JEE Main 2025 Apr 7 Shift 1, Chemistry Q9: Colligative Properties

JEE Main2025Apr 7, Shift 1Chemistry

At the sea level, the dry air mass percentage composition is given as nitrogen gas : 70.0, oxygen gas : 27.0 and argon gas : 3.0. If total pressure is 1.15 atm, then calculate the ratio of followings respectively :

(i) partial pressure of nitrogen gas to partial pressure of oxygen gas

(ii) partial pressure of oxygen gas to partial pressure of argon gas

(Given : Molar mass of N, O and Ar are 14, 16, and 40 g mol $^{- 1}$ respectively)

A
4.26, 19.3
B
2.59, 11.85
C
5.46, 17.8
D
2.96, 11.2

Solution

By Dalton's law, the ratio of partial pressures equals the ratio of mole fractions, which equals the ratio of moles. Take a 100 g sample of dry air:

$n_{N_{2}} = \frac{70}{28}, n_{O_{2}} = \frac{27}{32}, n_{Ar} = \frac{3}{40}$

(Molar masses: N $_{2}$ = 28, O $_{2}$ = 32, Ar = 40.)

Ratio (i):

$\frac{P _{N_{2}}}{P _{O_{2}}} = \frac{n _{N_{2}}}{n _{O_{2}}} = \frac{70/28}{27/32} = \frac{2.5}{0.84375} \approx 2.96$

Ratio (ii):

$\frac{P _{O_{2}}}{P _{Ar}} = \frac{n _{O_{2}}}{n _{Ar}} = \frac{27/32}{3/40} = \frac{0.84375}{0.075} \approx 11.25$

$∴$ The ratios are approximately 2.96 and 11.2 $\Rightarrow$ Option (4). The total pressure (1.15 atm) is a distractor.

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JEE Main 2025 Apr 7 Shift 1, Chemistry Q9: Colligative Properties

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