Class 9 Physical Science

Chapter 7 Some Properties Of Sound And Characteristics Of Sound Synopsis

Laws Of Reflection Of Sound:

1. The incident sound wave, the reflected sound wave, and the normal to the reflector at the point of incidence, all lie on the same plane.
2. The angle of incidence is equal to the angle of reflection.
3. Regular reflection of light waves occurs from a small-sized smooth reflector, whereas sound waves are not reflected by it. Again, regular reflection of sound waves occurs from a large-sized rough reflector.
4. If the frequency of sound is within the range of 20 Hz to 20 kHz, it produces a sensation of hearing in the human ear and is known as audible sound.
5. Sound having a frequency lower than 20 Hz is known as infrasonic sound.
6. Sound having frequency higher than 20 kHz is known as ultrasonic sound.
7. The wavelength of audible sound approximately ranges from 1.5 cm to 16 cm, hence a large reflector is required for the reflection of such sound waves.
8. If a sound produced from a source is reflected by a reflector and reaches the listener in addition to the original sound, then the reflected sound is called an echo of the original sound.
9. If we hear any sound, its trace remains in the brain for approximately 0.1s. During this time duration of 0.1s, any other sound entering the ears cannot be distinguished. This time duration is called persistence of hearing.
10. If the velocity of sound in air at 27°C is 348 m/s, the minimum distance between the source and the reflector for hearing an echo of an inarticulate sound is 17.4 m. The minimum distance between the source and the reflector for hearing the echoes of monosyllabic, bisyllabic, and trisyllabic sounds are 34.8m, 69.6m, and 104.4m respectively.
11. The persistence of sound within a closed space after its source ceases to produce sound, caused by multiple reflections of the sound from walls, roof, etc. is known as reverberation.
12. SONAR is an acronym for SOund Navigation And Ranging. With the help of ultrasonic sound, SONAR is used for the detection of ruins of a sunken ship or existence of a mountain deep inside an ocean.
13. When sound of a single frequency is emitted from a tuning fork or different musical instruments, then the emitted sound is known as tone. Sound due to a mixture of more than one frequency is called a note.
14. The sound which is produced by regular and periodic vibrations of the source and appears euphonious to our ears is known as musical sound. Sound of guitar, sitar, flute, tuning fork etc. are examples of musical sound.
15. The sound which is produced by irregular and non-periodic vibrations of the source and appears cacophonous to our ears is known as noise. Sound of crackers, automobiles, uproar of crowd etc. are examples of noise.

There Are Three Characteristics Of A Musical Sound. These Are:

1. Loudness,
2. Pitch and
3. Quality or timbre.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

1. Amount of sound energy passing through an unit area around a point in one second, perpendicular to the direction of sound is called intensity of the sound at that point.
2. The dimensional, formula of intensity of sound is MT^-3 and its unit in SI is W/m².
3. The characteristic property of a musical sound due to which a listener can differentiate between a high-pitched and a low-pitched sound is known as pitch of the sound.
4. Frequency is the cause, pitch is its effect. Frequency is a measurable physical quantity. But pitch is a sensation or feeling which is not measurable.
5. The characteristic property by which sound of same intensity and pitch emitted by different sources can be distinguished in known as the quality or timbre of the musical sound.

Chapter 7 Some Properties Of Sound And Characteristics Of Sound Short And Long Answer Type Questions

Question 1. Write down the laws of reflection of sound.

Answer:

The Laws Of Reflection Are:

1. The incident sound wave, the reflected wave and the normal to the reflector at the point of incidence, all lie on the same plane.
2. The angle of incidence is equal to the angle of reflection.

Question 2. Demonstrate with an experiment that during reflection of sound, angle of incidence = angle of reflection. Or, Demonstrate by an experiment that sound wave is reflected similar to a light wave.

Answer:

The Angle Of Incidence = Angle Of Reflection:-

Two hollow long pipes, open at both ends, are taken. Also, one table clock and one raised partition are taken. One long table is kept touching the wall and at a little distance from the wall, the partition is kept in a perpendicular way.

Now, the pipes are kept on the two sides of the partition in such a way that the axes of both the pipes intersect at the point O of the wall.

Now, a table clock is kept at the end A of the left pipe. The right pipe is now rotated slowly with respect to the point O so that a position is obtained, where tick-tock sound of the clock is heard loudly. Suppose, sound is heard loudly at the point D.

This is the reflected sound. The tick-tock sound of the clock passes through pipe AB, reflected at point O, and comes to the ear through the pipe CD and is heard. Now pencil marks are made at the points O, A, D, and E.

Next, pipe and partition are removed, and A,0; D,0; E,0 are connected. If ∠AOE and ∠DOE are measured, it is found that ∠AOE = ∠DOE, i.e., angle of incidence = angle of reflection.

So, it can be concluded that sound wave is reflected similar to a light wave.

Question 3. How is the property of reflection of sound utilised in a stethoscope?

Answer:

Property Of Reflection Of Sound Utilised In A Stethoscope:-

1. The stethoscope is a medical instrument using which doctors examine the heart or lungs of a patient. There are two rubber pipes in this instrument which are attached to a round metallic disc covered with a thin diaphragm.
2. If the two open ends of the rubber pipes are fitted in the ears and the disc is placed on the chest of the patient, sound of the heart of the patient is reflected repeatedly through the rubbers tubes and reaches the ears of the doctor.

Question 4. Why do we curve our palms by the side of our ear to clearly hear a faint sound from a distant place?

Answer:

We Curve Our Palms By The Side Of Our Ear To Clearly Hear A Faint Sound From A Distant Place:-

To clearly hear a faint sound from a distant place, we curve our palm by the side of our ear. In addition to the direct faint sound, sound reflected from the palm also enters the ear. As a result, loudness of the sound increases to some extent. Here, palm of the hand works like a concave reflector.

Question 5. Why is the roof of a modern auditorium curved like an arch?

Answer:

The Roof Of A Modern Auditorium Curved Like An Arch Because:-

The roof of a modern auditorium is curved like an arch so that it works as a concave reflector and if the source of sound is kept at its focus, sound spreads evenly throughout the auditorium. As a result, the loudness of the sound increases.

Question 6. How can one talk in a speaking tube?

Answer:

One Talk In A Speaking Tube As Follows:-

If we talk at one end of a long metallic tube, the produced sound advances through the tube without scattering and reaches the other end after repeated reflections. The listener on the other side hears this sound very clearly. This arrangement is called a speaking tube. A funnel is attached to the front side for the speaker.

Question 7. What do you mean by audible sound?

Answer:

Audible Sound Means:-

One can hear sound only when the number of vibrations per second, i.e., frequency of a vibrating body lies within the range of 20 Hz to 20000 Hz. This sound is called audible sound.

Question 8. What do you mean by infrasonic sound and ultrasonic sound? give examples.

Answer:

Infrasonic Sound And Ultrasonic Sound:-

1. Infrasonic sound is the sound whose frequency is less than 20 Hz.
2. Ultrasonic sound is the sound whose frequency is more than 20 kHz.

When a pendulum oscillates, its frequency remains less than 20 Hz. Sound produced in this case is infrasonic sound, hence inaudible. Again, when a bat flies, it produces a type of sound from its mouth whose frequency is more than 20kHz, that is, ultrasonic sound.

Question 9. A flying bat makes a sound from its mouth. What type of sound wave does it produce?

Answer:

Given

A flying bat makes a sound from its mouth.

The sound wave emitted by a flying bat from its mouth has a frequency more than 20 kHz. Hence, it is an ultrasonic sound. If there is any obstacle in front of the sound, sound waves get reflected. By hearing that reflected sound, the bat can locate the position of any obstacle. We can not hear this ultrasonic sound emitted from the mouth of a bat.

Question 10. Give some practical applications of ultrasonic sound.

Answer:

These Are Some Of The Practical Applications Of Ultrasonic Sound:

1. In the process of ultrasonography, ultrasound is used to detect the location of a tumour in a human body.
2. Ultrasonic sound is used to clean dirty clothes.
3. Ultrasonic sound is also used to measure depth of the sea, to locate submarine or a shoal of fish, or to understand the presence of aquatic animals inside sea.

Question 11. Mention two applications of ultrasonic sound in medical science.

Answer:

Applications Of Ultrasonic Sound In Medical Science:-

It has been confirmed by experiments that there are some bacteria which can be destroyed by the application of ultrasonic sound. Also in the process of ultrasonography or USG, ultrasonic sound is used to know more about the internal muscles, different joints, arteries, etc. of the body.

Question 12. Write about a few animals in nature who normally make use of ultrasonic sound in their manner of living.

Answer:

Animals In Nature Who Normally Make Use Of Ultrasonic Sound In Their Manner Of Living Are Given Below:-

It has been observed that animals like bats, dolphins, whales, etc. utilise ultrasonic sound. Bats emit ultrasonic sound of very high frequency (100 kHz – 200 kHz) from their mouth at the time of flying.

This sound is reflected back from some obstacle in front of it and this helps the bat in recognising the obstacle. It thus flies freely without any impediment. Whales, dolphins, etc. living deep inside the sea communicate with each other through the use of ultrasound.

Question 13. Light wave is reflected regularly from a small and smooth reflector whereas sound wave is not reflected. Why?

Answer:

LightWave Is Reflected Regularly From A Small And Smooth Reflector Whereas Sound Wave Is Not Reflected

Reflection of a wave is possible only when the size of the reflector is larger than its wavelength. Again, whether reflection of a wave is regular or scattered, depends on the roughness of the surface of the reflector. If the wavelength of the wave is greater than the roughness of the surface of the reflector, there is regular reflection, otherwise the reflection is scattered.

The wavelength of sound in the audible range is from 1.5 cm to 16 m and wavelength of light in the visible range is from 1000 A (or 4 x 10^-5 cm) to 8000 A (or 8 x 10^-5 cm ). The size of a small, smooth reflector is comparatively larger than the wavelength of visible light.

So, in this case, light is reflected and the value of roughness being less than the wavelength of light, the reflection is regular. But as the size of a reflector is smaller than the wavelength of a sound wave, there is no reflection of sound wave.

Question 14. What are the differences between reflection of sound and reflection of light?

Answer:

The Differences Between The Reflection Of Sound And Reflection Of Light Are:

Question 15. What are the differences between sound wave and light wave?

Answer:

The Differences Between Sound Wave And Light Waves Are:

Question 16. What is the persistence of hearing?

Answer:

Persistence Of Hearing:-

If we hear any sound, its trace remains in the brain for approximately 0.1s. During this time duration of 0.1 s, more than one sound entering the ears cannot be distinguished. This time duration is called persistence of hearing.

Question 17. Calculate the minimum distance between the source and the reflector for hearing an echo of a short-lived sound; taking time of persistence of hearing as 0.1 second. [Given, velocity of sound in air = 348 m/s.]

Answer:

Given, velocity of sound in air = 348 m/s.

1. As the velocity of sound in air is 348 m/s, it traverses a distance of 348 x 0.1 m = 34.8 m in 0.1s.
2. So, the minimum distance between the source and the reflector for hearing an echo of a short-lived sound = \(\frac{34.8}{2}\) = 17.4 m.

Question 18. Calculate the minimum distance between the source and the reflector to hear an echo for a syllabic and intelligible sound, taking time of persistence of sound as 0.1 s. [Given that velocity of sound in air is 348 m/s and a human being cannot pronounce more than five syllables in one second.]

Answer:

Given that velocity of sound in air is 348 m/s and a human being cannot pronounce more than five syllables in one second.

Intelligible sound consists of one or more than one syllable. A human being cannot pronounce more than five syllables in one second. So, sound has to reach the ear after 1/5 s = 0.20 s, in order to hear an echo of a monosyllabic word. As the velocity of sound in air is 348 m/s, sound traverses a distance of 348 x 0.2 m * 69.6 m during this time.

In this case, minimum distance between the source and the reflector = \(\frac{69.6}{2}\) m = 34.8 m

Similarly, minimum distances between the source and the reflector, to hear an echo for a bisyllabic and trisyllabic sound are 34.8 x 2 = 69.6 m and 34.8 x 3 = 104.4 m respectively.

Question 19. What do you mean by an echo?

Answer:

Echo:-

If a sound produced from a source is reflected by a reflector situated at some distance and reaches the listener in addition to the original sound, then the reflected sound is called an echo of the original sound.

Question 20. State two conditions for the echo to be heard.

Answer:

Echo Can Be Heard If:-

1. The minimum distance between the source of sound and the obstacle will be 17 m if velocity of sound in air be 340 m/s.
2. The obstacle will be rigid, hard, and big in size (compare to wavelength of sound wave).

Question 21. Echo can’t be heard in a small room. Explain.

Answer:

Echo Can’t Be Heard In A Small Room:-

The minimum distance between source of sound and the obstacle to hear an echo should be 17 m if the velocity of sound in air be 340 m/s. Therefore echo can’t be heard in a small room.

Question 22. How is the depth of a sea measured with the help of echo?

Answer:

Depth Of A Sea Measured With The Help Of Echo As Follows:-

The depth of a sea can be measured with the help of echo. From the bottom of the ship, a source of sound (S) and a hydrophone (H) (a hydrophone is a microphone designed to be used underwater for recording or listening to underwater sound) are hung from the two sides at the same depth no.

If an intense and short-lived sound is produced from the source, one sound traverses along the straight path SH in time t₁ to reach the hydrophone and another sound reaches the hydrophone in time t₂, after getting reflected from the bottom at P. The first sound is the original sound and the second one is the reflected sound. A perpendicular PQ is drawn on SH from P.

Now if V is the velocity of sound in water, then SH = Vt₁

or, SQ = \(=\frac{1}{2} S H=\frac{V t_1}{2}\)

Again, SP + PH = Vt₂

or, SP = \(\frac{V t_2}{2}\) [SP = PH]

∴ PQ = \(\sqrt{S P^2-S Q^2}\)

= \(\sqrt{\left(\frac{V t_2}{2}\right)^2-\left(\frac{V t_1}{2}\right)^2}=\frac{V}{2} \sqrt{t_2^2-t_1^2}\)

As h₀ is the depth of the source from the water surface, so depth of sea at that place from the surface of sea = \(h_0+P Q=h_0+\frac{v}{2} \sqrt{t_2^2-t_1^2}\)

Question 23. How is the height of an airplane measured with the help of echo?

Answer:

Height Of An Airplane Measured With The Help Of Echo:-

The height of an airplane can be measured from the surface of the earth with the help of echo. Suppose, an airplane is flying parallel to the earth with a velocity u.

When it comes to position A, an intense sound of very short duration is made and when the airplane reaches the position C, then an echo is heard. If the sound is reflected at point 6, a perpendicular BD is drawn from B to AC.

If f is the difference of time between creating of sound and hearing of echo, then AC = ut.

∴ AD = \(\frac{AC}{2}{/latex] = [latex]\frac{ut}{2}{/latex]

Again, time taken for sound to reach B from A = [latex]\frac{t}{2}{/latex]

∴ if V is the velocity of sound in air, AB = [latex]\frac{Vt}{2}{/latex]

height of the airplane from the surface of the earth,

h = BD = [latex]\sqrt{A B^2-A D^2}\)

= \(\sqrt{\left(\frac{V t}{2}\right)^2-\left(\frac{u t}{2}\right)^2}=\frac{t}{2} \sqrt{V^2-u^2}\)

Question 24. What is reverberation?

Answer:

Reverberation

The persistence of sound within a closed space after its source ceases to produce sound, caused by multiple reflections of the sound from walls, roof, etc. is known as reverberation.

Question 25. What measures are taken to prevent reverberation inside an auditorium?

Answer:

Measures To Take For The Prevention of Reverberation Inside An Auditorium:-

Reverberation takes place if a lot of reflectors are present within a small distance. For this reason, sound persists for a considerable amount of time inside an empty hall even after the original sound stops. This creates great difficulty for the audience as they cannot hear the original sound.

To remove this difficulty, walls of the auditorium are covered with soft pads or rubber-type materials. These work as sound absorbers. As a result, the problem of reverberation does not arise. Furniture and material of the seats are also made of materials which can absorb sound.

Question 26. What is SONAR?

Answer:

SONAR:-

SONAR is an acronym for Sound Navigation And Ranging. With the help of ultrasonic sound, SONAR is used for detection of ruins of a sunken ship or existence of a mountain deep inside an ocean.

Question 27. How can we measure the depth of a sea with the help of SONAR?

Answer:

We Can Measure The Depth Of A Sea With The Help Of SONAR As Follows:-

One sound emitting machine and a detector of SONAR are present in a ship. A very powerful ultrasonic sound wave is sent inside the sea with – the help of the sound emitter. This wave is reflected by the sea bed and comes back to the detector instrument.

Suppose, t is the time difference between the time of sending the wave and time of receiving the echo of the wave. If the velocity of the ultrasonic sound is V, then the depth of sea at that place is \(\frac{Vt}{2}\).

Question 28. What are the differences between a tone and a note?

Answer:

Differences Between A Tone And A Note:-

Tone is the sound of a single frequency, emitted from different musical instruments or tuning fork. In other words, it may be stated that the sound emitted from any musical instrument is a mixture of sounds of different frequencies. Each one of these frequencies is a tone. In the language of music, a summation of several tones is a note. Therefore, the note is a mixture of several frequencies.

Question 29. What do you mean by fundamental tone, overtone, and harmonic?

Answer:

Fundamental Tone, Overtone, And Harmonic:-

Fundamental tone: Fundamental tone is defined as the tone of minimum frequency present among the tones of a note.

Overtone: Overtone is defined as any tone present in a note other than the fundamental tone.

Harmonic: Harmonic is defined as any tone whose frequency is an integral multiple of the fundamental tone.

Question 30. The following tones with given frequencies are mixed in a sound emitted from a musical instrument: 256Hz, 512Hz, 1020Hz, and 1280Hz. Identify them as:

1. Fundamental tone
2. First harmonic
3. First overtone
4. Second overtone
5. Second harmonic.

Answer:

1. Fundamental tone: 256 Hz
2. First harmonic: 256 Hz
3. First overtone: 512 Hz
4. Second overtone: 1020 Hz
5. Second harmonic: 512 Hz

Question 31. What do you mean by musical sound and noise?

Answer:

Musical Sound And Noise:-

Musical sound: Musical sound is defined as that sound which is produced by regular and periodic vibrations of the source and appears euphonious to our ears. Sound of guitar, sitar and flute, tuning fork etc. are examples of musical sound.

Noise: Noise is defined as that sound which is produced by irregular and non-periodic vibration of the source and appears cacophonous to our ears. Sound of crackers, automobiles, uproar of crowd etc. are examples of noise.

Question 32. What are the characteristics of a musical sound? What is intensity of sound?

Answer:

Characteristics Of A Musical Sound:-

There are three characteristics of a musical sound. These are:

1. Loudness,
2. Pitch and
3. Quality or timbre.

Question 33. Amount of sound energy passing through an unit area around a point in one second, perpendicular to the direction of sound is called intensity of the sound at that point. what is the unit of intensity in SI? Calculate the dimensional formula of intensity.

Answer:

Given

Amount of sound energy passing through an unit area around a point in one second, perpendicular to the direction of sound is called intensity of the sound at that point.

Unity of intensity in SI is W • m^-2.

Dimensional formula of intensity = \(\frac{\text { dimensional formula of energy }}{\text { dimensional formula of area}}\) x dimensional formula of time

= \(\frac{M L^2 T^{-2}}{L^2 \times T}=M T^{-3}\)

Question 34. What do you understand by loudness of sound?

Answer:

Loudness Of Sound:-

Among the sounds we hear everyday, some are high-pitched and some are low-pitched. Loudness of sound is such a characteristic which defines the strength with which it reaches the ears of the listener. Loudness is a feeling only and is not measurable. Loudness of sound is measured by its intensity.

Question 35. Intensity is the cause, loudness is its effect—explain.

Answer:

Intensity Is The Cause, Loudness Is Its Effect

When intensity of sound is more, more amount of sound energy reaches our ears, and hence, greater loudness is felt by us. Again, when intensity is less, less amount of sound energy enters our ears, so loudness of sound is less. This is why it may be concluded that if intensity is the cause, then loudness is its effect.

Question 36. Intensity of sound depends on which factors?

Answer:

Intensity of sound depends on several factors

Amplitude of vibration of the source of sound:

Intensity of sound is directly proportional to the square of the amplitude of vibration of the source of sound.

Size of the source:

Larger the size of the source, more is the intensity of the sound.

Distance of the listener from the source:

Intensity of sound is inversely proportional to the square of the distance of the listener from the source.

Density of the medium:

Intensity is more if the density of the medium is more.

Flow of air: Intensity of sound increases if flow of air is in the direction of motion of the sound. But if flow of air is in the opposite direction of motion of sound, intensity gets reduced.

Question 37. On a particular day, wind is blowing from east to west. A song is played through a microphone. A man is standing slightly away from the microphone in the east. Now, this man crosses the microphone and moves to the west side of the microphone at an equal distance. He hears more intense sound in which case? Explain with reasons.

Answer:

Sound spreads in every direction with equal loudness. But if wind blows along the direction of propagation of sound, loudness increases in that direction. According to this question, wind blows from east to west. As a result, the man hears sound with greater intensity while standing in the west.

Question 38. What do you mean by pitch of a sound?

Answer:

Pitch Of A Sound:-

If a male, a female and a child pronounce a sound with the same intensity, this same sound is known to create three different feelings in our ears. The voice of a female appears to be shriller than that of a male, again the voice of a child appears to be shriller than that of a female. This characteristic of sound is called pitch.

Pitch is defined as that characteristic of a musical sound due to which a listener can differentiate between a high-pitched and a low-pitched sound.

Question 39. Explain the statement — frequency is the cause of a sound and pitch is its effect.

Answer:

Frequency Is The Cause Of A Sound And Pitch Is Its Effect

Pitch depends mainly on frequency. If the frequency of a tone is high, its pitch is also high. On the other hand, if frequency is low, pitch is also low. So, it is said that frequency is the cause and pitch is the effect.

But frequency is a physical quantity and is measurable whereas pitch is a feeling which is not measurable.

Question 40. Loudness and pitch of sound change mainly due to which physical quantities?

Answer:

Due to change of intensity of a musical sound, loudness of sound perceived by a person changes. It has been found by experiment that this sensation increases as intensity of sound increases.

On the other hand, due to change in frequency of a musical sound, the pitch of a sound perceived by ear changes. Generally, as the frequency of the source of sound increases, the sound appears to be more high-pitched in the ears of an individual.

Question 41. What do you mean by quality or timbre of a sound?

Answer:

Timbre Of A Sound:-

The characteristic property by which sound of same intensity and pitch emitted by different sources can be distinguished is called the quality or timbre of a musical sound. Due to this characteristic of sound, tones of the same intensity and pitch emitted from harmonium, violin and table, etc. can be easily differentiated by us.

Question 42. The quality or timbre of a note depends on which factors?

Answer:

The quality of a note depends on three factors. These are:

1. The number of overtones mixed with the fundamental tone;
2. Frequencies of the overtones and
3. Relative intensity of the overtones.

Chapter 7 Some Properties Of Sound And Characteristics Of Sound Very Short Answer Type Questions Choos The Correct Answer

Question 1. Due to which characteristic of sound, one can distinguish between the low-pitched and high-pitched sound of same intensity?

1. Loudness
2. Amplitude
3. Frequency
4. Pitch

Answer: 4. Pitch

Question 2. Which one is the first harmonic among the following frequencies: 500 Hz, 900 Hz, 550 Hz, 750 Hz, 300 Hz, 600 Hz?

1. 300 Hz
2. 500 Hz
3. 600 Hz
4. 900 Hz

Answer: 1. 300 Hz

Question 3. Out of 256 Hz, 512 Hz, 1020 Hz, and 1280 Hz, the frequency 1280 Hz is the

1. Last tone
2. Fourth overtone
3. Third harmonic
4. Second harmonic

Answer: 2. Fourth overtone

Question 4. Intensity of sound does not depend on

1. Shape of the source of sound
2. Amplitude of vibration of the source of sound
3. Density of the medium
4. Number of overtones in note

Answer: 4. Number of overtones in note

Question 5. The sound which contains more number of is more melodious.

1. Harmonics
2. Overtones
3. Fundamental tones
4. Beats

Answer: 1. Harmonics

Question 6. Characteristic of musical sound is

1. Loudness
2. Pitch
3. Quality
4. All of these

Answer: 4. All of these

Question 7. The quantity which does not produce any perception is

1. Loudness
2. Intensity
3. Pitch
4. Quality

Answer: 2. Intensity

Question 8. Among the following pairs, which one represents reciprocal quantities?

1. Frequency and wavelength
2. Frequency and time period
3. Velocity of wave and wavelength
4. Wavelength and loudness

Answer: 2. Frequency and time period

Question 9. Which of the following cannot reflect sound waves?

1. Range of mountains
2. High embankment of a river
3. A small mirror
4. Wall of an empty hall

Answer: 3. A small mirror

Question 10. Reverberation of sound may be heard

1. In a stethoscope
2. Inside an empty hall
3. In a distant wall
4. In vacuum

Answer: 2. Inside an empty hall

Question 11. Reverberation of sound is created

1. In one reflection
2. In two reflections
3. In three reflections
4. In multiple reflections

Answer: 4. In multiple reflections

Question 12. Rubber pads are fixed on the walls of a cinema hall in order to

1. Increase reflection of sound
2. Decrease reverberation of sound
3. Increase velocity of sound
4. Increase loudness of sound

Answer: 2. Decrease reverberation of sound

Question 13. If a tuning fork is struck, sound emitted from it has a frequency of 500 Hz. What is the frequency of the emitted sound when the tuning fork is struck with twice the previous force?

1. 1000 Hz
2. 500 Hz
3. 250 Hz
4. Insufficient data

Answer: 2. 500 Hz

Question 14. If the frequencies of fundamental tone, overtone, and harmonic are denoted by n₁, n₂, and n₃ respectively, then

1. n₁> n₂> n₃
2. n₁ > n₂ but n₂ > n₃
3. n₁< n₂> n₃
4. n₁< n₂ and n₁ ≤ n₃

Answer: 4. n₁< n₂ and n₁ ≤ n₃

Question 15. Which of the following is essential for a melodious sound?

1. The presence of more harmonics
2. The presence of more overtones
3. Presence of different frequencies
4. Presence of different musical instruments

Answer: 1. Presence of more harmonics

Question 16. Which of the following statements is correct?

1. Tone is a sound of single frequency
2. Note is a sound of single frequency
3. Tone is a sound composed of different frequencies
4. Tone is composed of several notes

Answer: 1. Tone is a sound of single frequency

Question 17. If the frequency of the source of sound increases, the sound heard by a person appears to be more

1. Loud
2. Discordant
3. Melodious
4. Shrill

Answer: 4. Shrill

Question 18. If the velocity of sound in air at 27°C is 348 m/s, then what is the minimum distance between the source and the reflector of sound to hear an echo of transient sound?

1. 17.4 m
2. 34.8 m
3. 69.6 m
4. 104.4 m

Answer: 1. 17.4 m

Question 19. If the velocity of sound in air at 27°C is 348 m/s, what is the minimum distance between the source and the reflector of sound to hear an echo of a monosyllabic sound?

1. 17.4 m
2. 34.8 m
3. 69.6 m
4. 104.4 m

Answer: 2. 34.8 m

Question 20. Bats detect the obstacles in their path by receiving the reflected

1. Electromagnetic waves
2. Radio waves
3. Infrasonic wave
4. Ultrasonic waves

Answer: 4. Ultrasonic waves

Question 21. Infrasonic sound can be heard by

1. Human beings
2. Dolphins
3. Rhinoceros
4. Monkey

Answer: 3. Rhinoceros

Question 22. Which of the following is used in ultrasonography?

1. Infrasonic sound
2. Ultrasonic sound
3. Audible sound
4. Radio waves

Answer: 2. Ultrasonic sound

Question 23. Which of the following is used in SONAR?

1. Infrasonic sound
2. Micro waves
3. Ultrasonic sound
4. Audible sound

Answer: 3. Ultrasonic sound

Question 24. Repeated reflection of sound produces

1. Reverberation
2. Stationary wave
3. Echo
4. Refraction

Answer: 1. Reverberation

Question 25. A mirror is a good reflector of light wave but poor reflector of sound wave because

1. Wavelength of sound is bigger than the mirror
2. Wavelength of light is bigger than the mirror
3. Velocity of sound is smaller than that of light
4. Sound can not travel in vacuum

Answer: 1. Wavelength of sound is bigger than the mirror

Question 26. Sound wave transfers a physical quantity ‘X’ from one place to another, where ‘X’ is

1. Mass
2. Velocity
3. Energy
4. Density

Answer: 3. Energy

Question 27. A sound wave strikes a wall perpendicularly. What is the angle of reflection of the sound wave?

1. 90°
2. 0°
3. 45°
4. 30°

Answer: 2. 0°

Question 28. A supersonic jet plane

1. Can fly faster than sound
2. Can fly at the highest velocity of sound
3. Cannot fly with the velocity of sound
4. Can fly with the velocity of light

Answer: 1. Can fly faster than sound

Question 29. A jet plane is flying with supersonic speed. Its Mach number is

1. Equal to one
2. More than one
3. Less than one
4. Equal to one hundred

Answer: 2. More than one

Chapter 7 Some Properties Of Sound And Characteristics Of Sound Answer In Brief

Question 1. What is the range of frequency of audible sound?

Answer: Range of frequency of audible sound is 20 Hz to 20000 Hz .

Question 2. What is the frequency of infrasonic sound?

Answer: Frequency of infrasonic sound is less than 20 Hz.

Question 3. What is the frequency of ultrasonic sound?

Answer: Frequency of ultrasonic sound is more than 20000 Hz.

Question 4. What is a tone?

Answer:

Tone

Sound of a single frequency is known as a tone.

Question 5. What is a note?

Answer:

Note

Note is a sound containing more than one frequency.

Question 6. Write the SI unit of intensity of sound.

Answer:

SI unit of intensity of sound

J • s^-1 • m^-2 or, W • m^-2.

Question 7. Give one application where principle of echo is utilised.

Answer: Hearing aid.

Question 8. 500 Hz is written on the body of tuning fork. Is the sound produced from this fork a tone or a note?

Answer: If the tuning fork is vibrated, the sound produced from it is a tone since sound of only 500 Hz frequency is produced.

Question 9. Which property of sound is applied in a doctor’s stethoscope?

Answer: Property of reflection of sound is applied in a doctor’s stethoscope.

Question 10. What type of sound is emitted when a tuning fork is struck?

Answer: Audible sound is emitted when a tuning fork is struck.

Question 11. What type of sound is used in the in a radar?

Answer: Ultrasound is used in a radar.

Question 12. Mention the name of an animal which makes use of ultrasound.

Answer: A bat uses ultrasound while flying.

Question 13. Which characteristic of a musical sound depends on its frequency?

Answer: Pitch of a musical sound depends on its frequency.

Question 14. Which characteristic of a musical sound depends on its intensity?

Answer: Loudness of a musical sound depends on its intensity.

Question 15. Which characteristic of a musical sound changes when the distance between the source of sound and the listener changes?

Answer: In this case, loudness of the musical sound changes.

Question 16. Sound of the same frequency and with the same intensity are emitted from different string instruments. Which characteristic of sound helps us in distinguishing the sound of one musical instrument from the other?

Answer: Quality or timbre of the musical sound helps us in distinguishing the sound of different musical instruments.

Question 17. Which is the first harmonic in a sound consisting of more than one frequency?

Answer: The sound with the minimum frequency is the first harmonic.

Question 18. What do you mean by the intensity of sound?

Answer:

Intensity of sound

Sound energy per unit time per unit area is known as the intensity of sound.

Question 19. What does SONAR stand for?

Answer: SONAR stands for SOund Navigation And Ranging.

Question 20. What is the basic principle with which SONAR works?

Answer: SONAR works on the principle of reflection of waves.

Question 21. State one important use of ultrasound in industries.

Answer: Ultrasounds are used to find faults and cracks in metal.

Question 22. State one important use of ultrasound for medical purpose.

Answer: Diagnosing the diseases in human body.

Question 23. What do you understand by the reverberation?

Answer: The phenomenon of prolongation of original sound due to the multiple reflection of sound wave even after the source stops producing sound is called reverberation.

Question 24. Why is the speed of sound grater in solid than in gas?

Answer: This is because particle of solids are closer than the particles of gases.

Question 25. Which wave property determines loudness?

Answer: Amplitude of the wave determines loudness.

Question 26. Does sound follow the same lows of reflection as light does?

Answer: Yes- Sound wave are reflected just like light wave

Question 27. Why is sound not heard when a bird flies?

Answer: When a bird flies, the frequency of its wings is less than 20 Hz which is in the infrasonic range and thus the sound cannot be heard.

Question 28. How does the velocity of sound in air change when temperature is increased?

Answer: Velocity of sound in air increases when temperature is increased.

Question 29. How does the velocity of sound in air change when humidity of air increases?

Answer: Velocity of sound in air increases when the humidity of air increases.

Question 30. When a pendulum oscillates, its frequency remains less than 20 Hz. What type of sound is produced in this case?

Answer: Infrasonic sound is produced in this case.

Question 31. Which type of sound waves are used to clean dirty clothes?

Answer: Ultrasonic sound waves are used to clean dirty clothes.

Question 32. Which type of sound waves are used to detect a shoal of fish deep inside sea water?

Answer: Ultrasonic sound waves are used to detect a shoal offish deep inside sea water.

Chapter 7 Some Properties Of Sound And Characteristics Of Sound Fill In The Blanks

Question 1. The perception of difference of pitch sound arises due to change of ______ of sound.

Answer: Frequency

Question 2. In the case of  ________ repeated reflection of sound occurs.

Answer: Reverberation

Question 3. Sound having frequency more than the frequency of the audible range is known as _______ sound.

Answer: Ultrasonic

Question 4. Sound having frequency less than the frequency of the audible range is known as ______ sound.

Answer: Infrasonic

Question 5. Stethoscope is the application of the property of ______ of sound.

Answer: Reflection

Question 6. Soft pads fixed to the walls of different auditoriums serve as __________

Answer: Sound absorbers

Question 7. SONAR uses ________ sound to determine the depth of an ocean.

Answer: Ultrasonic

Question 8. _______ of sound increases with the increase of size of the source of sound.

Answer: Intensity

Question 9. ________ of the sound increases with increase of frequency of the source of sound.

Answer: Pitch

Question 10. The intensity of sound __________ if air flows along the direction of motion of the sound.

Answer: Increases

Question 11. The difference in the voices of males and females is due to the ________ of sound.

Answer: Pitch

Question 12. Reflection of a wave is possible only when the size of the reflector is ________ compared to the wavelength of the wave.

Answer: Greater

Question 13. The trace of a sound remains in our brain for almost ________

Answer: 0.1 s

Question 14. The ratio of the velocities of a body and sound in a medium is called __________

Answer: Match number

Question 15. The velocity of a moving body is called ________ if the value of its Mach number is more than 1.

Answer: Super sonic

Chapter 7 Some Properties Of Sound And Characteristics Of Sound State Whether True Or False

Question 1. Echo and reverberation of sound refer to a single phenomenon.

Answer: False

Question 2. SONAR is an acronym for Sound Navigation and Ranging.

Answer: True

Question 3. The tone having the lowest frequency in a note is known as fundamental tone.

Answer: True

Question 4. Ultrasonic sound is used for the purpose of sterilizing of milk.

Answer: True

Question 5. Loudness of sound varies inversely to the square of the amplitude.

Answer: True

Question 6. Pitch of a note depends on its amplitude.

Answer: False

Question 7. A sound which is music to some one may be a noise to others.

Answer: False

Question 8. Sound of one frequency is tone.

Answer: True

Question 9. In a note the sound of least frequency is called the fundamental tone.

Answer: True

Question 10. A continuous noise of high level may cause migraine.

Answer: True

Chapter 7 Some Properties Of Sound And Characteristics Of Sound Numerical Examples

Useful Information

The impression of a sound persist in our brain for about \(\frac{1}{10}\) s.

If velocity of sound in air at a particular temperature = V then,

1. The minimum distance between listener and the reflector for hearing distinctly the echo of a short sound = \(\frac{1}{2}\) x V x \(\frac{1}{5}\) = \(\frac{V}{20}\)
2. Our ears can not recognise separately more than 5 syllables in one second, for monosyllabic sound the minimum distance between the listener and the reflector = \(\frac{1}{2}\) x V \(\frac{1}{5}\) = \(\frac{V}{10}\).

For disyllabic word the distance should be = \(\frac{1}{2}\) x V x \(\frac{2}{5}\) = 2 x \(\frac{V}{10}\) and so on.

Measurement of depth of a sea:

If, velocity of sound in water be V, the depth of the source and hydrophone from the surface of the sea be hQ, the depth of the sea with respect to the source be h, time taken by the sound to reach the hydrophone directly be t₁, and the time taken by the echo be t₂, then depth of the sea from its surface,

h = \(h_0+\frac{V}{2} \sqrt{t_2^2-t_1^2}\)

Measurement of the height of an airplane:

If, an airplane flying horizontally at a height h with a velocity u, produces a loud short sound and heard the echo after time t and velocity of sound in air be V then,

h = \(\frac{t}{2} \sqrt{V^2-u^2}\)

Question 1. A man fires a shot in front of a hill and hears the echo after 3 s. If the velocity of sound in air is 332 m/s, what is the distance between this man and the hill?

Answer:

Given

A man fires a shot in front of a hill and hears the echo after 3 s. If the velocity of sound in air is 332 m/s,

Let distance of the man from the hill is L m .

2L m is traversed in 3 s.

So, \(\frac{2L}{3}\) = 332 m 3

or, L = \(\frac{332 \times 3}{2}\) = 498 m

Question 2. A pilot fires a shot when the airplane is flying parallel to the earth’s surface at a speed of 72 km/h . If the echo is heard after 4 $ and the velocity of sound in air is 332 m/s, calculate the height of the airplane from the earth’s surface.

Answer:

Given

A pilot fires a shot when the airplane is flying parallel to the earth’s surface at a speed of 72 km/h . If the echo is heard after 4 $ and the velocity of sound in air is 332 m/s,

Velocity of the airplane, u = 72 km/h = 72 x \(\frac{5}{18}\) = 20 m/s

Velocity of sound, V = 332 m/s; time, t = 4 s

Suppose, height of the airplane from the surface of the earth =h

∴ h = \(\frac{t}{2} \sqrt{V^2-u^2}=\frac{4}{2} \sqrt{(332)^2-(20)^2}\) = 662.8

Question 3. A man standing between two parallel mountains hears the first echo after 2 s and the second echo after 5 s of firing a bullet from a gun. If the velocity of sound in air is 332 m/s, what is the distance between the two mountains?

Answer:

Given

A man standing between two parallel mountains hears the first echo after 2 s and the second echo after 5 s of firing a bullet from a gun. If the velocity of sound in air is 332 m/s,

Suppose the distance of the man from the nearer mountain is a m and its distance from the distant one is b m.

The distance between the two mountains = (o + b) m .

According to the question, sound traverses a distance 2a in 2 seconds and a distance 2b in 5 seconds.

∴ 2a = 2 x 332 or, a = 332 and 2b = 5 x 332 or, b = 830

∴ distance between the two mountains = 830 + 332 = 1162 m .

Question 4. An airplane is flying at a height of 1200 m with a velocity Of 125 m/s. A high- pitched sound of very short duration is emitted from the plane. How much time does the echo of that sound take to reach the pilot after being reflected from the earth’s surface? Velocity of sound in air = 325 m/s.

Answer:

Given

An airplane is flying at a height of 1200 m with a velocity Of 125 m/s. A high- pitched sound of very short duration is emitted from the plane.

Suppose, the time gap between emission of sound and its audibility as echo = t.

