First slide

Gaseous state

Question

A mixture of Ar (40) and N₂ gases have a density of 1.40 g L^-1 at STP. Hence, weight fraction of N₂ approximately in the mixture is

Moderate

A

0.3
B

0.7
C

0.4
D

0.5

Solution

$p = \frac{n}{V} RT = \frac{W}{mV} RT = \frac{dRT}{m} ∴ m (molar mass of mixture) = \frac{dRT}{p} = \frac{1 .40 \times 0 .0821 \times 273}{1} = 31 .38 g {mol}^{- 1} Let mixture contains x part N_{2} and (1 - x) part Ar. ∴ 31 .38 = \frac{m_{1} x + m_{2} (1 - x)}{x + (1 - x)} \Rightarrow 31 .38 = \frac{28 + 40 (1 - x)}{1} ∴ x = 0 .72$

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