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

(i) Calculate the total number of electrons present in one mole of methane. (ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C. (Assume that mass of a neutron = 1.675 × 10-27 kg). (iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP. Will the answer change if the temperature and pressure are changed ?
(Rev. 27-Oct-2023)

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### Question 2.2 NCERT Class XI Chemistry

(i) Calculate the total number of electrons present in one mole of methane. (ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C. (Assume that mass of a neutron = 1.675 × 10-27 kg). (iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP. Will the answer change if the temperature and pressure are changed ?

### Video Explanation(detailed solution given after this video)

Please watch this youtube video for a quick explanation of the solution:

### Solution in Detail

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

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.