Atoms and Molecules CBSE Class 9 Chemistry Worksheets Set 01

Questions with Answers

Question 1. Which of the following weighs the most: (i) 40 g of mercury, (ii) one mole of oxygen, (iii) 0.5 g atom of silver, (iv) 11.2 L of nitrogen at S.T.P.?
Answer:
(i) 40 g.
(ii) 1 mol O₂ = 32 g.
(iii) 0.5 g atom of Ag = 0.5 × 108 = 54 g.
(iv) 11.2 L N₂ = 0.5 mol = 14 g.
0.5 g atom of silver (54 g) weighs the most.
In simple words: Convert everything to grams.

Teacher's Note:
1. Mass = moles × molar mass.
2. A gram atom is one mole of atoms.

Question 2. The density of ammonia at S.T.P. is 0.758 g/L. Calculate the molar mass of ammonia.
Answer:
Molar mass = mass of 22.4 L at S.T.P. = 0.758 × 22.4 = 16.98 g.
Molar mass of NH₃ = 17 g/mol, which agrees with 14 + 3 = 17.
In simple words: Density multiplied by 22.4 gives the molar mass.

Teacher's Note:
1. This is how molar masses of gases are found.
2. Check with the formula.

Question 3. Calculate the number of moles of sulphur dioxide in 0.064 kg of the gas.
Answer:
Molar mass of SO₂ = 32 + 32 = 64 g/mol. Mass = 0.064 kg = 64 g.
Moles = 64/64 = 1 mole.
In simple words: Convert kg to g and divide by the molar mass.

Teacher's Note:
1. 1 kg = 1000 g.
2. Moles = mass / molar mass.

Question 4. Which of the following contains the least number of molecules and which the highest? (i) 33 g CO₂, (ii) 8 g O₂, (iii) 21 g N₂, (iv) 2.016 g H₂.
Answer:
(i) 33/44 = 0.75 mol; (ii) 8/32 = 0.25 mol; (iii) 21/28 = 0.75 mol; (iv) 2.016/2.016 = 1 mol.
Least: 8 g of O₂ (0.25 N_A molecules). Highest: 2.016 g of H₂ (N_A molecules).
In simple words: More moles means more molecules.

Teacher's Note:
1. Hydrogen has the smallest molar mass.
2. Use moles to compare.

Question 5. The atomic mass of helium is 4. Calculate the number of (i) atoms and (ii) molecules in 1 g of helium gas.
Answer:
Helium is monoatomic, so one molecule is one atom.
Moles = 1/4 = 0.25. Number = 0.25 × 6.022 × 10²³ = 1.5055 × 10²³.
So both the number of atoms and the number of molecules are 1.5055 × 10²³.
In simple words: A noble gas molecule contains only one atom.

Teacher's Note:
1. Atomicity of helium is 1.
2. Number = moles × N_A.

Question 6. Arrange the following quantities of gold in increasing order of mass: (i) 200 mg, (ii) 0.1 mole, (iii) 2.5 × 10²⁰ atoms (atomic mass of gold 197.2).
Answer:
(i) 200 mg = 0.200 g.
(ii) 0.1 mol = 0.1 × 197.2 = 19.72 g.
(iii) 2.5 × 10²⁰ atoms = (2.5 × 10²⁰/6.022 × 10²³) × 197.2 = 0.0819 g.
Increasing order: (iii) 0.0819 g < (i) 0.200 g < (ii) 19.72 g.
In simple words: Convert all to grams and then compare.

Teacher's Note:
1. 1 mole of gold = 197.2 g.
2. Do not compare the numbers without converting to the same unit.

Question 7. Calculate the mass of one molecule of (i) sulphuric acid and (ii) carbon monoxide.
Answer:
(i) Molar mass of H₂SO₄ = 2 + 32 + 64 = 98 g. Mass of one molecule = 98/(6.022 × 10²³) = 1.627 × 10⁻²² g.
(ii) Molar mass of CO = 28 g. Mass of one molecule = 28/(6.022 × 10²³) = 4.65 × 10⁻²³ g.
In simple words: Divide the molar mass by Avogadro's number.

Teacher's Note:
1. A mole contains 6.022 × 10²³ molecules.
2. Masses of molecules are very small.

Question 8. Calculate the volume at S.T.P. occupied by (i) 320 g of sulphur dioxide and (ii) 324 g of nitrogen pentoxide (N₂O₅).
Answer:
(i) Molar mass of SO₂ = 64 g. Moles = 320/64 = 5. Volume = 5 × 22.4 = 112 L.
(ii) Molar mass of N₂O₅ = 28 + 80 = 108 g. Moles = 324/108 = 3. Volume = 3 × 22.4 = 67.2 L.
In simple words: Find moles and multiply by 22.4 L.

