(i) Total number of electrons present in one mole of CH4
= 6 × 6.022 × 1023 + 4 × 6.022 × 1023 = 36.132 × 1023 + 24.088 × 1023
= 6.022 × 1024 electrons
(ii) We know that, p + n = A
6 + n = 14 [Since e = p = Z]
n = 14 – 6 = 8
Now, 1 mole of

14

‌

C = 14 g of C = 6.022 × 1023 atoms of

14

‌

C
= 6.022 × 1023 × 8 neutrons = 4.8176 × 1024 neutrons
(a) 14 g of

14

‌

C contains = 4.8176 × 1024 neutrons
7 mg (7 × 10–3 g) of

14

‌

C contains =

4.8176×7×10−3

14

neutrons
= 2.4088 × 1021 neutrons
(b) Mass of a neutron = 1.675 × 10–27 kg
Mass of 2.4088 × 1021 neutrons = 1.675 × 10–27 × 2.4088 × 1021
= 4.0347 × 10–6 kg
(iii) (a) 17 g of NH3 has 10 × 6.022 × 1023 electrons or protons = 6.022 × 1024
34 mg (34 × 10–3 g) of NH3 has protons =

6.022×1024

17

×34×10−3
= 1.2044 × 1022 protons
(b) Mass of a proton = 1.675 × 10–27 kg
Total mass of protons in 34 × 10–3 g = 1.2044 × 1022 × 1.675 × 10–27 kg
= 2.017 × 10–5 kg
The answer will not change if the temperature and pressure are changed.