Skip to content
Exercises · 2.2

Q.(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^{14}C. (Assume that mass of a neutron =1.675×10−27 kg= 1.675 \times 10^{-27}\ kg).
(iii) Find
(a) the total number and
(b) the total mass of protons in 34 mg of NH3NH_3 at STP. Will the answer change if the temperature and pressure are changed?
Tripura TbseTextbookSubjective· 3mImportance★★★★★est
14% · 20/140 Questions
✓ Free question

  1. 6.022×10246.022\times10^{24} electrons;
  2. 2.409×10212.409\times10^{21} neutrons of mass 4.035×10−6 kg4.035\times10^{-6}\ \text{kg};
  3. 1.204×10221.204\times10^{22} protons of mass 2.015×10−5 kg2.015\times10^{-5}\ \text{kg} — independent of TT and PP.

(i) Electrons in one mole of methane (CH4\text{CH}_4).

One CH4\text{CH}_4 molecule has 6 (C)+4×1 (H)=106\ (\text{C}) + 4\times1\ (\text{H}) = 10 electrons. In one mole there are NA=6.022×1023N_A = 6.022\times10^{23} molecules, so

Ne=10×6.022×1023=6.022×1024 electrons.N_{e} = 10 \times 6.022\times10^{23} = 6.022\times10^{24}\ \text{electrons}.

(ii) 7 mg of 14C^{14}\text{C}.

Each 14C^{14}\text{C} atom has 14−6=814 - 6 = 8 neutrons. Moles of 14C^{14}\text{C}:

n=7×10−3 g14 g mol−1=5×10−4 mol,atoms=5×10−4×6.022×1023=3.011×1020.n = \frac{7\times10^{-3}\ \text{g}}{14\ \text{g mol}^{-1}} = 5\times10^{-4}\ \text{mol},\quad \text{atoms} = 5\times10^{-4}\times6.022\times10^{23} = 3.011\times10^{20}.

  1. Total neutrons:

    Nn=8×3.011×1020=2.409×1021 neutrons.N_n = 8 \times 3.011\times10^{20} = 2.409\times10^{21}\ \text{neutrons}.

  2. Total mass of neutrons (mass of one neutron =1.675×10−27 kg= 1.675\times10^{-27}\ \text{kg}):

    m=2.409×1021×1.675×10−27=4.035×10−6 kg.m = 2.409\times10^{21} \times 1.675\times10^{-27} = 4.035\times10^{-6}\ \text{kg}.

    (iii) 34 mg of NH3\text{NH}_3. Each NH3\text{NH}_3 molecule has 7 (N)+3×1 (H)=107\ (\text{N}) + 3\times1\ (\text{H}) = 10 protons. Moles of NH3\text{NH}_3:

    n=34×10−3 g17 g mol−1=2×10−3 mol,molecules=2×10−3×6.022×1023=1.204×1021.n = \frac{34\times10^{-3}\ \text{g}}{17\ \text{g mol}^{-1}} = 2\times10^{-3}\ \text{mol},\quad \text{molecules} = 2\times10^{-3}\times6.022\times10^{23} = 1.204\times10^{21}.

(a) Total protons:

Np=10×1.204×1021=1.204×1022 protons.N_p = 10 \times 1.204\times10^{21} = 1.204\times10^{22}\ \text{protons}.

(b) Total mass of protons (mass of one proton =1.6726×10−27 kg= 1.6726\times10^{-27}\ \text{kg}):

m=1.204×1022×1.6726×10−27=2.015×10−5 kg.m = 1.204\times10^{22} \times 1.6726\times10^{-27} = 2.015\times10^{-5}\ \text{kg}.

Effect of temperature and pressure: the number of protons (and their mass) is fixed by the given mass of NH3\text{NH}_3 alone. It does not change with temperature or pressure — the "at STP" condition is irrelevant here.

✓Final answer

(i) 6.022×10246.022\times10^{24} electrons. (ii) (a) 2.409×10212.409\times10^{21} neutrons, (b) 4.035×10−6 kg4.035\times10^{-6}\ \text{kg}. (iii) (a) 1.204×10221.204\times10^{22} protons, (b) 2.015×10−5 kg2.015\times10^{-5}\ \text{kg}; the answer does not change with temperature or pressure.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.