Question

Question 2.2 (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?

Class XI Structure of Atom Page 65

Answer 1

(i) 1 mole of methane (CH4) = 1 mole of C + 4 × 1 mole of H
we know,
1 mole of C = 6 mole of electrons
1 mole of H = 1 mole of electron

so, 1 mole of CH4 = 6 + 4 = 10 mole of electrons
but 1 mole = 6.022 × 10²³
so, 1 mole of CH4 = 10 × 6.022 × 10²³ electrons
= 6.022 × 10²⁴ electrons


(ii) given weight of ¹⁴C = 7mg
atomic weight of ¹⁴C = 14 g/mol
so, no of mole = given weight/atomic weight
= 7mg/14g/mol = 0.5 × 10^-3 mol
we know,
number of neutron in one ¹⁴C = 14 - 6 = 8
number of mole of neutrons = no of mole of ¹⁴C × 8 = 0.5 × 10^-3 × 8 = 4 × 10^-3
total number of neutrons = no of mole of neutrons × Avogadro's constant
= 4 × 10^-3 × 6.022 × 10²³
= 24.088 × 10²¹

mass of 24.088 × 10²¹ neutrons = 24.088 × 10²¹ × 1.675 × 10^-27 Kg
= 4.0347 × 10^-6 Kg


(iii) given weight = 34 mg
molecular mass of NH3 = 17 g/mol
no of mole = 34mg/14g/mol = 2 × 10^-3 mol
we know,
1 mole of NH3 = (7 + 3) moles of protons
= 10 moles of protons
so, number of moles of NH3 in 34mg = 10 × 2 × 10^-3 = 2 × 10^-2

number of protons = no of mole of protons × Avogadro's constant
=2 × 10^-2 × 6.022 × 10²³
= 1.2044 × 10²² protons


mass of protons = no of protons × mass of 1 proton
= 1.2044 × 10²² × 1.6726 × 10^-27 Kg
= 2.01447 × 10^-5 Kg

Answer 2

Answer

1) 1 molecule of CH4 contains electron = 10

Therefore 1 mole (6.022 x 1023 atoms) contains electron = 6.022 x 1024

2) a) 1 g atom of 14C = 14g = 6.022 x 1023 atoms = 6.022 x 1024 x 8 neutrons

Thus 14 g or 14000 mg have 6.022 x 1024 x 8 neutrons

Therefore 7mg will have neutrons = 6.022 x 1024 x 8 /14000 x 7 = 2.4088 x 1022

b) mass of 1 neutron = 1.675 x 10-27kg

Therefore mass of 2.4088 x 1021 neutrons = 2.4088 x 1021 x 1.67x 10-27 = 4.0347 x 10-6 kg

3) a) 1 mol of NH3 = 17g NH3 = 6.022 x 1023 molecules of NH3 = (6.022x1023)(7 + 3) proton = 6.022x1024 protons

Therefore 34 mg i.e 0.034 g NH3 =6.022x1024 x 0.034/1 = 1.2044x1022protons

b) mass of 1 proton = 1.6726x10-27 kg

Therefore mass of 1.2044 x 1022 protons = (1.6726x 10-27)(1.2044x1022) kg = 2.0145x 10-5 kg

No,

The answer change if the temperature and pressure are changed...

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