NEET Foundation Chemistry Chapter 3 Atoms And Molecules Short Answer Questions

Chapter 3 Atoms And Molecules Short Answer Type Question And Answers

Question 1. Calculate the molar mass of HNO₃. [N = 14, O = 16, H = 1]
Answer.

Given:

Molar mass of HNO₃.

H = 1 × 1 = 01

N = 14 × 1 = 14

O = 16 × 3 = 48

Total mass = 63 grams

Molar mass of HNO₃ = 63 grams.

Question 2. Calculate the formula mass of CaCl₂. [Ca = 40, Cl = 35.5]
Answer.

Given:

1(Ca) + 2(Cl) 40 + 2 × (35.5) = 111 u

The formula mass of CaCl₂ is 111 u.

Question 3. A certain non-metal X forms two oxides 1 and 2. The mass percentage of oxygen in oxide 1 (X4O6) is 43.7, which is same as that of X in oxide 2. Find the formula of the second oxide.

Answer.

Now 43.7 parts of oxygen in I corresponds to = 6 oxygen atoms

Therefore, 56.3 parts of oxygen in II corresponds to

⇒ \(\frac{6 \times 56.3}{43.7}=7.730 \text { atoms }\)

Also 56.3 parts of X in 1 correspond to = 4 X atom

Therefore, 43.7 parts of X in 2 will correspond to

⇒ \(=\frac{4 \times 43.7}{56.3} \times 3.1 \times \text { atoms }\)

Now the atomic ration X : O in the second

Oxide = \(\frac{3.1}{3.1}: \frac{7.73}{3.1}\) or 1 : 25 or 2 : 5

The formula of the second oxide is X₂O₂.

Question 4. Calculate the mass of 0.2 moles of water (O = 16, H = 1).
Answer.

Given:

Gram molecular weight of H₂O = 2 × 1 + 16 = 18 g

1 mole of water weighs 18 g

Therefore, 0.2 moles of water weighs \(\frac{18}{1}\) × 0.2 = 3.6 g

Question 5. Calculate the volume of 7.1 g of chlorine (Cl = 35.5) at S.T.P.
Answer.

Given:

Gram Molecular Weight of Cl₂ (one mole) = 35.5 × 2 = 71 g.

71 g of Cl₂ at S.T.P occupies 22.4 litres

Therefore, 7.1 g of Cl₂ at S.T.P occupies

⇒ \(\frac{22.4}{71} \times 7.1=2.24 \text { litres }\)

Question 6. The reaction between aluminium carbide and water takes place according to the following equation:

Al₄C₃ + 12H₂O → 3CH₄ + 4Al(OH)₃

Calculate the volume of CH4 released from 14.4 g of Al₄C₃ by excess water at S.T.P. (C = 12, Al = 27)
Answer.

Given

CH4 released from 14.4 g of Al₄C₃ by excess water at S.T.P. (C = 12, Al = 27)

Molecular weight of Al₄C₃ is (27 × 4) + (12 × 3) = 144

144 g of Al₄C₃ produces 3 × 22.4 litres of CH₄ at S.T.P

Therefore, 14.4 g Al₄C₃ produces \(\frac{3 \times 22.4}{144} \times\) 14.4

⇒ \(=\frac{967.7}{144}=6.72 \text { litres }\)

Question 7. How many litres of ammonia are present in 3.4 kg of it? (N = 14, H = 1)
Answer.

Given:

Gram molecular weight of NH₃ = 14 + (1 × 3) = 17 g.

17 g of NH₃ = 22.4 litres

Therefore, 3.4 × 103g of NH₃ = \(\frac{22.4}{17} \times 3.4 \times 10^3\)

= \(\frac{76160}{17}\)

= 4480 litres.

Question 8. Define a mole.
Answer.

Mole: A mole is the amount of substance that contains as many elementary entities as there are atoms in 12 grams in carbon 12 isotopes .

Question 2. Find out the molar mass of sulphuric acid.
Answer.

Formula of sulphuric acid = H₂SO₄

No. of atoms:

H = 2

S = 1

O = 4

Atomic mass:

H = 1

S = 32

O = 16

Molar mass= (2 × 1) + (1 × 32) + (4 × 16) = 98 g/mol

Question 9. Calculate the number of moles and atoms in 240 grams of O (Oxygen) atom.
Answer.

Atomic Mass of O = 16 u

16 u of O = 1 atom of oxygen

16 g of O = 1 mole of oxygen

16 g of O = 6.022 × 10²³ atom of O

240 g of O = 15 moles of O

240 g of O = 15 × 6.022 × 10²³ atoms

Therefore, 240 g of O = 90.33 × 10²³ atoms.

Question 10. Find out the weight of 120.44 × 10²³ molecules of water.
Answer.

Molecular mass of H₂O = 18 u [(1 × 2) H + (16 × 1)O]

So, 1 mole = 18 g H₂O

6.022 × 10²³ = 18 g

120.44 × 10²³ = (120.44 × 10²³/6.022 × 10²³) × 18

= 20 × 18 = 360 g

Question 11. 6.6 g of CaCO₃on heating gave 2.98 g CaO and 3.62 g CO₂. Prove that these observations agree with law of conservation of mass.
Answer.

Given:

6.6 g of CaCO₃on heating gave 2.98 g CaO and 3.62 g CO₂.

CaCO₃ → CaO + CO₂

6.6g → 2.88 g + 3.52 g

Mass of the reactant = 6.4 g

Mass of the product = (2.98 + 3.62) g = 6.6 g

These results agree with the law of conservation of mass as the mass of reactants is equal to the mass of products.

Question 12. An 80.0 g sample of an unknown compound contains 16.4 g of hydrogen. What is the percent by mass of hydrogen in the compound?
Answer.

Given:

An 80.0 g sample of an unknown compound contains 16.4 g of hydrogen.

Mass of the compound = 80 g

Mass of hydrogen in the compound = 16.4 g

Therefore, the mass fraction of hydrogen in the unknown compound = (16.4/80) x 100% = 20.5%

Question 13. What is the importance of law of conservation of mass in everyday life?
Answer.

Importance of law of conservation of mass in everyday life: Law of conservation of mass is important to study to produce chemical reactions. Chemists can predict the amount of products that will be produced in a chemical reaction if they know the amount and identities of the reactants.

Question 14. What is the limitation of law of definite proportions?
Answer.

Limitation of law of definite proportions: The law of definite proportions does not hold good for those elements who are also present in different isotopic forms in a compound.