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Pascal's Law

The vapour de...

Question

The vapour density of a mixture consisting of $N O_{2}$ and $N_{2} O_{4}$ is $38.3$ at ${26.7}^{o} C$ . Calculate the number of moles of $N O_{2}$ in $100 g$ of the mixture.

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Solution

The vapour density of a gas is the density of that gas relative to that of hydrogen gas,
Thus,
Vapour density= $\frac{(d e n s i t y o f \frac{N O_{2}}{N_{2} O_{4}} m i x t u r e)}{(d e n s i t y o f H_{2} g a s)}$

Vapour density = $\frac{(m a s s o f \frac{N O_{2}}{N_{2} O_{4}} m i x t u r e)}{(m a s s o f H_{2} g a s)}$

Vapourdensity= $\frac{(m o l a r m a s s \frac{N O_{2}}{N_{2} O_{4}} m i x t u r e)}{(m o l a r m a s s o f H_{2} g a s)}$

Vapour density = $\frac{(m o l a r m a s s \frac{N O_{2}}{N_{2} O_{4}} m i x t u r e)}{2}$

Molar mass $\frac{N O_{2}}{N_{2} O_{4}}$ mixture= $V a p o u r d e n s i t y \times 2$

$= 38.3 \times 2$

$= 76.6 g {m o l}^{- 1}$

Let the percentage of $N O_{2}$ gas be x, the percentage of $N_{2} O_{4}$ gas be 1-x

$x (m o l a r m a s s o f N O_{2} g a s) + (1 - x) (m o l a r m a s s o f N_{2} O_{4}) = 76.6$

$x (46) + (1 - x) (92) = 76.6$
$x = 33.48$

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