According to the formula, h = \(=\frac{t}{2} \sqrt{v^2-u^2}\)

t = \(\frac{2 h}{\sqrt{V^2-u^2}}=\frac{2 \times 1200}{\sqrt{325^2-125^2}}\)

= \(\frac{2 \times 1200}{\sqrt{450 \times 200}}=\frac{2 \times 1200}{300}=8 \mathrm{~s}\)

So, pilot hears the echo after 8 s of its emission.

Question 5. An echo repeated four syllables. If the velocity of sound be 335 m/s , find the distance of the reflector.

Answer:

To hear echo for a word with four syllable, the minimum time required = 4 x 1/5 = 4/5 s

∴ The distance of the reflector = \(\frac{\text { distance travels }}{2}=\frac{4 \times 335}{5 \times 2}\) = 134 m

Question 6. Two children are at opposite ends of an aluminium rod. One strikes at one end of the rod with a hammer. Find the ratio of time taken by the sound waves in air and in the aluminium to reach the second child. (Speed of sound in aluminium is 6420 m/s and in air is 338 m/s.

Answer:

Given

Two children are at opposite ends of an aluminium rod. One strikes at one end of the rod with a hammer.

Let, l = length of the rod.

Time taken by sound to travel distance l in aluminium rod

⇒ \(t_1=\frac{\text { distance }}{\text { speed of sound in alumimium }}\)

= \(\frac{l}{64210} \mathrm{~s}\)

and time taken by sound to travel distance l in air

t₂ = \(\frac{l}{338}\) s

∴ The ratio of time taken by the sound wave in air and in the aluminium

= \(t_2: t_1=\frac{1}{338}: \frac{1}{6420}=3210: 169\)

Question 7. A man standing 12 m away from a high wall makes a short sound. If velocity of the sound in air is 330 m/s. Check whether he hear the echo or not.

Answer:

Given

A man standing 12 m away from a high wall makes a short sound. If velocity of the sound in air is 330 m/s.

The persistence of hearing = 0.1 s

∴ The minimum distance for hearing echo for short sound = \(\frac{0.1 \times 330}{2}\) = 16.5 m

According to the problem as the man standing at a distance 12 m from the wall, he can not hear the echo.

Question 8. Sound beyond the human audio range is produced at a place deep inside the sea. After 5 s, echo is detected by a SONAR at the same place. Calculate the depth of the sea from the place of production of sound. Velocity of sound in sea water is 1500 m/s.

Answer:

Given

Sound beyond the human audio range is produced at a place deep inside the sea. After 5 s, echo is detected by a SONAR at the same place.

Velocity of sound in sea water is 1500 m/s.

Suppose, depth of sea = h m

∴ sound travels 2h m in 5 seconds.

So, sound travels \(\frac{2h}{5}\) m in one second.

∴ \(\frac{2h}{5}\) = 1500 or, h = 1500 x \(\frac{5}{2}\) = 3750 m

Chapter 4 Matter Concept Of Mole Topic A Concept Of Mole And Importance Of Avogadro’s Number Synopsis

1. The literal meaning of the word ‘mole’ is heap.
2. One mole of a substance is the amount that contains 6.022 x 10²³ constituent particles (atoms, molecules, or ions) of the substance.
3. Chemical calculations can be done in a simpler and more convenient method by using the concept of mole.
4. The number of molecules present in one gram-mole of a substance, which may be either an element or a compound (solid, liquid, or gas) is known as Avogadro’s number (N_A).
5. The value of Avogadro’s number (N_A) is 6.22x 10²³. It is independent of both temperature and pressure.
6. It is very difficult to grasp the enormousness of Avogadro’s number. N_A = 602213670000000000000000.
7. The value of Avogadro’s constant is 6.022 x 10²³ mol^-23.
8. Avogadro’s number correlates with the microscopic and macroscopic world.
9. Avogadro’s number finds useful applications in calculations related to physics, chemistry, and biological science.

Class 9 Physical Science Chapter 4 Concept Of Mole

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

Chapter 4 Matter Concept Of Mole Topic A Concept Of Mole And Importance Of Avogadro’s Number Short And Long Answer Type Questions

Question 1. Define mole.

Answer:

Mole:

One mole of a substance is the amount of the substance (element or compound) which contains Avogadro’s number (6.022 x 10²³) of fundamental particles (atoms, molecules, or ions).

Question 2. What is Avogadro’s number?

Answer:

Avogadro’s Number:

The number of molecules present in one gram mole of a substance which may be either an element or a compound (solid, liquid, or gas) is known as Avogadro’s number.

Question 3. What is Avogadro’s constant? How is it different from Avogadro’s number?

Answer:

Avogadro’s Constant:

1. The number of particles present per mole of a substance is called Avogadro’s constant. Thus, Avogadro’s constant is Avogadro’s number/mole. Its value is 6.022 x 10²³ mol^-1. It is a universal constant.
2. Avogadro’s number is a pure number. It has no unit but the unit of Avogadro’s constant is mol^-1.

Class 9 Physical Science Chapter 4 Short And Long Answer Type Questions

Question 4. Why is it necessary to mention the corresponding fundamental particle while using the term ‘mole’?

Answer:

It/is necessary to mention the corresponding fundamental particle while using the term ‘mole’. This is because the amount of a substance entirely depends on the nature of particles present in it. For example, the fundamental particle of oxygen can be atom as well as molecule.

Thus, the term 1 mol oxygen1 does not clearly indicate the amount of oxygen present in that quantity, because 1 mol oxygen represents both 1 mol oxygen molecule and 1 mol oxygen atom.

Now, a 1 mol oxygen molecule contains twice the number of oxygen atoms present in a 1 mol oxygen atom, although the number of particles in both quantities are same.

Question 5. Avogadro’s number creates a correlation between the macroscopic and microscopic world— explain.

Answer:

Avogadro’s number creates a correlation between the macroscopic and microscopic world—

Molecules or atoms are too small to be seen. So, it is extremely difficult to count the number of atoms or molecules in a given mass of substance. However, by using the mole concept, scientists have been able to successfully calculate the number of atoms, molecules, or ions in a given mass of substance.

According to the definition of mole, 1 mol of a substance contains 6.022 x 10²³ number of fundamental particles. This number (6.022 x 10²³) is known as Avogadro’s number.

We cannot see 1 molecule of water but, 1 mol of water (which is equal to 18 g water) is visible to us. Now, 1 mol of water (i.e., 18 g water) contains 6.22x 10²³ molecules of water.

Thus it can be concluded that, Avogadro’s number creates a correlation between the macroscopic and microscopic world.

Class 9 Physical Science Chapter 4 Short And Long Answer Type Questions

Question 6. Discuss the significance of mole concept or Avogadro’s number.

Answer:

Significance Of Mole Concept:

The concept of mole is used in different fields of science. 1 mol of a substance contains Avogadro’s number (6.022 x 10²³) of fundamental particles (atoms, molecules or ions) this concept has simplified the chemical calculations to a large extent.

It has also become easier to correlate the different physical quantities with the help of Avogadro’s number and mole concept. Some important applications of Avogadro’s number are illustrated below

1. The number of atoms, molecules or ions present in a given mass of a substance is calculated by using mole concept. For example, 88g CO₂ = 88g/44g = 2 mol (as a gram-molecular mass of CO₂ is 44 g).

Hence, number of molecules in 88 g CO₂ = 2 x 6.022 x 10²³ = 12.044 x 10²³

2. The actual mass of an atom or a molecule is expressed in atomic mass unit (u).

Now, u = \(\frac{1}{\text { Avogradro’s number }}\) g

Thus, mass of one N₂ molecule = 28 u

= 28 x \(\frac{1}{6.022 \times 10^{23}}\) = 28 x 1.6605 x 10^-24

Class 9 Physical Science Chapter 4 Short And Long Answer Type Questions

3. Using Avogadro’s number, it is possible to calculate the number of molecules present in a definite volume of any gas at STP.

4. The atomic radii of solid metals can be calculated using Avogadro’s number.

5. The value of Boltzmann constant can be calculated using Avogadro’s number.

Question 7. Do 1 mol O₂ and 1 mol O represent the same quantity of oxygen?

Answer:

The molecular formula of oxygen is O₂ while O represents an atom of oxygen. Hence, 1 mol O₂ represents Avogadro’s number of oxygen molecules. On the other hand, 1 mol O₂ indicates Avogadro’s number of O-atoms.

The mass of Avogadro’s number of O₂ molecules is 32 g and that of Avogadro’s number of O-atoms is 16 g. Thus, 1 mol O₂ and 1 mol O do not represent the same quantity of oxygen.

Question 8. Can Avogadro’s constant be considered as universal constant?

Answer:

Yes, Avogadro’s constant may be considered as universal constant.

Reasons:

1. 22.4 L of any gas at STP contains Avogadro’s number of molecules.
2. Number of molecules in 1 g-mole or number of atoms in 1 g-atom of any element or compound are equal to Avogadro’s number.
3. Number of ions present in 1 g-ion of any sample is equal to Avogadro’s number.

Question 9. Write down the importance of Avogadro’s number in biology. Or, Mention an use of Avogadro’s number in biology.

Answer:

Importance of Avogadro’s number in biology

In biological science, Avogadro’s number can be used for quantitative calculations of solids, liquids, and gases.

Class 9 Physical Science Chapter 4 Short And Long Answer Type Questions

Example:

Chlorophyll contains 2.68% Mg. Using the concept of Avogadro’s number one can find the number of Mg atoms present in 1 g chlorophyll as 6.69 x 10²⁰.

Chapter 4 Matter Concept Of Mole Topic A Concept Of Mole And Importance Of Avogadro’s Number Very Short Answer Type Questions Choose The Correct Answer

Question 1. Which of the following correlates between the microscopic and macroscopic world?

1. Dalton’s number
2. Berzelius’ number
3. Avogadro’s number
4. Faraday’s number

Answer: 3. Avogadro’s number

Question 2. The meaning of the Latin word ‘moles’ is

1. Many
2. Having large volume
3. Having large mass
4. Heap

Answer: 4. Heap

Class 9 Physical Science Chapter 4 Very Short And Long Answer Type Questions

Question 3. The word ‘mole’ was first used by

1. Dalton
2. Ostwald
3. Avogadro
4. Millikan

Answer: 2. Ostwald

Class 9 Physical Science Chapter 4 Very Short And Long Answer Type Questions

Question 4. The value of Avogadro’s number is

1. 6.024 X 10²⁰
2. 0.6023 X 10²²
3. 0.6022 X 10²⁴
4. 0.623 x 10²³

Answer: 3. 0.6022 X 10²⁴

Question 5. The amount of Avogadro’s number of fundamental particles (example: electron, proton, atom, molecule, ion) is known as

1. Mole
2. Gram-mole
3. Gram-atom
4. Gram-ion

Answer: 1. Mole

Class 9 Physical Science Chapter 4 Very Short And Long Answer Type Questions

Question 6. The element for which the number of gram-moles and number of gram-atoms will be equal for any quantity is

1. Oxygen
2. Helium
3. Hydrogen
4. Chlorine

Answer: 2. Helium

Question 7. Number of atoms in 0.1 mol of a triatomic gas is

1. 6.022 x10²²
2. 1.806 X 10²³
3. 3.600 x10²³
4. 1.800 x 10²²

Answer: 2. 1.806 X 102²³

Class 9 Physical Science Chapter 4 Very Short And Long Answer Type Questions

Question 8. If 10^-5 mol electron flows per second through a wire, the number of electron flows per second through it is

1. 6.022 X10²⁸
2. 6.022 x 10²³
3. 6.022 X10^-23
4. 6.022 x 10¹⁸

Answer: 4. 6.022 x 10¹⁸

Question 9. 1 mol of oxygen atom means

1. 6.022 x 10²³ number of oxygen atoms
2. 2 x 6.022 x 10²³ number of oxygen atoms
3. 1/2 x 6.022 x 10²³ number of oxygen atoms
4. 6.022 x 10²³ number of oxygen molecules

Answer: 1. 6.022 x 10²³ number of oxygen atoms

Class 9 Physical Science Chapter 4 Very Short And Long Answer Type Questions

Question 10. Number of Cl^– ions obtained from 1 mol of CaCI₂ is

1. 2
2. 6.022 X 10²³
3. 12.046 x 10²³
4. 3 x 6.022 x 10²³

Answer: 3. 12.046 x 10²³

Question 11. Unit of Avogadro’s constant is

1. Mole
2. Per mole
3. (mole)²
4. Dobson

Answer: 2. Per mole

Class 9 Physical Science Chapter 4 Very Short And Long Answer Type Questions

Question 12. Number of molecules in 1 millimol ammonia is

1. 6.022 x 10²⁶
2. 6.022 x 10²³
3. 6.022 X 10²⁰
4. 6.0222 x 10^-3

Answer: 3. 6.022 X 10²⁰

Question 13. Which of the following cannot be calculated using Avogadro’s number?

1. Boltzman constant
2. Atomic radius of solid metals
3. Velocity of light in vacuum
4. Number of molecules in a definite volume

Answer: 3. Velocity of light in vacuum

Chapter 4 Matter Concept Of Mole Topic A Concept Of Mole And Importance Of Avogadro’s Number Answer In Brief

Question 1. Who determined the value of Avogadro’s number?

Answer: Robert Millikan .

Question 2. What is the unit of molar mass?

Answer: g • mol^-1 or kg • mol^-1.

Question 3. What is the effect of temperature and pressure on Avogadro’s number?

Answer: Mass and number of molecules do not depend on either temperature or pressure. Hence, temperature or pressure has no effect on Avogadro’s number.

Question 4. What is the SI unit of quantity of matter?

Answer: The SI unit of quantity of matter is mole.

Question 5. How many H-atoms are present in 1 mol hydrogen gas?

Answer: 1 mol hydrogen gas contains 6.022 x 10²³ molecules.

Hence, number of H-atoms in 1 mol of hydrogen gas = 2 x 6.022 x 10²³ = 12.044 x 10²³.

Question 6. How many Fe-atoms does 1 gram-atom of iron indicate?

Answer: 6.022 x 10²³ atoms of iron.

Question 7. How many moles of oxygen atoms do 6.22x 10²⁴ number of oxygen atoms indicate?

Answer: 6.02 2 x 10²⁴ number of oxygen atoms = 10 x 6.022 x 10²⁴ oxygen atoms = 10 mol of oxygen atoms.

Question 8. How many grams of CO₂ does 1 gram-mole of CO₂ indicate?

Answer: 1 gram-mole of CO₂ indicates 44 g of CO₂

Question 9. How many molecules are present in 1 millimole CO₂?

Answer: 6.022 x 10²⁰ number of molecules are present in 1 millimol CO₂.

Question 10. 2 balloons contain 1.8066 x 10²³ number of hydrogen molecules and 2 mol hydrogen gas respectively. Which balloon contains higher number of molecules?

Answer: The balloon having 2 mol of hydrogen gas.

Question 11. What is the relation between the number of constituent particles of a substance, Avogadro’s number, and number of moles?

Answer: Number of moles

= \(\frac{\text { Number of constituent particles of a substances }}{\text { Avogadro’s number }}\)

Question 12. What is the total charge of 1 mol electron?

Answer: The charge of 1 mol electron is 1 Faraday.

Chapter 4 Matter Concept Of Mole Topic A Concept Of Mole And Importance Of Avogadro’s Number Fill In The Blanks

Question 1. In physical science, the 6.022 x 10²³ is known as _______ number.

Answer: Avogradro’s

Question 2. In l mol oxygen gas, the number of oxygen atoms is _______

Answer: Fundamental

Question 3. 1 mol represents Avogadro’s number of _______ particles.

Answer: 2 x 6.022 x 10²³

Question 4. The unit of Avogadro’s constant is _______

Answer: mol^-1

Question 5. 6.22x 10²³ number of H⁺ ions will be produced from __________ gram-mol of H₂SO₄.

Answer: 0.5

Question 6. When the term ‘mole’ is used, it is necessary to mention the corresponding _______ particle.

Answer: Fundamental

Question 7. 1 millimol = _______ mol.

Answer: 10

Question 8. 12.044 x 10²³ number of protons = ________ mol protons.

Answer: 2

Question 9. 1 mol sugar = ________ g sugar.

Answer: 342

Question 10. 1 mol nitrogen molecule = ________ g nitrogen.

Answer: 28

Question 11. 1 millimol oxygen molecule indicates __________ number of oxygen molecules.

Answer: 6.022 x 10²⁰

Question 12. Number of moles = mass of the substance ÷ mass of _________ of the substance.

Answer: 1 mol

Chapter 4 Matter Concept Of Mole Topic A Concept Of Mole And Importance Of Avogadro’s Number State Whether True Or False

Question 1. The value of Avogadro’s constant changes with the change in both temperature and pressure.

Answer: False

Question 2. The given mass of any substance when divided by the number of moles gives the corresponding value of molar mass.

Answer: True

Question 3. The value of Avogadro’s number was determined by Millikan.

Answer: True

Question 4. The unit of Avogadro’s constant is mol^-1.

Answer: True

Question 5. The number of atoms in 1 gram-atom of oxygen is 6.022 x 10²³

Answer: True

Question 6. 1 mol N₂ and 1 mol N signify the same amount.

Answer: False

Question 7. 1 mol of CO₂ contains 1 mol oxygen atom.

Answer: False

Question 8. Avogadro’s number is not applicable for microscopic substances.

Answer: False

Question 9. Avogadro’s number is independent of the properties of matter, it only depends upon the volume, pressure, and temperature.

Answer: False

Chapter 4 Matter Concept Of Mole Topic A Concept Of Mole And Importance Of Avogadro’s Number Numerical Examples

Question 1. P and Q flasks contain 0.5 mol oxygen and 0.4 mol ozone gas respectively. In which flask number of oxygen atoms is higher?

Answer:

Given

P and Q flasks contain 0.5 mol oxygen and 0.4 mol ozone gas respectively

Number of oxygen atoms in P flask = 2 x 0.5 x 6.022 x 10²³ = 6.022 x 10²³

Number of oxygen atoms in Q flask = 3 X 0.4 X 6.022 X 10²³ = 1.2 x 6.022 x 10²³

∴ Q flask contains higher number of oxygen atoms than that of P flask.

Question 2. How many moles of C-atom and how many moles of O-atom are present in 44 gCO₂?

Answer:

Molecular mass of carbon dioxide = 12 + 2 x 16 =44

∴ 44 g CO₂ = 1 gram-mole of CO₂

Now, 1 molecule of CO₂ contains 1 C-atom and 2 O-atoms.

∴ 1 gram-mole or 44 g of C02 contains 1 mol of C-atom and 2 mol of O atoms.

Chapter 4 Matter Concept Of Mole Topic B Gram Atom Atomic Mass Unit Gram Mole Molar Volume Of Gases And Formula Mass Synopsis

In hydrogen (¹H) scale, the relative atomic mass of an element

= \(\frac{\text { mass of one atom of the element }}{\text { mass of one hydrogen atom }}\)

In ¹²C scale, the relative atomic mass of an element

= \(\frac{\text { mass of one atom of the element }}{\text { mass of one }{ }^{12} \mathrm{C} \text {-atom } \times \frac{1}{12}}\)

1. The relative atomic mass of an element is expressed as a ratio of the masses of two atoms. Hence, it has no unit.
2. The atomic mass of an element expressed in gram is known as the gram-atomic mass of the element.
3. The actual mass of an atom is expressed in atomic mass unit.
4. Atomic mass unit (1u) = 1/12 x actual mass of one C atom = 1.6605 x 10^-24 g
5. One gram atom of an element contains Avogadro’s number of atom.
6. Atomic mass is expressed as a ratio of two masses. So it does not have any unit.
7. The actual mass of an atom is very small. So, instead of actual mass, relative atomic mass of the element is used during chemical calculations.
8. The atomic mass of an element is the average of atomic masses of all the naturally occurring isotopes of that element. Consequently, the atomic mass of most of the elements is fractional.
9. The molecular mass of a substance expressed in gram is known as its gram-molecular mass.
10. One gram-mole of a substance (element or compound) contains Avogadro’s number of molecules.
11. Number of moles of a matter = \(\frac{\text { mass of the matter }}{\text { its molar mass }}=\frac{W}{M}\)
12. At a given temperature and pressure, the volume occupied by 1 mol of a substance (element or compound) is known as its molar volume.
13. At standard temperature and pressure, the molar volume of all the gases is 22.4 L.
14. Formula mass of a compound is the sum of the atomic masses of all the atoms present in a formula unit of the compound. In case of ionic compounds, the formula of the compound does not represent the molecule of the compound, so the use of formula mass is more appropriate.

Chapter 4 Matter Concept Of Mole Topic B Gram Atom Atomic Mass Unit Gram Mole Molar Volume Of Gases And Formula Mass Short Ans Long Answer Type Questions

Question 1. Define relative atomic mass with respect to the hydrogen scale.

Answer:

Relative atomic mass with respect to the hydrogen scale

Considering the mass of one hydrogen atom as unity (1), the number of times an atom of an element is heavier than a hydrogen atom indicates the relative atomic mass of the given element.

Thus, the relative atomic mass of an element is defined as the ratio of the mass of one atom of the element to the mass of one atom of hydrogen.

Relative atomic mass of an element = \(\frac{\text { mass of one atom of the element }}{\text { mass of one hydrogen atom }}\)

Question 2. Define relative atomic mass with respect to the carbon (¹²C) scale.

Answer:

Relative atomic mass with respect to the carbon (¹²C) scale

The number of times an atom of an element is heavier than 1/12 th part of the actual mass of an atom of ¹²C isotope indicates the relative atomic mass of the given element.

Thus, the relative atomic mass of an element is defined as the ratio of mass of one atom of the element to 1/12th part of the mass of an atom of ¹²C isotope.

Relative atomic mass of an element = \(\frac{\text { mass of one atom of the element }}{\frac{1}{12} \times \text { mass of one }{ }^{12} \mathrm{C} \text {-atom }}\)

= \(\frac{\text { mass of one atom of the element }}{\text { mass of one }{ }^{12} \mathrm{C} \text {-atom }} \times 12\)

Question 3. Why is ¹²C and not hydrogen considered As the standard element to determine as the standard element to determine the relative atomic mass?

Answer:

1. Hydrogen is the lightest element. So, while calculating atomic mass with respect to hydrogen, a very small error in measurement causes a large deviation in the actual result. This does not happen when ¹²C is taken as the standard element.
2. The atomic masses calculated with respect to hydrogen scale are found to be fractional for most of the elements. However, atomic masses of most of the elements are integers with respect to the ¹²C scale.

Due to these advantages, nowadays ¹²C is considered as the standard element in the determination of relative atomic mass.

Question 4. What is gram-atomic mass?

Answer:

Gram-Atomic Mass :-

The gram-atomic mass of an element is defined as the atomic mass of the element expressed in gram. For example, atomic mass of oxygen is 16. Therefore, the gram-atomic mass of oxygen is 16 g.

Question 5. What is gram-atom?

Answer:

Gram-Atom:-

1 gram-atom of an element is defined as the amount of the element expressed in gram, which contains 6.022 x 10²³ atoms of the element. For example, 1 gram-atom of nitrogen =14 g nitrogen, because 14g nitrogen contains 6.22x 10²³ number of atoms.

Question 6. Relative atomic mass of an element has no unit. Explain

Answer:

The Relative Atomic Mass Of An Element:-

= \(\frac{\text { mass of one atom of the element }}{\frac{1}{12} \times \text { mass of one }{ }^{12} \mathrm{C} \text {-atom }}\)

As relative atomic mass of an element is a ratio of the masses of two atoms, it has no unit.

Question 7. What is atomic mass unit?

Answer:

Atomic Mass Unit:-

Atomic mass unit may be defined as the unit in which the actual mass of an atom is expressed and which is equal to 1/12 th of the actual mass of an atom of ¹²C isotope.

1 atomic mass unit or lu = 1.6605 x 10^-24 g

Question 8. Why is the atomic mass of most elements fractional?

Answer:

The Atomic Mass Of Most Elements Fractional:-

Almost all the naturally occurring elements exist as a mixture of two or more isotopes. The isotopes have the same atomic number but different mass numbers. The relative atomic mass is calculated by taking the average of the mass numbers of different isotopes of that element.

Though the mass number is a whole number, yet the average is taken on the basis of percentage abundance of the isotopes which more or less have a fixed proportion in nature. This is why the atomic masses of most of the elements are found to be fractional.

For example, chlorine has two naturally occurring isotopes, \({ }_{17}^{35} \mathrm{Cl} \text { and }{ }_{17}^{37} \mathrm{Cl}\). The percentage abundance of these two isotopes are 75% and 25% respectively.

Hence, the average atomic mass of chlorine = \(\frac{35 \times 75+37 \times 25}{100}\) = 35.5 (fractional).

Question 9. Write the differences between atomic mass and mass of an atom of an element.

Answer:

Differences Between Atomic Mass And Mass Of An Atom Of An Element:-

1. The atomic mass of an element is the ratio of mass of one atom of the element to 1/12th part of the mass of an atom of ¹²C isotope. On the other hand, the mass of an atom of an element means the actual mass of the atom.
2. Atomic mass is a ratio. So, it has no unit. However, mass of an atom represents a definite mass. Hence, it has a unit of mass.
3. Mass of an atom = atomic mass of the element x atomic mass unit (u). For example, atomic mass of oxygen is 16 but, mass of 1 oxygen atom = 16 x 1.6605 x 10^-24 g = 26.656 x 10^-24 g

Question 10. What is gram-molecule or gram-mole?

Answer:

Gram-Molecule:-

1 gram-molecule or gram-mole of a substance (element or compound) is defined as the amount of the substance expressed in gram, which contains 6.022 x 10²³ molecules of the substance. For example, 1 gram-mole of oxygen =32 g oxygen, because, 32 g oxygen contains 6.22x 10²³ number of molecules.

Question 11. What is meant by molar mass?

Answer:

Molar Mass:-

Molar mass is the mass of 1 mol of a substance. Alternatively, it is defined as the mass of Avogadro’s number of constituent particles (atoms, molecules, or ions) of the substance. For example, the gram-atomic mass of helium is 4g.

Hence, molar mass of helium = 4g • mol^-1. Again, the gram-molecular mass of water is 18 g. Hence, molar mass of H₂O = 18 g • mol^-1.

Question 12. What is gram-molar mass?

Answer:

Gram-Molar Mass:-

The gram-molar mass of a substance (element or compound) is defined as the molar mass of that substance expressed in gram. For example, molar mass of carbon dioxide is 44 g • mol^-1. Therefore, gram-molar mass of CO₂ is 44 g.

Question 13. What is molar volume?

Answer:

Molar Volume:-

At a given temperature and pressure, the volume occupied by 1 mol of any substance (element or compound) is known as its molar volume at that temperature and pressure. The molar volume of any gas at standard temperature and pressure (STP) is 22.4 L.

Question 4. What can be known from ‘1 mol CO₂‘?

Answer:

From ‘1 mol CO₂‘ it can be known that

1. Mass of the CO₂ sample is 44 g.
2. Number of CO₂ molecules in the sample is 6.22 x 10²³.
3. Volume of the CO₂ gas at STP is 22.4 L.

Question 15. What is formula mass?

Answer:

Formula Mass:-

Formula mass of an ionic compound is the sum of the atomic masses of all the atoms present in a formula unit of the compound. For example, formula mass of NaCI = 23+ 35.5 = 58.5.

Question 16. What is gram-formula mass?

Answer:

Gram-Formula Mass:-

The gram-formula mass of an ionic compound is defined as that amount of the compound expressed in gram numerically equal to its formula mass. For example, formula mass of NaCI is 58.5. Therefore, the gram-formula mass of NaCI is 58.5g.

Question 17. What is meant by gram-formula unit?

Answer:

Gram-Formula Unit:-

The amount of a substance in gram calculated from its formula unit is called 1 gram-formula unit of the substance.

Thus, 1 gram-formula unit = \(\frac{\text { given mass of the substance in gram }}{\text { gram-formula mass of the substance }}\)

For example, 58.5 g of NaCI = \(\frac{58.5g}{58.5g}\)= 1 gram-formula unit of sodium chloride.

Question 18. Find the formula mass of CaCI₂.

Answer:

Formula Mass Of CaCI₂ 

Atomic mass of calcium = 40; atomic mass of chlorine = 35.5.

Hence, formula mass of CaCI₂ = 40 + 35.5 x 2 = 111

Question 19. Use of formula mass instead of molecular mass is more appropriate in case of ionic compounds. Explain.

Answer:

Use Of Formula Mass Instead Of Molecular Mass Is More Appropriate In Case Of Ionic Compounds

For an ionic compound, there is no existence of any discrete molecule. The formula of the compound represents the ratio of different ions in the compound.

For example, although sodium chloride is expressed by the formula NaCI, yet there is no existence of individual sodium chloride molecule. Actually, the compound contains Na⁺ ions and Cl^– ions in the ratio of 1:1 which is represented by the formula.

In the crystalline form of sodium chloride, each Na⁺ ion is surrounded by 6 Cl^– ions and each Cl^– ion is further surrounded by 6 Na⁺ ions forming a network structure called crystal lattice.

Thus, in the crystal lattice of sodium chloride, there is no presence of molecules. Hence, it is not appropriate to use the term molecular mass in case of ionic compounds rather, the term formula mass should be used.

Question 20. Molecular mass and formula mass are not always the same. Explain with examples.

Answer:

Molecular Mass And Formula Mass Are Not Always The Same

Molecular mass of an element or a compound is calculated from the formula of the substance. So, molecular mass is numerically equal to the formula mass. However, the two terms are not always the same.

The term molecular mass represents the relative mass of a molecule of a substance while formula mass represents the relative mass of 1 formula unit of the substance.

For substances which exist as molecules, molecular mass and formula mass are the same. However, there are compounds which do not exist as molecules. For such compounds, the term molecular mass is not applicable.

For example, CO₂ molecule actually exists. Hence, the term molecular mass is applicable for CO₂ and both molecular mass and formula mass of CO₂ is 44. However, NaCI molecule has no separate existence.

Its formula mass is 58.5 but the term molecular mass is not applicable for NaCI because it constitutes of Na⁺ and Cl^– ions.

Question 21. What is the relation between gram- molecular mass or gram-molecule and molar volume of a gas at STP?

Answer:

Relation Between Gram- Molecular Mass Or Gram-Molecule And Molar Volume Of A Gas At STP

From Avogadro’s law it can be proved that 1 mole of any gas at STP occupies 22.4 L. On the other hand, mass of 1 mol of molecules of any substance expressed in gram is called the gram- molecular mass of that substance.

Again, 1 gram- molecule of a substance is the amount of the substance expressed in gram, which is numerically equal to its molecular mass.

Thus, it can be concluded from these three relations that, at STP, volume of 1 gram-molecule of any gaseous substance is 22.4 L.

For example, 1 gram-molecule of oxygen = 32 g.

Hence, it can be said that 32 g of oxygen at STP occupies 22.4 L.

Similarly, the gram-molecular mass of carbon dioxide is 44 g. Hence, the volume of 44 g of CO₂ at STP is 22.4 L.

Chapter 4 Matter Concept Of Mole Topic B Gram Atom Atomic Mass Unit Gram Mole Molar Volume Of Gases And Formula Mass Very Short Answer Type Questions Choose The Correct Answer

Question 1. 1 amu is equal to

1. 1.66 X 10^-24 kg
2. 1.66 x 10^-25 g
3. 0.166 x 10^-23 g
4. 1.66 x 10^-20 g

Answer: 3. 0.166 x 10^-23 g

Question 2. The mass of 1 atom of ₁₇CI³⁵ is

1. 17 u
2. 34 u
3. 35 u
4. 70 u

Answer: 3. 35 u

Question 3. 1 u is equal to

Mass of one H-atom

Mass of one ¹²C-atom

Mass of one ¹⁶O-atom

1/12 th mass of one ¹²C-atom

Answer: 4. 1/12 th mass of one ¹²C-atom

Class 9 Physical Science Chapter 4  Concept of Mole

Question 4. Molar volume of a gas at STP is

1. 1 L
2. 11.1 L
3. 22.4 L
4. 5.6 L

Answer: 3. 22.4 L

Question 5. Molar mass of sulphuric acid is

1. 98 kg •mol^-1
2. 98 g •mol^-1
3. 49 u
4. 49 mg•mol^-1

Answer: 2. 98 g-mol^-1

Question 6. For which of the following gases, the mass of 22.4 L of the gas at STP is 44 g?

1. NH₃
2. H₂S
3. CO₂
4. CH₄

Answer: 3. CO₂

Class 9 Physical Science Chapter 4  Concept of Mole

Question 7. The formula mass of MgCI₂ (Mg = 24) is

1. 95 u
2. 65 u
3. 85 u
4. 75 u

Answer: 1. 95 u

Question 8. 117g NaCI is equal to how many gram- formula mass of NaCI?

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Question 9. The compound whose formula mass is equal to 106 is

1. CaCO₃
2. CaCI₂
3. NaCI
4. Na₂CO₃

Answer: 4. Na₂CO₃

Class 9 Physical Science Chapter 4  Concept of Mole

Question 10. The statement ‘molecular mass of sodium chloride is 58.5’ is wrong because

1. It is a radioactive substance
2. It is a solid
3. NaCI molecule has no separate existence
4. None of these

Answer: 3. NaCI molecule has no separate existence

Question 11. If atomic mass of an element is ‘x’ in hydrogen scale, its atomic mass in carbon scale will be

1. x
2. 1/12 x x
3. 12x
4. 12/x

Answer: 1. x

Question 12. Atomic mass of nitrogen in carbon scale is 14, its atomic mass in hydrogen scale will be

1. 14 x 1/3
2. 14 x 1/2
3. 14
4. 14 x 2

Answer: 3. 14

Class 9 Physical Science Chapter 4  Concept of Mole

Question 13. If mass and gram-molecular mass of a compound are ‘m’ g and ‘M’ g respectively, number of gram-mole of the compound is

1. m x M
2. m/M
3. M/m
4. \(\frac{(m)^2}{M}\)

Answer: 2. m/M

Chapter 4 Matter Concept Of Mole Topic B Gram Atom Atomic Mass Unit Gram Mole Molar Volume Of Gases And Formula Mass Answer In Brief

Question 1. Which element is considered as the standard for determining the relative atomic mass of an element?

Answer: To determine the relative atomic mass of a ¹²C element, the C isotope of carbon (\({ }_6^{12} C\)) is considered as the standard.

Question 2. Find the value of 1/12 th part of the mass of  one \({ }_6^{12} C\)-atom.

Answer: The value of 1/12th part of mass of one \({ }_6^{12} C\) atom = 1.6605 x 10^-24 g.

Question 3. How many grams of H₂O does 1 gram-mole of H₂O indicate?

Answer: 1 gram-mole of H₂O indicates 18 g of water.

Question 4. 1u = how many grams?

Answer: 1u = 1.6605 x 10^-24 g.

Question 5. How are mass, molar mass and number of moles related?

Answer: Given mass (m) = number of moles (n) x molar mass (M)

Class 9 Physical Science Chapter 4  Concept of Mole

Question 6. What is the mass of one molecule of water (H₂O) in amu?

Answer: 18 u.

Question 7. Find the actual mass of one atom of \({ }_8^{16} O\).

Answer: The actual mass of one atom of \({ }_8^{16} O\) = 16u

= 16 x 1.6605 x 10^-24 g = 26.568 X 10^-24 g

Question 8. What is meant by the statement ‘molecular mass of ammonia is 17 u’?

Answer: Molecular mass of ammonia is 17u means that the actual mass of one molecule of ammonia = 17 x 1.6605 x 10^-24 g = 28.2285 x 10^-24 g

Question 9. For which compound between CaCI₂ and H₂O, the use of formula mass is appropriate?

Answer: The use of formula mass is appropriate for CaCI₂ as it is an ionic compound.

Question 10. Find the percentage of carbon (by mass) in CO₂.

Answer: Molar mass of CO₂ = 12 + 16 x 2 = 44 So, percentage of carbon (by mass) in CO₂ = 12/44 x 100 =27.27%

Class 9 Physical Science Chapter 4  Concept of Mole

Question 11. Find the actual mass of a hydrogen atom. (1u =1.6605 x 10^-24 g)

Answer: Actual mass of a hydrogen atom = 1.008u = 1.008 x 1.6605 x 10^-24 g

Question 12. Atomic mass of sodium is 23—explain.

Answer: Atomic mass of sodium is 23 means that the mass of a sodium atom is 23 times to that of 1/12 th part of the mass of a C¹² isotope.

Question 13. What is the mass of a carbon atom unified atomic mass unit?

Answer: 12 u.

Question 14. Express the value of 1 amu in kg.

Answer: Value of 1 amu in kg is 1.6605 x 10^-27 kg

Question 15. What is Avogram?

Answer: 1/12th part of the mass of a C¹² -atom i.e.; 12 1.66 x 10^-24  g is termed as 1 Avogram. 1 amu = 1 Avogram.

Question 16. Name a gaseous substance whose atomic and molecular masses are equal to each other.

Answer: Helium (He).

Question 17. In which case gram-atomic mass and gram-molecular mass are of same meaning?

Answer: For monoatomic elements;

Example: noble gases.

Question 18. If atomic mass of sodium is 23, what will be the mass of a sodium atom in ‘amu’?

Answer: 23 u or 23 amu.

Class 9 Physical Science Chapter 4  Concept of Mole

Question 19. Name two substances for which both molar mass and atomic mass are same.

Answer: Neon and potassium.

Chapter 4 Matter Concept Of Mole Topic B Gram Atom Atomic Mass Unit Gram Mole Molar Volume Of Gases And Formula Mass Fill In The Blanks

Question 1. The actual mass of one ammonia molecule is _______ u.

Answer: 17

Question 2. The actual mass of an atom of an element = atomic mass of the element x ________ g.

Answer: 1.6605 x 10^-24

Question 3. Actual mass of a \({ }_7^{14} \mathrm{~N}\)-atom is ______ u.

Answer: 14

Question 4. Actual mass of an \({ }_8^{16} \mathrm{~O}\)-atom is _____ g.

Answer: 16 x 1.6605 x 10^-24

Question 5. At STP, the volume of 17 g NH₃ is _______ L.

Answer: 22.4

Question 6. At STP, the mass of 44.8 L of CO₄ (g) is ________ g.

Answer: 88

Class 9 Physical Science Chapter 4  Concept of Mole

Question 7. Formula mass of CaCI₂ is _______ u.

Answer: 111

Question 8. ______ part of the mass of an atom of a C¹² isotope is termed as atomic mass unit.

Answer: 1/12

Chapter 4 Matter Concept Of Mole Topic B Gram Atom Atomic Mass Unit Gram Mole Molar Volume Of Gases And Formula Mass State Whether True Or False

Question 1. The relative atomic mass of an element can be defined with respect to hydrogen (¹H) and carbon (¹²C) scales.

Answer: True

Question 2. The actual mass of one molecule of CO₂ is 7.31 x 10^-23 g.

Answer: True

Question 3. Formula mass is applicable for oxygen molecule.

Answer: False

Question 4. The formula mass of NaCI is 58.5.

Answer: True

Question 5. Atomic mass unit is denoted by ‘u’.

Answer: True

Class 9 Physical Science Chapter 4  Concept of Mole

Question 6. 1u =6.022 x 10²³ g.

Answer: False

Question 7. Atomic mass of an elemernt is generally fraction.

Answer: True

Chapter 4 Matter Concept Of Mole Topic B Gram Atom Atomic Mass Unit Gram Mole Molar Volume Of Gases And Formula Mass Numerical Examples

Question 1. Express in gram-atom

1. 46 g sodium,
2. 3 g carbon.

Answer:

1. Atomic mass of sodium = 23

∴ 23 g sodium = 1 gram-atom sodium

∴ 46 g sodium = 46/23 = 2 gram-atom sodium

2. Atomic mass of carbon = 12

∴ 12 g carbon = 1 gram-atom carbon

Class 9 Physical Science Chapter 4  Concept of Mole

∴ 3g carbon = 3/12 = 0.25 gram-atom carbon

Question 2. Express in gram

1. 0.25 mol nitrogen atom,
2. 1.5 mol sodium atom.

