A mixture of C3H8 and CH4 exerts a pressure of 320 mm Hg at temperature T K in a V litre flask. On complete combustion, gaseous mixture contains CO2 only and exerts a pressure of 448 mm Hg under identical conditions. Hence mole fraction of C3H8 in the mixture is

Question

A mixture of C3H8 and CH4 exerts a pressure of 320 mm Hg at temperature T K in a V litre flask. On complete combustion, gaseous mixture contains CO2 only and exerts a pressure of 448 mm Hg under identical conditions. Hence mole fraction of C3H8 in the mixture is (A) 0.2 (B) 0.8 (C) 0.25 (D) 0.75

Tardigrade Admin · Accepted Answer

Let V L at T K and 320 mmHg represents 1 mol Then V L at T K and 448 mmHg represents
=

\frac{448}{320}

mol = 1.4 mol

x mol C_{3} H_{8} (g) + S O_{2} (g) \to 3x mol 3 C O_{2} (g) + 4 H_{2} O (l)

(1-x) mol C H_{4} (g) + 2 O_{2} (g) \to (1-x) mol 3 C O_{2} (g) + 4 H_{2} O (l)

Let moles of

C_{3} H_{8}

= x

∴

Moles of

C H_{4}

= 1- x Moles of

C O_{2}

produced = 3x + 1 - x = 1 + 2x
1 + 2x = 1.4
2x = 0.4
x = 0.2
Mole fraction of

C_{3} H_{8} = \frac{x}{1}

= 0.2

Q. A mixture of C3​H8​ and CH4​ exerts a pressure of 320 mm Hg at temperature T K in a V litre flask. On complete combustion, gaseous mixture contains CO2​ only and exerts a pressure of 448 mm Hg under identical conditions. Hence mole fraction of C3​H8​ in the mixture is

0.2

0.8

0.25

0.75

Q. A mixture of $C_{3} H_{8}$ and $C H_{4}$ exerts a pressure of 320 mm Hg at temperature T K in a V litre flask. On complete combustion, gaseous mixture contains $C O_{2}$ only and exerts a pressure of 448 mm Hg under identical conditions. Hence mole fraction of $C_{3} H_{8}$ in the mixture is