Atoms and Molecules: Class 10 Science Chapter 7 Book Back Questions and Answers

Atoms and Molecules - Book Back Questions

I. Choose the best answer.

1. Which of the following has the smallest mass?

   a. 6.023 × 10²³ atoms of He

   b. 1 atom of He

   c. 2 g of He

   d. 1 mole atoms of He

2. Which of the following is a triatomic molecule?

   a. Glucose

   b. Helium

   c. Carbon dioxide

   d. Hydrogen

3. The volume occupied by 4.4 g of CO₂ at S.T.P

   a. 22.4 litre

   b. 2.24 litre

   c. 0.24 litre

   d. 0.1 litre

4. Mass of 1 mole of Nitrogen atom is

   a. 28 amu

   b. 14 amu

   c. 28 g

   d. 14 g

5. Which of the following represents 1 amu?

   a. Mass of a C – 12 atom

   b. Mass of a hydrogen atom

   c. 1/12^th of the mass of a C – 12 atom

   d. Mass of O – 16 atom

6. Which of the following statement is incorrect?

   a. One gram of C – 12 contains Avogadro’s number of atoms.

   b. One mole of oxygen gas contains Avogadro’s number of molecules.

   c. One mole of hydrogen gas contains Avogadro’s number of atoms.

   d. One mole of electrons stands for 6.023 × 10²³ electrons.

7. The volume occupied by 1 mole of a diatomic gas at S.T.P is

   a. 11.2 litre

   b. 5.6 litre

   c. 22.4 litre

   d. 44.8 litre

8. In the nucleus of $^{40}_{20}\text{Ca}$, there are

   a. 20 protons and 40 neutrons

   b. 20 protons and 20 neutrons

   c. 20 protons and 40 electrons

   d. 40 protons and 20 electrons

9. The gram molecular mass of oxygen molecule is

   a. 16 g

   b. 18 g

   c. 32 g

   d. 17 g

10. 1 mole of any substance contains ____ molecules.

    a. 6.023 × 10²³

    b. 6.023 × 10^-23

    c. 3.0115 × 10²³

    d. 12.046 × 10²³

II. Fill in the blanks

1. Atoms of different elements having same mass number, but different atomic numbers are called isobars.
2. Atoms of different elements having same number of neutrons are called isotones.
3. Atoms of one element can be transmuted into atoms of other element by artificial transmutation.
4. The sum of the numbers of protons and neutrons of an atom is called its mass number.
5. Relative atomic mass is otherwise known as standard atomic weight.
6. The average atomic mass of hydrogen is 1.0079 amu.
7. If a molecule is made of similar kind of atoms, then it is called Homo atomic molecule.
8. The number of atoms present in a molecule is called its atomicity.
9. One mole of any gas occupies 22400 ml at S.T.P.
10. Atomicity of phosphorous is 4.

III. Match the following

Column A	Column B (Correct Match)
1. 8 g of O2	0.25 moles
2. 4 g of H2	2 moles
3. 52 g of He	13 moles
4. 112 g of N2	4 moles
5. 35.5 g of Cl2	0.5 moles

IV. True or False: (If false give the correct statement)

1. Two elements sometimes can form more than one compound.

   Answer: True

2. Noble gases are Diatomic.

   Answer: False

   Correct statement: Noble gases are monoatomic.

3. The gram atomic mass of an element has no unit.

   Answer: False

   Correct statement: The gram atomic mass of an element is expressed in grams.

4. 1 mole of Gold and Silver contain same number of atoms.

   Answer: True

5. Molar mass of CO₂ is 42g.

   Answer: False

   Correct statement: Molar mass of CO₂ is 44g.

V. Assertion and Reason

Answer the following questions using the data given below:

• i) A and R are correct, R explains the A.
• ii) A is correct, R is wrong.
• iii) A is wrong, R is correct.
• iv) A and R are correct, R doesn’t explains A.

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1. Assertion: Atomic mass of aluminium is 27.

   Reason: An atom of aluminium is 27 times heavier than 1/12th of the mass of the C – 12 atom.

   Answer: i) A and R are correct, R explains the A.

2. Assertion: The Relative Molecular Mass of Chlorine is 35.5 a.m.u.

   Reason: The natural abundance of Chlorine isotopes are not equal.

   Answer: i) A and R are correct, R explains the A.

VI. Short answer questions

1. Define: Relative atomic mass.

   The Relative Molecular Mass of a molecule is the ratio between the mass of one molecule of the substance to 1/12th mass of an atom of Carbon-12.