Answer:

1. Atomic mass of nitrogen = 14

∴ 1 mol nitrogen atom = 14 g nitrogen .

∴ 0.25 mol nitrogen atom = 0.25 x 14g = 3.5 g nitrogen

2. Atomic mass of sodium = 23.

∴ 1 mol sodium atom = 23 g sodium

∴ 1.5 mol sodium atom = 1.5 x 23g = 34.5g sodium

Question 3. Express in gram-mole

1. 64 g oxygen,
2. 2.2 g carbon dioxide.

Class 9 Physical Science Chapter 4  Concept of Mole

Answer:

1. Molecular mass of oxygen = 32

∴ 32 g oxygen = 1 gram-mole oxygen

∴ 64 g oxygen = 64/32 = 2 gram-mole oxygen

2. Molecular mass of carbon dioxide (CO₂) = 12 + 2 x 16 = 44

∴ 44 g carbon dioxide = 1 gram-mole carbon dioxide

∴ 2.2 g carbon dioxide = 2.2/44 = 0.05 gram- mole carbon dioxide

Question 4. What is the mass of l gram-atom of oxygen? How many atoms are there in this quantity of oxygen?

Answer:

Given

mass of l gram-atom of oxygen

Atomic mass of oxygen = 16

∴ Mass of 1 gram-atom of oxygen = 16 g

Again, 16 g oxygen = 1 mol oxygen atom = 6.022 x 10²³ number of oxygen atoms

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Synopsis

If Avogadro’s number = N_A and gram-molecular mass = M, then

1. Mass of 1 atom of an element = \(\frac{\text { gram-atomic mass }}{N_A}\)

2. Mass of 1 molecule of any substance = \(\frac{M}{N_A}\)

3. Number of molecules in W gram of any substance = \(\frac{W \times N_A}{M}\)

4. Number of molecules in VI of gas at STP = \(=\frac{V \times N_A}{22.4}\)

5.

 

6. Number of molecules in n mol of any substance = n x N_A.

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Short And Long Answer Type Questions

Question 1. How will you determine the number of moles and number of molecules In a given mass of a substance?

Answer:

Number of moles in a given mass of a substance

(n) = \(\frac{\text { given mass of the substance in gram }}{\text { gram-molecular mass of the substance }}\)

= \(\frac{m g}{M g \cdot \mathrm{mol}^{-1}}=\frac{m}{M} \mathrm{~mol}\)

Number of molecules in a given mass of substance (N) = Number of moles x Avogadro’s number = n x N_A = m/M x N_A.

Class 9 Physical Science Chapter 4  Concept of Mole

Question 2. What is the actual mass of 1 molecule of NH₃?

Answer:

Actual mass of 1 molecule of NH₃

Molecular mass of NH₃ = 14 + 1 x 3 = 17

Hence, actual mass of one NH₃ molecule =17u

= 17 X 1.6605 X 10^-24 g = 28.2285 x 10^-24 g

Question 3. Explain whether 0.1 g of diamond will contain the same number of carbon atoms as present in 0.1 g of graphite.

Answer:

Both graphite and diamond are the allotropes of carbon.

Hence, equal amount (by mass) of both substances will contain same number of carbon atoms.

Thus, 0.1 g of diamond and 0.1 g of graphite contain equal number of carbon atoms.

Question 4. How many gram-atoms and gram-moles of oxygen are present in 32 g oxygen?

Answer:

Atomic mass of oxygen = 16.

∴ 16 g oxygen = 1 gram-atom oxygen.

∴ 32 g oxygen = 32/16 = 2 gram-atom oxygen.

Now, molecular mass of oxygen = 32.

∴ 32 g oxygen = 1 gram-mole oxygen.

Question 5. Among equal volumes of Cl₂  and O₂ at STP, which one has more mass?

Answer:

Given

Among equal volumes of Cl₂  and O₂ at STP

We know, the volume of 1 g-mole of any gaseous substance at STP is 22.4 L.

Now 1 g-mole Cl₂ = 2 X 35.5 g = 71 g Cl₂

1 g-mole O₂ = 2 x 16 g = 32 g O₂

∴ At STP, mass of 22.4 L of Cl₂ gas = 71 g and mass of 22.4 L of O₂ gas = 32 g

∴ Cl₂ gas will have more mass than O₂ at given conditions.

Class 9 Physical Science Chapter 4  Concept of Mole

Question 6. An isotope of chlorine is 35 times heavier than 1/12 th of the mass of a C¹² isotope. What is the actual mass of the Cl-leading isotope?

Answer:

Given

An isotope of chlorine is 35 times heavier than 1/12 th of the mass of a C¹² isotope.

1/12th of the mass of a C12 isotope = 1u = 1.6605 x 10^-24  g

∴ The actual mass of the chlorine isotope = 35 x 1.6605 x 10^-24 g

= 58.1175 x 10^-24 g = 5.81175 X 10^-23 g

Question 7. Prove that 1 gram-molecule of any substance contains the same number of molecules.

Answer:

Let, the molecular mass of a substance be M.

∴ Its molecule will be M times heavier than that of a hydrogen atom.

If the actual mass of a hydrogen atom be xg, then the actual mass of a molecule of that substance = M x xg.

Now, the mass of 1 gram-molecule of the substance is Mg.

∴ Number of molecules constituting 1 gram- molecule of the substance = \(\frac{M}{M \times x}=\frac{1}{x}\)

Since value of x is definite, so value of 1/x is also definite.

For example, number of molecules in 1 molecule oxygen gas

= \(\frac{32}{32 x}=\frac{1}{x} \quad\left[because M_{0_2}=32\right]\)

Similarly number of molecules in1 gram-molecule nitrogen gas.

= \(\frac{28}{28 x}=\frac{1}{x} \quad\left[because M_{N_2}=28\right]\)

Similarly, number of molecules in W g of B \(\frac{W}{M_B} \times 6.022 \times 10^{23}\)

∴ Ratio of number of molecules in Wgof A and

B = \(\frac{W}{M_A} \times 6.022 \times 10^{23}: \frac{W}{M_B} \times 6.022 \times 10^{23}\)

= \(M_B: M_A\)

Thus, the ratio of the number of molecules present in equal masses of two substances having different molecular masses is equal to the reciprocal of the ratio of their molecular masses.

Class 9 Physical Science Chapter 4  Concept of Mole

Question 9. Molecular mass and atomic element are x and y respectively. Express the mass of1 molecule and 1 atom of that element In gram.

Answer:

Given

Molecular mass and atomic element are x and y respectively.

Since atomic mass of the element is y, hence mass of 1 gram-atom = yg

∴ Total mass of N_A number of atoms = yg [N_A = Avogadro’s number]

∴ Actual mass of 1 atom = \(\frac{y}{N_A}\) g

Again, molecular mass of the element =x

∴ Mass of 1 gram-molecule of the element = xg

∴ Total mass of N_A number of molecules = xg

∴ Mass of a molecule of the element = \(\frac{x}{N_A}\)g

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Very Short Answer Type Questions Choose The Correct Answer

Question 1. Number of molecules in 36 g water is

1. 6.022 X 10²³
2. 12.044 X 10²³
3. 3.011 x 10²³
4. 9.0345 X 10²³

Answer: 2. 12.044 X 10²³

Question 2. Which of the following contains maximum number of moles?

1. 18 g water,
2. 34 g ammonia,
3. 71 g chlorine,
4. 8 g hydrogen.

Answer: 4. 8 g hydrogen.

Question 3. Which of the following contains maximum number of atoms?

1. 28g N₂,
2. 18g H₂O,
3. 17g NH₃,
4. 16g CH₄.

Answer: 4. 16gCH₄.

Class 9 Physical Science Chapter 4  Concept of Mole

Question 4. Number of H-atoms in 2 mol water is

1. 1 mol
2. 2 mol
3. 3 mol
4. 4 mol

Answer: 4. 4 mol

Question 5. Number of moles of H⁺ ions obtained from 98 g H₂SO₄ is

1. 1 mol
2. 2 mol
3. 3 mol
4. 4 mol

Answer: 2. 2 mol

Question 6. Amount of Ca(OH)₂ required to produce 1 mol of OH^– ions is

1. 74 g
2. 148 g
3. 37 g
4. 60 g

Answer: 3. 37 g

Question 7. Which of the following has a volume of 22.4 L at STP?

1. 36 g water
2. 64 g oxygen
3. 71 g chlorine
4. 4 g hydrogen

Answer: 3. 71 g chlorine

Question 8. Which of the following contains maximum number of molecules?

1. 44 g CO₂
2. 48 g O₃
3. 8 g H₂
4. 64 g SO₂

Answer: 3. 8 g H₂

Class 9 Physical Science Chapter 4  Concept of Mole

Question 9. At STP, the mass of 44.8 L of a gas is 88 g. The molecular formula of the gas is

1. N₂
2. CO₂
3. NH₃
4. O₂

Answer: 2. CO₂

Question 10. Which of the following contains maximum number of molecules?

1. 10 g O₂ (g)
2. 15 L H₂ (g) at STP
3. 5 L N₂ (g) at STP
4. 0.5 g H₂ (g)

Answer: 2. 15 L H₂ (g) at STP

Question 11. Number of atoms in 4.25 g of NH₃ is

1. 6.022 x 10²³
2. 4 x 6.022 x 10²³
3. 1.7 x 10²⁴
4. 4.25 X 6.022 x 10²³

Answer: 1. 6.022 x 10²³

Question 12. How many grams of CO₂ will contain the same number of molecules as that in 44.8 L of NH₃ at STP?

1. 44 g
2. 66 g
3. 88 g
4. 132 g

Answer: 3. 88 g

Question 13. Which of the following contains maximum number of molecules?

1. 1 g CO₂
2. 1 g H₂
3. 1 g O₂
4. 1 g CH₄

Answer: 2. 1 g H₂

Question 14. Mass of 44.8 L of a gas at STP is 92 g. Formula of the gas is

1. N₂
2. NO₂
3. NH₂
4. O₂

Answer: 2. NO₂

Class 9 Physical Science Chapter 4  Concept of Mole

Question 15. Mass of 2 mol of ammonia is

1. 12 g
2. 51 g
3. 34 g
4. 17 g

Answer: 3. 34 g

Question 16. The volume of 34 g ammonia at STP is

1. 11.2 L
2. 22.4 L
3. 44.8 L
4. 42.2 L

Answer: 3. 44.8 L

Question 17. If the mass of 6.022 x 10²⁰ atoms of an element is 0.012 g, its atomic mass will be

1. 11
2. 12
3. 13
4. 24

Answer: 2. 12

Question 18. The volume of 2 g-mol CO₂ at STP is

1. 22.4 L
2. 44.8 L
3. 11.2 L
4. 2.24L

Answer: 2. 44.8 L

Question 19. Number of carbon atoms in 1.71 g of sugar is

1. 3.6 x 10²²
2. 7.2 x 10²²
3. 6.6 x 10²²
4. None of the above

Answer: 1. 3.6 x 10²²

Question 20. Amount of carbon present in 0.5 mol potassium ferrocyanide (K₄[Fe(CN)₆]) is ®

1. 1.5 mol
2. 36 g
3. 18 g
4. 3.6 g

Answer: 2. 36 g

Class 9 Physical Science Chapter 4  Concept of Mole

Question 21. Which of the following contains least number of molecules?

1. 1 g H₂
2. 1 g N₂
3. 1.1 g O₂
4. 1.5 g O₂

Answer: 3. 1.1 g O₂

Question 22. Number of moles of oxygen atoms present in 6.022 x 10²⁴ molecules of CO is

1. 10
2. 5
3. 1
4. 0.5

Answer: 1. 10

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Answer In Brief

Question 1. Which of the given substances has maximum number of molecules 28 g N₂, 32 g O₂,18 g H₂O, and 100 g CaCO₂?

Answer: 28g N₂ = 1 mol N₂;

32g O₂ = 1 mol O₂;

18g H₂O = 1 mol H₂O;

100 g CaCO₃ = 1 mol CaCO₃

As all the substances contain the same number of moles, the number of molecules in each of the substances will be equal i.e., 6.22x 10²³.

Question 2. What is the mass of 112 mL hydrogen gas at NTP?

Answer:

Mass of 112 mL hydrogen gas at NTP

0.01 g.

Question 3. A gas jar contains 17 g NH₃ and another contains 44.8 L NH₃ at STP. Which gas jar contains more molecules?

Answer: 17g NH₃ = 17/17 mol = 1 mol NH

Again, 44.8L NH₃ at STP = \(\frac{44.8}{22.4}\) mol NH₃ = 2 mol NH₃

Hence, the second gas jar contains more number of molecules.

Class 9 Physical Science Chapter 4  Concept of Mole

Question 4. What is the volume of 7g of nitrogen at NTP?

Answer:

The volume of 7g of nitrogen at NTP

At NTP, volume of 28 g of nitrogen = 22.4 L

Hence, at NTP, volume of 7g of nitrogen = \(\frac{22.4}{28}\) X 7 L = 5.6 L.

Question 5. What is the volume of 88 g CO₂ (g) at STP?

Answer:

Volume of 88 g CO₂ (g) at STP

88gCO₂ = \(\frac{88}{44}\)mol CO₂ = 2mol CO₂

Hence, volume of 88g of CO₂ at STP = 2 X 22.4L = 44.8 L.

Question 6. Between 42g nitrogen and 64 g oxygen, which has more gram-atoms?

Answer: 42 g nitrogen = \(\frac{42}{14}\)gram-atoms of nitrogen = 3 gram-atoms of nitrogen.

64g oxygen = \(\frac{64}{16}\) gram-atoms of oxygen = 4 gram-atoms of oxygen.

Hence, 64g of oxygen contains more number of gram-atoms.

Question 7. Which of the following has maximum number of molecules at STP? 100m³ CO₂, 200cm³ NH₃, 150cm³ O₂.

Answer: At STP, 200 cm³ of NH₃ will contain the maximum number of molecules.

Question 8. How many atoms are there in 67.2L of a diatomic gas at STP?

Answer: 67.2 L of a gas at STP = \(\frac{67.2}{22.4}\) mol of the gas = 3 mol of the gas

So, number of molecules of the gas = 3 x 6.022 x 10²³

As the gas is diatomic, the number of atoms = 2 x 3 X 6.022 x 10²³ = 36.132 x 10²³

Class 9 Physical Science Chapter 4  Concept of Mole

Question 9. How many molecules are present in 109.5 g HCI?

Answer: Number of molecules in 109.5 g HCI

= \(\frac{109.5}{36.5}\) x 6.022 x 10²³ = 18.066 x 10²³ = 1.8066 X 10²⁴

Question 10. Are the masses of 1 mol sodium and 1 mol oxygen are same?

Answer: The masses of 1 mol sodium and 1 mol oxygen are not same.

Question 11. How many atoms are there in 11.5 g sodium?

Answer: Number of atoms in 11.5 g sodium

= \(\frac{11.5}{23}\) X 6.022 x 10²³ = 3.011 x 10²³

Question 12. What will be the volume of hydrogen gas at STP produced from the reaction of 23 g of Na with water?

Answer: 2Na + 2H₂O → 2NaOH + H₂ ↑

∴ Volume of hydrogen gas at STP = 11.2 L

Question 13. What is the total number of electrons present in 1 mol of water?

Answer: Number of electrons in 1 mol of water = 10 mol

= 10 x 6.022 x 10²³ = 6.022 x 10²⁴

Question 14. What is the mass of 0.5 mol CO₂ gas?

Answer: Mass of 0.5 mol of CO₂ = 44 X 0.5 g = 22 g

Class 9 Physical Science Chapter 4  Concept of Mole

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Fill In The Blanks

Question 1. Number of CO₂ molecules in 88g CO₂ is _______

Answer: 2 x 6.022 x 10²³

Question 2. 16mol protons will be obtained from ______ mol oxygen.

Answer: 1

Question 3. Number of electrons present in 14 g nitrogen is _______

Answer: 7 x 6.022 X 10²³

Question 4. Number of protons in 18 g water is _______

Answer: 10 X 6.022 X 10²³

Question 5. 1 mol of OH^– ions is obtained from ______ g NaOH.

Answer: 40

Question 6. At STP, the volume of __________ mol of any gas (elemental or compound) is 22.4 L.

Answer: 1

Question 7. The number of H-atoms in 10 mol water is _________ mol.

Answer: 20

Question 8. 98 g H₂SO₄ = _______ mol H₂SO₄.

Answer: 1

Question 9. Total mass of 3.011 x 10²³ number of oxygen atoms is ________ g.

Answer: 8

Question 10. At STP, the volume of a gas (in L) = number of moles of the gas molecules x _______ L.

Answer: 22.4

Class 9 Physical Science Chapter 4  Concept of Mole

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass State Whether True Or False

Question 1. The molar volume of 256 g of sulphur dioxide gas at STP is 89.6 L.

Answer: True

Question 2. The percentage of calcium in CaCO₃ is 60%.

Answer: False

Question 3. 2mol OH^– ions can be obtained from 110 g KOH.

Answer: False

Question 4. The number of atoms present in 3.2 g CH₄ is 6.022 x 10²³.

Answer: True

Question 5. The number of chlorine atoms in 71 g HCI is 0.5 x 6.022 x 10²³.

Answer: False

Question 6. 022 x 10²³ number of molecules are present in 36 g of water.

Answer: False

Question 7. Volume of 32 g SO₂ at STP is 11.2 L.

Answer: True

Question 8. 4 mol CO₂ = 88 g CO₂.

Answer: False

Question 9. Number of molecules present in 2 millimol of chlorine gas is 2 x 6.022 x 10²⁰.

Answer: True

Question 10. The volume of 64 g oxygen at STP is 10 L.

Answer: False

Question 11. 44 g of CO₂ contains 6.022 x 10²³ number of molecules.

Answer: True

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Numerical Examples

Question 1. Find the mass of 1 molecule of CO₂.

Answer:

Molar mass of carbon dioxide = 12 + 2 x 16 = 44

Now, 1 mol of CO₂ contains 6.022 x 10²³ molecules.

Hence, mass of 6.022 x 10²³ CO₂ molecules = 44 g

∴ Mass of one CO₂ molecule = \(\frac{44}{6.022 \times 10^{23}} g\) = 7.3 x 10²³ g

Question 2. Between 100 g calcium and 100 g iron, which one will contain more number of atoms? (Given: atomic masses of calcium and iron are 40 u and 56 u respectively)

Answer:

Number of moles in 100 g calcium = \(\frac{\text { given mass }}{\text { gram-atomic mass }}=\frac{100 \mathrm{~g}}{40 \mathrm{~g}}=2.5\)

∴ Number of atoms in 100 g calcium = 2.5 x 6.022 x 10²³ = 1.5055 X 10²⁴

Again, number of moles in 100 g iron

= \(\frac{\text { given mass }}{\text { gram-atomic mass }}=\frac{100 \mathrm{~g}}{56 \mathrm{~g}}=1.78\)

∴ Number of atoms in 100 g iron = 1.78 X 6.022 X 10²³ = 1.07 x 10²⁴

Therefore, 100 g calcium contains more number of atoms than 100 g iron.

Question 3. How many grams of H₂SO₄ are required to produce 1 gram-ion of H⁺?

Answer:

\(\mathrm{H}_2 \mathrm{SO}_4(a q) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+\mathrm{SO}_4^{2-}(a q)\)

 

1 molecule of H2S04 gives two H⁺ ions in aqueous solution.

Hence, 1 gram-mole of H₂SO₄  gives 2 gram-ions of H⁺ ion.

∴ 1 gram-ion of H⁺ ion will be produced from 1/2 gram-mole of H₂SO₄= 98/2 = 49g H₂SO₄ [as 1 gram-mole of H₂SO₄ = 98 g]

Thus, 49 g H₂SO₄ is required to produce 1 gram- ion of H⁺ ions.

Question 4. Which of the following has maximum number of molecules? 0.5 mol H₂(g), 10g O₂(g) and 15 L Cl₂ at STP.

Answer:

Number of molecules present in a substance is proportional to its number of moles.

Amount of H₂ = 0.5 mol

\(10 \mathrm{~g} \mathrm{O}_2 =\frac{\text { given mass }}{\text { gram-molecular mass }}=\frac{10 \mathrm{~g}}{32 \mathrm{~g}}=0.31 \mathrm{~mol} \mathrm{O}_2\)

 

22.4 L Cl₂ at STP = 1 mol Cl₂

∴ 15 L Cl₂ = \(\frac{15L}{22.4L}\) = 0.67 mol Cl₂

Hence, 15 L Cl₂ at STP will have maximum number of molecules.

Question 5. Arrange the gases in increasing number of atoms present in given amounts

1. 18gH₂O,
2. 18gCO₂
3. 18gCO₂,
4. 18gCH₄.

Answer:

1. 18gH₂O = 18/18 = 1 mol H₂O = 6.022 x 10²³ molecules of H₂O

As 1 molecule of H₂O have 3 atoms, number of atoms = 3 x 6.022 x 10²³

2. 18 g O₂= 18/32 =0.5625 mol O₂ = 0.5625 x 6.022 x 10²³ molecules of O₂.

As 1 molecule of O₂ have 2 atoms, number of atoms = 2 X 0.5625 X 6.022 X 10²³ = 1.125 x 6.022 x 10²³

3. 18 g CO₂ = 18/44 = 0.409 mol CO₂ = 0.409 x 6.022 x 10²³ molecules of CO₂

As 1 molecule of CO₂ have 3 atoms, number of atoms = 3 x 0.409 x 6.022 x 10²³ = 1.227 x 6.022 x 10²³

4. 18 g CH₄ = 18/16 = 1.125 mol CH₄ = 1.125 x 6.022 x 10²³ molecules of CH₄

As 1 molecule of CH₄ has 5 atoms, number of atoms = 5 X 1.125 x 6.022 x 10²³ = 5.625 X 6.022 X 10²³

Hence, correct order is 18g O₂ < 18g CO₂ < 18g HO₂ < 18g CH₄

Question 6. 3.42 g sucrose is dissolved in 18 g water. Find the number of oxygen atoms in the formed solution.

Answer:

Molecular mass of sucrose (C₁₂H₂₂O₁₁) = 12 x 12 + 1 x 22 + 16 x 11 = 342

∴ 3.42 g sucrose = 3\(\frac{3.42}{342}\) = 0.01 mol sucrose = 0.01 x 6.022 x 1023 molecules of sucrose

As 1 molecule of sucrose has 11 oxygen atoms, number of O-atoms in 3.42 g sucrose = 11 X 0.01 x 6.022 x 10²³

Molecular mass of water (H₂O) = 1 x 2 + 16 =18

∴ 18 g water = 1 mol water = 6.022 x 10²³ molecules of water.

As each molecule of water contains 1 O-atom, number of O-atoms in 18 g water = 6.022 x 10²³

∴ Total number of O-atoms in formed solution = 11 x 0.01 x 6.022 X 10²³ + 1 x 6.022 x 10²³ = 1.11 X 6.022 x 10²³ = 6.68 X 10²³

Question 7. Calculate the number of H-atoms present in 9 g water.

Answer:

Molecular mass of H₂O = 1 x 2 + 16 = 18

Hence, 9 g water = 9/18 = 0.5 mol water = 0.5 x 6.022 x 10²³ molecules of water

Each molecule of water contains two H-atoms.

Hence, number of H-atoms in 9 g water = 2 x 0.5 x 6.022 x 10²³ = 6.022 x 10²³

Question 8. State whether number of atoms in

1. 1 g oxygen atom,
2. 1 g oxygen molecule and
3. 1 g ozone molecule will be same or different? Justify your answer.

Answer:

1. 16 g O-atom = N_A no. of O-atoms

∴ 1 g O-atom = \(\frac{N_A}{16}\) no. of O-atoms

2. 32 g O₂ molecule = N_A no. of O₂ molecules

∴ 1 g O₂ molecule = \(\frac{N_A}{32}\)no. of O₂ molecules

= \(\frac{N_A}{32}\) x 2 no. of O-atoms = \(\frac{N_A}{16}\)no. of O-atoms.

3. 48 g O₃ molecule = N_A no. of O₃ molecules

∴ 1 g O₃ molecule = \(\frac{N_A}{48}\) no. of O₃ molecules

= \(\frac{N_A}{48}\) x 3 no. of O-atoms = \(\frac{N_A}{16}\) no. of O-atoms.

Hence, number of atoms in each of the substances will be equal.

Question 9. Can you drink Avogadro’s number of water molecules? (Given: Density of water = 1g • m^-1)

Answer:

Molecular mass of water = 2 x 1 + 16 = 18 Gram-molecular mass of water = 18 g and molar mass of water = 18 g • mol^-1

Thus, the mass of Avogadro’s number (6.022 x 10²³) of water molecules = 18 g

Density of water = 1g • mL^-1.

Hence, volume of 18 g water = 18 x 1 = 18mL

Thus, one can easily drink 18 mL water (approximately one test tube of water) or Avogadro’s number (6.022 x 10²³) of water molecules.

Question 10. What is the volume of 4.4 g CO₂ at STP?

Answer:

Gram-molecular mass of CO₂ = 12 + 16 x 2 = 44 g

Hence, at STP, volume of 44 g of CO₂ = 22.4 L

∴ Volume of 4.4 g of CO₂ at STP = \(\frac{22.4}{44}\) x 4.4L = 2.24 L

Question 11. At STP, the volume of 0.44 g of a gas is 224 cm³. Find the molecular mass of the gas.

Answer:

At STP, mass of 224 cm³ of the gas = 0.44 g

∴ At STP, the mass of 22.4L or 22400 cm³ of the gas = \(\frac{0.44}{224}\) x 22400g = 44g

Hence, the molecular mass of the gas = 44

Question 12. What is the volume of 8 g of oxygen at standard temperature and pressure?

Answer:

Molecular mass of oxygen = 32

∴ 1 gram-mole oxygen = 32 g oxygen

Hence, at STP, volume of 32 g oxygen is 22.4 L.

∴ At STP, volume of 8 g oxygen = \(\frac{22.4}{32}\) x 8L = 5.6 L

Question 13. Find the mass (in gram) of 5.6 L ammonia at STP.

Answer:

Gram-molecular mass of ammonia (NH₃) = 14+ 1×3 = 17 g

Hence, at STP, the mass of 22.4 L of ammonia = 17 g

∴ At STP, the mass of 5.6 L of ammonia = \(\frac{17}{22.4}\) x 5.6 g = 4.25 g

Question 14. Find the volume of 4 g SO₂ at standard temperature and pressure.

Answer:

Molecular mass of SO₂ = 32 + 16 x 2 = 64

Hence, at STP, the volume of 64 g SO₂ = 22.4L

∴ At STP, the volume of 4 g SO₂ = \(\frac{22.4}{64}\) x 4L= 1.4 L

Question 15. What is the mass of 4 gram-moles of oxygen? What will be the volume of this quantity of oxygen at STP?

Answer:

Molecular mass of oxygen = 32

Hence, 1 gram-mole of oxygen = 32 g oxygen

∴ 4 gram-moles of oxygen = 4 x 32 g = 128 g oxygen

Volume of 1 gram-mole of oxygen at STP = 22.4 L

Therefore, the volume of 4 gram-moles of oxygen at STP = 4 X 22.4L = 89.6L

Question 16. State whether, at STP, both 22.4 L NH₃ and 22.4 L CO₂ will contain

1. same number of molecules?
2. same number of atoms?

Answer:

1. At STP, 22.4 L NH₃ = 1 mol NH₃ Now, 1 mol NH₃ = 6.022 x 10²³ molecules NH₃

At STP, 22.4 L carbon dioxide = 1 mol carbon dioxide = 6.022 x 10²³ molecules of CO₂

Hence, both NH₃ and carbon dioxide contain same number of molecules.

2. 1 molecule of NH₃ contains 4 atoms.

Hence, number of atoms in 22.4 L NH₃ at STP = 4 X 6.022 X 10²³ = 24.088 X 10²³

Again, 1 molecule of CO₂ contains 3 atoms.

Hence, number of atoms in 22.4 L CO₂ at STP = 3 X 6.022 X 10²³ = 18.066 X 10²³

Therefore, 22.4 L NH₃ will contain more atoms than that in 22.4 L CO₂.

Question 17. Find the number of constituent ions in 11.1 g calcium chloride.

Answer:

Gram-formula mass of calcium chloride (CaCI₂) = 40 + 35.5 X 2 = 111 g

∴ Number of formula units in 111 g CaCl₂ = 6.022 x 10²³

∴ Number of formula units in 11.1 g CaCI₂

= \(\frac{6.022 \times 10^{23}}{111}\) x 11.1 = 6.022 x 10²²

Each formula unit of calcium chloride contains Ca²⁺ ions and 2CI^– ions.

Hence, number of Ca²⁺ ions in 11.1 g CaCI₂ = 1 x 6.022 X 10²² = 6.022 x 10²²

Number of Cl^– ions in 11.1 g CaCI₂ = 2 X 6.022 X 10²² = 12.044 x 10²²

Question 18. What is the gram-formula mass of NaCI? Calculate the number of gram-formula units present in 234 g of NaCI.

Answer:

Atomic masses of sodium and chlorine are 23 and 35.5 respectively.

Hence, formula mass of NaCI = 23 + 35.5 = 58.5

∴ Gram-formula mass of NaCI = 58.5g

Number of gram-formula units in 234 g NaCI = \(\frac{234 \mathrm{~g}}{58.5 \mathrm{~g}}\) = 4

Question 19. A mixture of hydrogen and oxygen contains 20% hydrogen by weight. Calculate total number of molecules present per gram of the mixture.

Answer:

Given

A mixture of hydrogen and oxygen contains 20% hydrogen by weight.

Mass of hydrogen in 1 g of the mixture = \(\frac{1 \times 20}{100} \mathrm{~g}\) 0.2 g

∴ Mass of oxygen in 1 g of mixture = (1 – 0.2) g = 0.8 g

Now, number of molecules of hydrogen in 0.2 g of hydrogen gas

= \(\frac{0.2}{2}\) X 6.022 X 10²³ = 6.022 x 10²³

Number of molecules of oxygen in 0.8 g of oxygen gas

= \(\frac{0.8}{32}\) x 6.022 x 10²³ = 1.5055 x 10²²

∴ Total number of molecules per gram of the mixture

= (6.022 + 1.5055) x 10²² = 7.5275 x 10²²

Question 20. Calculate the total charge in 6 g of \(\mathrm{CO}_3^{2-}\) ion.

Answer:

Formula mass of \(\mathrm{CO}_3^{2-}\) ion = (12 + 3 x 16) = 60

∴ Number of moles of \(\mathrm{CO}_3^{2-}\) ion = \({6}{60}\) = 0.1

Now, total negative charge of a \(\mathrm{CO}_3^{2-}\) ion is equal to the negative charge of 2 electrons.

Again, total charge of 1 mol electron = 96500 C

Total charge of 6 g or 0.1 mol of \(\mathrm{CO}_3^{2-}\) ion = 2 X 0.1 X 96500 C = 19300 C

Question 21. Calculate the number of Na^⊕ and Cl^Θ ions in 117 g of sodium chloride. [Na = 23, Cl = 35.5]

Answer:

Formula mass of sodium chloride = (23+ 35.5) = 58.5 117

∴ 117 g of NaCI = \({117}{58.5}\)= 2gram-formula mass of NaCI.

Now in NaCI, number of Na^⊕ ions = number of Cl^Θ ions

∴ Number of Na® ions in 2 gram-formula mass of NaCI = number of Cl^Θ ions in 2 gram-formula mass of NaCI

= 2 x 6.022 x 10²³ = 12.044 x 10²³ = 1.2044 X 10²⁴

Question 22. You have been given a glass of water mixed with 10 g glucose (C₆H₁₂O₆). If you drink the entire solution, calculate the number of glucose molecules you have consumed.

Answer:

Molecular mass of glucose = 6 x 12 + 12 x 1 + 6 x 16 = 180

∴ Number of moles of glucose in 10 g = \(\frac{10}{180}\)= 0.0555

∴ Number of molecules of glucose in 10 g = 0.0555 X 6.022 X 10²³ = 3.345 x 10²²

∴ 3.345 x 10²² number of glucose molecules will be consumed.

Question 23. Calculate the number of electrons, protons and neutrons in 1 mol O^2- ion.

Answer:

Number of electrons in an O^2- ion = (8 + 2) = 10

Number of protons in an O^2- ion = 8

∴ Number of neutrons in an O^2- ion = (16 – 8) = 8 (assuming ₈O¹⁶16 isotope)

Number of electrons in 1 mol O^2- ion = 1 x 10 x 6.022 x 10²³ = 6.022 x 10²⁴

∴ Number of protons in 1 mol O^2- ion = 1 X 8 x 6.022 X 10²³ = 4.817 X 10²⁴

∴ Number of neutrons in 1 mol O^2- ion = 1 x 8 x 6.022 x 10²³ = 4.817 x 10²⁴

Question 24. Calculate the number of atoms and molecules in 124 g phosphorus.

Answer:

Atomic mass of phosphorus = 31

∴ 1 g-atom phosphorus =31 g phosphorus

Molecular mass of phosphorus =31 x 4 = 124

∴ 1 g-molecule phosphorus = 124 g phosphorus

∴ 124 g phosphorus =  \(\frac{124}{31}\)g-atom phosphorus

= 4 g-atom phosphorus

= 4 x 6.022 x 10²³ atoms of phosphorus

= 2.4088 x 10²⁴ atoms of phosphorus 124 g phosphorus

= \(\frac{124}{124}\) g-molecule phosphorus

= 1 x 6.022 x 10²³ molecules of phosphorus = 6.022 x 10²³ molecules of phosphorus

Question 25. Calculate the number of hydrogen atoms in 1 L of water (at 4°C temperature).

Answer:

Density of water at 4°C = 1 g • mL^-1

∴ Mass of 1 L or 1000 mL of water at 4°C = 1000 g

Now molecular mass of water = 1 x 2 + 16 = 18

∴ Number of molecules in 1000 g of water = \(\frac{6.022 \times 10^{23}}{18} \times 1000\) = 3.345 x 10²⁵

Again, number of H-atoms in 1 molecule of water = 2

∴ Number of H-atoms in 1 L water = 2 X 3.345 X 10²⁵ = 6.69 X 10²⁵

Question 26. Calculate the number of hydrogen atoms present in 51 g ammonia gas. What will be the volume of that amount of gas at STP?

Answer:

Gram-molecular mass of NH₃ = 17 g

∴ 51 g ammonia = \(\frac{51}{17}\) mol = 3 mol ammonia

Now, the number of H-atoms per molecule of ammonia= 3

∴ Number of H-atoms in 51 g or 3 mol ammonia gas = 3 x 3 x 6.022 x 10²³ = 5.4198 x 10²⁴

The volume of 51 g or 3 mol of ammonia gas at STP = 3 X 22.4 L = 67.2 L.

Question 27. Which one is heavier between 1 mol NO molecule and 0.5 mol NO₂ molecule?

Answer:

Molecular mass of NO = 14 + 16 = 30

∴ Mass of 1 mol NO molecules = 30 g

Again, molecular mass of NO₂ = 14 + 16 X 2 = 46

∴ Mass of 0.5 mol NO₂ molecules = 46 X 0.5 = 23 g

∴ 1 mol NO molecules will be heavier than that of 0.5 mol NO₂ molecules.

Question 28. Calculate the total number of atoms in 80 u helium.

Answer:

Mass of a helium atom = 4 u

∴ Number of atoms in 80 u helium = \(\frac{80u}{4u}\) = 20.

Question 29. Calculate the number of molecules left when 10²¹ molecules of CO₂are removed from 200 mg of CO₂•

Answer:

200 mg of CO₂ = 0.2 g of CO₂

Now, 44g of CO₂₃ contains 6.022 x 10²³ molecules

∴ 0.2 g of CO₂ contains

⇒ \(\frac{6.022 \times 10^{23} \times 0.2}{44}=2.7372 \times 10^{21}\) molecules

∴ On removing 10²¹ molecules, number of CO₂ molecules left

= 2.7372 X 10²¹ – 10²¹ = 1.7372 X 10²¹

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Miscellaneous Type Questions

Match The Columns

1.

Answer: 1. B, 2. A, 3. D, 4. C

2.

Answer: 1. D, 2. A, 3. B, 4. C

Chapter 4 Matter Solution Topic A True Solution Colloid And Suspension Dissolution Of Smaller Ion Or Molecules And Larger Molecules In Water Synopsis

1. A solution is a homogeneous mixture of two or more substances in which the amount of each constituent can be varied within a certain limit.
2. The component of a solution which is present in greater amount is called the solvent. The solution has the same physical state as that of the solvent.
3. On the other hand, the component of the solution which is present in lesser amount is called the solute.
4. A colloidal solution is a stable heterogeneous system of two immiscible phases in which one phase (solid, liquid or gas) is dispersed as particles with diameter ranging from 10^-7-10^-5 cm, in another phase (solid, liquid or gas).
5. Colloid is not a special type of substance, rather it is a state of the substance.
6. The medium in which the colloidal particles remain uniformly dispersed is known as the dispersion medium and the substance whose particles remain dispersed in a colloidal solution is called the dispersed phase.
7. When light rays are passed through a colloidal solution, the rays get scattered by the colloidal particles and as a result tye path of light becomes clearly visible. This phenomenon is known as Tyndai. effect.
8. Tyndall effect helps to distinuguish between true solutions and colloidal solutions.
9. On the basis of the physics states of the re dispersion medium and dispersed phase, colloids areclassified into eight groups.
10. Colloids are classified into eight groups. They are sol, solid, sol, gel, emulsion, solid aerosol, liquid aerosol, solid foam and foam.
11. An emulsion is a colloidal solution in which both the dispersed phase and dispersion medium are liquid.
12. Emulsions are of two types: oil-in-water type and water-in-oil type emulsion.
13. The chemical substances used to enhanse the stability of an emulsion are called emulsifiers. Soaps and detergents are well-known emulsifiers.
14. A suspension is a heterogeneous and unstable system in which particles having diameter greater than 10^-5 cm remain dispersed in the solvent. On standing, the particles slowly separate out from the mixture and settle at the bottom.

Chapter 4 Matter Solution Topic A True Solution Colloid And Suspension Dissolution Of Smaller Ion Or Molecules And Larger Molecules In Water Short And Long Answer Type Questions

Question 1. What is a solution?

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

Answer:

Solution:-

If a homogeneous mixture of two or more substances (solid, liquid or gas) have uniform properties (in terms of constituents and structure) throughout the mixture and the amounts of the constituents can be varied within certain limits, then the homogeneous mixture is said to be a solution.

Question 2. What do you mean by solvent and solute?

Answer:

Solvent And Solute:-

Solvent: The component of a solution which is generally present in greater amount and whose physical state is the same as that of the solution is known as the solvent.

Solute: The component of a solution which is present in lesser amount and which remains dissolved in another solid, liquid or gaseous substance to form a homogeneous mixture (solution) is called a solute. For example, sugar dissolves in water to form a homogeneous mixture. Here, sugar is the solute.

In some solutions, where both solvent and solute are in the same phase, the terms solvent and solute is defined with respect to their relative quantities in the solution.

For example, when 70 parts of alcohol mix with 30 parts of water, alcohol is considered as the solvent and water as the solute. On the other hand, in a mixture of 70 parts of water and 30 parts of alcohol, water is the solvent and alcohol is the solute.

Question 3. All solutions are mixtures but all mixtures are not solutions. Justify.

Answer:

All Solutions Are Mixtures But All Mixtures Are Not Solutions:-

If a mixture of two or more substances is homogeneous in nature and their composition can be varied only within a certain limit, then the mixture is called a solution. For example, common salt dissolved in water forms a solution.

However, a mixture may also be heterogeneous in nature, in which the composition is not uniform throughout and can be changed in any proportion. This type of mixtures cannot be termed as solution. For example, a mixture of sand and sugar cannot be called a solution.

