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Elevation in Boiling Point

The vapour de...

Question

The vapour density (hydrogen $= 1$ ) of a mixture containing ${N O}_{2}$ and $N_{2} O_{4}$ is $38.3$ at ${26.7}^{o} C$ . Calculate the number of moles of ${N O}_{2}$ in $100 g r a m s$ of the mixture.

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Solution

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
At equi. $(1 - x)$ $2 x$
$x$ (degree of dissociation) $= \frac{D - d}{(n - 1) d}$
Given, $d = 38.3; D = \frac{M o l . m a s s o f N_{2} O_{4}}{2} = \frac{92}{2} = 46; n = 2$
So, $x = \frac{46 - 38.3}{38.3} = 0.2$
At equilibrium, amount of $N_{2} O_{4} = 1 - 0.2 = 0.8 m o l$
and amount of ${N O}_{2} = 2 \times 0.2 = 0.4 m o l$
Mass of the mixture $= 0.8 \times 92 + 0.4 \times 46$
$= 92.0 g$
Since, $92$ gram of themixture contains $= 0.4 m o l$ ${N O}_{2}$
$100$ gram of the mixture contains $== \frac{0.4 \times 100}{92} = 0.43$ mol ${N O}_{2}$

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