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Charles' Law

The vapour de...

Question

The vapour density of a mixture containing $N O_{2}$ and $N_{2} O_{4}$ is 27.6. The moles of $N_{2} O_{4}$ in 10 moles of the mixture is:

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Solution

We know,
$m o l a r m a s s = v a p o r d e n s i t y \times 2$
molar mass = $27.6 \times 2 = 55.2 g / m o l$
Molar mass of the mixture = $mole fraction of NO2×molar mass of NO2+mole fraction of N2O4×molar mass of N2O4$
Let 1 mole of mixture has $x$ mole $N_{2} O_{4}$
$2 \times 27.6 = x (92) + (1 - x) 46$
$\Rightarrow 55.2 = 92 x + 46 - 46 x$
$\Rightarrow x = 0.2$

Hence 0.2 mole of $N_{2} O_{4}$ is present in 1 mole of the mixture. Hence 2 mole of $N_{2} O_{4}$ will be present in 10 moles of the mixture.

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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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