Thus, all solutions are mixtures but all mixtures are not solutions.

Question 4. Define homogeneous and heterogeneous mixtures.

Answer:

Homogeneous And Heterogeneous Mixtures:-

Homogeneous Mixtures: A homogeneous mixture is a solid, liquid or gaseous mixture that has the same proportions of its components throughout any given sample.

Example: air or sugar solution.

Heterogeneous Mixture: A heterogeneous mixture is simply a mixture that is not uniform in composition. It has components in which proportions vary throughout the sample.

Example: mixture of sand and sulphur powder.

Question 5. Why is an aqueous solution of sugar called a true solution?

Answer:

An Aqueous Solution Of Sugar Called A True Solution:-

When sugar is added to water, the sugar molecules occupy the intermolecular spaces between the water molecules and dissolve in it. As a result, the sugar molecules form a homogeneous mixture with water and cannot be separated easily. This is why an aqueous solution of sugar is called a true solution.

Question 6. Discuss the properties of a true solution.

Answer:

The Properties Of A True Solution:-

1. A true solution possesses the following properties
2. A true solution is a homogeneous mixture of solvent and solute. It has uniform properties (in terms of components, physical properties and structural features) throughout the solution.
3. The solute particles present in a true solution are so small (particle diameter is less than or equal to 10^-8 cm) that they cannot be seen even under an ultramicroscope.
4. In a true solution, the solvent and solute may exhibit some changes in their physical properties, but their chemical properties remain unchanged.
5. The relative amount of the components in a solution can be increased or decreased within a certain limit.
6. The solute and solvent in a true solution cannot be separated by filtration or by gravitational separation. Even if the solution is kept undisturbed for a long period of time, the solute does not settle down.
7. The components of a true solution can be separated by physical methods such as, evaporation, distillation or crystallisation.
8. Heat may or may not be evolved or absorbed during formation of a true solution.

Question 7. What is a colloidal solution?

Answer:

Colloidal Solution:-

A colloidal solution is said to be a stable heterogeneous system of two immiscible phases in which one phase (solid, liquid or gas) comprises of the particles with diameter ranging from 10^-7 – 10^-5 cm, is dispersed into another phase (solid, liquid or gas).

For example, freshly precipitated ferric hydroxide when shaken with water and small amount of ferric chloride, it forms a colloidal solution.

Question 8. What is meant by dispersion medium and dispersed phase of a colloidal solution?

Answer:

Dispersion Medium: The medium in which the colloidal particles remain uniformly dispersed is called the dispersion medium.

Dispersed Phase: The component of a colloidal solution which remain uniformly dispersed in the dispersion medium and consists of particles with diameter ranging from 10~7-10-5 cm is called the dispersed phase.

Example: In a gold sol, water is the dispersion medium while gold particles form the dispersed phase.

Question 9. State whether colloidal particles can be separated from a solution by filtration using a filter paper.

Answer:

1. The pores of the filter paper are generally larger than 10^-5 cm. So, the colloidal particles having diameter of 10^-7 to 10^-5 cm can easily pass through a filter paper. So, colloidal particles cannot be separated from a solution using a filter paper.
2. On the other hand, colloidal particles cannot pass through a parchment paper. So, parchment paper is effective in separating colloidal particles from a solution.

Question 10. State the imporatnt properties of a colloidal solution.

Answer:

The Important Properties Of A Colloidal Solution:-

A colloidal solution has the following properties

1. Heterogeneity:

Colloidal solutions are heterogeneous in nature. The dispersed phase neither dissolves completely in the dispersion medium nor does it separate out from the dispersion medium. The colloidal particles remain dispersed in the dispersion medium.

2. Tyndall Effect:

When a beam of light is allowed to pass through a colloidal solution, scattering of light by the colloidal particles occurs and the path of the beam through the solution gets illuminated and clearly visible. This phenomenon is called Tyndall effect.

3. Brownian Motion:

The colloidal particles in the dispersion medium are in continuous random motion moving in zigzag paths. This motion is known as Brownian motion.

4. Electrophoresis:

The colloidal particles are charged particles. If electricity is passed through a colloidal solution, then the colloidal particles move towards the oppositely charged electrodes. This movement of colloidal particles towards a specific electrode, under the influence of an electric field is called electrophoresis.

5. Passage Through Filter Paper And Parchment Paper:

Colloidal particles can pass through filter paper but cannot pass through parchment paper.

Question 11. What are the diameters of the solute particles in true solutions, colloidal solutions and suspensions?

Answer:

In true solutions, diameter of solute particles is less than or equal to 10^-8 cm or 0.1 nm.

In colloidal solutions, the diameter of solute particles ranges from 10^-7 -10^-5 cm or 1-100 nm. In suspensions, the diameter of solute particles is greater than 10^-5 cm or 100 nm.

Lyophilic And Lyophobic Sol:

If the dispersed phase of a sol has high affinity for the molecules of dispersion medium, then it is called a lyophilic sol. Sols of starch, gum, gelatin, glue etc. are examples of such sol. If the dispersion medium of such sols is water, then these are called hydrophilic sols.

If the dispersed phase of a sol has a very little or almost no affinity for the molecules of the dispersion medium, then it is called a lyophobic sol. Some common examples are ferric hydroxide [Fe(OH)₃] sol, arsenius sulphide [As₂S₃] sol, etc. If water is the dispersion medium, then such sols are called hydrophobic sols.

Question 12. Discuss the significance of Brownian motion.

Answer:

The Significance Of Brownian Motion Are As Follows:-

1. Brownian motion is an inherent evidence of the incessant motion of particles in a solution.
2. The value of Avogadro’s number can be determined by measuring Brownian motion.
3. Due to Brownian motion, the colloidal particles are in continuous motion. This motion prevents the particles to settle down even under gravitational force. Thus, Brownian motion plays a very significant role in stabilising a colloidal solution.

Question 13. What is a suspension?

Answer:

Suspension:-

A heterogeneous and unstable system in which particles of a substance (usually a solid) of diameter greater than 10^-5 cm remain suspended in another substance (usually a liquid) is called a suspension. On standing, the particles slowly separate out from the mixture and settle down at the bottom of the vessel.

Example:

When finely powdered barium sulphate (BaSO₄) is shaken vigorously with water, it is observed that the particles get evenly distributed throughout the solvent and remain suspended in water to form a suspension. If the beaker is kept undisturbed for some time, then barium sulphate particles begin to settle down at the bottom of the beaker.

Question 14. State the properties of a suspension.

Answer:

Properties Of A Suspension:-

A suspension has the following properties

1. A suspension is a heterogeneous mixture.
2. When kept undisturbed for some time, the suspended particles gradually settle down under gravity.
3. The solution is turbid and hence light cannot pass through a suspension. However, a suspension may sometimes exhibit the Tyndall effect.
4. The suspended particles are visible under ordinary microscopes and sometimes even to the naked eyes.
5. The suspended particles cannot pass through a parchment paper or a filter paper.

Question 15. Colloidal particles do not settle down at the bottom of the beaker, but particles of suspensions do. Explain with reason.

Answer:

Colloidal particles do not settle down at the bottom of the beaker, but particles of suspensions do.

Due to Brownian motion of the colloidal particles, they are always in a continuous random motion. The possibility of aggregation of colloidal particles is thus very low. So, colloidal particles do not settle down at the bottom of the beaker.

On the other hand, particles of a suspension do not exhibit Brownian motion. Hence, on standing, the particles come closer to each other and aggregate to form larger particles which eventually settle down under gravity.

Question 16. Explain why turbid water collected from ponds or rivers do not become clear on prolong standing.

Answer:

Mud particles act as colloid particles in the turbid water of ponds or rivers. These colloid particles can not be precipitated out naturally. That is why those water do not become clear on prolong standing.

Question 17. Compare the properties of a true solution, a colloidal solution and a suspension.

Answer:

The properties of a true solution, a colloidal solution and a suspension are compared in the following table

Question 18. What is the Tyndall effect?

Answer:

Tyndall Effect:-

When a beam of light is allowed to pass through a colloidal solution, scattering of light by the colloidal particles occurs and the illuminated path of light can be visualised from a direction at right angles to the incident light. This phenomenon is called Tyndall effect and it has been named after the Irish physicist John Tyndall.

Question 19. Mention some practical applications of Tyndall effect.

Answer:

Some Practical Applications Of Tyndall Effect:-

1. Ultramicroscope was developed on the basis of Tyndall effect. It is used to determine the size of particles in aerosols by using the principle of Tyndall effect.
2. Tyndall effect is used to determine the presence of particulate matter (which destroys the efficiency of catalyst) in air during the reaction of sulphur dioxide and air in the preparation of sulphuric acid by contact process.

Question 20. Cigarette smoke sometimes light blue in colour. Why?

Answer:

Cigarette Smoke Sometimes Light Blue In Colour Because:-

Cigarette smoke is a colloidal system which belongs to the category of solid aerosol. It consists of fine particles of carbon dispersed in air (dispersion medium).

Due to its colloidal nature, it can scatter light beam passing through it, which is known as Tyndall effect. As blue light is scattered to the maximum extent, cigarette smoke sometimes appears light blue in color.

Question 21. Classify colloids on the basis of the physical states of dispersed phase and dispersion medium and give examples for each category.

Answer:

On the basis of the physical states of the dispersed phase and dispersion medium, colloids can be classified into eight categories as described below

Question 22. Why do colloidal solutions formed by different methods exhibit different colours?

Answer:

The colour of a colloidal solution depends on the wavelength of visible light scattered by the colloid particles. The scattering of light, on the other hand, depends on the size of colloidal particles in the solution.

The size of colloidal particles formed is different for different methods. So, a colloidal solution exhibits different colours when formed by different methods.

Question 23. What is a sol? Give example.

Answer:

Sol:-

Colloidal solutions in which the dispersed phase is a solid while the dispersion medium is a liquid is called a sol.

Example: Gold sol, sulphur sol, arsenius sulphide sol etc. In each of these sols, gold, sulphur and arsenius sulphide particles respectively form the dispersed phase while water acts as the dispersibn medium.

Question 24. What is an aerosol? How are they categorised?

Answer:

Aerosol:-

An aerosol is a colloidal system in which the dispersion medium is a gas or air. p If the dispersed phase in an aerosol is solid, then it is called solid aerosol such as smoke, dust particles floating in air etc. On the other hand, if the dispersed phase of an aerosol is liquid, then it is called liquid aerosol such as fog, cloud etc.

Question 25. What are alcohol and hydrosol?

Answer:

Alcohol And Hydrosol:-

Colloids with alcohol as dispersion medium and having solids as dispersed phase are termed as alcohols.

Again, colloids whose dispersed phases and dispersion medium are solid and water respectively are termed as hydrosol.

Question 26. State some important applications of emulsions.

Answer:

Some Important Applications Of Emulsions Are Given Below:-

1. Milk, butter, margarine, vanishing cream, and cold cream used in our everyday life are all emulsions.
2. Cleansing action of soap is due to the emulsification of grease with water. Water and soap together forms a colloidal solution and removes grease along with dirt and dust from clothes by forming emulsion.
3. A wide variety of medicines like cod liver oil, and vitamin B-complex are oil-in-water type emulsions. These are easily absorbed by our digestive system and provide quick relief.

Question 27. Classify emulsions and define each of them with examples.

Answer:

Emulsions Are Of Two Types:-

1. Oil-in-water type emulsion and
2. Water-in-oil type emulsion.

Here, the term oil represents any liquid that is immiscible with water.

1. Oil-In-Water Type Emulsion:

When small quantity of oil is agitated with an excess amount of water, an oil-in-water type emulsion is formed.

Here, water is the dispersion medium and oil is the dispersed phase.

Example: Milk, vanishing cream etc. When small amount of nitrobenzene is agitated with water, this type of emulsion is produced.

2. Water-In-Oil Type Emulsion:

When small quantity of water is agitated with an excess amount of oil, a water-in-oil type emulsion is formed. Here, oil is the dispersion medium and water is the dispersed phase.

Examples: Butter, cod liver oil, cold cream etc.

Question 28. Write the differences between oil-in-water type emulsions and water-in-oil type emulsions.

Answer:

The major differences between oil-in-water type emulsions and water-in-oil type emulsions are given below

Question 29. What is Brownian motion? How does this motion arise?

Answer:

Brownian Motion:-

1. The continuous, random motion of particles of the dispersed phase in a colloidal solution in zigzag paths is known as Brownian motion.
2. Brownian motion arises due to random collisions of the particles of the dispersion medium with the colloidal particles in dispersed phase.

Question 30. What is an emulsifier? Give example.

Answer:

Emulsifier:-

While preparing an emulsion of two immiscible liquids, a third substance is generally added to the mixture. The mixture is then vigorously agitated. The third substance which is added in small quantity to stabilise the emulsion is known as an emulsifier or emulsifying agent.

Example: In case of milk which is a colloid (emulsion), a protein named casein acts as the emulsifier.

Question 31. How will you prepare colloidal sulphur?

Answer:

Preparation Of Colloidal Sulphur:-

Dilute hydrochloric acid reacts with sodium thiosulphate to produce sulphur. When the concentration of sulphur in the solution reaches a certain level, the sulphur particles instead of getting precipitated, form colloidal particles and remain in the colloidal solution as the dispersed phase.

\(\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+2 \mathrm{HCl} \rightarrow 2 \mathrm{NaCl}+\mathrm{S}+\mathrm{SO}_2+\mathrm{H}_2 \mathrm{O}\)

This reaction is called clock reaction.

Question 32. Name a substance that is used as an emulsifying agent. Mention its role as an emulsifying agent in our daily life.

Answer:

1. Soaps, detergents etc. act as emulsifying agents.
2. Soaps and detergents help in the formation of emulsion between water and grease or oil. When soap or sodium stearate is dissolved in water, it dissociates to form a stearate ion.

The ion has two different parts—the long hydrocarbon part (C₁₇H₃₅ —) which is hydrophobic and the ionic part ( — COO^–) which is hydrophilic. Thus, the hydrocarbon chain dissolves in oil or grease of the clothes while the ionic part dissolves in water.

In this way, an emulsion is formed and oil or grease, along with dirt and dust is easily removed from clothes by the action of soap.

Question 33. Water from a tube well sometimes becomes turbid and brownish on standing. Explain with reason.

Answer:

Water From A Tube Well Sometimes Becomes Turbid And Brownish On Standing:-

Water collected from tube wells sometimes contains ferrous bicarbonate [Fe(HCO₃)₂] salt dissolved in it. This compound is oxidised by aerial oxygen to produce brown ferric hydroxide [Fe(OH)₃]. This ferric hydroxide forms colloidal solution with water. As a result, the water becomes turbid and brownish on standing.

4Fe(HCO₃)₂ + 2H₂O + O₂ → 4Fe(OH)₃  ↓+ 8CO₂

Question 34. Proteins, DNA, starch etc. form colloidal solution with water. Why?

Answer:

Proteins, DNA, starch etc. form colloidal solution with water.

Proteins, starch, DNA etc. are polymeric biomolecules. These are large molecules known as macromolecules. The diameter of the molecules of these substances is larger than that of the solute particles of a true solution (diameter ≤ 10^-8 cm).

Hence, these molecules cannot occupy the intermolecular spaces between water molecules. So, these substances cannot form true solutions with water.

However, macromolecules have great affinity towards water molecules and when these substances are added to water, they do not separate out from the mixture. The dimension of the particles is in the range of ≤10^-7 – ≤10^-5 cm. Therefore, they form a colloidal solution with water.

Chapter 4 Matter Solution Topic A True Solution Colloid And Suspension Dissolution Of Smaller Ion Or Molecules And Larger Molecules In Water Very Short Answer Type Questions Choose The Correct Answer

Question 1. Which of the following is a true solution?

1. Aqueous solution of sugar
2. Cod liver oil
3. Mixture of sand and water
4. Milk

Answer: 1. Aqueous solution of sugar

Question 2. Which of the following is a colloidal solution?

1. Aqueous solution of sugar
2. Mixture of BaS04 and water
3. Mixture of sand and water
4. Milk

Answer: 4. Milk

Question 3. Which of the following is an emulsion?

1. Curd
2. Milk
3. Milk of magnesia
4. Soda water

Answer: 2. Milk

Question 4. The solution in which scattering of light can be observed is

1. Blood
2. Mixture of sand and water
3. Aqueous solution of copper sulphate
4. Aqueous solution of sodium chloride

Answer: 1. Blood

Question 5. The correct order of stability is

1. Suspension < colloidal solution < true solution
2. Colloidal solution < true solution < suspension
3. True solution < colloidal solution < suspension
4. Colloidal solution < suspension < true solution

Answer: 1. Suspension < colloidal solution < true solution

Question 6. For which of the following solutions, the solute particles can pass through parchment paper?

1. Milk
2. Cod liver oil
3. Aqueous solution of sugar
4. Sand mixed with water

Answer: 3. Aqueous solution of sugar

Question 7. Which of the following solutions is homogeneous in nature?

1. Milk
2. Cod liver oil
3. Aqueous solution of sugar
4. Mixture of sand and water

Answer: 3. Aqueous solution of sugar

Question 8. Which of the following acts as dialyzer in human body?

1. Lungs
2. Kidneys
3. Liver
4. Stomach

Answer: 2. Kidneys

Question 9. In which of the following, the dispersion medium is not a liquid?

1. Fog
2. Foam
3. Sulphur sol
4. Cream

Answer: 1. Fog

Question 10. Milk is a type of

1. Gel
2. Foam
3. Oil-in-water type emulsion
4. Water-in-oil type emulsion

Answer: 3. Oil-in-water type emulsion

Question 11. An example of a solid sol is

1. Metal alloy
2. Gold sol
3. Paneer
4. Pumice stone

Answer: 1. Metal alloy

Question 12. An example of gel is

1. Milk of magnesia
2. Paneer
3. Cream
4. Lather of soap

Answer: 2. Paneer

Question 13. A colloidal solution of a liquid dispersed in another liquid is known as

1. Sol
2. Gel
3. Foam
4. Emulsion

Answer: 4. Sol

Question 14. The easiest way to identify a colloidal solution is by

1. Observing Tyndall effect
2. Observing Brownian motion
3. Electrodialysis

Answer: 1. Observing Tyndall effect

Question 15. The molecules of which of the following substances will not occupy the intermolecular spaces between water molecules when it is dissolved in water?

1. Ethanol
2. Sugar
3. Glycerin
4. Protein

Answer: 4. Protein

Question 16. In presence of colloidal particles, light rays get

1. Reflected
2. Refracted
3. Scattered
4. Deviated

Answer: 3. Scattered

Question 17. For which of the following dispersion medium is not liquid?

1. Fog
2. Lather
3. Sulphur sol
4. Cream

Answer: 1. Fog

Question 18. Descending order of the size of particles of true solution (A), Colloidal sol (B) and suspension (C) is

1. A >B > C
2. B > A > C
3. C >B > A
4. A < B > C

Answer: 3. C >B > A

Question 19. Particle with which diameter, can fit in the intermolecular space of water

1. 10^-9 cm
2. 10^-6 cm
3. 10^-5 cm
4. 10^-4 cm

Answer: 1. 10^-9 cm

Question 20. Example of liquid aerosol

1. Cloud
2. Smoke
3. Milk
4. Cream

Answer: 1. Cloud

Question 21. For which of the following dispersed phase and dispersion medium are liquid and gas respectively?

1. Emulsion
2. Sol
3. Gel
4. Liquid Aerosol

Answer: 4. Liquid Aerosol

Question 22. Example of emulsifier

1. Arrowroot
2. Gelatine
3. Dettol
4. Cod liver oil

Answer: 2. Gelatine

Question 23. Which one is suspension?

1. Milk
2. Milk of magnesia
3. Cake
4. Shaving cream

Answer: 4. Shaving cream

Question 24. Which one is not an example of emulsion?

1. Butter
2. Shampoo
3. Cake
4. Shaving cream

Answer: 4. Shaving cream

Chapter 4 Matter Solution Topic A True Solution Colloid And Suspension Dissolution Of Smaller Ion Or Molecules And Larger Molecules In Water Answer In Brief

Question 1. The diameter of a particle ‘O’ present in water is 80 nm. What is the nature of the solution?

Answer: Q will form a colloidal solution.

Question 2. Name a colloidal solution which is used in our daily life.

Answer: Milk.

Question 3. Mention the type of solutions formed when sulphur is separately dissolved in water and alcohol.

Answer: Sulphur remains dispersed in water to form a colloidal solution. On the other hand, it completely dissolves in alcohol to form a true solution.

Question 4. Give some examples of solid sol.

Answer:

Some examples of solid sol

Coloured glass, ruby glass (Au/glass), gemstones, metal alloys etc.

Question 5. The particulate matter in air belongs to which type of colloid?

Answer: Solid aerosol.

Question 6. What is meant by gel?

Answer:

Gel meaning:

Gel is a colloidal solution in which a liquid is dispersed in a solid. So, the dispersion medium is solid and dispersed phase is liquid. Jelly, gelatin etc. are some examples of gel.

Question 7. Give some examples of liquid aerosol.

Answer: Fog and cloud are some examples of liquid aerosol.

Question 8. Give an example of a colloid in which the dispersion medium is solid and the dispersed phase is gas.

Answer: The colloids in which the dispersion medium is solid and the dispersed phase is gas are known as solid foams. Some common examples are, cake, pumice stone (air dispersed in silicate compounds) etc.

Question 9. Give some examples of hydrophilic colloids.

Answer: Starch, gelatin, protein, cellulose, soap etc. are some examples of hydrophilic colloids.

Question 10. Give some examples of hydrophobic colloids.

Answer: Silver sol, gold sol, arsenius sulphide sol, ferric hydroxide sol etc. are some examples of hydrophobic colloids.

Question 11. Which substance acts as the emulsifying agent in milk?

Answer: In milk, casein (which is a protein) acts as the emulsifying agent.

Question 12. What is the type of colloid in which both the dispersion medium and dispersed phase are liquid known as?

Answer: The colloids in which both the dispersion medium and dispersed phase are liquid are known as emulsions.

Question 13. Name some covalent compounds which are soluble in water.

Answer: Some covalent compounds which are soluble in water are sugar, alcohol, HCl etc.

Question 14. Name a water soluble organic compound.

Answer: Sugar is a water-soluble organic compound.

Question 15. Name an electrovalent compound which is insoluble in water.

Answer: An electrovalent compound which is insoluble in water is barium sulphate (BaSO₄).

Question 16. Give an example of a colloidal solution in which the colloidal particles are large molecules.

Answer: Starch remains dispersed in water to form a colloidal solution. Here, large starch molecules are present as colloidal particles.

Question 17. What is the diameter of solute particles in a true solution?

Answer: ≤ 10^-8 cm.

Question 18. What is the diameter of solute particles in a colloidal solution?

Answer: 10^-7-10^-5 cm.

Question 19. What is the diameter of solute particles in a suspension?

Answer: 10^-5 cm.

Question 20. Which method is used for the separation of crystalloids from colloids?

Answer: electrodialysis.

Question 21. Name the random, continuous motion exhibited by colloidal particles in a colloidal solution

Answer: Brownian motion.

Question 22. Why gas mixtures cannot form colloidal systems?

Answer: Gas mixtures always form homogeneous mixtures irrespective of the ratio in which they are mixed. Thus, they cannot form colloidal systems.

Question 23. The suspended particulate matters belong to which type of colloids?

Answer: SPM belong to the solid aerosol class of colloid.

Question 24. By which property colloids and True solutions can be distinguished?

Answer: Tyndall Effect.

Question 25. Name an emulsifier used in daily life.

Answer: Soap or detergent.

Question 26. Name the process by which crystalloid can be separated from colloid using diffusion through a semi-permeable membrane.

Answer: Dialysis.

Question 27. Which solution is used with HSO to prepare colloidal Sulphur?

Answer: Hypo solution or the dilute solution of Sodium thiosulphate.

Question 28. Name two diseases caused due to solid aerosol.

Answer: Asthma and silicasis.

Question 29. Which protective colloid is used in icecream?

Answer: Gelatine.

Question 30. Given an example of a solution where salute and solvent both are solid.

Answer: Brass.

Question 31. Give an example of a colloid which is an essential component of our body.

Answer: Blood (colloidal solution of Albumin).

Question 32. Give an example of a colloidal system where finer particles of water are dispersed in fat.

Answer: Butter.

Question 33. Mention the dispersion phase in O/W type Emulsion.

Answer: Oil.

Question 34. Give an example of colloid causing air pollution.

Answer: Solid aerosol.

Question 35. Why Gelatine is used to prepare ice cream?

Answer: Gelatine is used to produce ice cream to give the stability of the colloidal system (ice cream).

Question 36. Give an example of a colloidal system where the dispersed phase and the dispersion medium are gas and liquid respectively.

Answer: Soda-water.

Question 37. Which type of colloid is the soap-lather? Mention the nature of dispersed phase and dispersion medium of soap lather.

Answer:

1. Soap-lather belong to the ‘Foam’ family of colloid.
2. The dispersed phase and dispersion medium here are of gaseous and liquid respectively.

Question 38. Which type of solution will formed by dissolving starch in water?

Answer: Colloidal solution.

Question 39. Which type of solution is formed by dissolving barley in hot water.

Answer: Colloidal solution.

Question 40. Name the property of scattering of light by colloid particles.

Answer: Tyndall effect.

Question 41. Give an example of a colloid whose dispersed phase and dispersion medium, both are solid.

Answer: Curd.

Question 42. Give an example of a colloid which is used as disinfectant.

Answer: Dettol.

Question 43. How solute and solvent are termed in colloidal system?

Answer: Solute and solvent are termed as dispersed phase and dispersion medium respectively in colloidal system.

Question 44. Name the solvent and solutes of air.

Answer: Solvent of air is nitrogen and oxygen, carbon dioxide and other gases are the solutes of air.

Question 45. Name the solutes and solvents in—brass and sodium amulgum.

Answer: In brass the solvent is copper and the solute is zinc.ln sodium amulgum solute and solvents are mercury and sodium respectively.

Chapter 4 Matter Solution Topic A True Solution Colloid And Suspension Dissolution Of Smaller Ion Or Molecules And Larger Molecules In Water Fill In The Blanks

Question 1. True solutions are _______ in nature.

Answer: Homogeneous

Question 2. A Colloidal solutions are ____ in nature.

Answer: Heterogeneous

Question 3. Milk is a colloid in which ______ is the dispersed phase and ______ is the dispersion medium.

Answer: Far, water

Question 4. ________ solutions show Tyndall effect.

Answer: Colloidal

Question 5. The colloidal particles in a colloidal solution can be observed by using ______

Answer: Ultramicroscope

Question 6. When sodium chloride is dissolved in water, it forms a ________ solution.

Answer: True

Question 7. When starch is dissolved in water, it forms a _________ solution.

Answer: Colloidal

Question 8. A ________ is a is a colloidal solution in which the dispersed phase is solid.

Answer: Sol

Question 9. A _________ is a colloidal system in which the dispersed phase is liquid and the dispersion medium is solid.

Answer: Gel

Question 10. An example of solid foam is ________

Answer: Cake

Question 11. A colloidal system is known as _______ if the dispersed phase is a gas and the dispersion medium is solid.

Answer: Solid foam

Question 12. Dispersed phase and dispersion medium, in case of pumice stone, are _______ and _______ respectively.

Answer: Gas, solid

Question 13. The solutes in case of colloidal solution are termed as _______

Answer: Dispersed phase

Question 14. Dispersed phase and dispersion medium for emulsions are _______ and _______ respectively.

Answer: Liquid, liquid

Chapter 4 Matter Solution Topic A True Solution Colloid And Suspension Dissolution Of Smaller Ion Or Molecules And Larger Molecules In Water State Whether True Or False

Question 1. The component of a solution, which is present in a lesser amount is known as the solute.

Answer: True

Question 2. The amount of CaCI₂ present in 500 mL of 25% CaCI₂ solution is 125 g.

Answer: True

Question 3. The solute particles of a colloidal solution can pass through parchment paper.

Answer: False

Question 4. The solute particles of a true solution are visible under an ultramicroscope.

Answer: False

Question 5. A true solution does not exhibit Tyndall effect.

Answer: True

Question 6. If the diameter of solute particles is 10^-4 cm, the corresponding solution is a true solution.

Answer: False

Question 7. Colloidal system whose dispersed phase and dispersion medium both are gaseous, does not exist.

Answer: True

Question 8. Particles of a true solution can not be observed even with the help of a powerful microscope.

Answer: True

Question 9. Example of water-in-oil type emulsion is butter.

Answer: True

Question 10. Colloidal solution are homogeneous.

Answer: False

Question 11. Colloidal solution are transparent.

Answer: False

Chapter 4 Matter Solution Topic A True Solution Colloid And Suspension Dissolution Of Smaller Ion Or Molecules And Larger Molecules In Water Numerical Examples

Question 1. 8 g of NaOH is dissolved in 2 L of its aqueous solution, Calculate the strength of the solution in mole/litre unit.

Answer:

Given

8 g of NaOH is dissolved in 2 L of its aqueous solution

Molecular mass of NaOH = (23 + 16 + 1) = 40

mass of 1 mol NaOH = 40 g

∴ 8g NaOH = 8/40 mol =0.2 mol

∴ Amount of NaOH in 2 L solution =0.2 mol

∴ Amount of NaOH in 1 L solution = 0.2/2 mol = 0.1 mol

∴ Strength of the solution =0.1 mol/L

Question 2. 300 g saturated solution of KNO₃ at 100°C is cooled down to 10°C. Calculate the amount of salt that should be crystallised out from the solution (Solubility of KNO₃ at 100°C and 10°C are 250 and 20 respectively).

Answer:

Given

300 g saturated solution of KNO₃ at 100°C is cooled down to 10°C.

At 100°C, amount of KNO₃ dissolved in 100 g H₂O = 250 g

∴ (100 + 250)g saturated solution contains 100 g solvent

∴ 300 g saturated solution contains \(\frac{100 \times 300}{350}\)g solvent = 85.71 g solvent

∴ Amount of solute in 300 g solution (300 – 85.71)g = 214.29g

Now at 10°C,

Amount of KNO₃ that can be dissolved in 100 g solvent = 20 g

∴ Amount of KNO₃ that can be dissolved in 85.71 g solvent = \(\frac{20 \times 85.71}{100} \mathrm{~g}\) g = 17.142 g

∴ Amount of KNO₃ that should be crystallised out = (214.29 – 17.142)g = 197.148g

Question 3. 4 g of NaOH (molar mass 40) is dissolved in 500 ml of a solution. Calculate the strength of the solution in

1. g • L^-1,
2. % strength (w/v) and
3. mol • L^-1.

Answer:

Given

4 g of NaOH (molar mass 40) is dissolved in 500 ml of a solution.

1. Amount of NaOH in 500 ml solution = 4 g

∴ Amount of NaOH in 1000 ml or 1 L

= \(\frac{4 \times 1000}{500} \mathrm{~g}\) = 8g

∴ Strength of solution = 8 g • L^-1

2. Amount of NaOH in 500 ml solution = 4 g

∴ Amount of NaOH in 100 ml solution

= \(\frac{4 \times 100}{500} g\) = 0.8 g

∴ Strength of the solution is 0.8% (w/v)

3. Again, 4 g NaOH = 4/40 mol = 0.1 mol

∴ Strength of the solution = 0.1/500 x 1000 = 0.2 mol • L^-1

Question 4. Strength of a glucose solution is 0.8 mol • L^-1. Express the strength in % (w/v).

Answer:

Molar mass of glucose is = 180.

1 L or 1000 ml. solution contains 0.8 mol glucose = 0.8 x 180 g = 144 g of glucose

∴ 100 mol of solution contains = \(\frac{144 \times 100}{1000}\) g of glucose = 14.4 g glucose.

∴ % strength (w/v) of the solution is 14.4%.

Question 5. Calculate the amount of salt required to prepare 200 ml 2% sodium bicarbonate solution. Express the strength in gram per litre.

Answer:

100 ml of 2% solution contains 2 g of sodium bicarbonate.

∴ Amount of salt in 200 ml solution = 2/100 x 200 g = 4g

∴ Amount of salt in 1000 ml solution = 4/100 X 1000 = 20 g

Strength of the solution is 20 g • L^-1.

Question 6. Among 120 ml 10 % NaCI solution and 120 ml, 1 mol • L^-1 NaCI solution, which contains a greater amount of NaCI?

Answer:

Amount of NaCI present in 120 ml of 10% NaCI solution = 10/100 x 120 g = 12 g

Amount NaCI present in 120 ml of 1 mol • L^-1 solution = \(\frac{(23+35.5) \times 120}{1000}\) g = 7.02 g

∴ 120 ml 18% NaCI solution contains greater amount of sodium chloride.

Chapter 4 Matter Solution Topic B Solubility Strength Of Solution And Its Unit Synopsis

1. Solubility of a substance at a given temperature is the maximum number of gram units of the substance that can be dissolved in 100g of a solvent at that temperature.
2. The curve obtained by plotting the solubility of a substance in a given solvent along the y-axis against temperature along x-axis is called the solubility curve of that solute.
3. Solubility of solids in liquids generaily increases with increase in temperature.
4. Example: KNO₃, NaCI, NH₄CI. sugar, blue vitriol etc.
5. Solubility of CaSO₄, Ca(OH)₂, etc. in water decreases with incease in temperature.
6. Solubility of gases in liquid decreases with increase in temperature and increases with decrease in temperature.
7. At constant temperature solubility of gaseous solutes in liquid solvents increases with increase in pressure and decreases with decreases in pressure.

Chapter 4 Matter Solution Topic B Solubility Strength Of Solution And Its Unit Short And Long Answer Type Questions

Question 1. Why It necessary to mention temperature in the definition of solubility?

Answer:

The solubility of a substance in a given solvent varies with temperature. Hence, the amount of a particular solute required to saturate 100 g of a solvent will be different at different temperatures. Thus, it is necessary to mention temperature in the definition of solubility.

Question 2. Why is it necessary to mention a definite amont of solvent in the definition of solubility?

Answer:

At a particular temperature, the amount of solute required to make a saturated solution depends on the amount of solvent. If greater amount of solvent is taken, then it can dissolve more amount of solute whereas, lesser amount of solvent dissolves lesser amount of solute.

Hence, we cannot determine the amount of solute required to make a saturated solution, if the amount of solvent is not specified. Thus, it is necessary to specify the amount of solvent in the definition of solubility.

Question 3. What are the units used to express the strength of a solution? Comment on their dependence on temperature.

Answer:

Units used to express the strength of a solution

Different units used to express the concentration or strength of a solution are volume percentage (% W/V), moles per litre or molarity (mol • L^-1) and gram per litre (g • L^-1).

Although mass or number of moles of a substance is independent of temperature, volume of a solution changes with temperature. As all the units are dependent on volume of the solution, these are also dependent on temperature.

Question 4. What is meant by molarity or moles per litre strength of a solution? Give example.

Answer:

Molarity or moles per litre strength of a solution

At a given temperature, the number of gram-moles of a solute dissolved in 1 litre (or 1000mL) of a solution is known as the molarity or moles per litre strength of the solution.

Example: 5 gram-moles of HN03 is dissolved in 1 litre aqueous solution. Hence, the molarity of the solution will be 5. The molar strength is expressed as mol • L^-1 or (M).

Question 5. What is meant by gram per litre strength of a solution? Give example.

Answer:

Gram per litre strength of a solution

The amount of solute in gram dissolved per litre of the solution is expressed in gram per litre unit.

Example: If 5.0g of sodium chloride is dissolved in 1 L of solution, then the strength of the sodium chloride solution in gram per litre unit will be 5. The unit is expressed as g • L^-1.

Question 6. What changes will be observed in the (W/V) concentration of a solution if the temperature of the solution is raised?

Answer:

The following changes will be observed in the (W/V) concentration of a solution if the temperature of the solution is raised

1. With the increase in temperature, the volume of the solution increases. Hence, the ratio of mass of solute to the volume of solution decreases. As a result, concentration of the solution also decreases.
2. If some amount of solvent vapourises due to increase in temperature, then the volume of solution also decreases. Hence, the concentration of the solution also increases.

Question 7. Why molarity depends on temperature?

Answer:

We know molarity of solution = \(\frac{\text { number of } \mathrm{g} \cdot \text { mol of solute }}{\text { volume in } \mathrm{L} \text { of the solution }}\)

Now volume of the solution changes with temperature. Thus molarity of the solution also changes with temperature, although the number of g • mol of the solute remains unaltered. Hence molarity depends on temperature.

Question 8. If the amount of the dissolved solute in a gram of saturated solution is b gram, calculate the solubility of the solute at that temperature.

Answer:

We know, solubility of a solute at t°C = \(\frac{\text { mass of the solute in saturated solution }}{\text { mass of the solvent in-the solution }} \times 100\)

= \(\frac{b \mathrm{~g}}{(a-b) g} \times 100=\frac{100 b}{(a-b)}\)

Question 9. What are the factors on which the solubility of a substance depends?

Answer:

The solubility of a substance in a given solvent depends on the following factors

1. Nature of solute:

Different solutes dissolve to various extents in a given solvent at a particular temperature.

2. Temperature:

For most solid substances, solubility in a given liquid increases with the rise in temperature. However, there are some gases whose solubility in a given liquid decreases with the rise in temperature.

3. Atmospheric pressure:

At a particular temperature, the solubility of a gas in a liquid generally increases with the increase in pressure above the liquid and vice-versa. This is why C02 dissolved in water under very high pressure comes out as bubbles when a sodawater bottle is opened.

Question 10. What is a solubility curve? Draw the solubility curve of Glauber’s salt in water and explain its solubility with the help of the curve.

Answer:

Solubility curve

The curve obtained by plotting the solubility of a solute in a given solvent along the Y-axis against temperature along X-axis, is called the solubility curve of that solute. This curve easily demonstrates the relation between temperature and solubility of a substance.

The solubility curve of Glauber’s salt in water (hydrated sodium sulphate) rapidly rises up with the rise in temperature upto 32.38°C. After 32.38°C, if the temperature is further increased, the solubility gradually starts decreasing.

Above 32.38°C, hydrated crystals of sodium sulphate (Na₂SO₄ – 10H₂O) give up ten water molecules to form anhydrous sodium sulphate (Na₂SO₄). So, above 32.38°C, the solubility curve actually represents the solubility of anhydrous sodium sulphate.

Question 11. Draw the solubility curves of KNO₃ and NaCI in water and mention the effect of temperature on solubility of these salts.

Answer:

The solubility curve of KNO₃ rises steeply with rise in temperature. This means that the solubility of KNO₃ increases rapidly with the increase in temperature.

The solubility curve of NaCI is almost parallel to the X-axis. This indicates that the solubility of NaCI in water remains almost unchanged over a wide range of temperature.

Question 12. How does the solubility of Ca(OH)₂ in water changes with the rise in temperature? Explain with the help of its solubility curve.

Answer:

From the solubility curve, it can be said that the solubility of Ca(OH)₂ in water decreases with the rise in temperature.

Question 13. State the significance of solubility curve.

Answer:

Significance of solubility curve

1. Without actually performing an experiment, the solubility of a substance at different temperatures can be determined with the help of the solubility curve.
2. It helps to compare the solubilities of different substances at a given temperature.
3. The solubility curve helps us to determine which substance will preferentially crystallise out if a solution of a mixture of substances is subjected to cooling or evaporation.

Question 14. Discuss the effect of temperature on the solubility of a gas in a liquid.

Answer:

The effect of temperature on the solubility of a gas in a liquid

In general, the solubility of a gas in a liquid decreases with the rise in temperature. For example, CO₂ is soluble in water. At ordinary temperature, some amount of C02 gas dissolves in water. However on heating, the dissolved CO₂ fizzles out from the solution.

Question 15. Discuss the effect of pressure on the solubility of a gas in a liquid.