Teacher's Note:
1. Molar volume at S.T.P. is 22.4 L.
2. Mass → moles → volume.

Question 9. Calculate the molar mass of a compound if 1.5055 × 10²⁴ molecules weigh 70 g.
Answer:
Moles = 1.5055 × 10²⁴ / 6.022 × 10²³ = 2.5.
Molar mass = 70/2.5 = 28 g/mol.
In simple words: Number of molecules gives moles; mass divided by moles is the molar mass.

Teacher's Note:
1. The compound could be CO or N₂.
2. Molar mass = mass / moles.

Question 10. Calculate the number of atoms in (i) 52 mole of He, (ii) 52 amu of He, (iii) 52 g of He.
Answer:
Helium is monoatomic.
(i) 52 × 6.022 × 10²³ = 3.13 × 10²⁵ atoms.
(ii) One He atom has a mass of 4 amu, so 52 amu is 52/4 = 13 atoms.
(iii) 52 g is 52/4 = 13 mol, so 13 × 6.022 × 10²³ = 7.83 × 10²⁴ atoms.
In simple words: Each part uses a different unit; convert carefully.

Teacher's Note:
1. amu is the mass of one atom, g is the mass of many.
2. Atoms = moles x N_A.

Question 11. When 0.48 g of magnesium is burnt in excess oxygen, 0.80 g of magnesium oxide is obtained. A sample of pure magnesium oxide from another source contains 60% magnesium and 40% oxygen by mass. Show that the data illustrate the law of constant proportions.
Answer:
First sample: oxygen = 0.80 − 0.48 = 0.32 g, so Mg : O = 0.48 : 0.32 = 3 : 2 = 1.5 : 1.
Second sample: Mg : O = 60 : 40 = 3 : 2 = 1.5 : 1.
The ratio is the same, so the data illustrate the law of constant proportions.
In simple words: The ratio of masses of magnesium to oxygen is 3 : 2 in both.

Teacher's Note:
1. Law of constant proportions: a compound always has the same elements in the same mass ratio.
2. The ratio does not depend on the source.

Question 12. Dry hydrogen is passed over 1.6 g of hot cupric oxide and 1.28 g of copper is obtained. 0.64 g of copper is dissolved in nitric acid, the solution is evaporated and strongly heated, and 0.8 g of cupric oxide is obtained. Show that cupric oxide has a definite composition.
Answer:
First sample: oxygen = 1.60 − 1.28 = 0.32 g, so Cu : O = 1.28 : 0.32 = 4 : 1.
Second sample: oxygen = 0.80 − 0.64 = 0.16 g, so Cu : O = 0.64 : 0.16 = 4 : 1.
Both ratios are 4 : 1, so cupric oxide has a definite composition.
In simple words: Copper and oxygen combine in the ratio 4 : 1 by mass in both preparations.

Teacher's Note:
1. Both methods give the same compound.
2. Proust's law.

Question 13. Analysis of 1.171 g of sodium chloride gave 0.46 g of sodium and 0.712 g of chlorine. Another sample showed 39.0% sodium and 61.0% chlorine. Do the results agree with the law of constant proportions?
Answer:
First sample: Cl : Na = 0.712 : 0.46 = 1.55 (sodium = 1.171 − 0.712 = 0.459 g).
Second sample: Cl : Na = 61.0 : 39.0 = 1.56.
The ratios agree (within experimental error), so the results agree with the law of constant proportions.
In simple words: Both samples have chlorine and sodium in about 1.56 : 1.

Teacher's Note:
1. Small differences come from measurement.
2. Percent composition is the same.

Question 14. Determine the number of water molecules in one drop of water of diameter 2 mm (density of water 0.9 g/mL).
Answer:
Radius = 1 mm = 0.1 cm. Volume = (4/3)π(0.1)³ = 4.19 × 10⁻³ cm³.
Mass = 0.9 × 4.19 × 10⁻³ = 3.77 × 10⁻³ g.
Moles = 3.77 × 10⁻³/18 = 2.09 × 10⁻⁴.
Molecules = 2.09 × 10⁻⁴ × 6.022 × 10²³ = 1.26 × 10²⁰.
In simple words: Volume → mass → moles → molecules.

Teacher's Note:
1. 1 mL = 1 cm³.
2. The number is huge although the drop is tiny.

Question 15. Calculate the volume of a single molecule of water (density of water 1 g/mL).
Answer:
One mole of water (18 g) occupies 18 mL and has 6.022 × 10²³ molecules.
Volume of one molecule = 18/(6.022 × 10²³) = 2.99 × 10⁻²³ mL (about 3 × 10⁻²³ cm³).
In simple words: Divide the molar volume of water by N_A.

Teacher's Note:
1. The molecule is very small.
2. Treat the liquid as closely packed.