(i) Total number of electrons present in one mole of CH4 = 6 × 6.022 × $10^{23}$ + 4 × 6.022 × $10^{23}$ = 36.132 × $10^{23}$ + 24.088 × $10^{23}$ = 6.022 × $10^{24}$ electrons (ii) We know that, p + n = A 6 + n = 14 [Since e = p = Z] n = 14 – 6 = 8 Now, 1 mole of $\,{}^{14}\mathrm{C}$ = 14 g of C = 6.022 × $10^{23}$ atoms of $\,{}^{14}\mathrm{C}$ = 6.022 × $10^{23}$ × 8 neutrons = 4.8176 × $10^{24}$ neutrons (a) 14 g of $\,{}^{14}\mathrm{C}$ contains = 4.8176 × $10^{24}$ neutrons 7 mg (7 × $10^{-3}$ g) of $\,{}^{14}\mathrm{C}$ contains = $\frac{4.8176 \times 7 \times 10^{-3}}{14}$ neutrons = 2.4088 × $10^{21}$ neutrons (b) Mass of a neutron = 1.675 × $10^{-27}$ kg Mass of 2.4088 × $10^{21}$ neutrons = 1.675 × $10^{-27}$ × 2.4088 × $10^{21}$ = 4.0347 × $10^{-6}$ kg (iii) (a) 17 g of $\mathrm{NH}_3$ has 10 × 6.022 × $10^{23}$ electrons or protons = 6.022 × $10^{24}$ 34 mg (34 × $10^{-3}$ g) of $\mathrm{NH}_3$ has protons = $\frac{6.022 \times 10^{24}}{17} \times 34 \times 10^{-3}$ = 1.2044 × $10^{22}$ protons (b) Mass of a proton = 1.675 × $10^{-27}$ kg Total mass of protons in 34 × $10^{-3}$ g = 1.2044 × $10^{22}$ × 1.675 × $10^{-27}$ kg = 2.017 × $10^{-5}$ kg The answer will not change if the temperature and pressure are changed.