Answer:

The effect of pressure on the solubility of a gas in a liquid

At a particular temperature, the solubility of a gas in a liquid generally increases with the increase in pressure and decreases with the decrease in pressure. So, at a higher pressure relatively large amount of gas dissolves in a liquid. This principle is used in the preparation of soda water where excess CO₂ is dissolved in water by applying pressure.

Question 16. Why does soda water fizz when the cap of the bottle is removed?

Answer:

At a particular temperature, the solubility of a gas in a liquid increases with increase in pressure. In soda water bottles, carbon dioxide is dissolved in water applying high pressure.

When the cap of the bottle is removed, the pressure suddenly decreases and as a result the solubility of CO₂ in water decreases. Consequently, excess carbon dioxide comes out from water and eventually causes the fizz.

Question 17. A bottle containing liquor ammonia must be cooled before opening. Why?

Answer:

A bottle containing liquor ammonia must be cooled before opening.

A saturated aqueous solution of ammonia is called liquor ammonia. With rise in temperature, solubility of NH₃ in water decreases and hence some NH₃ gas comes out of the solution and causes excess pressure inside the bottle.

If the bottle is opened under this condition, then excess NH₃ gas along with some dissolved NH₃ gas may spurt out from the bottle causing serious accident. However, if the bottle is cooled down, the excess NH₃ again dissolves in water and no such accident occurs when the bottle is opened.

Question 18. Why bubbles are formed when water is heated?

Answer:

Some amount of air generally remains dissolved in water at room temperature. Now solubility of gases in liquid decreases with increase in temperature. That is why solubility of air decreases when we heat the water and hence the excess air comes out of water forming bubbles.

Chapter 4 Matter Solution Topic B Solubility Strength Of Solution And Its Unit Very Short Answer Type Questions Choose The Correct Answer

Question 1. HCI (g) dissolves in water, because HCI

1. Is a gaseous substance
2. Reacts with water
3. Is an ionic compound
4. Is a polar covalent compound

Answer: 4. Is a polar covalent compound

Question 2. Water can dissolve ionic compounds, because water is a/an

1. Covalent compound
2. Ionic compound
3. Polar covalent compound
4. Liquid

Answer: 3. Polar covalent compound

Question 3. The temperature upto which the solubility curve for the solubility of Glauber’s salt in water rises is

1. 25.5°C
2. 32.4°C
3. 35.2°C
4. 37.24°C

Answer: 2. 32.4°C

Question 4. At a constant temperature, the solubility of a gas in a given volume of liquid is directly proportional to the pressure of the gas. This statement is known as

1. Boyle’s law
2. Charles’ law
3. Raoult’s law
4. Henry’s law

Answer: 4. Henry’s law

Question 5. 01 Unit of solubility is

1. g • cm^-3
2. kg • m^-3
3. g
4. Unitless

Answer: 4. Unitless

Question 6. With rise in temperature, the solubility of Glauber’s salt (Na₂SO₄ • 10H₂O) in water

1. Decreases
2. Increases
3. Remains almost unchanged
4. Initially increases and then decreases

Answer: 4. Initially increases and then decreases

Question 7. With rise in temperature, the solubility of potassium nitrate (KNO₃) in water

1. Decreases
2. Increases
3. Remains almost unchanged
4. Initially increases and then decreases

Answer: 1. Decreases

Question 8. Which of the following units of concentration is not affected by any change in temperature?

1. Mass-volume percentage (% W/V)
2. Mass percentage (% W/W)
3. Moles per litre
4. Gram per litre

Answer: 2. Mass percentage (% W/W)

Question 9. With rise in temperature, the solubility of a gas in a liquid

1. Increases
2. Decreases
3. Remains almost unchanged
4. Initially increases and then decreases.

Answer: 2. Decreases

Question 10. 10% Na₂CO₃ solution means

1. 10g Na₂CO₃ is dissolved in 100 mL solution
2. 100g Na₂CO₃ is dissolved in 10 mL solution
3. 10g Na₂CO₃ is dissolved in lOmL solution
4. 10g Na₂CO₃ is dissolved in 100g solution

Answer: 1. 10g Na₂CO₃ is dissolved in 100 mL solution

Question 11. The amount of NaCI present in 200 mL of 50% NaCI solution is

1. 50 g
2. 75 g
3. 100 g
4. 150 g

Answer: 3. 100 g

Question 12. The concentration of a solution prepared by dissolving 40 g NaOH in 1000 mL water will be

1. 1 mol • L^-1
2. 2 mol • L^-1
3. 3 mol • L^-1
4. 4 mol • L^-1

Answer: 1. 1 mol • L^-1

Question 13. If xg of a solute is required to make yg saturated solution at f°C temperature; solubility at that temperature will be

1. x/y x 100
2. y
3. \(\frac{x}{y-x} \times\) x 100
4. \(\frac{y}{x-y} \times\) x 100

Answer: 3. \(\frac{x}{y-x} \times\) x 100

Question 14. With increase in temperature, the solubility of slaked lime in water

1. Increases
2. Decreases
3. Remains same
4. Increases firstly then decreases

Answer: 2. Decreases

Question 15. Solubility does not depend upon

1. Temperature
2. Pressure
3. Nature of solvent
4. Gravitational force of the earth

Answer:  4. Gravitational force of the earth

Question 16. The gas which comes out as effervescence from a bottle of soda water when we open it is

1. NH₃
2. H₂
3. He
4. CO₂

Answer: 4. CO₂

Question 17. The solubility of which one decreases with increase in temperature?

1. KNO₃
2. Ca(OH)₂
3. NaCI
4. NH₄CI

Answer: 2. Ca(OH)₂

Question 18. Solubility of NaCI in water at 273K is 37. Amount of water required to make a saturated solution containing 100g of NaCI at that temperature will be

1. 270.27g
2. 245 g
3. 222.22g
4. 145 g

Answer: 1. 270.27 g

Question 19. Solubility of which one in water increases reading with increase in temperature?

1. Ca(OH)₂
2. NaNO₃
3. NaCI
4. NH₄CI

Answer: 2. NaNO₃

Question 20. Solubility of which one increases slightly with increase in temperature?

1. ZnSO₄
2. CaSO₄
3. NaCI
4. Na₂SO₄-10H₂O

Answer: 3. NaCI

Question 21. With increase in pressure, the solubility of a solid in a liquid

1. Increases
2. Decreases
3. Firstly increases and then decreases
4. Remains same

Answer: 4. Remains same

Question 22. Amount of Na₂CO₃ in 100 ml of 25% Na₂CO₃ solution

1. 25 g
2. 2.5 g
3. 225 g
4. 250g

Answer: 4. 250g

Question 23. If we express the strength of 10% Na₂CO₃ solution as mol/litre, it will be

1. 2.00
2. 1.00
3. 0.74
4. 0.94

Answer: 4. 0.94

Chapter 4 Matter Solution Topic B Solubility Strength Of Solution And Its Unit Answer In Brief

Question 1. What is the unit of solubility?

Answer: Solubility is a unitless physical quantity. It has no unit.

Question 2. Name some compounds whose solubility in water decreases with the increase in temperature.

Answer: The compounds whose solubility in water decreases with the increase in temperature are calcium sulphate (CaSO₄), calcium nitrate [Ca(NO₃)₂], slaked lime or calcium hydroxide [Ca(OH)₂], cerium sulphate [Ce₂(SO₄)₃] etc.

Question 3. Name a solid whose solubility in water remains almost constant with change in temperature.

Answer: Solubility of common salt or sodium chloride (NaCI) in water remains almost constant with change in temperature

Question 4. Name a solid whose solubility in water initially increases with the rise in temperature but starts to decrease when the temperature is further increased.

Answer: The solubility of Glauber’s salt (Na₂SO₄-10H₂O) in water initially increases with the rise in temperature up to 32.4°C and then the solubility begins to decrease when the temperature is further increased.

Question 5. How is the solubility of a solid solute in a liquid solvent affected by pressure?

Answer: There is no effect of pressure on the solubility of a solid solute in a liquid solvent.

Question 6. State Henry’s law.

Answer:

Henry’s law:

At constant temperature, the solubility of a given gas in a given volume of liquid solvent is directly proportional to the pressure of that gas.

Question 7. What will happen if a saturated solution of sugar prepared at room temperature is cooled down to 5°C?

Answer: When a saturated solution of sugar prepared at room temperature is cooled down to 5°C, the solubility of sugar decreases and some amount of sugar settles down at the bottom.

Question 8. How much NaCI is present in 50 mL of 20% aqueous solution of NaCI?

Answer: 50 mL of 20% aqueous solution of NaCI will contain 10 g of NaCI.

Question 9. What change in W/V concentration of a • solution will be observed with the rise in temperature?

Answer: If the solvent does not evaporate on heating, then the W/V concentration of a solution generally decreases if the temperature is increased.

Question 10. Why does molarity of a solution depend on temperature?

Answer: Molarity of a solution is the number of moles, of solute present per litre of a solution. As the volume of solution depends on temperature (although mass of solute is independent of temperature), the molarity of a solution also depends on temperature.

Question 11. Why the statement—“Solubility of NaCO₃ in water is 20“—is erroneous?

Answer: The temperature is not mentioned here which makes the statement erroneous.

Question 12. What do you mean by the statement— Solubility curve of NaCI is almost parallel to x-axis?

Answer: It means that the solubility of NaCI in water remains almost unaltered with change in temperature.

Question 13. State whether the solubility of gas in water increases or decreases with decrease in pressure.

Answer: Solubility of gas in water decreases with decrease in pressure.

Question 14. How the solubility of Ca(OH)₂ changes with increase in temperature.

Answer: Solubility of Ca(OH)₂ in water decreases with increase in temperature.

Question 15. Which of the following solutions of higher concentration: 100 g/L NaOH and 10 molar NaOH.

Answer: Concentration of NaOH in 10 molar NaOH solution is higher.

Question 16. What do you mean by 15% Na₂CO₃ solution.

Answer: 15% Na₂CO₃ solution means that 15 g of Na₂CO₃ is dissolved in 100 ml solution.

Chapter 4 Matter Solution Topic B Solubility Strength Of Solution And Its Unit Fill In The Blanks

Question 1. If 5 g of a salt is dissolved in 50 mL water, then the strength of the solution will be _______ % (W/V).

Answer: 10

Question 2. The solubility of those substances which absorb heat during dissolution in water, _______ with the rise in temperature.

Answer: Increase

Question 3. The chemical substance added to a mixture of two immiscible substances to increase the stability of the emulsion is called a/an _______ agent.

Answer: Emulsifying

Question 4. The solubility-curve for the solubility of NaCI in water is almost ________ to the temperature-axis.

Answer: Parallel

Question 5. The solubility curve of ________ rises up with the rise in temperature and then starts decreasing after reaching a maximum.

Answer: Glauber’s salt

Question 6. The solubility of calcium hydroxide in water ________ with rise in temperature.

Answer: Decreases

Question 7. The solubility of Glauber’s salt at 32.4°C as obtained from its solubility curve actually represents the solubility of ________

Answer: Anhydrous Na₂SO₄

Question 8. CO₂ is dissolved in water at ________ pressure to produce soda water.

Answer: High

Question 9. With rise in temperature, the %(W/V) strength of a solution _________

Answer: Decreases

Question 10. To express the solubility of a substance, mentioning of ________ is necessary.

Answer: Temperature

Question 11. If 10 g of NaOH is dissolved in 100 ml of NaOH solution, strength of the solution is ________

Answer: 10%

Question 12. Solubility of CO₂ in water _______ with increase in temperature.

Answer: Decreases

Question 13. Solubility of solids in liquid depends on _______

Answer: Temperature

Chapter 4 Matter Solution Topic B Solubility Strength Of Solution And Its Unit State Whether True Or False

Question 1. The amount of CaCI₂ present in 500 mL of 25% CaCI₂ solution is 125 g.

Answer: True

Question 2. The unit of solubility is gram per litre.

Answer: False

Question 3. The solubility of cerium sulphate decreases with the rise in temperature.

Answer: True

Question 4. Molarity of a solution is independent of temperature.

Answer: False

Question 5. The solubility of a gas in a liquid usually decreases with the increase in pressure above the liquid.

Answer: False

Question 6. W/V concentration of a solution increases with the rise in temperature.

Answer: False

Chapter 4 Matter Solution Topic B Solubility Strength Of Solution And Its Unit Numerical Examples

Question 1. At 50°C, 100g of a saturated solution on complete evaporation gives 5O g of residue. Find the solubility of the substance at that temperature.

Answer:

Given

At 50°C, 100g of a saturated solution on complete evaporation gives 5O g of residue.

Amount of solvent = Amount of solution – Amount of solute = (100 – 50) g = 50 g

∴ At 50°C, 50 g of solvent dissolves 50 g of solute

∴ 1 g of solvent dissolves 50/50 g of solute

∴ 100 g of solvent dissolves 50/50 x 100 g of solute = 100 g of solute.

Hence, solubility of the substance at 50°C is 100.

Question 2. At 30°C, the solubility of a salt in water is 40. Find the amount of water required to make a saturated solution of 30 g of the salt at that temperature.

Answer:

Given

At 30°C, the solubility of the salt in water is 40.

∴ 40 g of the solute dissolves in 100 g of water.

∴ 30 g of the solute dissolves in x 30 g of water = 75 g of water.

Thus, 75 g of water will be required to make a saturated solution of 30 g of the salt at 30°C.

Question 3. The solubility of a salt in water at 60 30°C are 140 and 75 respectively. 50 g of the saturated solution of that salt at 60°C is cooled down to 30°C. How much of the salt will separate out from the solution?

Answer: 

At 60° C, 100g of water disisolves 140g salt.

∴ 50g of water dissolves (\(\frac{140}{100} \times 50\))g =70g salt.

At 30°C, 100g of water dissolves 75 g salt.

∴ 50g of water dissolves (\(\frac{75}{100} \times 50\)) g = 37.5g salt.

∴ When the solution is cooled down to 30°C, the amount of solute that will separate out from the solution = (70 – 37.5)g = 32.5g

Question 4. At 40°C, 20g of an unsaturated solution on evaporation gives 4g residue. What amount of salt has to be dissolved in 100 g of solution to make a saturated solution? [Solubility of the salt in water at 40°C is 40]

Answer:

Given

20 g of solution has 4g of salt dissolved in it.

∴ 100g of solution will contain 4/20 x 100 g of the salt = 20g of the salt

∴ 100g of the solution contains = (100 – 20)g = 80 g water

∴ At 40°C, 100 g of water is saturated by \(\frac{40 \times 80}{100}\) g salt = 32 g of the salt

Hence, amount of salt required to make the solution saturated = (32 – 20) g = 12g

Question 5. Concentration of a solution containing NaOH is 100 g • L^-1. Express the concentration in mass-volume percentage (% W/V).

Answer:

1000 mL of the solution contains 100g NaOH.

∴ 100 mL of the solution contains \(\frac{100 \times 100}{1000}\) g of NaOH = 10 g of NaOH.

Hence, concentration in mass-volume percentage (% W/V) = 10%

Question 6. 1.575 g oxalic acid is dissolved in 250 mL water. Find the cone, of the solution in g-L^-1 & %W/V.

Answer:

In 250 mL water, the amount of oxalic acid dissolved is 1.575 g.

∴ In 1000 mL water, the amount of oxalic acid dissolved will be = \(\frac{1.575 \times 1000}{250} \mathrm{~g}\) = 6.3 g

Hence, concentration of the solution is 6.3g • L^-1.

Again, 1000 mL of water dissolves 6.3 g oxalic acid.

∴ 100 mL of water dissolves \(\frac{6.3 \times 100}{1000}\) g = 0.63 g

oxalic acid. Hence, concentration of the solution in % W/V = 0.63%

Question 7. 60g NaOH is dissolved in 500 mL water. Express the cone, of the solution in mol • L^-1.

Answer:

In 500 mL water, mass of NaOH dissolved = 60g

∴ In 1000 mL water, mass of NaOH dissolved = 60/500 x 1000 = 120g

Now, molecular mass of NaOH = 23 + 16 + 1 = 40.

∴ 1 mol NaOH = 40 g NaOH.

∴ 120 g NaOH = 120/40 mol Na0H = 3 mo1 NaOH

Hence, 1 L of the solution contains 3 mol of NaOH. So, the concentration of the solution is 3mol • L^-1.

Question 8. How many grams of 50% (W/W) H₂SO₄ solution will be required to prepare 500 mL 2 (M) H₂SO₄ solution?

Answer:

Molecular mass of H₂SO₄ = 98.

∴ Mass of H₂SO₄ in 1000mL 2 (M) H₂SO₄ = 2 x 98 = 196g

Mass of H₂SO₄ dissolved in 500 mL 2 (M) H₂SO₄= 2 x 98/1000 x 5008 = 98g

Now, in 50% (W/W) H₂SO₄ solution, 100 g of the solution has 50g of H₂SO₄ dissolved in it.

∴ 98g H₂SO₄is dissolved in 100/50 x 98g = 196 g solution.

Hence, 196 g of H₂SO₄ solution will be required.

Question 9. Mass percentage & molar concentration of an HNO₃ solution are 31.5 and 4 (M) respectively. Express the density in g • mL^-1.

Answer:

Mass percentage of HNO₃ solution is 31.5. Hence, 100g of solution has 31.5 g HNO₃.

Let, the density of the solution be d g • mL^-1.

∴ Volume of 100 g of the solution = 100/d mL

Now, 31.5g HN03 = 31.5/63 mol = 0.5 mol HNO₃

Hence, the molar concentration of the solution = \(\frac{0.5}{100 / d} \times 1000\)  = 5  x d(M)

Now, 5 x d = 4; Hence, d = 0.8

Thus, density of the solution is 0.8 g • mL^-1.

Chapter 4 Matter Solution Topic C Saturated Unsaturated Supersaturated Solution Synopsis

1. The solution in which the maximum amount of solute remains dissolved in the solvent at a given temperature is called a saturated solution.
2. An unsaturated solution is one in which, the solvent has the capacity to dissolve more amount of solute at a particular temperature.
3. A supersaturated solution at a particular temperature is defined as the solution in which the solvent, under special conditions, contains more amount of solute than that required to form a saturated solution.
4. A supersaturated solution is highly unstable. It readily precipitates the excess solute, when the solution is disturbed or shaken.
5. The strength of a solution is generally expressed in mass percentage (% W/W), mass-volume percentage (% W/V), volume percentage (% V/V), mol • L^-1 or g • L^-1 units.
6. x% W/V aqueous solution of a substance means that, at a given temperature x g of the substance is dissolved in 100mL water.

Chapter 4 Matter Solution Topic C Saturated Unsaturated Supersaturated Solution Short And Long Answer Type Questions

Question 1. What is a saturated solution?

Answer:

Saturated solution

A solution in which, the maximum amount of solute remains dissolved in the solvent at a given temperature, is called a saturated solution. If more amount of solute is further added to the solution at that temperature, the excess solute settles at the bottom, but the concentration of the solution remains unchanged

Question 2. What is an unsaturated solution?

Answer:

Unsaturated solution

An unsaturated solution is one in which, at a particular temperature, the solvent has the capacity to dissolve more amount of solute. On adding more solute, the concentration of the solution increases and after some time the solution becomes saturated.

Question 3. Differentiate between saturated and unsaturated solutions.

Answer:

The differences between saturated and unsaturated solutions are as follows

Question 4. How will you convert a saturated solution into an unsaturated solution?

Answer:

A saturated solution can be converted to an unsaturated solution by the following methods

1. At a particular temperature, a saturated solution can be made unsaturated by adding some more amount of solvent.
2. In most cases, on heating, a saturated solution becomes unsaturated at a higher temperature.

Question 5. How will you convert an unsaturated solution into a saturated solution?

Answer:

An unsaturated solution can be converted to a saturated solution by the following methods

1. At a particular temperature, an unsaturated solution can be made saturated by adding some more amount of solute.
2. In most cases, on cooling, an unsaturated solution becomes saturated at a lower temperature.
3. Evaporating some amount of solvent from a solution by heating and then subsequently cooling the solution to its initial temperature will decrease the amount of solvent in the solution. The solution thus obtained is a saturated solution at that temperature.

Question 6. Why the saturated solution of glauber salt cannot be unsaturated by heating the solution?

Answer:

The solubility of glauber salt (Na₂SO₄ – 10H₂O) increases rapidly with increase in temperature upto 32.4°C. At a higher temperature than 32.4°C, the salt looses its water of crystallisation and transform to anhydrous sodium sulphate.

On increasing the temperature above 32.4°C solubility of anhydrous sodium sulphate decreases with increase in temperature. As a consequence, the excess salt gets precipitated out from the saturated solution. Hence the above solution cannot be unsaturated by heating.

Question 7. What is a supersaturated solution?

Answer:

Supersaturated solution

A supersaturated solution at a particular temperature is defined as the solution in which the solvent contains more solute at a specific condition than that required to form a saturated solution at that temperature.

Question 8. State two important characteristics of a supersaturated solution.

Answer:

Two important characteristics of a supersaturated solution are

A supersaturated solution is highly unstable. Excess solute from the solution settles down at the bottom of the beaker on slightly shaking the solution, by adding a small crystal of the solute or in presence of dust particles.

On adding more solute to a supersaturated solution at a particular temperature, the concentration of the solution decreases.

Question 9. How will you prepare the super saturated solution of sodium thiosulphate?

Answer:

A small amount of hypo or sodium thiosulphate crystal is taken in a test tube and closed at top with cotton. The test tube is then immersed partially in boiling water bath. Crystals lose their water of crystallisation and get dissolved in that water forming a clear solution.

The test tube is then cooled, without stirring, to room temperature. The solution thus obtained is the super saturated solution of sodium thiosulphate.

Question 10. Differentiate between saturated and supersaturated solutions.

Answer:

The differences between saturated and supersaturated solutions are as follows

Question 11. With the help of a simple experiment, how will you determine whether a solution is saturated, unsaturated or supersaturated?

Answer:

A solution of a given solute in a given solvent is taken. Some more amount of the solute is added to the solution. The following observations will help us to identify whether this solution is saturated, unsaturated or supersaturated

1. If the solute completely or partially dissolves in the solution on stirring at the same temperature, then the solution is an unsaturated solution.
2. If the added solute deposits at the bottom of the beaker instead of dissolving and its concentration remains unchanged, then it is a saturated solution.
3. If on addition of excess solute, the added solute separates out from the solution and settles at the bottom of the beaker instead of dissolving and the amount of deposited solute is more than the added amount of solute, then the solution is a supersaturated solution.

Question 12. You have been given a saturated solution of common salt at room temperature. If the temperature of the solution has been reducedto5°Cby keeping it in an ice block, will the solution remain saturated? What will bethe probable observations?

Answer:

1. The solution will remain saturated.
2. Solubility of common salt changes slightly with temperature. So by cooling the solution only a little amount of salt will be observed to get precipitated out.

Question 13. what is the reason behind the damage of health of the paint workers?

Answer:

Many volatile substances like methylated spirit, tarpene oil, naptha, acetone etc. are used as solvent of paints during painting. These solvents evaporated at room temperature and enter easily to the body of the paint workers through inhalation. These can easily be in contact with the skin of the workers also.

These solvents cause allergy and several bronchiole or lung diseases which causes harm to the health of the workers involved in painting.

Chapter 4 Matter Solution Topic C Saturated Unsaturated Supersaturated Solution Very Short Answer Type Questions Choose The Correct Answer

Question 1. A supersaturated solution in water can be prepared by

1. CuSO₄ • 5H₂O
2. Na₂SO₄ • 10H₂O
3. FeSO₄ • 7H₂O
4. Na₂S₂O₃ • 5H₂O

Answer: 4. Na₂S₂O₃ • 5H₂O

Question 2. The amount of dissolved solute present in a supersaturated solution compared to the maximum amount of solute that can be present in a saturated solution at a given temperature is

1. Less
2. More
3. Equal
4. None of these

Answer: 2. More

Question 3. Formula of hypo

1. Na₂SO₄  • 10H₂O
2. Na₂S₂O₃
3. Na₂S₂O₃  • 5H₂O
4. NaHSO₄

Answer: 3. Na₂S₂O₃  • 5H₂O

Question 4. What happens when the temperature of the KNO₃ solution saturated at 80°C reduces to 40°C?

1. No change
2. Solution will become unsaturated
3. Some amount of KNO₃ will settle down
4. Solution will become supersaturated

Answer: 3. Some amount of KNO₃ will settle down

Question 5. The solution in which more solute can be dissolved, is

1. Saturated solution
2. Unsaturated solution
3. Supersaturated solution
4. Colloid

Answer: 2. Unsaturated solution

Question 6. If some amount of solute is added to a saturated solution at a definite temperature, density of the solution will

1. Increase
2. Reduce
3. Remain same
4. None of the above

Answer: 3. Remain same

Chapter 4 Matter Solution Topic C Saturated Unsaturated Supersaturated Solution Answer In Brief

Question 1. Classify solutions on the basis of their concentration.

Answer: On the basis of their concentration, solutions can be classified into three types—saturated solution, unsaturated solution and supersaturated solution.

Question 2. Among saturated, unsaturated and supersaturated solutions, which one is the least stable?

Answer: A supersaturated solution is the least stable.

Question 3. Name a solid which can form a supersaturated solution.

Answer: Sodium thiosulphate pentahydrate (Na₂S₂O₃ 5H₂O), commonly known as hypo, can form a supersaturated solution.

Question 4. State whether the solubility of common salt dissolved in a cup of water will be equal in Kolkata and Sikkim.

Answer: The solubility of a substance salt depends on temperature. As the temperature of the two places are different, the solubility of common salt in a cup of water will also be different.

Question 5. What amount of of acid (in g) is present per litre of 2 (M) H₂SO₄ solution? (Molecular mass of H₂SO₄ is 98)

Answer: Per litre of 2 (M) H₂SO₄ solution will contain 98 x 2 = 196 g of acid.

Question 6. Which type of solutions cannot form a saturated solution?

Answer: Solutions consisting of two miscible liquids can never form a saturated solution. For example, a solution of water and ethyl alcohol cannot form a saturated solution.

Question 7. On cooling a saturated solution, whether it remains saturated or not?

Answer: If we cool a saturated solution, some amount of solute gets precipitated out. But the remaining solution remains saturated.

Chapter 4 Matter Solution Topic C Saturated Unsaturated Supersaturated Solution Fill In The Blanks

Question 1. A _______  solution is one, whose concentration increases if some more solute is added to the solution.

Answer: Unsaturated

Question 2. In a ________ solution, the concentration remains unchanged even if some amount of solute is added to it.

Answer: Saturated

Question 3. A saturated solution dissolves more solute under a specific condition to produce _______ solution.

Answer: Supersaturated

Question 4. At a particular temperature, the dissolved solute particles in a saturated solution is in _______ with the precipitated particles.

Answer: Equilibrium

Question 5. At a given temperature, if some more solute is added to a supersaturated solution, its solubility ________

Answer: Decreases

Question 6. Unsaturated solution can absorb more _______

Answer: Solute

Chapter 4 Matter Solution Topic C Saturated Unsaturated Supersaturated Solution State Whether True Or False

Question 1. An unsaturated aqueous solution of ammonia is called liquid ammonia.

Answer: False

Question 2. Supersaturated solutions are stable in nature.

Answer: False

Question 3. A saturated solution remains saturated at any temperature.

Answer: False

Question 4. When more solute is added to a saturated solution, the excess solute will be sedimented.

Answer: True

Question 5. Supersaturated solution cannot be prepared by heating copper sulphate crystal in water- bath.

Answer: True

Chapter 4 Matter Solution Topic D Crystallisation Motion Of ParticlesIn Solution And Different Solvents Other Than Water Synopsis

1. The continuous random motion of colloidal particles in a colloidal solution is known as Brownian motion.
2. Colloidal particles exhibit Brownian motion in a colloidal solution. However, Brownian motion is not observed in case of particles of a suspension.
3. The process by which crystals of a substance are formed from its saturated solution or its molten state either by cooling the solution or by sublimation is known as crystallisation.
4. The phenomenon of spontaneous mixing of two or more substances irrespective of their nature without any external help to form a homogeneous mixture is called diffusion.
5. Apart from water, compounds such as ethyl alcohol, methyl alcohol, acetone, chloroform, kerosene etc, are also used as solvents. These are known as aprotic solvents.
6. Use of volatile solvents may cause serious health hazards.

Chapter 4 Matter Solution Topic D Crystallisation Motion Of ParticlesIn Solution And Different Solvents Other Than Water Short And Long Answer Type Questions

Question 1. What are crystals? On which factors does the shape of a crystal depend?

Answer:

Crystals:-

1. Homogeneous solid substances having definite geometrical shape and bounded by symmetrically arranged plane surfaces that meet up at sharp edges are known as crystals. The substances formed of crystalline particles are known as crystalline substances. Some common examples are alum, diamond, sugar, copper sulphate etc.
2. The shape of a crystal entirely depends on the number of particles (atoms, molecules or ions) present in the crystal and their geometrical arrangement.

Question 2. What do you mean by crystallisation and residue?

Answer:

Crystallisation And Residue:-

Crystallisation: The process by which crystals of a substance are formed from its saturated solution or molten state either by cooling the solution or by sublimation is known as crystallisation.

Residue: The remaining solution that is left behind in the beaker after a solid is separated out as crystals from its saturated solution on cooling is called residue.

Question 3. What is the primary condition for the purification of a solid by crystallisation?

Answer:

The Primary Condition For The Purification Of A Solid By Crystallisation:-

To purify a solid by crystallisation method, the solid must be soluble in a specific solvent while the impurities present in the solid must be insoluble or have different solubilities in that particular solvent.

Question 4. Mention two important applications of crystallisation.

Answer:

Two Important Applications Of Crystallisation:-

1. Sea water contains different impurities along with dissolved salts. The method of crystallisation is applied for the removal of those impurities and extraction of salt from sea water.
2. Impure samples of potash alum [K₂SO₄ – AI₂(SO₄)₃– 24H₂O] and nitre (KNO₃) are purified by crystallisation.

Question 5. Crystallisation is a more effective method for purifying a solid than evaporation. Explain.

Answer:

Crystallisation Is A More Effective Method For Purifying A Solid Than Evaporation:-

1. Crystallisation is a more effective method for purifying a solid than evaporation because of the following reasons
2. During evaporation, the solvent is evaporated from the solution by applying heat. However on heating, some solids may decompose to produce new substances or substances like sugar may also get burnt to form a black mass of carbon particles.
3. During evaporation, the solid impurities present in the solution do not separate out from the solution. So, the residue may contain large amount of impurities along with the desired substance. Thus, the solid obtained by evaporation may not be pure. However, during crystallisation, the impurities remain in the solution, while the solid crystallises out in pure form.

Question 6. What is overgrowth? What is seed crystal? Give examples.

Answer:

Overgrowth:-

1. When a small crystal of the solute is dipped into its saturated solution with the help of a thread, the crystal grows bigger in size due to deposition of more solute crystals on its surface. This phenomenon is known as overgrowth of crystal.
2. Seed crystal is a small crystal of the solute, dipped into its saturated solution with the help of a thread. The crystal becomes larger due to deposition of more solute crystals on its surface.
3. If a small crystal of potash alum is dipped into a saturated solution of alum, alum starts depositing on the surface of the crystal and it grows bigger in size. Here, the small crystal is the seed crystal and the phenomenon is called overgrowth.

Question 7. Mention some uses of seed crystals in different industries.

Answer:

Uses Of Seed Crystals In Different Industries:-

1. Seed crystals find important applications in many industries, such as
2. During production of cane sugar (sucrose) from sugarcane juice, pure crystals of sugar are used as seed crystals.
3. Seed crystals are used to accelerate the crystallisation of precious stones (like diamond), gemstones (like sapphire) and semiconductors (like silicone, germanium etc.)
4. In Baeyer’s process of extraction of alumina, hydrated crystals of Al₂O₃ are used as seed crystals to accelerate precipitation of alumina.

Question 8. What is diffusion?

Answer:

Diffusion:-

The phenomenon of spontaneous mixing of two or more substances irrespective of their nature and. molar mass, without any external help to form a homogeneous mixture is called diffusion. In a solution, the solute particles move from a region of higher concentration to a region of lower concentration due to diffusion until a homogeneous mixture is formed.

Question 9. How does diffusion take place?

Answer:

Diffusion Take Place As Follows:-

In a solution, both solvent and solute molecules move incessantly due to their high kinetic energy.

Consequently, the solute molecules collide among themselves as well as with the solvent molecules leading to continuous and random motion (Brownian motion). As a result, the solute particles move from a region of higher concentration to a region of lower concentration by the process of diffusion until an equilibrium is reached.

Question 10. “Like dissolves like”— explain the statement with examples.

Answer:

“Like Dissolves Like”:-

When a solute is added to a solvent, three different types of interactions take place in the solution

1. Solvent-solvent interaction,
2. Solute-solute interaction and
3. Solvent-solute interaction.

When the nature and extent of these three interactions are similar, then the solute dissolves in the solvent. For example, polar solutes like sodium chloride dissolve easily in polar solvents like water.

In this solution, the interaction between two water molecules is almost identical to the interaction between sodium chloride molecules as well as that between water and sodium chloride molecules. Similarly, a non-polar solute is found to be highly soluble in non-polar organic solvents.

Question 11. Write some uses of solvent.

Answer:

Some Uses Of Solvent:-

In different industries, methyl alcohol is extensively used as a solvent for colours, varnishes, celluloid substances, cements, fats etc.

Question 12. Write some uses of ethyl alcohol as a solvent.

Answer:

Uses Of Ethyl Alcohol As A Solvent:-

Ethyl alcohol is a very important solvent. It is used in different industries to dissolve resins, soaps, varnishes, colours, rayons, lax, synthetic rubbers, synthetic fibres, dyes etc. Ethyl alcohol is also used extensively as a solvent in pharmaceutical industries.

Question 13. Write some uses of acetone as a solvent.

Answer:

Uses Of Acetone As A Solvent:-

Acetone is a widely used organic solvent in chemical industries. It is used as a solvent for acetylene and semi-synthetic polymers like nitrocellulose, cellulose acetate etc. It is also used to dissolve varnishes and polishing materials.

Question 14. Write some uses of kerosene as a solvent.

Answer:

Uses Of Kerosene As A Solvent:-

Kerosene is an important organic solvent. It is used to dissolve different organic substances as well as colours. It can also remove colour stains from clothes.

Question 15. Write some uses of chloroform as a solvent.

Answer:

Uses Of Chloroform As A Solvent:-

Chloroform is an excellent organic solvent. In different industries, it is used to dissolve substances like fat, rubber, wax, resin etc. It is also widely used in the extraction of oils, gums and alkaloids.

Question 16. Why are volatile solvents used to dissolve colours and varnishes?

Answer:

Volatile Solvents Used To Dissolve Colours And Varnishes:-

Colours and varnishes are generally dissolved in volatile solvents because of the following reasons

1. When the colour or varnish is applied on a surface, the volatile solvent evaporates easily. As a result, a smooth layer of colour or varnish is obtained.
2. When a colour is sprayed by dissolving it in a volatile solvent, less amount of colour is required thus, making the process more cost-effective.

Question 17. Discuss some harmful effects of using volatile solvents.

Answer:

Harmful Effects Of Using Volatile Solvents:-

In our everyday life, we come across a number of volatile solvents. Volatile organic solvents like methyl alcohol, methylated spirit, turpentine or tarpin oil, naphtha, acetone etc. are widely used as solvents for oil paints, paint thinners, spray paints, varnishes etc.

Being volatile in nature, these solvents evaporate very fast even at ordinary temperature and can enter our body through respiratory system and get absorbed in the blood through our lungs.

In fact, these solvents may directly come in contact with our skin. Thus exposure to these solvent may cause serious damage to our body. The harmful effects caused by volatile solvents can be categorised into two groups

1. Harmful effects due to short-term exposure and
2. Harmful effects due to long-term exposure.

Harmful effects due to short term exposure to volatile solvents include

1. Allergies,
2. Headache,
3. Asthma,
4. Nausea and vomiting tendency,
5. Drowsiness,
6. Burning sensation in respiratory tract eyes, nose etc.

Harmful effects due to long term exposure to volatile solvents include

1. Increased risk of cancer,
2. Damage of kidneys and liver,
3. Delayed childhood development and
4. Damage of central nervous system.

Question 18. Mention the harmful effects of ethyl alcohol, methyl alcohol and acetone when they are used as solvents.

Answer:

Harmful Effects Of Ethyl Alcohol:

It is used as a solvent for different medicines. On entering the human body, it affects the nervous system. Thus, transmission of nervous impulse across the body gets disrupted and the person suffers from dizziness, Ethyl alcohol also reacts with some medicines and cause adverse side effects.

Harmful Effects Of Methyl Alcohol:

Methyl alcohol (CH₃OH) is a volatile toxic compound and it decomposes to form formaldehyde (HCHO). If 10 mL CH₃OH enters into the body, it may cause blindness due to the effect of HCHO, if 30 mL enters, the person may become unconscious and if 100 mL enters the body, the person may even die.

Harmful Effects Of Acetone:

If excess acetone enters into the body through respiration, then it may lead to temporary nervous breakdown.

Question 19. Mention the harmful effects of chloroform and kerosfene when they are used as solvents.

Answer:

Harmful Effects Of Chloroform:

Excess chloroform on entering the body through respiration, paralyses the central nervous system. It can even cause cardiac problems leading to death. In presence of light and air, chloroform gets oxidised to form poisonous carbonyl chloride or phosgene (COCI₂) gas.

Harmful Effects Of Kerosene:

Kerosene has a suffocating and choking odour and it adversely affects living organisms.

Chapter 4 Matter Solution Topic D Crystallisation Motion Of ParticlesIn Solution And Different Solvents Other Than Water Very Short Answer Type Questions Choose The Correct Answers

Question 1. Brownian motion of colloidal particles is caused due to

1. Convection
2. Attraction between the particles of dispersed phase and dispersion medium
3. Unequal collision between the particles of dispersed phase and dispersion medium
4. Change in temperature

Answer: 3. Unequal collision between the particles of dispersed phase and dispersion medium

Question 2. The correct order for diffusion is

1. Solid > liquid > gas
2. Liquid > solid > gas
3. Gas > solid > liquid
4. Gas > liquid > solid

Answer: 4. Gas > liquid > solid

Question 3. The Brownian motion of colloidal particles indicates

1. Linear motion
2. Circular motion
3. Spiral motion
4. Random motion

Answer: 4. Random motion

Question 4. Which of the following is not an organic solvent?

1. Liquid ammonia
2. Benzene
3. Chloroform
4. Acetone

Answer: 1. Liquid ammonia

Question 5. Which of the following substances cannot form crystals by sublimation?

1. Iodine
2. Ammonium chloride
3. Sodium chloride
4. Benzoic acid

Answer: 3. Sodium chloride

Question 6. Just before crystallisation, the solution somewhat becomes

1. Saturated
2. Unsaturated
3. Supersaturated
4. Precipitated

Answer: 3. Supersaturated

Question 7. An example of a crystalline solid is

1. Glass
2. Wax
3. Pitch
4. Common salt

Answer: 4. Common salt

Question 8. An example of an amorphous solid is

1. Sugar
2. Salt
3. Glass
4. Blue vitriol

Answer: 3. Glass

Question 9. Which of the following is a method of crystallisation?

1. Sublimation
2. Filtration
3. Fractional distillation
4. Decantation

Answer: 1. Sublimation

Question 10. Mixing of the solute and solvent particles due to continuous, random motion of the particles is known as

1. Osmosis
2. Diffusion
3. Effusion
4. Distillation

Answer: 2. Diffusion

Question 11. An example of an organic solvent is

1. Water
2. Liquid ammonia
3. Acetone
4. Liquid SO₂

Answer: 3. Acetone

Question 12. An inorganic solvent except water is

1. Chloroform
2. Kerosene
3. Liquid ammonia
4. Ethyl alcohol

Answer: 3. Liquid ammonia

Question 13. An example of lyophilic sol is

1. Gold sol
2. Silver sol
3. Sulphur sol
4. Starch sol

Answer: 4. Sulphur sol

Question 14. An example of a lyophobic sol is

1. Ferric hydroxide sol
2. Starch sol
3. Gelatin
4. Soap

Answer: 1. Ferric hydroxide sol

Question 15. In which of the following diffusion does not take place?

