Question

A mixture of $C_3H_8$ and $CH_4$ exerts a pressure of 320 mm Hg at temperature T K in a V litre flask. On complete combustion, gaseous mixture contains $CO_2$ only and exerts a pressure of 448 mm Hg under identical conditions. Hence mole fraction of $C_3H_8$ in the mixture is

• A. 0.2
• B. 0.8
• C. 0.25
• D. 0.75

Answer 1

Answer: (A)

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
$\underset{\text{x mol}}{{C_3H_8(g) }}+ SO_2(g) \rightarrow \underset{\text{3x mol}}{{3CO_2(g) }} + 4H_2O(l)$
$\underset{\text{(1-x) mol}}{{CH_4(g) }}+ 2O_2(g) \rightarrow \underset{\text{(1-x) mol}}{{3CO_2(g) }} + 4H_2O(l)$
Let moles of $C_3H_8$ = x
$\therefore $ Moles of $CH_4$ = 1- x Moles of $CO_2$ produced = 3x + 1 - x = 1 + 2x
1 + 2x = 1.4
2x = 0.4
x = 0.2
Mole fraction of $C_3H_8 = \frac{x}{1}$ = 0.2