2. Write the different types of isotopes of oxygen and its percentage abundance.

   Oxygen exists as a mixture of three stable isotopes in nature.

   Isotope	Mass (amu)	% abundance
   $^{16}_8\text{O}$	15.9949	99.757
   $^{17}_8\text{O}$	16.9991	0.038
   $^{18}_8\text{O}$	17.9992	0.205

3. Define: Atomicity

   The number of atoms present in the molecule is called as its atomicity.

4. Give any two examples for heterodiatomic molecules.

   Examples of heterodiatomic molecules are HCl, CO.

5. What is Molar volume of a gas?

   (i) The volume occupied by one mole of any gas at S.T.P is called molar volume.

   (ii) One mole of any gas occupies = 22.4 litre or 22400 ml at S.T.P.

6. Find the percentage of nitrogen in ammonia.

   The percentage of nitrogen in ammonia ($\text{NH}_3$) is calculated as:

   $$ \% \text{ of Nitrogen in NH}_3 = \frac{\text{Mass of Nitrogen}}{\text{Molar Mass of NH}_3} \times 100 $$ $$ = \frac{14}{17} \times 100 \approx 82\% $$

VII. Long answer questions

1. Calculate the number of water molecule present in one drop of water which weighs 0.18 g.

   Answer:

   Mass of water = 0.18 g

   Molecular Mass of water ($\text{H}_2\text{O}$) = 18 g/mol

   $$ \text{No of moles} = \frac{\text{Given Mass}}{\text{Molecular Mass}} = \frac{0.18}{18} = 0.01 \text{ mole} $$ $$ \text{Number of molecules} = \text{No of moles} \times \text{Avogadro number} $$ $$ = 0.01 \times 6.023 \times 10^{23} $$ $$ = 6.023 \times 10^{21} \text{ molecules} $$

2. N₂ + 3 H₂ → 2 NH₃

   (The atomic mass of nitrogen is 14, and that of hydrogen is 1)

   1 mole of nitrogen (____ g) + 3 moles of hydrogen (____ g) → 2 moles of ammonia (____ g)

   Answer: 1 mole of nitrogen (28 g) + 3 moles of hydrogen (6 g) → 2 moles of ammonia (34 g)

3. Calculate the number of moles in

   i) 27g of Al ii) 1.51 × 10²³ molecules of NH₄Cl

   Answer:

   (i) 27 g of Al

   $$ \text{No of moles} = \frac{\text{Mass}}{\text{Atomic Mass}} = \frac{27}{27} = 1 \text{ mole} $$

   (ii) 1.51 x 10²³ molecules of NH₄Cl

   $$ \text{No of moles} = \frac{\text{Number of molecules}}{\text{Avogadro number}} $$ $$ = \frac{1.51 \times 10^{23}}{6.023 \times 10^{23}} = 0.25 \text{ mole} $$

4. Give the salient features of “Modern atomic theory”.

   Answer:

   The main postulates of modern atomic theory are as follows:

   • (i) An atom is no longer indivisible.
   • (ii) Atoms of the same element may have different atomic mass. (e.g., Isotopes $^{35}_{17}\text{Cl}$, $^{37}_{17}\text{Cl}$).
   • (iii) Atoms of different elements may have same atomic masses. (e.g., Isobars $^{40}_{18}\text{Ar}$, $^{40}_{20}\text{Ca}$).
   • (iv) Atoms of one element can be transmuted into atoms of other elements.
   • (v) Atoms may not always combine in a simple whole number ratio. (e.g., Glucose C₆H₁₂O₆).
   • (vi) Atom is the smallest particle that take part in a chemical reaction.
   • (vii) Mass of an atom can be converted into energy, according to the equation $E = mc^2$.

5. Derive the relationship between Relative molecular mass and Vapour density.

   Answer:

   Relative molecular mass (Hydrogen scale): The Relative Molecular Mass of a gas or vapour is the ratio between the mass of one molecule of the gas or vapour to mass of one atom of Hydrogen.