1. Hydrogen and nitrogen
2. Oxygen and water
3. NaCI and Na₂SO₄
4. Sugar and water

Answer: 3. NaCI and Na₂SO₄

Question 16. In which of the following, there is no water of crystallisation?

1. Blue vitriol
2. Borax
3. Washing soda
4. Sugar

Answer: 4. Sugar

Question 17. Anhydrous copper sulphate, by nature, is

1. Deliquescent
2. Hygroscopic
3. Supersaturated
4. None of these

Answer: 2. Hygroscopic

Question 18. The residual solution after separating the crystal by crystallisation is

1. Saturated
2. Unsaturated
3. Supersaturated
4. Colloid

Answer: 1. Saturated

Question 19. Inorganic, non-aqueous solvent is

1. Chloroform
2. Liquid CO₂
3. Ethanol
4. Kerosene

Answer: 2. Liquid CO₂

Question 20. Universal solvent is

1. Alcohol
2. Water
3. Ether
4. All of the above

Answer: 2. Water

Question 21. Which solvent is used to remove nail polish?

1. Ethyl alcohol
2. Methyl alcohol
3. Acetone
4. Tarpene oil

Answer: 3. Acetone

Question 21. Crystal without water of crystallisation is

1. Common salt
2. Blue vitriol
3. Alum
4. Epsom salt

Answer: 1. Common salt

Chapter 4 Matter Solution Topic D Crystallisation Motion Of ParticlesIn Solution And Different Solvents Other Than Water Answer In Brief

Question 1. Define water of crystallisation.

Answer: One or more than one water molecule attached to the crystals of certain compounds on which the colour and shape of the crystal largely depend is known as water of crystallisation.

Question 2. Write the formula of common alum.

Answer: The formula of common alum is K₂SO₄ – AI₂(SO₄)₃– 24H₂O.

Question 3. Why does diffusion take place in a solution?

Answer: Diffusion takes place due to continuous random motion of solute and solvent particles in a solution.

Question 4. What is the effect of temperature on diffusion of a liquid?

Answer: With the rise in temperature, diffusion in a liquid occurs at a faster rate.

Question 5. Give some examples of organic solvents.

Answer: Ethyl alcohol, methyl alcohol, acetone, chloroform, kerosene, carbon tetrachloride etc., are some examples of organic solvents.

Question 6. Give some examples of inorganic solvents other than water.

Answer: Some examples of inorganic solvents other than water are liquid ammonia, liquid nitrogen dioxide, liquid sulphur dioxide etc.

Question 7. Name a poisonous organic solvent.

Answer: Methyl alcohol

Question 8. Name an organic solvent which is actually a mixture.

Answer: Kerosene is an organic solvent which is actually a mixture of hydrocarbons.

Question 9. Which volatile solvent acts as a anaesthetic?

Answer: Chloroform is used as anaesthetic.

Question 10. Give an example of a solvent which can dissolve both of sulphur and phosphorous.

Answer: Carbon disulphide (CS₂).

Question 11. Name two non-aqueous solvent.

Answer: Ethyl alcohol and chloroform.

Question 12. Name a crystal containing water of crystallisation.

Answer: Hypo or sodium thiosulphate (Na₂S₂O₃ – 5H₂O).

Question 13. Name the solvent used in paint and varnish.

Answer: Methylated spirit, tarpene oil, naptha, acetone etc.

Question 14. Give an example of efflorescent crystal.

Answer: Blue vitriol or blue stone i.e. CuSO₄, 5H₂O is an example of efflorescent crystal.

Question 15. Give an example of a crystal which control water of crystallisation but is not an efflorescent one.

Answer: Common alum [K₂SO₄ – AI₂(SO₄)₃– 24H₂O].

Question 16. What happens when blue vitriol is heated over 250°C?

Answer: Water of crystallisation of blue vitriol crystal evaporated upon heating over 250°C to produce white powder of amorphous copper sulphate.

Question 17. Mention the number of water of crystallisation in common alum.

Answer: 24

Chapter 4 Matter Solution Topic D Crystallisation Motion Of ParticlesIn Solution And Different Solvents Other Than Water Fill In the Blanks

Question 1. The size of a crystal depends upon the _______ and _______ of the constituent atoms or particles present in the crystal.

Answer: Size, arrangement

Question 2. Glass is a/an _______ solid.

Answer: Amorphous

Question 3. Acetone is an example of ________

Answer: Organic

Question 4. Number of water of crystallisation in glauber Salt is ______

Answer: 10

Question 5. Example of a crystalline compound without water of crystallisation is _________

Answer: Sodium chloride

Chapter 4 Matter Solution Topic D Crystallisation Motion Of ParticlesIn Solution And Different Solvents Other Than Water State Whether True Or False

Question 1. Brownian motion of colloidal particles is continuous and random in nature.

Answer: True

Question 2. In diffusion, the solute particles move from a region of higher concentration to lower concentration.

Answer: True

Question 3. Methyl alcohol on entering the body decomposes to form acetaldehyde.

Answer:  False

Question 4. Chloroform is a volatile solvent which is oxidised to form the toxic compound, phosgene.

Answer: True

Question 5. Seed crystals can accelerate the process of crystallisation of gemstones.

Answer: True

Question 6. Sublimation is a method of crystallisation.

Answer: True

Chapter 4 Matter Solution Topic D Crystallisation Motion Of ParticlesIn Solution And Different Solvents Other Than Water Miscellaneous Type Questions

Match The Column

1.

Answer: 1.D, 2. C, 3. A, 4. B

2.

Answer: 1. B, 2. D, 3. A, 4. C

Chapter 4 Matter Acids Bases And Salts Topic A Concept Of Acid And Bases Synopsis

1. According to Arrhenius theory of electrolytic dissociation, an acid is a hydrogen-containing compound which dissociates in aqueous solution to produce H⁺ ion (or H₃O⁺ ion) as the only cation. On the other hand, compounds that yield OH^– ions as the only anion in their aqueous solutions are called bases.
2. Acids like H₂SO₄, HCI, HNO₃ and bases like NaOH, KOH find important applications in different industries.
3. Metals do not produce hydrogen on reacting with cold and very dilute HNO₃. Only Mg and Mn can displace hydrogen from cold and very dilute nitric acid.
4. Hydrochloric acid is identified by using aqueous solution of silver nitrate. Aqueous solution of barium chloride is used to identify sulphuric acid. Nitric acid is identified by the ring test.
5. Strong acids and strong bases are highly corrosive in nature. So, proper precautionary measures are adopted while using them.
6. When concentrated nitric acid is refluxed With small amount of starch, a yellow-coloured solution of nitric acid containing different oxides of nitrogen is obtained. It is called fuming nitric acid.

Chapter 4 Matter Acids Bases And Salts Topic A Concept Of Acid And Bases Short And Long Answer Type Questions

Question 1. Define acids in the light of Arrhenius’s theory of electrolytic dissociation.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

Answer:

Arrhenius’s Theory Of Electrolytic Dissociation :-

According to Arrhenius theory of electrolytic dissociation, an acid is defined as a hydrogen-containing compound that dissociates in; water to produce H⁺ ions (or H₃O⁺ ions) as the only cation.

For example, HCI ionises in water to produce H⁺ (or H₃O⁺) ions as the only cation. Hence, HCI is an acid.

\(\mathrm{HCl} \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-} ; \mathrm{H}^{+}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}\)

Question 2. Mention the general properties of acids.

Answer:

General Properties Of Acids Are

1. Acids are ionised in aqueous solution forming H^⊕ ion (H₃O^⊕ ion actually) and thus can conduct electricity.
2. Acids are sour in taste.
3. Acids turn blue litmus paper red.
4. Acids generally produce hydrogen gas in reaction with metal.
5. Acids generally produce carbon dioxide in reaction with carbonates and bicarbonates.
6. Acids produce salt and water in reaction with alkali.

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Question 3. Why do free H⁺ Ions not exist in water or aqueous solution?

Answer:

H⁺ ions combine with electronegative oxygen atoms in water to produce hydroxonium or hydronium ions (H₃O⁺). So, free H⁺ ions do not exist in water or aqueous solution.

Question 4. What are inorganic or mineral acids and give some examples.

Answer:

Inorganic Acids :

Acids which are obtained from minerals or produced from inorganic substances are known as mineral acids. Hydrochloric acid (HCI), sulphuric acid (H₂SO₄), nitric acid (HNO₃) are some examples of mineral acids.

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Question 5. What are organic acids and give some examples.

Answer:

Organic Acids :

Acids (containing carbon atoms) which are obtained from plants or animals are called organic acids. For example, formic acid is found in ant venom, lemon contains citric acid, lactic acid is present in curd etc.

Question 6. What is meant by ionisation of an acid in aqueous solution and give example.

Answer:

Ionisation Of An Acid In Aqueous Solution

In aqueous solution, an acid dissociates to form H⁺ ions which are unstable in nature. So these ions combine with water molecules to form hydronium ions (H₃O⁺). This decomposition of an acid in its aqueous solution is known as the ionisation of an acid.

\(\mathrm{HCl}(a q)+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

Question 7. What are strong acids and give examples.

Answer:

Strong Acids :

Acids which almost completely dissociate in aqueous solution to produce large amount of H₃O⁺ ions are called strong acids. Some examples are hydrochloric acid (HCI), sulphuric acid (H₂SO₄), nitric acid (HNO₃), perchloric acid (HCIO₄) etc.

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Question 8. What are weak acids and give examples.

Answer:

Weak Acids :

Acids which partially dissociate in aqueous solution to produce small amount of H₃O⁺ ions and most of the molecules remain undissociated in the solution are called weak acids. Some examples are acetic acid (CH₃COOH), carbonic acid (H₂CO₃), citric acid etc.

Question 9. Give definition of a base according to Arrhenius’s theory.

Answer:

Arrhenius’s Theory :

According to Arrhenius theory, a base can be defined as a compound (mainly the metal oxides and metal hydroxides) which on dissociating in water produces hydroxyl ions (OH^–) as the only anion.

Na₂O, NaOH, CaO are some examples of bases.

Question 10. Mention the general properties of bases.

Answer:

General Properties Of Bases Are :

1. Bases form OHe ion as anion when they are dissolved in water.
2. Bases taste bitter and are generally soapy in nature.
3. Bases turn red litmus paper blue.
4. Bases form salt and water in reaction with acids.

Question 11. NaOH is a strong base but NH₄OH is a weak base. Explain.

Answer:

Difference Between Strong And Weak Base :

NaOH dissociates completely into Na^⊕ and OH^Θ ions in aqueous solution.

⇒ \(\mathrm{NaOH} \rightleftharpoons \mathrm{Na}^{\oplus}+\mathrm{OH}^{\ominus}\)

But NH₄OH dissociates partially into NH^⊕ and OH^Θ ion in aqueous solution.

⇒ \(\mathrm{NH}_4 \mathrm{OH} \rightleftharpoons \mathrm{NH}_4^{\oplus}+\mathrm{OH}^{\ominus}\)

Depending of the nature of ionic dissociation in their respective aqueous solutions, NaOH acts as a strong base and NH₄OH acts as a weak base.

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Question 12. Mention the limitations of Arrhenius acid-base concept.

Answer:

Limitations Of Arrhenius Acid-Base Concept :

The limitations of Arrhenius acid-base concept are as follows

1. According to this concept, an acid or a base is defined on the basis of their dissociation in aqueous solutions. Thus, the concept is limited to aqueous medium only. It cannot explain the acidic or basic property of compounds in solvents other than water (non-aqueous solvents) like liquid ammonia, benzene etc.

2. According to this concept, a base must yield OH^– ions in aqueous medium. Thus, the theory fails to explain the basic nature of compounds like ammonia (NH₃), methyl amine (CH₃NH₂), aniline (C₆H₅NH₂) etc. Similarly acidic nature of compounds like BF₃ or AICI3 cannot be explained by this theory as these compounds do not produce H⁺ ions in aqueous medium.

Question 13. All acids are hydrogen-containing compounds, but all hydrogen-containing compounds are not acids—justify the statement with example.

Answer:

All Acids Are Hydrogen-Containing Compounds, But All Hydrogen-Containing Compounds Are Not Acids :

According to Arrhenius theory, an acid must contain H-atom. However, a compound containing hydrogen will be called an acid only if it has the following properties

1. It will dissociate in aqueous solution to produce H₃O⁺ (hydroxonium) ions.
2. It will produce hydrogen gas on reacting with metals.
3. It will react with a base or an alkali to produce salt and water.
4. Its aqueous solution will turn blue litmus red. However there are compounds containing hydrogen, which do not exhibit these properties.

For Example

1. Compounds like PH₃, NH₃, CH₄, sugar (C₁₂H₂₂O₁₁) do not H₃O⁺ ions in aqueous solution. Neither do they react with metals to produce hydrogen gas nor do they react with bases to form salt and water.
2. Sodium reacts with H₂O to produce hydrogen gas but the reaction produces NaOH instead of a salt.
3. Al, Zn etc. react with compounds like sodium hydroxide, potassium hydroxide etc. to produce hydrogen gas but these compounds do not yield H₃O⁺ ions in their aqueous solution. So, these compounds are not acids.

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Question 14. Why are CaH₂ and CH₄ not acids?

Answer:

CaH₂ and CH₄ Are Cannot Be Termed As An Acids Because :

CaH₂ produces H^⊕ ion or hydride ion. That is why it cannot be termed as an acid.

\(\mathrm{CaH}_2 \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{H}^{\ominus}\)

CH₄ is a covalent compound and cannot form H^⊕ ion in aqueous solution. Hence CH₄ cannot be termed as acid.

Question 15. All alkalis are bases, but all bases are not alkalis—explain.

Answer:

All Alkalis Are Bases, But All Bases Are Not Alkalis :

A base dissolves in water to form an alkali.

However, all bases are not soluble in water and these bases do not form alkalis. Thus, all alkalis are bases, but bases insoluble in water are not alkalis.

For example, Na₂O is a base. It dissolves in water to form NaOH. Sodium hydroxide (NaOH), potassium hydroxide (KOH), calcium hydroxide [Ca(OH)₂] dissolve in water to give OH^– ions. So these are alkalis.

⇒ \(\mathrm{NaOH} \rightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-} ; \mathrm{KOH} \rightarrow \mathrm{K}^{+}+\mathrm{OH}^{-}\)

⇒ \(\mathrm{Ca}(\mathrm{OH})_2 \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-}\)

On the other hand, aluminium hydroxide [AI(OH)₃], zinc hydroxide [Zn(OH)₂], ferric hydroxide [Fe(OH)₃] etc. react with acids to form salt and water. So they are bases. However, as they are insoluble in water, they cannot be called alkalis.

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Question 16. Sodium hydroxide (NaOH) is a base, but  CIOH is an acid. Why?

Answer:

Sodium Hydroxide (NaOH) Is A Base, But  CIOH Is An Acid Because :

NaOH dissociates in water to produce Na⁺ and OH^– ions. As it yields OH^– ions in water, it is called a base. On the other hand, CIOH dissociates in water to give CIO^– and H⁺ ions.

H⁺ ions further combine with water molecules to form H₃O⁺ ions. As it produces H₃O⁺ ions in aqueous solution, it is an acid.

Question 17. State whether AI(OH)₃ is a base or an alkali. Justify your choice.

Answer:

Aluminium Hydroxide AI(OH)₃ Is Not An Alkali.

Water soluble metal hydroxides are known as alkalis. Aluminium hydroxide [AI(OH)₃] is a base as it reacts with acid to produce salt and water but it is insoluble in water. So, aluminium hydroxide is not an alkali.

Question 18. Give an example of a reaction where two gases react to form a solid.

Answer:

Example Of A Reaction Where Two Gases React To Form A Solid :

Ammonia gas reacts with hydrogen chloride gas to form fine particles of solid ammonium chloride which floats in air forming white fumes.

Equation: \(\mathrm{NH}_3+\mathrm{HCl} \rightarrow \mathrm{NH}_4 \mathrm{Cl}\)

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Question 19. Write down equations for the reaction of HCI with NaOH, Mg(OH)₂ and AI(OH)₃ respectively.

Answer:

Equations For The Reaction Of HCI With NaOH, Mg(OH)2 And AI(OH)3 Respectively : 

HCI forms corresponding salts and water in reaction with the mentioned bases:

⇒ \(\mathrm{NaOH}+\mathrm{HCl} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{Mg}(\mathrm{OH})_2+2 \mathrm{HCl} \rightarrow \mathrm{MgCl}_2+2 \mathrm{H}_2 \mathrm{O}\)

∴ \(\mathrm{Al}(\mathrm{OH})_3+3 \mathrm{HCl} \rightarrow \mathrm{AlCl}_3+3 \mathrm{H}_2 \mathrm{O}\)

Question 20. Give examples of chemical reactions of hydrochloric acid with carbonates and bicarbonate compounds.

Answer:

Chemical Reactions Of Hydrochloric Acid With Carbonates And Bicarbonate Compounds :-

Hydrochloric acid reacts with carbonate and bicarbonate compounds to liberate carbon dioxide and produce corresponding chloride salts.

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{HCl} \rightarrow 2 \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

⇒ \(\mathrm{NaHCO}_3+\mathrm{HCl} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Question 21. Under what condition does Cu react with HCI? Give equation.

Answer:

Condition For The Reaction Of Cu With HCI

Copper slowly reacts with hot and concentrated HCI in the presence of air or oxygen O₂(g) to produce cupric chloride (CuCI₂) and water.

Question 22. Give an example of the reducing property of hydrochloric acid (HCI).

Answer:

Reducing Property Of Hydrochloric Acid (HCI) :

Hydrochloric acid reduces manganese dioxide (MnO₂) to manganous chloride (MnCI₂) and itself gets oxidised to evolve greenish-yellow chlorine gas.

⇒ \(\mathrm{MnO}_2+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_2+\mathrm{Cl}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}\)

Question 23. How will you identify hydrochloric acid? Give corresponding equation.

Answer:

Identification Of Hydrochloric Acid:

When silver nitrate (AgNO₃) solution is added to an aqueous solution of hydrochloric acid (HCI), a curdy white precipitate of silver chloride (AgCI) is obtained. The precipitate is insoluble in nitric acid but dissolves readily in excess ammonium hydroxide (NH₄OH) solution.

⇒ \(\mathrm{AgNO}_3+\mathrm{HCl} \rightarrow \mathrm{AgCl} \downarrow+\mathrm{HNO}_3
(white ppt.)\)

Question 24. Mention some uses of hydrochloric acid in industries.

Answer:

Uses Of Hydrochloric Acid In Industries.

Some important uses of hydrochloric acid are as follows

1. In the preparation of chlorine and different metal chlorides.
2. Preparation of aqua regia and glucose (from starch) and in removing scales from boilers
3. In dye industries, tanneries, pharmaceutical industries and during galvanisation and tinplating of iron.

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Question 25. What happens when carbonate and bicarbonate compounds react with nitric acid (HNO₃)? Give equations.

Answer:

Nitric acid reacts with carbonate and bicarbonate compounds to liberate carbon dioxide and produce corresponding nitrate salts.

\(\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{HNO}_3 \rightarrow 2 \mathrm{NaNO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\) \(\mathrm{NaHCO}_3+\mathrm{HNO}_3\rightarrow \mathrm{NaNO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

Question 26. Write with equation what happens when copper turnings are heated with hot and concentrated nitric acid.

Answer:

When copper turnings are heated with hot and concentrated nitric acid, brown fumes of nitrogen dioxide (NO₂) gas is evolved. Copper is oxidised by nitric acid to copper nitrate [Cu(NO₃)₂] which makes the solution bluish-green in colour.

\(\mathrm{Cu}(s)+4 \mathrm{HNO}_3(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2(a q)+2 \mathrm{NO}_2(g) \uparrow+2 \mathrm{H}_2 \mathrm{O}(l)\)

Question 27. Give An example of the oxidising property of nitric acid.

Answer:

Example Of The Oxidising Property Of Nitric Acid :

Hot and concentrated nitric acid oxidises metallic zinc to zinc nitrate [Zn(NO₃)₂] and itself gets reduced to nitrogen dioxide.

\(\mathrm{Zn}+4 \mathrm{HNO}_3 \rightarrow \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}\)

Question 28. Which acid is termed as ‘aqua fortis’ or ‘strong water’? How can hydrogen be produced form that acid?

Answer:

1. Nitric acid is termed as ‘aqua fortis’.
2. In the reaction with Magnesium (Mg), very dilute (1%) nitric acid produces hydrogen.

⇒ \(\mathrm{Mg}+2 \mathrm{HNO}_3 \rightarrow \mathrm{Mg}\left(\mathrm{NO}_3\right)_2+\mathrm{H}_2 \uparrow\)

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Question 29. What is fuming nitric acid?

Answer:

Fuming Nitric Acid :

When concentrated nitric acid (HNO₃) is refluxed with starch, a brownish-yellow solution is obtained containing different oxides of nitrogen (mainly NO₂) dissolved in it. This solution is called fuming nitric acid. It has been so named because the dissolved NO₂ continuously comes out from the solution as brown fumes.

Question 30. Why does concentrated nitric acid turn yellow in presence of sunlight?

Answer:

In presence of sunlight, concentrated nitric acid decomposes to form brown-coloured NO₂ gas. This gas dissolves in nitric acid and turns the solution yellow.

⇒ \(4 \mathrm{HNO}_3 \rightarrow 4 \mathrm{NO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \uparrow\)

Question 31. Prove that oxygen is present in nitric acid.

Answer:

Oxygen Is Present In Nitric Acid :

When concentrated HNO₃ is added drop-wise on red-hot pumice stone, a brown gas mixture evolves. When this gas mixture in passed through a ‘Ll’ tube dipped in freezing mixture, a colourless gas is evolved through ‘U’ tube which ignites a flameless burning stick.

The colourless gas gets absorbed by the alkaline potassium pyrogallate solution and turns the colour of the solution brown. So the gas is oxygen which comes from nitric acid.

\(4 \mathrm{HNO}_3 \rightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{NO}_2+\mathrm{O}_2\)

This reaction proves the presence of oxygen in nitric acid.

Question 32. Concentrated nitric acid can not be kept in copper vessel but the same can be kept in aluminium vessel. Why?

Answer:

Nitric acid produces brown fumes of nitrogen dioxide and copper nitrate if it is kept in a copper vessel.

⇒ \(\mathrm{Cu}+4 \mathrm{HNO}_3 \rightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O}\)

That is why concentrated HNO₃ can not be kept in copper vessel.

If concentrated nitric acid is kept in an aluminium vessel, nitric acid reacts with aluminium to form aluminium oxide. This aluminium oxide forms a thin layer which is insoluble in HNO₃.

So the concentrated HNO₃ can not come in contact with aluminium furthermore. That is why concentrated HNO₃ can be kept in aluminium vessel.

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Question 33. Describe ring test performed for identification of nitric acid. Give relevant equations.

Answer:

Ring Test Performed For Identification Of Nitric Acid :

A sample of the given solution is taken in a test tube and freshly prepared ferrous sulphate (FeSO₄) solution is added to it. Then the solution is cooled and concentrated H₂SO₄ is gradually poured into the solution along the side of the test tube.

Sulphuric add being denser moves down and a brown ring is formed at the junction of the two liquid layers (acid + solution consisting of FeSO₄ and sample). This test confirms the presence of nitric acid or any other nitrate salt.

HNO₃ reacts with FeSO₄ in the presence of concentrated H₂SO₄ to produce nitric oxide (NO). The formed nitric acid then combines with FeSO₄ to form a complex [Fe(H₂O)₅NO]SO₄ [pentaaquanitrosyliron (II) sulphate] which is responsible for the formation of the brown ring.

⇒ \(2 \mathrm{HNO}_3+3 \mathrm{H}_2 \mathrm{SO}_4+6 \mathrm{FeSO}_4 \longrightarrow 3 \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+2 \mathrm{NO}+4 \mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{FeSO}_4+6 \mathrm{H}_2 \mathrm{O} \rightarrow\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{SO}_4\)

⇒ \({\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{SO}_4+\mathrm{NO} \longrightarrow}{\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}_3\right] \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}}\)

Question 34. Why is freshly prepared ferrous sulphate solution used in the ring test for the identification of nitric acid?

Answer:

The Reason For Freshly Prepared Ferrous Sulphate Solution Used In The Ring Test For The Identification Of Nitric Acid Is : 

In the presence of atmospheric oxygen, ferrous sulphate solution is rapidly oxidised to ferric sulphate. Ferric sulphate cannot form any complex with NO.

Hence, ferrous sulphate solution stored in laboratory cannot form the brown ring under the given reaction conditions. Thus, freshly prepared ferrous sulphate is used in the ring test for identification of nitric acid.

Question 35. The brown ring formed in the ring test disappears when the test tube is shaken. Why?

Answer:

The Brown Ring Formed In The Ring Test Disappears When The Test Tube Is Shaken Because:

When the test tube is shaken, concentrated H₂SO₄ combines with water and evolves large amount of heat which decomposes the complex compound, [Fe(H₂O)₅NO]SO₄ and liberates nitric oxide (NO) gas.

NO comes out as bubbles from the test tube. As the complex responsible for ring formation decomposes, the brown ring disappears.

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Question 36. Mention some uses of nitric acid in industries.

Answer:

Uses Of Nitric Acid In Industries:

Some important uses of nitric acid in industries are as follows

1. It is used in the manufacture of different explosives like nitroglycerine, picric acid, TNT (trinitrotoluene) etc.
2. It is used in the production of celluloids, synthetic colours, synthetic silk etc.
3. HNO₃ is used in the production of electrical cells, nitrate salts, fertilisers, aqua regia etc. with a glass rod. As a result, the produced heat is evenly distributed, thereby reducing the chance of any accident.

Question 37. What happens when carbonate and bicarbonate compounds react with sulphuric acid (H₂SO₄)? Give equations.

Answer:

Carbonate and bicarbonate compounds react with sulphuric acid to liberate carbon dioxide and produce corresponding sulphate salts.

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

⇒ \(\mathrm{NaHCO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{NaHSO}_4+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

Question 38. Give an example of the oxidising property of sulphuric acid.

Answer:

Example Of The Oxidising Property Of Sulphuric Acid:

Hot and concentrated sulphuric acid oxidises metallic zinc to zinc sulphate (ZnSO₄) and itself gets reduced to sulphur dioxide.

⇒ \(\mathrm{Zn}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{ZnSO}_4+\mathrm{SO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}\)

Question 39. Give suitable conditions and equations of the reaction between copper and sulphuric acid.

Answer:

Copper reacts with hot and concentrated sulphuric acid to produce blue-coloured copper sulphate and sulphur dioxide gas which has a smell of burnt sulphur.

⇒ \(\mathrm{Cu}(s)+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CuSO}_4(a q)+\mathrm{SO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}(l)\)

Question 40. Give an example of hygroscopic property of concentrated sulfuric acid.

Answer:

When cone. H₂SO₄ is added to blue vitriol crystals, white powder of anhydrous copper sulfate is formed. This is an example of the hygroscopic property of cone. H₂SO₄ .

⇒ \(\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O} \stackrel{\text { Conc. } \mathrm{H}_2 \mathrm{SO}_4}{-5 \mathrm{H}_2 \mathrm{O}} \mathrm{CuSO}_4\)

Question 41. What happens when a few drops of cone. H₂SO₄ are added to sugar?

Answer:

Sugar turns to black carbon when a few drops of cone. H₂SO₄ are added to it.

⇒ \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \frac{\text { Conc. } \mathrm{H}_2 \mathrm{SO}_4}{-11 \mathrm{H}_2 \mathrm{O}} 12 \mathrm{C} \text { (black) }\)

 

Question 42. Fe reacts with dilute H₂SO₄ to liberate hydrogen, but Cu does not—explain.

Answer:

Only metals present above hydrogen in the metal activity series can replace hydrogen from a dilute acid. Iron is placed above hydrogen while copper is placed below hydrogen in the metal activity series.

So iron reacts with dilute sulphuric acid to liberate hydrogen but copper is not able to do so. It must be noted that, copper does not react with dilute sulphuric acid but it is oxidised by hot and concentrated sulphuric acid.

Question 43. How will you identify sulphuric acid? Give equation.

Answer:

When barium chloride (BaCI₂) solution is added to an aqueous solution of sulphuric acid (H₂SO₄), white barium sulphate (BaSO₄) is precipitated which is insoluble in hydrochloric acid (HCI) or nitric acid (HNO₃).

⇒ \(\mathrm{H}_2 \mathrm{SO}_4+\mathrm{BaCl}_2 \rightarrow \mathrm{BaSO}_4 \downarrow+2 \mathrm{HCl}\)

Question 44. Three bottles contain equal volumes of concentrated HCI, concentrated HNO₃ and concentrated H₂SO₄ How will you identify the bottle containing H₂SO₄  without conducting any chemical test?

Answer:

The specific gravity of concentrated H₂SO₄ is much greater than that of concentrated HCI and concentrated HNO₃. Hence, the bottle which is heaviest contains concentrated H₂SO₄.

Question 45. What is oleum?

Answer:

Oleum:

When an excess amount of SO₃ gas is passed through 98% cone. H₂SO₄, the gas gets absorbed by the acid and an oily liquid substance of formula H₂S₂O₃ is formed. This substance is termed as oleum.

⇒ \(\mathrm{H}_2 \mathrm{SO}_4+\mathrm{SO}_3 \rightarrow \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7\)

Question 46. Prove the presence of hydrogen in sulphuric acid.

Answer:

Presence Of Hydrogen In Sulphuric Acid :

At room temperature, dil. H₂SO₄ and granulated zinc react to produce a colourless, odourless gas in which an ignited splint is extinguished and the gas burns with blue flame in air forming a colourless liquid.

This colourless liquid turns the white anhydrous copper sulphate blue. So this colourless liquid is water which was formed by the combustion of the colourless, odourless gas.

Therefore, the said gas is hydrogen which was formed by the reaction of granulated zinc and dil. H₂SO₄. Hence the pressence of hydrogen in H₂SO₄ is proved.

⇒ \(\mathrm{Zn}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{ZnSO}_4+\mathrm{H}_2\)

Question 47. Mention some uses of sulphuric acid in industries.

Answer:

Some important uses of sulphuric acid are as follows

1. H₂SO₄is used in the preparation of inorganic acids like hydrochloric acid and nitric acid.
2. H₂SO₄ is used in the preparation of fertilisers like super phosphate, ammonium sulphate.
3. Sulphuric acid finds application in the production of alum, glucose, ethers, alcohols, dyes, explosives etc.
4. It is used for processes such as, petroleum refining and galvanisation.
5. H₂SO₄ is also used in lead storage batteries, extraction of metals and manufacture of films and synthetic silk.

Question 48. Acids are not stored in metal containers. Explain.

Answer:

Acids are not stored in metal containers.

Metals placed above hydrogen in the activity series react with acids to produce salts and hydrogen gas. So, acids cannot be kept in contact with active metals. For example, Fe reacts with dilute H₂SO₄ to produce FeSO₄ salt and hydrogen gas.

Question 49. Nowadays, the contept of nascent hydrogen is considered to be obsolete— explain with an example.

Answer:

When hydrogen gas is passed through an aqueous solution of ferric chloride acidified with dilute H₂SO₄, no change is observed. Flowever when, zinc dust is added to the solution, ferric chloride (FeCl₃) is reduced to ferrous chloride (FeCI₂) and the solution turns colourless.

Earlier, it was considered that nascent hydrogen (which is more reactive in nature) produced due to the reaction between zinc dust and dilute sulphuric acid was responsible for the reduction of ferric chloride to ferrous chloride.

However, it has been proved that the reduction is due to the electrons released by the metals in the acid solution. Nascent hydrogen has no role in reducing the compounds, rather the metal acts as the reductant.

Zn atoms release electrons and get converted to Zn²⁺ ions. Fe³⁺ ions accept those electrons and are reduced to Fe²⁺ ions. Again H⁺ ions present in the solution are also reduced to hydrogen gas simultaneously by accepting the electrons released by zinc atoms.

⇒ \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 e \text { (oxidation) }\)

⇒ \(2 \mathrm{Fe}^{3+}+2 e \rightarrow 2 \mathrm{Fe}^{2+} \text { (reduction) }\)

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

⇒ \(\mathrm{H}^{+}+e \rightarrow \mathrm{H} \text { (reduction); } \mathrm{H}+\mathrm{H} \rightarrow \mathrm{H}_2 \uparrow\)

It should also be noted that, nowadays the concept of nascent oxygen is also considered to be obsolete.

Question 50. Write with equation what happens when zinc dust is added to hot and concentrated caustic soda solution.

Answer:

Zinc dust reacts with hot and concentrated caustic soda solution (NaOH) to produce sodium zincate (Na₂ZnO₂) salt and hydrogen gas.

⇒ \(\mathrm{Zn}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2 \mathrm{ZnO}_2+\mathrm{H}_2 \uparrow\)

Question 51. Concentrated sodium hydroxide (NaOH) solution should not be boiled in aluminium vessels. Why?

Answer:

Aluminium reacts with sodium hydroxide solution (NaOH) to form soluble sodium aluminate salt along with the evolution of hydrogen gas. As a result, the vessels get corroded.

⇒ \(2 \mathrm{Al}+2 \mathrm{NaOH}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaAlO}_2+3 \mathrm{H}_2 \uparrow\)

So, concentrated NaOH solution should not be boiled in aluminium vessels.

Question 52. Mention some uses of sodium hydroxide in industries.

Answer:

Sodium hydroxide is a widely used chemical in different industries. Some of its uses are as follows

1. Sodium hydroxide is used in the manufacture of soaps, paper and synthetic fibres.
2. It is used in the production of organic dyes.
3. It is used in the refining of petroleum-based products.
4. Caustic soda (NaOH) is used as a reagent in the industrial preparation of different chemicals.

Question 53. Concentrated acids are diluted by adding the acid into water and not by adding water into the acid. Why?

Answer:

Large amount of heat is evolved when water is added to acid and this causes the acid to boil. As a result, the acid may spurt out from the beaker and cause serious accident.

Apart from this, the glass beaker may break due to evolution of excess amount of heat. So, the acid is gradually added to large amount of water with continuous stirring.

Question 54. Why should one be cautious while working with concentrated acids or Concentrated alkalis?

Answer:

Concentrated acids or concentrated alkalis are highly corrosive substances. They cause blister-like wounds on human skin. Burns caused due to acids and alkalis are known as acid burns and alkali burns respectively.

Concentrated acids are extremely harmful for eyes. Apart from these, acids may cause burning sensation, blackening of skin and also damage the living tissues.

Acid vapours if inhaled may cause breathing troubles. Concentrated alkalis also produce similar problems. Thus, one should, be careful while working with concentrated acids or concentrated alkalis.

Question 55. What do you mean by acid burns and alkali bums?

Answer:

Acid burns and alkali bums

When concentrated mineral acids like HCI, H₂SO₄, HNO₃ etc., and concentrated alkalis like NaOH, KOH etc., come in contact with our skin, these cause blisters on our skin and even burn it.

The burns caused by the corrosive action of acids are known as acid burns while the burns caused by the corrosive action of strong alkalis ate known as alkali burns.

Question 56. What are the steps that should be immediately taken in case of an acid burn or an alkali burn?

Answer:

Concentrated acids or alkalis cause blisters on our skin. In such cases, the affected part should be immediately washed with plenty of water repeatedly.

In case of an acid burn, the affected part should be treated with 1% dilute sodium bicarbonate solution and in case of an alkali burn, the affected part should be washed with 1% acetic acid solution. After the preliminary treatment is done, one should consult a doctor.

Question 57. Human skin becomes yellowish when it mcomes in contact with dilute nitric acid. Why?

Answer:

Dilute nitric acid reacts with proteins present in the human skin to form yellow-coloured xanthoproteic acid. This makes the skin yellowish.

Question 58. Human skin blackens when it is exposed to concentrated sulphuric acid. Why?

Answer:

Concentrated sulphuric acid is strongly hygroscopic and corrosive in nature. When it comes in contact with human skin, it absorbs water from the skin causing the skin to blacken.

Question 59. What changes will be observed in the colour of a litmus paper when it comes in contact with dry HCI gas and an aqueous solution of hydrogen chloride gas?

Answer:

No change will be observed in the colour oflitmus paper when it comes in contact with dry HCI gas. However, the litmus paper turns red in aqueous solution of hydrogen chloride gas.

Question 60. HCI gas is not an acid, but aqueous solution of HCI is strongly acidic. Why?

Answer:

Hydrogen chloride is a covalent compound. It does not dissociate in gaseous state or in liquid state to produce H⁺ ion. Hence, hydrogen chloride gas is not an acid but is a neutral compound.

However, in aqueous solution, polar water molecules separate the partially positively charged H-atom from partially negatively charged Cl-atom in the polar H — Cl molecule.

Consequently, HCI molecules ionise to form H⁺ and Cl^– ions. H⁺ ions further combines with water molecules to form H₃O⁺ ions and thus, aqueous solution of HCI shows acidic property.

For example, if a dry mixture of sodium bicarbonate and tartaric acid is taken in a glass and the mouth of the glass is covered with a deflated balloon, no change is observed.

However, if some water is added to the glass and the deflated balloon is placed at the mouth of the flask, it is seen that the balloon gets inflated due to the liberation of CO₂ gas.

Question 61. A mixture of tartaric acid crystals and sodium bicarbonate undergo chemical reaction only in presence of water. Explain with reason.

Answer:

Tartaric acid and sodium bicarbonate do not react with each other in dry state because they do not behave as acids or bases in dry state. However, when water is added to the mixture, tartaric acid ionises to produce H⁺ ions which in turn react with bicarbonate ions (HCO₃) produced due to dissociation of sodium bicarbonate to liberate CO₂ gas.

⇒ \(\mathrm{H}^{+}+\mathrm{HCO}_3^{-} \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

Question 62. HCI is added to silver nitrate solution.

Answer:

When silver nitrate (AgNO₃) solution is added to hydrochloric acid, insoluble silver chloride (AgCI) is produced which forms a curdy white precipitate.

⇒ \(\mathrm{AgNO}_3+\mathrm{HCl} \underset{\text { (curdy white) }}{\rightarrow \mathrm{AgCl} \downarrow}+\mathrm{HNO}_3\)

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Question 63. Dilute hydrochloric acid is added to calcium carbonate.

Answer:

Calcium carbonate or marble reacts with dilute hydrochloric acid to produce calcium chloride (CaCI₂) and carbon dioxide gas.

\(\mathrm{CaCO}_3+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

Question 64. A mixture of concentrated HCI and Mno2 is heated.

Answer:

When a mixture of manganese dioxide (MnO₂) and concentrated HCI is heated, HCI reduces MnO₂ to manganous chloride (MnCl₂) and itself gets oxidised to evolve greenish-yellow chlorine gas.

⇒ \(\mathrm{MnO}_2+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_2+\mathrm{Cl}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}\)

Question 65. Ammonia is allowed to react with hydrogen chloride gas.

Answer:

Ammonia gas reacts with hydrogen chloride (HCI) gas to produce dense white fumes of ammonium chloride (NH₄CI).

⇒ \(\mathrm{NH}_3(g)+\mathrm{HCl}(g) \rightarrow \mathrm{NH}_4 \mathrm{Cl}(s)\)

Question 66. Dilute hydrochloric acid (HCI) is kept in an aluminium vessel.

Answer:

If dilute HCI is kept in an aluminium vessel, then aluminium reacts with the dilute acid to produce aluminium chloride (AlCI₃) along with evolution of hydrogen gas. Thus, the vessel gets corroded.

