(solved)Question 2.2 of NCERT Class XI Chemistry Chapter 2

$\displaystyle \underline{\underline{\text{Part (i)}}}$

1 molecule $\displaystyle \text{CH}_4 = ^{12}_{\:6}\text{C} + \:^1_1\text{H} \times 4$

$\displaystyle \therefore$ electrons $\displaystyle = 6 + (1 \times 4) = 10$

electrons in 1 mol = $\displaystyle 10 \times 6.02 \times 10^{23}$

$\displaystyle = 6.02 \times 10^{24}\:\underline{Ans}$

$\displaystyle \underline{\underline{\text{Part (ii)}}}$

we know 14 g of $\displaystyle ^{14}_{\:6}\text{C}\equiv 1 $ mol

$\displaystyle \implies 7 \text{ g} \equiv 1/2$ mol

$\displaystyle \implies 7 \text{ mg} \equiv 1/2000$ mol

$\displaystyle = \frac{1}{2000} \times 6.02 \times 10^{23}$ atoms

$\displaystyle = 3.01 \times 10^{20}$ atoms . . . (1)

but $\displaystyle ^{14}_{\:6}\text{C} \implies \text{A} = 14\text{, Z} = 6$

$\displaystyle \therefore$ 1 atom contains A - Z = 8 neutrons

$\displaystyle \therefore 3.01 \times 10^{20} \text{ atoms } \equiv 8 \times 3.01 \times 10^{20}$

$\displaystyle = 2.41 \times 10^{21} \text{ neutrons} \:\underline{Ans}$

total mass = $\displaystyle 2.41 \times 10^{21} \times 1.675 \times 10^{-27}$ kg

$\displaystyle = 4.04 \times 10^{-6}\text{ kg}\:\underline{Ans}$

$\displaystyle \underline{\underline{\text{Part (iii)}}}$

Given 34 mg NH3 i.e., 0.034 g NH3

now, molar mass NH3 = 14 + 3 x 1 = 17 g/mol

∴ 17 g NH3 = 1 mol

∴ 0.034 g NH3 = $\displaystyle \frac{1}{17} \times 0.034 = \frac{1}{500}$ mol

$\displaystyle = \frac{1}{500} \times 6.02 \times 10^{23}$

$\displaystyle = 1.204 \times 10^{21}$ molecules . . . (1)

no. of protons in an atom = Z

∴ protons in N = 7, H = 1

∴ 1 molecule NH3 = 7 + (1 x 3) = 10 protons

∴ $\displaystyle 1.204 \times 10^{21}$ contain $\displaystyle 10 \times 1.204 \times 10^{21}$

$\displaystyle = 1.204 \times 10^{22}$ protons Ans

$\displaystyle \underline{\underline{\text{Part (iii) - will ans change?}}}$

Will the answer change if the temperature and pressure are changed ?

Ans: If temperature and pressure are changed in such a way that the number of moles of the gas remain same, then the answer will not change.