   Vapour Density: Vapour density is the ratio of the mass of a certain volume of a gas or vapour, to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.

   $$ \text{Vapour Density (VD)} = \frac{\text{Mass of a given volume of gas or vapour at S.T.P}}{\text{Mass of the same volume of hydrogen}} $$

   According to Avogadro's law, equal volumes of all gases contain an equal number of molecules. Let this number be 'n'.

   $$ \text{VD} = \frac{\text{Mass of 'n' molecules of gas or vapour}}{\text{Mass of 'n' molecules of hydrogen}} $$

   Cancelling 'n', we get:

   $$ \text{VD} = \frac{\text{Mass of 1 molecule of gas or vapour}}{\text{Mass of 1 molecule of hydrogen}} $$

   Since hydrogen is diatomic (H₂), one molecule of hydrogen contains 2 atoms.

   $$ \text{VD} = \frac{\text{Mass of 1 molecule of gas or vapour}}{2 \times \text{Mass of 1 atom of hydrogen}} $$

   From the definition of relative molecular mass:

   $$ \text{Relative Molecular Mass} = \frac{\text{Mass of 1 molecule of gas or vapour}}{\text{Mass of 1 atom of hydrogen}} $$

   Substituting this into the V.D. equation:

   $$ \text{VD} = \frac{\text{Relative Molecular Mass}}{2} $$

   Therefore, the relationship is:

   $$ \text{Relative Molecular Mass} = 2 \times \text{Vapour Density} $$

VIII. HOT question

1. Calcium carbonate is decomposed on heating in the following reaction:

   $$ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 $$

   i. How many moles of Calcium carbonate are involved in this reaction?

   ii. Calculate the gram molecular mass of calcium carbonate involved in this reaction.

   iii. How many moles of CO₂ are there in this equation?

   Answer:

   (i) The number of moles of CaCO₃ involved is 1 mole.

   (ii) Gram molecular mass of CaCO₃ = (Mass of Ca) + (Mass of C) + (3 × Mass of O)

   = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100 g/mol.

   (iii) The number of moles of CO₂ in the reaction is 1 mole.

IX. Solve the following problems

1. How many grams are there in the following?

   i. 2 moles of hydrogen molecule, H₂

   ii. 3 moles of chlorine molecule, Cl₂

   iii. 5 moles of sulphur molecule, S₈

   iv. 4 moles of phosphorous molecule, P₄

   Answer:

   (i) 2 moles of hydrogen molecule (H₂)

   Mass = No. of moles × Molecular mass = 2 × 2 = 4 g

   (ii) 3 moles of Chlorine molecule (Cl₂)

   Mass = No. of moles × Molecular mass = 3 × 70.9 = 212.7 g

   (iii) 5 moles of sulphur molecule (S₈)

   Mass = No. of moles × Molecular mass = 5 × (32 × 8) = 5 × 256 = 1280 g

   (iv) 4 moles of phosphorous molecule (P₄)

   Mass = No. of moles × Molecular mass = 4 × (31 × 4) = 4 × 124 = 496 g

2. Calculate the % of each element in calcium carbonate. (Atomic mass: C-12, O-16, Ca-40)

   Answer: Molar mass of CaCO₃ = 100 g/mol.

   $$ \% \text{ of Ca} = \frac{40}{100} \times 100 = 40\% $$ $$ \% \text{ of C} = \frac{12}{100} \times 100 = 12\% $$ $$ \% \text{ of O} = \frac{3 \times 16}{100} \times 100 = \frac{48}{100} \times 100 = 48\% $$

3. Calculate the % of oxygen in Al₂(SO₄)₃. (Atomic mass: Al-27, O-16, S-32)

   Answer:

   Molar mass of Al₂(SO₄)₃ = (2 × 27) + (3 × 32) + (12 × 16) = 54 + 96 + 192 = 342 g/mol.

   $$ \% \text{ of Oxygen} = \frac{\text{Mass of Oxygen in compound}}{\text{Total Molar Mass}} \times 100 $$ $$ = \frac{192}{342} \times 100 \approx 56.1\% $$

4. Calculate the % relative abundance of B-10 and B-11, if its average atomic mass is 10.804 amu.

   Answer:

   Let the % abundance of B-10 be $x$ and B-11 be $y$.

   $ x + y = 100 \implies y = 100 - x $

   Atomic mass of B-10 = 10.01294 amu

   Atomic mass of B-11 = 11.00931 amu (using a more standard value)

   $$ \text{Average Atomic Mass} = \frac{( \text{Mass}_1 \times x ) + ( \text{Mass}_2 \times y )}{100} $$ $$ 10.804 = \frac{(10.01294 \times x) + (11.00931 \times (100 - x))}{100} $$ $$ 1080.4 = 10.01294x + 1100.931 - 11.00931x $$ $$ 1080.4 = 1100.931 - 0.99637x $$ $$ 0.99637x = 1100.931 - 1080.4 $$ $$ 0.99637x = 20.531 $$ $$ x = \frac{20.531}{0.99637} \approx 20.6\% $$

   So, $ y = 100 - 20.6 = 79.4\% $

   The relative abundance of B-10 is approximately 20.6% and B-11 is 79.4%.