⇒ \(2 \mathrm{Al}+6 \mathrm{HCl} \rightarrow 2 \mathrm{AlCl}_3+3 \mathrm{H}_2 \uparrow\)

Question 67. Aqueous sodium carbonate is treated with hydrogen chloride (HCI) gas.

Answer:

Aqueous sodium carbonate reacts with hydrogen chloride (HCI) gas to produce sodium chloride, carbon dioxide and water.

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{HCl} \rightarrow 2 \mathrm{NaCl}+\mathrm{CO}_2 \uparrow+\mathrm{H}_2 \mathrm{O}\)

Question 68. Metallic copper Is treated with concentrated HCI in presence of oxygen.

Answer:

Metallic copper slowly reacts with hot and concentrated HCI in presence of oxygen to produce cupric chloride (CuCI₂) and water.

⇒ \(2 \mathrm{Cu}(s)+\mathrm{O}_2(g)+4 \mathrm{HCl}(a q) \rightarrow \stackrel{\Delta}{2 \mathrm{CuCl}_2(a q)+2 \mathrm{H}_2 \mathrm{O}(I)}\)

Question 69. Concentrated HNO₃ is added dropwiseon hot charcoal.

Answer:

When concentrated nitric acid is added dropwise to hot charcoal (carbon), carbon gets oxidised to carbon dioxide (CO₂) gas while nitric acid gets reduced to form reddish-brown nitrogen dioxide (NO₂) gas.

⇒ \(\mathrm{C}+4 \mathrm{HNO}_3 \rightarrow \mathrm{CO}_2 \uparrow+4 \mathrm{NO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}\)

Question 70. Copper is heated with concentrated nitric acid.

Answer:

When copper is heated with concentrated nitric acid, copper is oxidised by nitric acid to form cupric nitrate. [Cu(NO₃)₂]. Nitric acid itself gets reduced to nitrogen dioxide.

⇒ \(\mathrm{Cu}+4 \mathrm{HNO}_3 \rightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}\)

Question 71. Hot and concentrated nitric acid is kept in an iron vessel.

Answer:

Hot and concentrated nitric acid initially reacts with iron to produce ferric nitrate [Fe(NO₃)₃], NO₂ and water.

⇒ \(\mathrm{Fe}+6 \mathrm{HNO}_3 \rightarrow \mathrm{Fe}\left(\mathrm{NO}_3\right)_3+3 \mathrm{NO}_2 \uparrow+3 \mathrm{H}_2 \mathrm{O}\)

Ferric nitrate then decomposes to form ferric oxide which forms a layer over the iron vessel and prevents further reaction of iron with the acid.

⇒ \(4 \mathrm{Fe}\left(\mathrm{NO}_3\right)_3 \rightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3+12 \mathrm{NO}_2+3 \mathrm{O}_2 \uparrow\)

Question 72. Metallic magnesium is treated with very dilute nitric add.

Answer:

Metallic magnesium reacts with very dilute nitric acid to form magnesium nitrate [Mg(NO₃)₂] and hydrogen gas.

⇒ \(\mathrm{Mg}+2 \mathrm{HNO}_3 \rightarrow \mathrm{Mg}\left(\mathrm{NO}_3\right)_2+\mathrm{H}_2 \uparrow\)

Question 73. Cold and dilute nitric acid is added to copper.

Answer:

Copper reacts with cold and dilute nitric acid to form cupric nitrate [Cu(NO₃)₂], nitrous oxide (N₂O) and water.

⇒ \(4 \mathrm{Cu}+10 \mathrm{HNO}_3 \rightarrow 4 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+\mathrm{N}_2 \mathrm{O} \uparrow+5 \mathrm{H}_2 \mathrm{O}\)

Question 74. Hot and concentrated nitric acid is added dropwise on red-hot pumice stone.

Answer:

When hot and concentrated nitric acid is added dropwise on red-hot pumice stone, the acid decomposes to form nitrogen dioxide (NO₂) gas along with oxygen and water vapour.

⇒ \(4 \mathrm{HNO}_3 \rightarrow 4 \mathrm{NO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O} \uparrow+\mathrm{O}_2 \uparrow\)

Question 75. Nitric acid vapour is passed over metallic copper.

Answer:

Metallic copper reacts with nitric acid vapour to produce cupric oxide (CuO) and water along with evolution of nitrogen gas.

⇒ \(5 \mathrm{Cu}+2 \mathrm{HNO}_3 \rightarrow 5 \mathrm{CuO}+\mathrm{N}_2 \uparrow+\mathrm{H}_2 \mathrm{O}\)

Question 76. Copper is treated with 1:1 HNO₃ at room temperature.

Answer:

Blue-coloured cupric nitrate, nitric oxide gas and water are formed by the reaction of cccper 1 : 1 HNO₃ at room temperature.

⇒ \(3 \mathrm{CU}+8 \mathrm{HNO}_3 \rightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+4 \mathrm{H}_2 \mathrm{O}+2 \mathrm{NO} \uparrow\)

Question 77. A mixture of potassium nitrate and concentrated H₂SO₄ is heated around 200ºC.

Answer:

When a mixture of potassium nitrate and concentrated H₂SO₄ is heated around 200ºC, potassium hydrogen sulphate KHSO₄ and nitric acid are produced.

⇒ \(\mathrm{KNO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{KHSO}_4+\mathrm{HNO}_3\)

Question 78. Dilute sulphuric acid is added to barium chloride solution.

Answer:

When dilute sulphuric acid is aaced to barium chloride (BaCI₂) solution, white barium su pnate is precipitated which is insoluble in hytrochiorc acid (HCI) and nitric acid (HNO₃).

⇒ \(\mathrm{H}_2 \mathrm{SO}_4+\mathrm{BaCl}_2 \underset{\text { (white) }} \rightarrow \mathrm{BaSO}_4 \downarrow+2 \mathrm{HCl}\)

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Question 79. Concentrated H₂SO₄ is heated with copper turnings.

Answer:

When concentrated H₂SO₄ is heated with copper turnings, copper sulphate is produced and ‘ sulphur dioxide gas having a smell of burnt S sulphur is liberated.

⇒ \(\mathrm{Cu}(s)+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CuSO}_4(a q)+\mathrm{SO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}(I)\)

Question 80. Concentrated sulphuric acid H₂SO₄ is added to formic acid.

Answer:

When concentrated H₂SO₄ is added to formic acid, sulphuric acid absorbs water molecules from formic acid and produces carbon monoxide (CO).

⇒ \(\mathrm{HCOOH}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CO} \uparrow+\left[\mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{SO}_4\right]\)

Chapter 4 Matter Acids Bases And Salts Topic A Concept Of Acid And Bases Very Short Answer Type Questions Choose The Answer

Question 1. A covalent compound whose aqueous solution is acidic is

1. CH₄
2. CCI₄
3. HCI
4. NH₃ 

Answer: 3. HCI

Question 2. A covalent compound whose aqueous solution is alkaline

1. CH₄
2. CCI₄
3. HCI
4. NH₃ 

Answer: 4. NH₃ 

Question 3. One of the limitations of Arrhenius acid-base theory is that

1. It is applicable only for organic solvents
2. It is applicable only for inorganic solvents
3. Water is essentially required as the solvent
4. No solvent is required

Answer: 3. Water is essentially required as the solvent

Question 4. In aqueous solution H⁺ ion is present as

1. H⁺  ion
2. H₂⁺ ion
3. H₃O⁺ ion
4. H₃O^– ion

Answer: 3. H₃O⁺ ion

Question 5. Which of the following is used to prepare soaps and detergents?

1. HNO₃
2. HCI
3. H₂SO₄
4. NaOH

Answer: 4. NaOH

Question 6. Which of the following is known as ‘aqua fortis’?

1. NaOH
2. HNO₃
3. HCI
4. H₂SO₄  

Answer: 2. HNO₃

Question 7. Which of the following is called the ‘king of chemicals’?

1. NaOH
2. HNO₃
3. HCI
4. H₂SO₄  

Answer: 4. H₂SO₄  

Question 8. The acid used in the preparation of starch from glucose is

1. HCI
2. HNO₃
3. H₂SO₄  
4. H₃PO₄  

Answer: 1. HCI

Question 9. Aqua regia is a

1. 3:1 mixture of concentrated HCI and concentrated HNO₃
2. 1:3 mixture of concentrated HCI and concentrated H₂SO₄
3. 2:3 mixture of HNO₃ and H₂SO₄
4. 1:1:1 mixture of HCI, HNO₃ and H₂SO₄

Answer: 1. 3:1 mixture of concentrated HCI and concentrated HNO₃

Question 10. Which of the following does not produce hydrogen on reacting with dilute acids?

1. Fe
2. Zn
3. Cu
4. Mg

Answer: 3. Cu

Question 11. Which of the following can produce hydrogen by reacting with cold and very dilute HNO₃?

1. Fe
2. Zn
3. Cu
4. Mg

Answer: 4. Mg

Question 12. The compound used to identify H₂SO₄ is

1. BaSO₄
2. BaCI₂
3. Ba(OH)₂
4. BaCO₃

Answer: 2. BaCI₂

Question 13. Which of the following is identified by using an aqueous solution of AgNO₃?

1. HCI
2. HNO₃
3. H₂SO₄
4. NaOH

Answer: 1. HCI

Question 14. NaOH reacts with HCI in

1. Dry condition
2. Aqueous solution
3. Benzene solution
4. CCL solution

Answer: 2. Aqueous solution

Question 15. Which of the following is not a mineral acid?

1. Hydrochloric acid
2. Citric acid
3. Sulphuric acid
4. Nitric acid

Answer: 2. Citric acid

Question 16. Which of the following is not a base?

1. NaOH
2. KOH
3. NH₄OH
4. C₂H₅OH

Answer: 4.C₂H₅OH

Question 17. Which of the following is a strong hygroscopic substance?

1. Cone. H₂SO₄
2. Cone. HNO₃
3. Cone. HCI
4. CH₃COOH

Answer: 1. Cone. H₂SO₄

Question 18. Which of the following can be identified through ring test?

1. HCI
2. H₂SO₄
3. NaOH
4. HNO₃

Answer: 4. HNO₃

Question 19. \(\mathrm{Zn} \underset{\mathrm{HNO}_3}{\stackrel{\text { dil, cold }}{\longrightarrow}} \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+X+\mathrm{H}_2 \mathrm{O}\); Here ‘X’ is

1. N₂O
2. NO
3. NO₂
4. NH₄NO₃

Answer: 4. NH₄NO₃

Question 20. The brown ring compound formed in the ring test is

1. \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5(\mathrm{NO})_2\right] \mathrm{SO}_4\)
2. \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4\)
3. \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right]\left(\mathrm{SO}_4\right)_2\)
4. \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4\)

Answer: 2. \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4\)

Question 21. Brown fumes is seen to be coming out of the surface of which acid?

1. HNO₃
2. H2CO₃
3. H₂SO₄
4. HCI

Answer: 1. HNO₃

Question 22. The gas formed by heating Cu-turnings with I cone. HN03 is

1. CO₂
2. NO₂
3. NO
4. N₂O

Answer: 2. NO₂

Question 23. According to Arrhenius Theory, an acid produces (in aqueous medium)

1. H^⊕ ion as only anion
2. H^⊕ ion as only cation
3. OH^⊕ ion as only anion
4. OH^⊕ ion as only cation

Answer: 2. H^⊕ ion as only cation

Question 24. The metal which produces H2 in reaction with acid as well as with base is

1. Copper
2. Magnesium
3. Zinc
4. Iron

Answer: 3. Zinc

Question 25. If acid falls on any body part, which one of the following should be used to wash it off?

1. Sodium bicarbonate solution
2. Lemon juice
3. Lime water
4. Caustic soda solution

Answer: 1. Sodium bicarbonate solution

Question 26. Which one of the following is formed by the reaction of copper with cold and dil. HNO₃?

1. NO
2. NO₂
3. N₂O
4. N₂

Answer: 3. N₂O

Question 27. Which one of the following is hydracid?

1. H₂SO₄
2. HNO₃
3. HCI
4. HCOOH

Answer: 3. HCI

Question 28. The acid which is secreted inside the stomach is

1. HNO₃
2. H₂SO₄
3. HCI
4. H₂CO₃

Answer: 3. HCI

Question 29. Cone. H₂SO₄ is a

1. Deliquescent substance
2. Efflorescent substance
3. Crystalline substance
4. Hygroscopic substance

Answer: 4. Hygroscopic substance

Question 30. The gas produced by the reaction of MnO₂ with dil. hydrochloric acid is

1. H₂
2. O₂
3. CI₂
4. N₂

Answer: 3. CI₂

Chapter 4 Matter Acids Bases And Salts Topic A Concept Of Acid And Bases Answer In Brief

Question 1. For which property of water, an acid ionises in water to produce H₃O⁺ ions?

Answer: Due to the polar nature of water, an acid ionises in water to produce H₃O⁺ ions.

Question 2. Give some examples of organic acid.

Answer: Acetic acid (CH₃COOH), formic acid (HCOOH) etc., are some examples of organic acid.

Question 3. Name an inorganic acid which is synthesised in human body.

Answer: Hydrochloric acid (HCl).

Question 4. Show the reaction for dissociation of HNO₃ and CH₃COOH in aqueous medium.

Answer:

⇒ \(\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{\oplus}+\mathrm{NO}_3^{\ominus}\)

⇒ \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}^{\ominus}+\mathrm{H}_3 \mathrm{O}^{\oplus}\)

Question 5. Give an example (name and formula) of a weak acid.

Answer: Acetic acid (CH₃COOH).

Question 6. Name two acids used in daily life.

Answer: Examples of acids used in daily life are

1. Acetic acid (vinegar is used in cooking),
2. Hydrochloric acid (muriatic acid is used to clean bathrooms).

Question 7. Name a base which is not a metallic oxide or hydroxide.

Answer: Ammonia (NH₃).

Question 8. Name two basic substances used in daily life.

Answer: Washing soda (Na₂CO₃) and baking soda (NaHCO₃) are two basic substances used in daily life.

Question 9. Can BCI₃ be considered as an acid according to Arrhenius theory?

Answer: BCI₃ cannot be considered as an acid as it does notyield any H⁺ ion in aqueous solution.

Question 10. Name a metal which cannot produce hydrogen gas by reacting with dilute mineral acids.

Answer: Copper.

Question 11. Give example of a volatile mineral acid.

Answer: Hydrochloric Acid (HCI).

Question 12. Name a covalent gaseous compound which ionises in water and acts as a strong acid.

Answer: Hydrogen chloride (HCI) is a gaseous covalent compound which ionises in water and acts as a strong acid.

Question 13. What is muriatic acid?

Answer: Muriatic acid is hydrochloric acid (HCI).

Question 14. Which metals can displace hydrogen from cold and dilute nitric acid?

Answer: Metals such as magnesium (Mg) and manganese (Mn) can displace hydrogen from cold and dilute nitric acid.

Question 15. Which acid is used to remove impurities from gold?

Answer: Nitric acid (HNO₃).

Question 16. State the major use of aqua regia.

Answer: Aqua regia is used to dissolve noble metals like gold, silver, platinum etc.

Question 17. Which acid is used in the preparation of TNT?

Answer: Nitric acid (HNO₃).

Question 18. Which gas is excreted from jewellery shops?

Answer: Nitrogen dioxide (NO₂).

Question 19. Which acid is used more in jewellery industry?

Answer: HNO₃ (nitric acid).

Question 20. Which acid is used to prepare nitroglycerine?

Answer: Nitric acid is used to prepare nitroglycerine.

Question 21. Which acid is used in the ring test for identification of nitric acid?

Answer: Sulphuric acid.

Question 22. Write down the name and formula of the compound which forms the brown ring in the ring test for identification of nitric acid.

Answer: Pentaaquanitrosyliron (II) sulphate and its formula is [Fe(H₂O)₅(NO)]SO₄.

Question 23. Which acid is used in the preparation of inorganic fertiliser, ammonium sulphate?

Answer: Sulphuric acid (H₂SO₄).

Question 24. Name the acid which is used in the preparation of HCI(g).

Answer: Sulphuric acid (H₂SO₄).

Question 25. Give example of a hygroscopic acid.

Answer: Cone. H₂SO₄

Question 26. Which salt is produced when aluminium reacts with hot and concentrated aqueous solution of NaOH?

Answer: Sodium aluminate (NaAIO₂).

Question 27. What is oil of vitriol?

Answer: Sulfuric acid is termed as oil of vitriol.

Chapter 4 Matter Acids Bases And Salts Topic A Concept Of Acid And Bases Fill In The Blanks

Question 1. The formula of hydronium ion is ________

Answer: H₃O⁺

Question 2. A base is a compound which dissociates in water to produce _______ ions.

Answer: OH^–  

Question 3. Ammonia is a base because it reacts with acids to produce _______

Answer: Salts

Question 4. Fuming nitric acid is a strong ______ agent.

Answer: Oxidising

Question 5. ______ is used in the preparation of picric acid.

Answer: HNO₃

Question 6. ________ is used to absorb water vapour.

Answer: H₂SO₄

Question 7. The base used to prepare synthetic fibre or rayon is _______

Answer: NaOH

Question 8. Copper reacts with hydrochloric acid in presence of _______

Answer: O₂

Question 9. The metal which is not used in the preparation of hydrogen is ________

Answer: Cu

Question 10. Freshly prepared solution of _________ is required for the ring test.

Answer: FeSO₄

Question  11. The water pollutant which is produced when Cu reacts with hot and cone. HNO₃ is ______

Answer: Cu(NO₃)₂

Question 12. The gas which is produced when Cu reacts with hot and cone. H₂SO₄ is _______

Answer: SO₂

Question 13. A concentrated acid is diluted by gradually adding_______ into _______

Answer: Acid, water

Question 14. A mixture of tartaric acid and sodium bicarbonate undergoes chemical reaction when ________ is added to it.

Answer: Water

Question 15. Hydrochloric acid (HCI) ionises in ________ solution and exhibits its acidic property.

Answer: Aqueous

Question 16. Hydrochloric acid (HCI) is identified by using ______ solution.

Answer: Silver nitrate

Question 17. Hydrochloric acid (HCI) reacts with Zn to produce hydrogen and ________

Answer: ZnCI₂

Question 18. The basicity of an acid is the number of replaceable __________ atoms present in one molecule of the acid.

Answer: Hydrogen

Question 19. An example of a tribasic acid is _________

Answer: H₃PO₄

Question 20. ________ solution is used to treat alkali burns.

Answer: 1% acetic acid

Question 21. Excess secretion of acid causes _________ in stomach.

Answer: Ulcer

Question 22. If oxygen atom is present in acid, the acid is termed as ________ acid.

Answer: Oxy

Question 23. Aluminium hydroxide reacts with NaOH to produce _______ and ________

Answer: Sodium aluminate, water

Chapter 4 Matter Acids Bases And Salts Topic A Concept Of Acid And Bases State Whether True Or False

Question 1. Water soluble metal hydroxides are called alkalis.

Answer: True

Question 2. The compound responsible for the formation of brown ring in the ring test is pentaaquanitrosyliron(ll) sulphate.

Answer: True

Question 3. Prolonged secretion of acid in the stomach causes ulcer.

Answer: True

Question 4. Hydrochloric acid exhibits reducing property while nitric acid exhibits oxidising property.

Answer: True

Question 5. The basicity of H₂SO₄ is 1.

Answer: False

Question 6. Nitric acid is extensively used in the manufacture of explosives.

Answer: True

Question 7. HCI is the organic acid produced in human body.

Answer: False

Question 8. BaCI₂ solution is used to identify HNO₃.

Answer: False

Chapter 4 Matter Acids Bases And Salts Topic B Qualitative Concept Of pH Synopsis

1. The H⁺ ion concentration of a dilute solution is expressed in terms of pH.
2. pH is the negative logarithm of H⁺ ion concentration of a solution. pH of a neutral solution is 7 at 25°C. Any acidic solution has pH less than 7 and alkaline solution has pH greater than 7.
3. pH-paper is used as a universal indicator to get an idea about the pH value of a solution. The paper shows different colours in solutions of different pH values.
4. The decay of tooth enamel occurs when the pH level of mouth cavity falls below 5.5.
5. The pH level of soil or water bodies is increased by addition of quick lime.

Chapter 4 Matter Acids Bases And Salts Topic B Qualitative Concept Of pH Short And Long Answer Type Questions

Question 1. What is meant by pH of a solution?

Answer:

pH of a solution

pH of a solution is defined as the negative logarithm (to the base 10) of the molar concentration of H₃O⁺ ions in the solution. pH comes from the German word Potenz de Hydrogen where Potenz means power.

⇒ \(p H=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\log _{10} \frac{1}{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}\)

where [H₃O⁺ ] = molar concentration of H₃O⁺ ions in the solution.

Question 2. What idea is obtained about the nature of a solution from its pH value?

Answer:

At 25°C, if the pH of a solution is less than 7, then the solution is acidic in nature. If the pH is greater than 7, then the solution is alkaline in nature whereas if the pH is equal to 7, then the solution is a neutral solution.

Question 3. What Is a pH scale? Mention the range of a pH scale.

Answer:

pH scale

The scale used to express the acidic or alkaline nature of a solution with respect to the pH of the solution is called a pH-scale.

In general, the H₃O⁺ ion concentration or OH^–   ion concentration of a dilute solution does not exceed 1mol • L^-1.

If the H₃O⁺ ion concentration of a solution is 1 mol • L^-1 then its pH will be = \(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right] =-\log _{10^1} 1=0\)

On the other hand, if the OH^– ion concentration of a solution is 1 mol • L^– 1, then its H₃O⁺ ion concentration will be 10^-14mol • L^-1 (at 25°C \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \times\left[\mathrm{OH}^{-}\right]=10^{-14}\)).

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Hence its pH will be = \(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log _{10} 10^{-14}=14\)

Thus, pH of a dilute aqueous solution at 25°C ranges from 0 to 14.

Question 4. The H₃O⁺ ion concentration in a solution at 25°C is 10^-4  mol • L^-1. Find the pOH of the solution.

Answer:

The H₃O⁺ ion concentration in the solution, [H₃O⁺] = 10^-4 mol • L^-1

∴ pH of the solution = \(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\) = \(-\log \left(10^{-4}\right)=-(-4)=4\)

At 25°C, pH + pOH = 14

∴ pOH of the solution = 14 – pH = 14 – 4 = 10

Question 5. If the H₃O⁺ ion concentration in a solution is 1 x 10^-9 mol• L^-1, then find the pH of the solution.

Answer:

The H₃O⁺ ion concentration in the solution, [H₃O⁺ ] = 1 x 10^-9 mol • L^-1

∴ pH of the solution = \(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\) = \(-\log \left(1 \times 10^{-9}\right)=-(-9)=9\)

Question 6. The H₃O⁺ ion concentration of four solutions M, N, O and P are 10^-2, 10^-3, 10^-4 and 10^-5 • mol • L^-1 respectively. Find the pH of the solutions and arrange the solutions in order of their increasing acid strength.

Answer:

Given

The H₃O⁺ ion concentration of four solutions M, N, O and P are 10^-2, 10^-3, 10^-4 and 10^-5 • mol • L^-1 respectively.

The H₃O⁺ ion concentration of four solutions

M, N, O and P are 10^-2, 10^-3, 10^-4 and 10^-5 mol • L^-1 respectively.

∴ \(pH of solution M =-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\)

= \(-\log _{10}\left(10^{-2}\right)=-(-2)=2\)

∴ \(pH of solution N =-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\)

= \(-\log _{10}\left(10^{-3}\right)=-(-3)=3\)

∴ \(\mathrm{pH} of solution O =-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\)

= \(-\log _{10}\left(10^{-4}\right)=-(-4)=4\)

∴ \(pH of solution P =-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\)

= \(-\log _{10}\left(10^{-5}\right)=-(-5)=5\)

We know that the acidity of a solution increases as the pH of the solution decreases. Thus, increasing order of acid strength of the solutions is P < O < N < M.

Question 7. The pH of a solution increases by 1 unit. What will be the change in H₃O⁺ ion concentration of the solution?

Answer:

Let us consider that the pH of the given solution is 4. Therefore, the H₃O⁺ ion concentration of the solution will be [H₃O⁺] = 10^-pH = 10^-5 mol • L^-1.

Now, the pH of the solution is increased by 1 unit. So, the pH of the solution now becomes 5.

Hence, the H₃O⁺ ion concentration of the solution will be [H₃O⁺] = 10^-pH = 10^-5 mol • L^-1.

Class 9 Physical Science Chapter 4 Matter Short And Long Answer Type Questions

Thus, concentration of a dilute aqueous solution becomes l/10th of its initial concentration when pH is increased by 1 unit.

Similarly, the concentration of a dilute aqueous solution becomes 10 times of its initial concentration if the pH is decreased by 1 unit.

Question 8. State whether the pH of a solution can be less than zero or greater than 14. Justify your answer.

Answer:

At 25°C, the pH of a dilute aqueous solution generally ranges from 0 to 14 because the concentration of H₃O⁺ or OH^–  ions in dilute aqueous solution usually does not exceed 1 mol • L^-1.

However, if the concentration of H₃O⁺ ions in a solution exceeds 1 g-ion L^-1, then the pH of the solution becomes less than zero. Similarly, if the concentration of OH^– ions in a solution exceeds 1 g-ion L^-1, then the pH of the solution becomes greater than 14.

Question 9. pH of the solutions P, Q and R are 13, 6 and 2 respectively.

1. which one of these solutions will form NH₃ from (NH₄)₂SO₄?
2. which one is strongly acidic?
3. which solution will contain both molecules and ions?

Answer:

1. Solution P will form NH3 from (NH₄)₂SO₄ as it is strongly basic in nature.
2. Solution R is strongly acidic as depicted by the given pH value.
3. Solution Q of pH 6 will contain both molecules and ions.

Question 10. What is pH+-paper? How is it used to determine the pH of a solution?

Answer:

pH+-paper

1. A pH-paper is a paper coated with universal indicator which is used to get an idea about the pH of a solution.
2. To determine the pH of a solution, the pH-paper is dipped into the solution. The paper attains a definite colour depending on the pH of the solution. The colour of the pH-paper is then compared with the standard pH colour strip and consequently, the pH of the solution is determined.

Question 11. Mentionb with reason whether the pH of the aqueous solutions of vinger and soda will be grater or less than 7.

Answer:

We know, pH of acidic and basic solution are less and greater than 7 respectively.

Now the main component of vinegar is acetic acid (CH₃COOH) which dissociates partially in the aqueous solution forming H₃O^⊕ and CH₃COO^Θ ion.

\(\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{H}^{\oplus}+\mathrm{CH}_3 \mathrm{COO}^{\ominus}\)

 

Hence the aqueous solution of vinegar will have pH less than 7.

On the other hand the aqueous solution of soda (Na₂CO₃) becomes alkaline due to hydrolysis..

 

Class 9 Physical Science Chapter 4 Matter 

\(\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{CO}_3\)

 

Hence the aqueous solution of soda will have pH greater than 7.

Question 12. Why does dental erosion take palce?

Answer:

The enamel of our tooth is made of calcium phosphate. During consumption of food, mainly sweets, the food particles get stuck between our teeth. These food particles are decomposed by bacteria present in our mouth to produce acids which decrease the pH level of the mouth cavity.

If the pH level falls below 5.5, then the acids react with calcium phosphate and cause erosion of the tooth enamel.

Question 13. Explain why brushing of teeth is essential after having meals?

Answer:

Food particles, especially the carbohydrate particles get stuck to the dental gap and form acids by the bacterial decomposition. As a consequence, pH inside the mouth decreases.

Dental enamel made of calcium phosphate, starts decaying at a pH less than or equal to 5.5. Most of the toothpastes have a pH around 9.0 which neutralise the formed acid at the dental gap and prevents enamel decay. That is why brushing of teeth after meal is essential.

Question 14. Sometimes lime is added to the water bodies during piscicultyure. Why?

Answer:

Sometimes lime is added to the water bodies during piscicultyure.

If the acidity of a water body increases, it affects the flora and fauna of the water body. The plants and fishes of the water body die due to excess acid. So, alkaline substances like quick lime or calcium oxide is added to water.

It neutralises the excess acids and maintains the acid-base balance of the water body. However, it should also be noted that excess use of lime may increase the alkali level of the water body which also adversely affects its ecosystem.

Question 15. Why is lime added to agricultural lands?

Answer:

Due to acid rain and excessive use of inorganic fertilisers like ammonium sulphate, the acidity of soil increases. As a result, the fertility of soil decreases and it becomes unfit for the growth of crops.

To revive the fertility of soil, alkaline substances like lime is added to the soil. Lime neutralises the excess acid present in the soil and makes the soil fertile.

Chapter 4 Matter Acids Bases And Salts Topic B Qualitative Concept Of pH Very Short Abswer Type Questions Choose The Correct Answers

Question 1. The hardest part of our body is

1. Enamel of teeth
2. Bones
3. Skin
4. Nails

Answer: 1. Enamel of teeth

Question 2. The pH of two solutions, A and B are 3 and 6 respectively. It means that

1. Solution A is twice as acidic as solution B
2. Solution B is twice as acidic as solution A
3. Solution A is 1000 times more acidic than solution B
4. Solution B is 1000 times more acidic than solution A

Answer: 3. Solution A is 1000 times more acidic than solution B

Class 9 Physical Science Chapter 4 Matter Very Short Abswer Type Questions Choose The Correct Answers

Question 3. The chemical nature of commonly used toothpastes is

1. Acidic
2. Neutral
3. Alkaline
4. Amphoteric

Answer: 3. Alkaline

Question 4. For which of the following solutions, pH will be maximum?

1. 1 mol • L^-1 HCI solution
2. 0.1 mol • L^-1 HCI solution
3. 0.01 mol • L^-1 HCI solution
4. 0.001 mol • L^-1 HCI solution

Answer: 4. 0.001 mol • L^-1 HCI solution

Question 5. pH of lemon juice is

1. > 7
2. = 7
3. < 7
4. > 14

Answer: 3. < 7

Question 6. pH of blood is

1. 4
2. 5
3. 7
4. 7.35
5. 7.45

Answer: 4. 7.45.

Class 9 Physical Science Chapter 4 Matter Very Short Abswer Type Questions Choose The Correct Answers

Question 7. pH of milk is

1. 5.5-5.6
2. 3.4-3.7
3. 6.5-6.7
4. 8.5-9.5

Answer: 3. 6.5-6.7

Question 8. pH of a solution is expressed as

1. -log[H₃O⁺]
2. -log[OH^–]
3. -log[H ]
4. log [OH^–]

Answer: 1. -log[H₃O⁺]

Question 9. If the H⁺ ion concentration of a solution is 10^-5 g-ion/litre, then the pH of the solution will be

1. 10
2. 5
3. 3
4. 7

Answer: 2. 5

Question 10. The pH at which the enamel of teeth begins to decay is

1. 3.5
2. 4.5
3. 5.5
4. 6.5

Answer: 3. 5.5

Question 11. Due to acid rain, the pH of agricultural lands

1. Increases
2. Decreases
3. Remains unchanged
4. Increases in some places and decreases in other places

Answer: 2. Decreases

Question 12. The range of the pH scale at 25°C is

1. 0-10
2. 0-14
3. 0-12
4. 1-14

Answer: 2. 0-14

Question 13. If acid is added to pure water, pH of the medium will

1. Increase
2. Decrease
3. Remain same
4. Be zero

Answer: 2. Decrease

Class 9 Physical Science Chapter 4 Matter Very Short Abswer Type Questions Choose The Correct Answers

Question 14. If a litmus paper is dipped in an alkaline solution, its colour will be

1. Blue
2. Red
3. Yellow
4. Green

Answer: 1. Blue

Question 15. Which of the solutions has the highest pH ?

1. Caustic soda
2. Milk of magnesia
3. Vinegar
4. Formic acid

Answer: 1. Caustic soda

Question 16. If a solution turns reddish-pink on addition of phenolphthalein, its pH may be

1. 4
2. 7
3. 10
4. 5

Answer: 3. 10

Question 17. Concept of pH scale was given by

1. Arrhenius
2. Sorensen
3. Rutherford
4. Lewis

Answer: 2. Sorensen

Question 18. pH of an alkaline solution may be

1. 5.4
2. 6.2
3. 7
4. 9.2

Answer: 4. 9.2

Question 19. Which is the correct order of pH?

1. Tomato juice < water < toothpaste
2. Water < toothpaste < tomato juice
3. Water < tomato juice < toothpaste
4. Toothpaste < tomato juice < water

Answer: 1. Tomato juice < water < toothpaste

Question 20. pH of pure water at 25°C is

1. 5.7
2. 6.2
3. 7
4. 7.5

Answer: 3. 7

Question 21. pH of an acidic solution may be

1. 9.7
2. 7
3. 14
4. 5.1

Answer: 4. 5.1

Question 22. pH will be lowest for

1. NaHCO₃ solution
2. HCI solution
3. KOH solution
4. Na₂CO₃ solution

Answer: 2. HCI solution

Class 9 Physical Science Chapter 4 Matter Very Short Abswer Type Questions Choose The Correct Answers

Question 23. pH of pure water is 7. Which of the following is true for water?

1. \(\mathrm{H}^{\oplus}+\mathrm{OH}^{\ominus} \rightleftharpoons \mathrm{H}_2 \mathrm{O}\)
2. \(\mathrm{H}_2 \mathrm{O}+\mathrm{H}^{\oplus} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{\oplus}\)
3. \(\mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{\oplus}+\mathrm{OH}^{\ominus}\)
4. \(2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_4 \mathrm{O}_2\)

Answer: 3. \(\mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{\oplus}+\mathrm{OH}^{\ominus}\)

Question 24. Favourable pH of water for piscicullture should be

1. 5.5-6.5
2. > 8
3. 7.5
4. 7

Answer: 1. 5.5-6.5

Question 25. Which of the solution has the highest pH?

1. 1 mol • L^-1 HCI
2. 0.1 mol • L^-1 HCI
3. 0.01 mol • L^-1 HCI
4. 0.001 mol • L^-1 HCI

Answer: 4. 0.00lmol • L^-1 HCI

Question 26. A solution turns yellow with addition of methyl orange. pH of the solution is

1. 7
2. 4
3. > 7
4. < 7

Answer: 3. > 7

Question 27. pH of vinegar is

1. 4
2. 5
3. 7
4. 7.35-7.45

Answer: 1. 4

Class 9 Physical Science Chapter 4 Matter Very Short Abswer Type Questions Choose The Correct Answers

Question 28. To reduce the acidity of soil which one of the following should be added?

1. Bleaching powder
2. Blue vitriol
3. Lime
4. Compost fertiliser

Answer: 3. Lime

Chapter 4 Matter Acids Bases And Salts Topic B Qualitative Concept Of pH Answer In Brief

Question 1. What is the concentration of H^⊕ ion in pure water at 25°C?

Answer: 10 mo1. L

Question 2. In which solvent between water and benzene, an acid does not undergo ionisation?

Answer: An acid does not undergo ionisation in benzene because it is a non-polar solvent.

Question 3. What is the nature of an aqueous solution having pH = 4?

Answer: Acidic in nature.

Question 4. What is the nature of an aqueous solution having pH = 7?

Answer: Neutral in nature.

Question 5. What is the nature of an aqueous solution having pH = 9?

Answer: Alkaline in nature.

Question 6. What is the value of pH of pure water at 25°C?

Answer: The pH of pure water at 25°C is 7.

Question 7. What is the pH of rain water?

Answer: The pH of rain water is generally 5.6.

Question 8. Odd one out: litmus paper, pH paper, methyl orange, phenolphthalein.

Answer: pH paper.

Question 9. What is the major component of tooth enamel?

Answer: Calcium phosphate [Ca₃(PO₄)₂].

Question 10. Why does acid form at the base of the tooth?

Answer: The residual food particles, especially the sugar particles get stuck to the space between the teeth and are dissociated by bacteria to form acid. Hence the statement.

Question 11. Which substance should be used to make alkaline agricultural land suitable for farming?

Answer: Compost fertilizer is generally used

Chapter 4 Matter Acids Bases And Salts Topic B Qualitative Concept Of pH Fill In The Blanks

Question 1. Scientist _______ introduced the pH scale.

Answer: Sorensen

Question 2. The term pH originates from the German word ________

Answer: Potenz de Hydrogen

Question 3. At 25°C, the maximum pH value of a dilute solution is _______

Answer: 14

Question 4. pH is the negative logarithm of _______ ion concentration in a solution.

Answer: H₃O⁺

Question 5. Blood is slightly ______ in nature.

Answer: Alkaline

Chapter 4 Matter Acids Bases And Salts Topic B Qualitative Concept Of pH State Whether True Or False

Question 1. The pH of an aqueous solution is 3. The solution is alkaline in nature.

Answer: False

Question 2. If the H₃O⁺ ion concentration of an aqueous solution is 1 x 10^-5 mol • L^-1, then the pH of the solution will be 5.

Answer: True

Question 3. Saliva is acidic in nature.

Answer: True

Question 4. pH of pure water is 7 at any temperature.

Answer: True

Chapter 4 Matter Acids Bases And Salts Topic C Acidic Basic Amphoteric Oxides And Acid Rain Synopsis

1. Metallic oxides and hydroxides that react with acids to form salt and water are called bases. Water  soluble metallic hydroxides are called . alkalis. Ail metallic oxides and hydroxides are not water soluble. Hence, they are bases but are not termed as alkalis.  Thus, we can say that all alkalis are bases but all bases are not alkalis.
2. The mixture consisting of 3 volumes of concentrated HCI and 1 volume of concentrated HNO₃ is commonly known as aqua regia. It is exclusively used to dissolve noble metals like gold, silver, platinum etc.
3. Oxides which react with bases to produce salt and water are known as acidic oxides. All non-metallic oxides are generally considered as acidic oxides.
4. Oxides which react with acids to produce salt and water are known as basic oxides. In general, all metallic oxides are basic oxides.
5. An amphoteric oxide reacts with both acids and bases to produce salt and water.
6. When the acid level of rain water exceeds a certain value, it is called acid rain. Acid rain increases the acidity of soil and thus, decreases its fertility. It also adversely affects changing the aquatic plants and animals by increasing the acid level of water bodies.
7. Due to acid rain, surfaces of buildings, sculptures and monuments made of marbles get covered with layers of CaSO₄ making them appear lustreless. With time, the layers come off and produce scars and pits on the marble buildings. These scars are known as stone cancer.

Chapter 4 Matter Acids Bases And Salts Topic C Acidic Basic Amphoteric Oxides And Acid Rain Short And Long Answer Type Questions

Question 1. Chromium trioxide is an acidic oxide though it is a metallic oxide. Explain.

Answer:

Chromium trioxide is an acidic oxide though it is a metallic oxide.

Chromium trioxide (CrO₃) dissolves in water to produce chromic acid (H₂CrO₄). Hence, it is an acidic oxide.

\(\mathrm{CrO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{CrO}_4\)

Question 2. What are acidic oxides and give examples.

Answer:

Acidic oxides

The oxides which react with bases to form salt and water are known as acidic oxides. Generally oxides of non-metals are acidic oxides. Some examples of acidic oxides are carbon dioxide (CO₂), sulphur dioxide (SO₂), phosphorus pentoxide (P₂O₅) etc.

⇒ \(\mathrm{S}+\mathrm{O}_2 \rightarrow \mathrm{SO}_2 ; \mathrm{SO}_2+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2 \mathrm{SO}_3+\mathrm{H}_2 \mathrm{O}\)

Question 3. Define basic oxides with examples.

Answer:

Basic oxides

The oxides which react with acids to form salt and water are known as basic oxides. Generally oxides of metals are basic oxides. Some examples re calcium oxide (CaO), ferrous oxide (FeO), magnesium oxide (MgO) etc.

\(2 \mathrm{Ca}+\mathrm{O}_2 \rightarrow 2 \mathrm{CaO} ; \mathrm{CaO}+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}\)

Question 4. Define amphoteric oxides with examples.

