A mixture contains $N_{2} O_{4}$ and $N O_{2}$ in the ratio $2 : 1$ by volume. The vapour denstiy of the mixture is

45.4

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49.8

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32.6

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38.3

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Solution

The correct option is A 38.3
Vapour density $N_{2} O_{4} = \frac{m o l a r - m a s s - N_{2} O_{4}}{m o l a r - m a s s - H_{2}} = \frac{92.011}{2.016} = 45.64$

Vapour density $N O_{2} = \frac{m o l a r - m a s s - N O_{2}}{m o l a r - m a s s - H_{2}} = \frac{46.0055}{2.016} = 22.82$

The ratio is 2:1 in favour $N_{2} O_{4}$ , Therefore

v.d. of $N_{2} O_{4} = 45.64 \times 2 = 91.28$

v.d. of $N O_{2} = 22.82 \times 1 = 22.82$

sum is 114.1

Average is $\frac{114.1}{3} = 38.33$

Hence , option D is correct .

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The vapour density of a mixture containing NO2 and N2O4 is 38.3 at 27'C. Calculate moles of NO2 in 100 g of mixture.

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