Answer:

Amphoteric oxides

The oxides which react with both acids and bases to form salt and water are called amphoteric oxides. Some examples are, zinc oxide (ZnO), aluminium oxide (Al2O₃) etc.

⇒ \(\mathrm{ZnO}+2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_2+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{ZnO}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2 \mathrm{ZnO}_2+\mathrm{H}_2 \mathrm{O}\)

Question 5. why is aluminium oxide called an amphoteric oxide?

Answer:

Aluminium oxide reacts with both acids and bases separately to form salt and water.

So, aluminium oxide is known as an amphoteric oxide.

Question 6. Classify the following oxides: \(\mathrm{ZnO}, \mathrm{SO}_2 \text {, }\mathrm{P}_2 \mathrm{O}_5, \mathrm{~K}_2 \mathrm{O}, \mathrm{Al}_2 \mathrm{O}_3\).

Answer:

Acidic oxides: \(\mathrm{SO}_2, \mathrm{P}_2 \mathrm{O}_5\)

Basic oxides: \(\mathrm{Na}_2 \mathrm{O}, \mathrm{K}_2 \mathrm{O}\)

Amphoteric oxides: \(\mathrm{ZnO}, \mathrm{Al}_2 \mathrm{O}_3\)

Question 7. What are acid anhydrides?

Answer:

Acid anhydrides

The oxide of a non-metal (acidic oxide) which dissolves in water to form its corresponding acid is called an acid anhydride. For example, phosphorus pentoxide (P₂O₅) dissolves in water to form phosphoric acid (H₃PO₄). Thus, phosphorus pentoxide is the anhydride of phosphoric acid.

⇒ \(\mathrm{P}_2 \mathrm{O}_5+3 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}_3 \mathrm{PO}_4\)

Question 8. Define mixed acid anhydride with example.

Answer:

Mixed acid anhydride

If a non-metallic oxide dissolves in water to produce more than one acid, then the oxide is considered as the anhydride of both the acids and is called a mixed acid anhydride.

For example, NO₂ dissolves in cold water to produce nitrous acid (HNO₂) and nitric acid (HNO₃). Hence, NO₂ is a mixed acid anhydride.

⇒ \(2 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{HNO}_2+\mathrm{HNO}_3\)

Question 9. How is HNO₃, formed by lightning?

Answer:

Formation of HNO₃ by lightning

N₂ and O₂ of the atmosphere reacts with each other during lightning to form nitric oxide.

⇒ \(\mathrm{N}_2+\mathrm{O}_2 \stackrel{\text { lightning }}{\longrightarrow} 2 \mathrm{NO}\)

This nitric oxide then combines with oxygen to form nitrogen dioxide. Nitrogen dioxide then reacts with rain water to form nitric acid.

⇒ \(2 \mathrm{NO}+\mathrm{O}_2 \rightarrow 2 \mathrm{NO}_2\)

⇒ \(3 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{HNO}_3+\mathrm{NO}\)

Question 10. What is acid rain?

Answer:

Acid rain

Normal rainwater is slightly acidic. However, if rain water contains excess amount of acids like H₂SO₄ , HNO₃ and HCI, then the pH of rain water ranges between 3.5-5.6. This is known as acid rain.

Question 11. What are the causes of acid rain?

Answer:

Causes of acid rain

Different causes of acid rain can be classified into two categories—natural and man-made causes. Acidic oxides emitted due to natural causes and different human activities mix with the atmospheric water vapour to cause acid rain.

1. Natural Causes:

1. SO₂ gas is released into the atmosphere during volcanic eruptions
2. Nitrogen and oxygen present in air combine to form different oxides of nitrogen NO_x during lightning discharges.
3. Bacterial decomposition of ammonium salts present in soil also produces oxides of nitrogen (NO_x) that are released into the atmosphere.

2. Man-Made Causes:

1. Different oxides of sulphur and nitrogen are released into the atmosphere due to combustion of fossil fuels like coal and petroleum used in automobiles, thermal power plants, oil refineries, metal extraction industries etc.
2. Huge amounts of HCI gas is released from factories where hydrochloric acid is extensively used.
3. These gases react with atmospheric oxygen, ozone and water vapour to produce different acids which float in air in the form of aerosols and come down on the earth’s surface along with rain water.

Question 12. What is stone cancer?

Answer:

Stone Cancer:

Scars and pits formed on the surface of buildings, sculptures, memorials and monuments made of marble due to acid rain are collectively called stone cancer. The scars are formed due to the reaction of the acid with marble (calcium carbonate, CaCO₃).

Question 13. Discuss the harmful effects of acid rain.

Answer:

Harmful Effects Caused By Acid Rain

The harmful effects of acid rain are as follows

1. Effect On Soil And Vegetation:

Acid rain increases acidity of soil. The increased acidity changes the solubilities of different metal salts in the soil. Consequently, fertility of the soil decreases which reduces the agricultural productivity. Photosynthetic activities of the plants get disrupted due to acid rain. Soil microorganisms and other living organisms of the soil are also adversely affected due to acid rain.

2. Effect On Aquatic Plants And Animals:

Acid rain decreases normal pH of different water bodies which in turn decreases reproductive capacity of fish leading to decresed production of spawns. Excessive increase in the acid level of water kills the flora and fauna of the water body and disturbs the marine ecosystem.

3. Effect On Human Beings:

Due to acid rain, some metals dissolve in rain water to form toxic salts which on entering the human body cause harmful effects. Acid rain damages our skin, hair and body cells. Acids like H₂SO₄ and HNO₃ present in acid rain enter the human body and adversely affect our nervous system, respiratory system and digestion process.

4. Effect on sculptures and monuments: Acid rain causes extensive damage to buildings, memorials, monuments and sculptures made of marble. Marble (CaCO₃) reacts with acids and forms insoluble calcium sulphate that deposits on the surfaces of the buildings. This causes scars, pits on the surface and erodes it.

⇒ \(\mathrm{CaCO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4 \downarrow+\mathrm{CO}_2 \uparrow+\mathrm{H}_2 \mathrm{O}\)

Chapter 4 Matter Acids Bases And Salts Topic C Acidic Basic Amphoteric Oxides And Acid Rain Vert Short Answer Type Questions Choose The Correct Answer

Question 1. An example of an acidic oxide is

1. CaO
2. Na₂O₂
3. CO₂
4. MgO

Answer: 3. CO₂

Question 2. An example of a basic oxide is

1. CO2
2. No
3. CaO
4. SO

Answer: 3. CaO

Question 3. An example of an amphoteric oxide is

1. ZnO
2. CaO
3. MgO
4. FeO

Answer: 1. ZnO

Question 4. Acid rain has adversely affected

1. Taj Mahal
2. Victoria Memorial
3. Sculptures made of marbles
4. All of these

Answer: 4. All of these

Question 5. As a consequence of acid rain, pH of soil

1. Increases
2. Decreases
3. Remains unaltered
4. Sometime increases, sometime decreases

Answer: 2. Decreases

Question 6. Which of the following can form each of acidic, basic and amphoteric oxide?

1. S
2. Pb
3. Ca
4. Cr

Answer: 4. Cr

Question 7. The gas released from Mathura oil refinery which is the reason behind the weathering of the Taj Mahal is

1. CO₂
2. CH₄
3. CO
4. SO₂

Answer: 4. SO₂

Question 8. Number of amphoteric oxides among Cr₂O₃, ZnO, PbO, Al₂O₃ is

1. 1
2. 2
3. 3
4. 4

Answer: 4. 4

Question 9. The gas responsible for acid rain is

1. CH₄
2. NO₂
3. N₂O
4. O₃

Answer: 2. NO₂

Question 10. Which of the following elements form acidic oxide?

1. Na
2. Mg
3. P
4. AI

Answer: 3. P

Question 11. N₂O₅ is

1. Acidic oxide
2. Basic oxide
3. Amphoteric oxide
4. Neutral oxide

Answer: 1. Acidic oxide

Question 12. Which one is not an acid anhydride?

1. CO₂
2. CO
3. N₂O₅
4. SO₃

Answer: 2. CO

Question 13. Which one of water, carbon dioxide, calcium oxide and nitrogen dioxide is a neutral oxide?

1. Water
2. Carbon dioxide
3. Calcium oxide
4. Nitrogen dioxide

Answer: 1. Water

Question 14. Depict the number of acidic oxides among \(\mathrm{SO}_2, \mathrm{~N}_2 \mathrm{O}_5, \mathrm{NO}_2, \mathrm{SO}_3, \mathrm{O}_3, \mathrm{CO}_2, \mathrm{H}_2 \mathrm{O},\mathrm{CO}, \mathrm{BaO}_2 and \mathrm{P}_2 \mathrm{O}_5\)?

1. 6
2. 8
3. 4
4. 7

Answer: 1. 6

Question 15. Which one of the following is a bicompound of oxygen, but not an oxide?

1. OF₂
2. NO₂
3. SO₂
4. CO₂

Answer: 1. OF₂

Question 16. Which one of the following is an acidic oxide?

1. MnO
2. Mn₂O₃
3. MnO₂
4. Mn₂O₇

Answer: 4. Mn₂O₇

Chapter 4 Matter Acids Bases And Salts Topic C Acidic Basic Amphoteric Oxides And Acid Rain Answer In Brief

Question 1. Name some metallic oxides which exhibit acidic property in water.

Answer: Manganese heptoxide (Mn₂O₇), chromium trioxide (CrO₃) etc.

Question 2. Give some examples of acidic oxide.

Answer: Some examples of acidic oxide are carbon dioxide (CO₂), sulphur trioxide (SO₃) etc.

Question 3. Name two acidic metallic oxide.

Answer: Manganese heptoxide (Mn₂O₇) and chromium trioxide (CrO₃).

Question 4. Which type of oxide is P₂O₅?

Answer: P₂O₅ is an acidic oxide.

Question 5. Which class of oxides does calcium oxide belong to?

Answer: Calcium oxide belongs to the class of basic oxide.

Question 6. Give some examples of amphoteric oxide.

Answer: Some examples of amphoteric oxide are zinc oxide (ZnO), aluminium oxide (Al₂O₃) etc.

Question 7. Which acid is formed when carbon dioxide (CO₂) gas dissolves in water?

Answer: Carbonic acid (H₂CO₃).

Question 8. Which acid is formed when nitrogen dioxide (NO₂) gas dissolves in water?

Answer: Nitric acid (HNO₃) and nitrous acid (HNO₂)

Question 9. Which acid is formed when sulphur dioxide (SO₂) gas dissolves in water?

Answer: Sulphurous acid (H₂SO₃).

Question 10. Which gas dissolves in water to produce sulphuric acid?

Answer: Sulphur trioxide (SO₃).

Question 11. Give an example of neutral oxide.

Answer: Water (H₂O).

Question 12. Give example of a base which is neither metallic oxide nor metallic hydroxide.

Answer: Ammonia (NH₃).

Question 13. What is the pH range of acid rain?

Answer: pH range of acid rain is 3.5 to 5.6;

Question 14. Write down the names of two gases responsible for acid rain.

Answer: SO₂ and NO₂ gas.

Chapter 4 Matter Acids Bases And Salts Topic C Acidic Basic Amphoteric Oxides And Acid Rain Fill In the Blanks

Question 1. The corrosion of marble buildings and monuments caused by acid rain is known as ______

Answer: Stone cancer

Question 2. An amphoteric oxide reacts with both ______ and _______ to produce salts and water.

Answer: Acids, bases

Question 3. ________ oxides are generally acidic in nature.

Answer: Non-metallic

Question 4. ________ oxides are generally basic in nature.

Answer: Metallic

Question 5. An acid present in acid rain is _______

Answer: H₂SO₄ or HNO₃

Question 6. SO₂ dissolves in water to produce ______ acid.

Answer: Sulphurous

Question 7. The product formed due to oxidation of SO₂ dissolves in water to produce ________ acid.

Answer: Sulphuric

Question 8. ________ is an example of mixed acid is an anhydride.

Answer: NO

Question 9. Aluminium oxide is an example of ________ oxide.

Answer: Amphoteric

Question 10. Fe₃O₄ is an example of _______ oxide.

Answer: Mixed

Question 11. Example of neutral oxide is ________

Answer: H₂O

Chapter 4 Matter Acids Bases And Salts Topic C Acidic Basic Amphoteric Oxides And Acid Rain State Whether True Or False

Question 1. Nitrogen dioxide dissolves in water to form nitric acid only.

Answer: False

Question 2. Acid rain mainly consists of the acids like HCI, H₂SO₄ and CH₃COOH .

Answer: False

Question 3. CO is an acidic oxide.

Answer: False

Question 4. Example of an amphoteric oxide is Al₂O₃.

Answer: True

Question 5. PbO is an amphoteric oxide.

Answer: True

Question 6. CO is dissolved in acid rain.

Answer: False

Question 7. NO is an acidic oxide.

Answer: False

Chapter 4 Matter Acids Bases And Salts Topic D Neutralisation Indicator Antacids Salts And Their Classification Synopsis

1. The substances which indicate the end point of an acid-base neutralisation reaction by changing their colour are known as acid-base indicators.
2. Equivalent amount of an acid reacts quantitatively with equivalent amount of a base to neutralise each other and produce salt and water. This reaction is known as neutralisation reaction.
3. Antacids are chemical compounds which are basic in nature and neutralise excess hydrochloric acid secreted in the stomach.
4. The salt formed when all the replaceble hydrogen atoms of an acid are completely replaced by a metal or some other basic radical is known as a normal salt.
5. The salt formed when the replaceable hydrogen atoms of a dibasic or polybasic acid are partially replaced by a metal or some other basic radical is known as an acid salt.
6. The salt formed when the hydroxyl group (OH^–) of a base or an alkali is partially replaced by an acid radical is known as a basic salt.

Chapter 4 Matter Acids Bases And Salts Topic D Neutralisation Indicator Antacids Salts And Their Classification Short And Long Answer Type Questions

Question 1. What are add-base indicators? Name two acid-base indicators.

Answer:

Add-base indicators

The chemical substances which indicate the end point of an acid-base neutralisation reaction by changingtheir colour are called acid-base indicators.

Two acid-base indicators

Two widely used acid-base indicators are methyl orange and phenolphthalein.

Question 2. What changes will be observed in the colour of a litmus paper when it comes in contact with dry ammonia gas and aqueous solution of ammonia?

Answer:

No change will be observed in the colour of litmus paper when it comes in contact with dry ammonia gas. However, the litmus paper turns blue in aqueous solution of ammonia.

Question 3. What is acid-base neutralisation reaction?

Answer:

Acid-base neutralisation reaction

Equivalent amount of an acid reacts quantitatively with equivalent amount of a base to produce salt and water and as a result, both the acid and the base lose their individual properties.

This reaction is known as acid-base neutralisation reaction. For example, 1 gram-equivalent HCI reacts with 1 gram-equivalent NaOH to produce salt and water. Both HCI and NaOH lose their individual properties in course of this reaction.

⇒ \(\begin{gathered}
\mathrm{HCl}(a q) \rightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q) \\
\mathrm{NaOH}\left((a q) \rightarrow \mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q)\right) \\
\hline \mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaCl}(a q)+\mathrm{H}_2 \mathrm{O}(l)
\end{gathered}\)

Question 4. What is neutralisation point? How is neutralisation point determined?

Answer:

Neutralisation point

In an acid-base reaction, the point at which the acid and the base quantitatively react with each other to produce salt and water and as a result, the solution becomes neutral is known as neutralisation point.

The neutralisation point of an acid-base reaction is identified by using an indicator. An indicator indicates the neutralisation point by changing its colour.

Question 5. How wilt you differentiate between Na₂CO₃ and NaHCO₃ solution with one drop of an indicator?

Answer:

Both Na₂CO₃ and NaHCO₃ hydrolyse in their aqueous solutions to form NaOH thereby making the resulting solution alkaline. However, the pH value of Na₂CO₃ solution is greater than that of NaHCO₃.

Hence, on addition of one drop of phenolphthalein to Na₂CO₃ solution, the solution turns pink. On the other hand, if one drop of phenolphthalein is added to NaHCO₃ solution, the solution remains colourless.

Hence one drop of phenolphthalein is sufficient to differentiate between Na₂CO₃ solution and NaHCO₃ solution.

Question 6. Mention the characteristic features of indicators used in neutralisation reactions. Mention the criteria for choosing an indicator for neutralisation reaction with example.

Answer:

Characteristic features of indicators used in neutralisation reactions

An acid-base indicator should be such that it can exhibit different colours in acidic, basic and neutral solutions. For example, methyl orange is yellow in alkaline solution, pinkish-red in acidic solution and orange in neutral solution.

An acid-base indicator is chosen for a particular neutralisation reaction on the basis of the change in pH-value of the solution at the equivalence point or end point.

The criteria for choosing an indicator for neutralisation reaction with example.

The selected indicator must exhibit a sharp change in colour in the same pH range as required around the end point. For example, in the neutralisation of HCI and NaOH, a sharp change in pH of the solution from 3.34 to 9.7 is observed around the end point.

The pH range of methyl orange is 3.4-4.4. Hence, methyl orange can be used for the neutralisation reaction of HCI and NaOH.

Question 7. Tabulate the colours exhibited by the following indicators in acidic, basic and neutral medium—methyl orange, litmus, phenolphthalein and methyl red.

Answer:

The colours exhibited by the given indicators in acidic, basic and neutral medium are tabulated below

Question 8. How is an indicator chosen for an acid- base neutralisation reaction?

Answer:

At the end point of an acid-base neutralisation reaction, there is a rapid and sharp change in the pH of the solution. The indicators which can change their colour in that pH-range are selected for those acid-base neutralisation reactions.

For example, in the neutralisation of a strong acid and a strong base, the range of pH change is maximum near the end point. The pH range of almost all the indicators lies in this range.

Hence, any indicator is suitable for titration of a strong acid and a strong base. On the other hand, in the neutralisation of a weak acid and a weak base, the range of pH change is minimum near the end point.

No suitable indicator having pH range in this region has been found. Hence, the titration of a weak acid and a weak base is not possible.

Question 9. Mention the colour of the solutions if phenolphthalein is added to them.

1. NaCl
2. Ca(OH)₂
3. HCI
4. MgCI₂.

Answer:

Question 10. What are antacids? Give some examples.

Answer:

Antacids

The substances which are used to neutralise the excess hydrochloric acid secreted in the stomach and maintain the pH of the gastric juice at an optimum level, are known as antacids.

Gelusil, Digene, Diovol are some of the commonly used antacids.

Question 11. Write the compositions of common antacids, Gelusil and Diovol.

Answer:

The compositions of common antacids, Gelusil and Diovol are

Gelusil is composed of aluminium hydroxide, magnesium trisilicate and siloxane.

Diovol is composed of aluminium hydroxide, magnesium carbonate and magnesium hydroxide.

Question 12. What are systemic land non-systemic antacids? Give examples of each.

Answer:

Systemic antacids: The antacids which get readily dissolved and absorbed in blood and hence, disturb the acid-base balance of the body are called systemic antacids.

They follow a definite mechanism to reduce acidity. Common examples include sodium bicarbonate, sodium citrate etc.

Non-systemic antacids: Non-systemic antacids are insoluble and are very poorly absorbed in the blood. Hence, they do not disturb the acid-base balance of the body.

They do not follow any definite mechanism of functioning as antacids. Some common examples include aluminium hydroxide, different calcium salts etc.

Question 13. What is milk of magnesia? Write its uses and side effects.

Answer:

Milk of magnesia: Aqueous suspension of magnesium hydroxide is known as milk of magnesia. It contains 7-8% magnesium hydroxide.

Uses:

1. Magnesium hydroxide is an effective antacid.  It rapidly reacts with HCI produced in stomach and neutralises the acid. Hence, it provides quick relief from acidity.
2. It is often used along with AI(OH)₃ as antacid. This provides quick relief from acidity and eliminates any possibility of constipation due to AI(OH)₃.

Side Effects:

1. HCI reacts with Mg(OH)₂ to produce MgCl₂ which may cause diarrhoea.
2. Deposition of magnesium in the kidneys can cause toxic effect in the body.

Question 14. Discuss the uses and side effects of sodium bicarbonate as antacid.

Answer:

Uses: Sodium bicarbonate is highly soluble in water and hence, it works rapidly as an antacid. However, it acts for a relatively short duration.

Side Effects:

1. Sodium bicarbonate produces CO₂ in the stomach which may cause uneasiness and infection in the bile duct.
2. It disturbs the acid-base balance of the body.
3. It increases the Na⁺ ion concentration in blood which may cause high blood pressure.

Question 15. Discuss the uses and side effects of aluminium hydroxide as antacid.

Answer:

Uses of aluminium hydroxide as antacid: Aluminium hydroxide is used to decrease the hyperacidity of stomach. It works slowly and hence, takes time to relieve the pain. Large amount of acid is produced in the stomach due to peptic ulcer, dyspepsia etc.

These result in chest pain, uneasiness, acidity etc. Aluminium hydroxide readily reacts with the excess acid in the stomach and provides relief.

Side Effects of aluminium hydroxide as antacid:

1. Excess use of AI(OH)₃ may cause constipation.
2. Prolonged and continuous use of AI(OH)₃ may decrease the phosphate level of blood. As a result, the excretion of Cu through urine increases which results in renal rickets.

Question 16. State whether the aqueous solution of all normal salts are neutral in nature.

Answer:

A normal salt is produced due to complete neutralisation of an acid and a base. So, the aqueous solution of a normal salt should be neutral in nature.

However, there are a number of normal salts whose aqueous solutions are either acidic or alkaline in nature. The cations and anions produced from these salts in their aqueous solutions react with water molecules thereby increasing the concentration of H₃O⁺ or OH^–  ions. As a result, the solution becomes acidic or alkaline accordingly.

Apart from a salt of strong acid and strong base, the solutions of all other normal salts are either acidic or alkaline.

For example, NH₄CI is a normal salt of strong acid, HCI and weak alkali, NH₄OH . The hydrolysis of this salt produces excess H₃O⁺ ions in water and the aqueous solution becomes acidic.

On the other hand, Na₂CO₃ is a salt of weak acid, H₂CO₃ and strong base, NaOH. It hydrolyses to produce excess OH^– ions in the solution thus, making it alkaline in nature.

Question 17. Baking soda was used in earlier days as antacid. But nowadays Mg(OH)₂, AI(OH)₃ are used instead. State the reason behind it.

Answer:

Baking soda can disturb the acid-base equilibrium of the body is it can be absorbed by blood. Baking soda can also interfere with how the body absorbs some medications.

Now-a-days, Mg(OH)₂ and AI(OH)₃ are used instead. These antacids do not destroy the acid- base equilibrium inside the body as they do not get absorbed by blood. These antacids form corresponding salt and water to decrease the acidity by neutralising the excess amount of HCI present in stomach.

Question 18. What is meant by basicity of an acid?

Answer:

Basicity of an acid

The number of replaceable hydrogen atoms present in 1 molecule of an acid or the number of H⁺ (or H₃O⁺) ions furnished by 1 molecule of the acid in aqueous solution is called the basicity of the acid.

For example, 1 molecule of HCI dissociates in aqueous solution to produce one H⁺ ion while 1 molecule of H₂SO₄ dissociates to give two H⁺ ions. So, the basicity of HCI and H₂SO₄ are 1 and 2 respectively.

Question 19. What is meant by acidity of a base?

Answer:

Acidity of a base

The number of OH^– ions produced by 1 molecule of a base is called the acidity of the base. In other words, it may be defined as the. number of molecules of a monobasic acid required to neutralise 1 molecule of the base quantitatively.

For example, 1 molecule of NaOH furnishes one OH^– ion in aqueous solution and 1 molecule of monobasic acid, HCI is required for complete neutralisation of 1 molecule of NaOH. Hence, acidity of NaOH is 1.

Question 20. Define normal salt with examples.

Answer:

Normal salt with examples

The salt produced when all the replaceable hydrogen atoms of an acid are completely replaced by a metal or basic radical is known as normal salt. Some examples are sodium chloride (NaCI), zinc sulphate (ZnSO₄) etc.

\(\mathrm{Zn}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{ZnSO}_4+\mathrm{H}_2\)

 

A molecule of H₂SO₄ has two replaceable hydrogen atoms. Zn reacts with sulphuric acid and replaces both the hydrogen atoms forming ZnSO₄. Thus, ZnSO₄ is a normal salt.

Question 21. Define acidic salts with examples.

Answer:

Acidic Salts :-

The salt produced when the replaceable hydrogen atoms of a polyprotic acid is partially replaced by a metal or basic radical is known as acidic salt. For example, sodium bisulphate (NaHS04), sodium bicarbonate (NaHCO₃) etc.

⇒ \(\mathrm{H}_2 \mathrm{SO}_4+\mathrm{NaOH} \rightarrow \mathrm{NaHSO}_4+\mathrm{H}_2 \mathrm{O}\)

H₂SO₄ has two replaceable hydrogen atoms. NaHSO₄ (sodium bisulphate) is produced when one hydrogen atom of sulphuric acid is replaced by Na-atom. Hence, NaHSO₄ is an acidic salt.

Question 22. Define basic salts with examples.

Answer:

Basic Salts :-

The salt produced when the replaceable hydroxyl radicals of a base or an alkali are partially replaced by an acid radical is known as basic salt. Some examples of basic salts are basic lead nitrate [Pb(OH)NO₃], basic lead chloride [Pb(OH)CI] etc.

Example 23. Explain ‘sodium bicarbonate is an acid salt’.

Answer:

Sodium Bicarbonate Is An Acid Salt :-

Sodium bicarbonate is the salt of strong alkali, NaOH and weak acid.H₂CO₃. Now H₂CO₃ is a dibasic acid (\(\mathrm{H}_2 \mathrm{CO}_3 \rightleftharpoons 2 \mathrm{H}^{\oplus}+\mathrm{CO}_3^{2-}\)) and of the two replaceable hydrogens of H₂CO₃, one H-atom is replaced by Na-atom to form NaHC03. That is why NaHCO₃ is an acid salt.

Question 24. HCI and HNO₃ always form normal salts but H₂SO₄ can form both normal and acid salts—Explain why.

Answer:

HCI and HNO₃ always form normal salts but H₂SO₄ can form both normal and acid salts

HCI and HNO₃ are monobasic acids. By the replacement of the only replacable H-atom, corresponding normal salts can only be formed. On the other hand, H₂SO₄ is a dibasic acid. So there is a probability of formation of both acid salt and normal salt.

When one of the two replacable ‘H’ is replaced by metal ion, acid salt is formed and when both of the replaceable hydrogen atoms are replaced by metal atoms, normal salt is formed.

Question 25. Mention the name and formula of a normal salt and an acid salt which are used in everyday life.

Answer:

Normal Salt And An Acid Salt :-

A normal salt used in everyday life is sodium chloride (NaCI) or common salt.

An acid salt used in everyday life is baking powder or sodium bicarbonate (NaHCO₃).

Question 26. Aqueous solution of sodium bicarbonate is alkaline although it Is an acid salt—Explain.

Answer:

Aqueous solution of sodium bicarbonate is alkaline although it Is an acid salt

Sodium bicarbonate dissociates into Na^⊕ and HCO₃^Θ  of ion in aqueous solution.

⇒ \(\mathrm{NaHCO}_3 \rightleftharpoons \mathrm{Na}^{\oplus}+\mathrm{HCO}_3^{\ominus}\)

Now in aqueous solution the tendency of loosing a proton (H⁺) from HCO₃^Θ ion is less than the tendency to accept a proton. So HCO₃ accepts a proton to form undissociated H₂CO₃.

⇒ \(\mathrm{HCO}_3^{\ominus}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_2 \mathrm{CO}_3+\mathrm{OH}^{\ominus}\)

So the concentration of OH^Θ ion increases in the aqueous solution of NaHCO₃. As a result the solution becomes alkaline.

Chapter 4 Matter Acids Bases And Salts Topic D Neutralisation Indicator Antacids Salts And Their Classification Very Short Answer Type Questions Choose The Correct Answer

Question 1. An example of an acid-base indicator is

1. Methyl alcohol
2. Methyl orange
3. Methyl chloride
4. Methylamine

Answer: 2. Methyl orange

Question 2. In neutral solution, the colour of litmus is

1. Violet
2. Red
3. Orange
4. Yellow

Answer: 1. Violet

Question 3. An indicator which shows pink colour in alkaline solution is

1. Litmus
2. Methyl orange
3. Phenolphthalein
4. Methyl red

Answer: 3. Phenolphthalein

Question 4. In neutral solution, the colour of methyl red indicator is

1. Red
2. Blue
3. Yellow
4. Orange

Answer: 4. Orange

Question 5. In acidic solution, the colour of litmus is

1. Blue
2. Red
3. Yellow
4. Violet

Answer: 2. Red

Question 6. An aqueous solution of sodium carbonate (Na₂CO₃) is alkaline because it is a salt consisting of

1. Strong acid and strong base
2. Weak acid and weak base
3. Strong acid and weak base
4. Weak acid and strong base

Answer: 4. Weak acid and strong base

Question 7. An acid salt is produced when sodium hydroxide (NaOH) reacts with

1. HCI
2. HNO₃
3. H₂SO₄
4. CH₃COOH

Answer: 3. H₂SO₄

Question 8. The indicator used in the titration of a strong acid and a weak base is

1. Methyl orange
2. Litmus
3. Phenolphthalein
4. Any indicator

Answer: 1. Methyl orange

Question 9. The indicator which is used in the titration of a weak acid and a strong base is

1. Methyl orange
2. Litmus
3. Phenolphthalein
4. Any indicator

Answer: 3. Phenolphthalein

Question 10. The indicator which is used to differentiate between aqueous solutions of Na₂CO₃ and NH₄CI is

1. Methyl orange
2. Litmus
3. Phenolphthalein
4. Any indicator

Answer: 3. Phenolphthalein

Question 11. The indicator used in the titration of a strong acid and a strong base is

1. Methyl orange
2. Methyl red
3. Phenolphthalein
4. Any indicator

Answer: 4. Any indicator

Question 12. The indicator suitable for titration of a weak acid and a weak base is

1. Methyl orange
2. Phenolphthalein
3. Any indicator
4. No indicator is suitable

Answer: 4. No indicator is suitable

Question 13. Omeprazole is used as a/an

1. Pain killer
2. Antipyretic
3. Drug to decrease acid secretion in the stomach
4. Analgesic

Answer: 3. Drug to decrease acid secretion in the stomach

Question 14. An example of a systemic antacid is

1. NaHCO₃
2. AI(OH)₃
3. NaOH
4. Ca(OH)₂

Answer: 1. NaHCO₃

Question 15. An example of a non-systemic antacid is

1. NaHCO₃
2. Al(OH)₃
3. NaOH
4. Ca(OH)₂

Answer: 2. Al(OH)₃

Question 16. An example of a normal salt is

1. Na₂HPO₄
2. NaH₂PO₄
3. NaH₂PO₃
4. Na₂SO₄

Answer: 4. Na₂SO₄

Question 17. Which of the following is the major component of milk of magnesia?

1. Fe(OH)₃
2. Mn(OH)₂
3. Mg(OH)₂
4. Ca(OH)₂

Answer: 3. Mg(OH)₂

Question 18. An example of an acidic salt is

1. Na₂SO₄
2. NaH₂PO₂
3. NaCl
4. Na₂HPO₄

Answer: 4. Na₂HPO₄

Question 19. An example of a basic salt is

1. Pb(OH)CI
2. PbCl₂
3. Pb(OH)₂
4. PbO

Answer: 1. Pb(OH)CI

Question 20. The aqueous solution of which of the following salts is acidic?

1. NH₄CI
2. NaCl
3. Na₂SO₄
4. CH₃COONa

Answer: 1. NH₄CI

Question 21. The compound used as an antacid

1. Ca(OH)₂
2. Mg(OH)₂
3. ZnO
4. NaOH

Answer: 2. Mg(OH)₂

Question 22. Which olne of the following is a tribasic acid?

1. H₃PO₃
2. H₃PO₄
3. CH₃COOH
4. H₃PO₂

Answer: 2. H₃PO₄

Question 23. Baking powder is

1. Acid
2. Base
3. Acid salt
4. Salt

Answer: 3. Acid salt

Question 24. To neutralise the excess acid of stomach, which one is used

1. Antibiotic
2. Antacid
3. Analgesic
4. Antiseptic

Answer: 2. Antacid

Question 25. To form basic salt, minimum acidity of a base should be

1. 1
2. 2
3. 3
4. None of the above

Answer: 2. 2

Question 26. Number of neutral salts among NaH₂PO₂, Na₂HPO₃, Na₂SO₄, Ca(OH)CI is

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Question 27. A component of digestive juices is

1. H₂SO₄
2. HNO₃
3. HCI
4. HCOOH

Answer: 3. HCI

Question 28. The colour observed by adding a drop of phenolphthalein to an aqueous solution of washing soda is

1. Red
2. Blue
3. Colourless
4. Pink

Answer: 4. Pink

Question 29. Which of the following compounds is not used as antacid?

1. Sodium bicarbonate
2. Calcium carbonate
3. Sodium hydroxide
4. Aluminium silicate

Answer: 3. Sodium hydroxide

Question 30. Which of the following compound is present in baking powder?

1. Na₂CO₃
2. H₂CO₃
3. NaHCO₃
4. CaCO₃

Answer: 3. NaHCO₃

Chapter 4 Matter Acids Bases And Salts Topic D Neutralisation Indicator Antacids Salts And Their Classification Answer In Brief

Question 1. Name two acid-base indicators.

Answer: Methyl orange and phenolphthalein are two widely used acid-base indicators.

Question 2. How do indicators indicate the end point of an acid-base neutralisation reaction?

Answer: An indicator indicates the end point of an acid-base neutralisation reaction by changing its colour.

Question 3. Name an indicator which exhibits orange colour in neutral solution.

Answer: Methyl orange.

Question 4. Name two indicators which exhibit yellow colour in alkaline solution.

Answer: Methyl orange and methyl red exhibit yellow colour in alkaline solution.

Question 5. Give an example of an acid which can produce two types of salts by reacting with a base.

Answer: Sulphuric acid (H₂SO₄) reacts with a base to produce two types of salts (acid salt and normal salt). It partially reacts with NaOH to produce an acid salt, sodium bisulphate (NaHSO₄) and on complete neutralisation it produces a normal salt, sodium sulphate (Na₂SO₄).

Question 6. Name an indicator that can be used for the neutralisation reaction of CH₃COOH and NaOH.

Answer: Phenolphthalein.

Question 7. Name an indicator that can be used for the neutralisation reaction of H₂SO₄ and NaOH.

Answer: Any indicator.

Question 8. Name an indicator that can be used for the neutralisation reaction of CH₃COOH and NH₄OH.

Answer: No indicator is suitable for the neutralisation reaction of weak acid, CH3COOH and weak base, NH4OH.

Question 9. what will be the colour of the Na₂CO₃ solution on addition of methyl orange?

Answer: Yellow.

Question 10. Neutralisation reaction is mainly the reaction of which ions?

Answer: H^⊕ and OH^Θ.

Question 11. State the number of molecules of caustic soda required to neutralise 1 molecule of sulfuric acid.

Answer: 2 molecules of caustic soda are required.

Question 12. What is the chemical name of widely used antacid, Zantac?

Answer: The chemical name of widely used antacid Zantac is ranitidine.

Question 13. How does the stomach get affected if acidity continues for a prolonged period of time?

Answer: If acidity continues for a prolonged period of time, it causes ulcer in the stomach.

Question 14. Name the acid salts produced when orthophosphoric acid reacts with NaOH.

Answer: Disodium hydrogen phosphate (Na₂HPO₄) and sodium < dihydrogen phosphate (NaH₂PO₄).

Question 15. Name an acid which always forms normal salt when it reacts with bases.

Answer: Hydrochloric acid (HCI).

Question 16. Give the name and formula of an acid salt which is formed due to partial neutralisation of a monobasic acid.

Answer: An acid salt which is formed due to partial neutralisation of a monobasic acid such as, hydrogen fluoride (HF) is potassium bifluoride (KHF₂).

Question 17. What is the reason behind ‘alkalosis disease’?

Answer: Alkalosis disease Can be caused by taking excess NaHCO₃ as antacid.

Question 18. Give an example of a salt without a metal atom.

Answer: Ammonium chloride (NH₄CI).

Question 19. State about the nature of the aqueous solution of NaHSO₄.

Answer: The aqueous solution of NaHSO₄ is acidic.

Question 20. Write down the names of two main components of antacid.

Answer: Aluminium hydroxide [AI(OH)₃] and magnesium hydroxide [Mg(OH)₂].

Question 21. Write down the formula of the acid salt obtained from the reaction of phosphoric acid and magnesium hydroxide.

Answer: MgHPO_4, (magnesium hydrogen phosphate).

Question 22. What is ‘milk of magnesia’?

Answer: Aqueous suspension of magnesium hydroxide.

Chapter 4 Matter Acids Bases And Salts Topic D Neutralisation Indicator Antacids Salts And Their Classification Fill In The blanks

Question 1. The indicator which turns colourless in acidic solution is ___________

Answer: Phenolphthalein

Question 2. The colours of methyl orange and methyl red change in _______ solution.

Answer: Acidic

Question 3. ________ indicator turns blue in alkaline medium.

Answer: Litmus

Question 4. Universal indicator is actually a _______ of some selective indicators.

Answer: Mixture

Question 5. _______ types of salts are produced when NaOH reacts with H₂SO₄.

Answer: Two

Question 6.  HNO₃ reacts with NaOH to produce _______ type of salt.

Answer: One

Question 7. The name of the salt, CuSO₄  • Cu(OH)₂ is ________

Answer: Basic copper sulphate

Question 8. ______ and _______ are produced in an acid-base neutralisation reaction.

Answer: Salt, water

Question 9. The indicators used in acid-base titrations are generally very weak organic _______ or organic _______

Answer: Acid, bases

Question 10. _______ is an example of a systemic antacid.

Answer: Sodium bicarbonate

Question 11. _______ is an example of a non-systemic antacid.

Answer: Aluminium hydroxide

Question 12. _______ is a normal salt of an inorganic base and an organic acid.

Answer: CH₃COONa

Question 13. [Cu(NH₃)₄]SO₄ is a _______ salt.

Answer: Complex

Question 14. The colour of the KCI solution will be _________ phenolphthalein is added to it.

Answer: Colourless

Question 15. NaH₂PO₂ is a ________ salt.

Answer: Neutral

Question 16. The colour of methyl orange in acid solution is _________

Answer: Pink

Question 17. NaHSO₄ is an example of _________ salt.

Answer: Acid

Question 18. Sodium bicarbonate is an example of _________ salt.

Answer: Acid

Question 19. Methyl orange turns _______ in neutral solution.

Answer: Orange

Question 20. Acidity of H₃PO₄ is ________

Answer: 3

Chapter 4 Matter Acids Bases And Salts Topic D Neutralisation Indicator Antacids Salts And Their Classification State Whether True Or False

Question 1. Methyl orange indicator turns colourless in basic solution.

Answer: False

Question 2. Phenolphthalein can be used as an indicator in the neutralisation reaction of CH₃COOH and NH₄OH.

Answer: False

Question 3. Sodium chloride and zinc sulphate belong to the category of normal salt.

Answer: True

Question 4. Excessive use of aluminium hydroxide as antacid for a long period of time may cause renal rickets.

Answer: True

Question 5. Excessive use of chemical fertilisers on agricultural lands for a long period of time makes the soil alkaline.

Answer: False

Question 6. Sodium carbonate is an alkali.

Answer: F

Question 7. Oxalic acid is a dibasic acid.

Answer: True

Question 8. Sodium surface is an acidic salt.

Answer: False

Chapter 4 Matter Acids Bases And Salts Topic D Neutralisation Indicator Antacids Salts And Their Classification Miscellaneous Type Questions

Match The column

1.

Answer: 1. C, 2. A, 3. D, 4. B

2.

Answer: 1. B, 2. D, 3. A, 4. C