The density of the vapour of a substance at $1$ atm pressure and $500$ K is $0.36$ kg $m^{- 3} .$ The vapour effuses through a small hole at a rate of $1.33$ times faster than oxygen under the same condition.If its molar volume in $m^{3} / m o l e$ is $x$ , then value of $100 x$ is:

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Solution

Density of the vapour in given conditions $0.36 k g m^{- 3} = 360 g m^{- 3}$

Now, by Graham's law of diffusion,

$\frac{r_{vap.}}{r_{oxygen}} = \sqrt{\frac{M_{oxygen}}{M_{vap.}}}$

$A l s o, 1.33 \approx \frac{4}{3}$

Hence, $\frac{r_{vap.}}{r_{oxygen}} = \frac{4}{3}$

Solving the above equation.

$M_{vap.} = 18 g m o l^{- 1}$

Hence, molar volume $= \frac{M o l . m a s s}{d e n s i t y}$

$\Rightarrow M o l a r v o l u m e = \frac{18}{360} = 0.05 m^{3} m o l^{- 1}$
$∴ x = 0.005$

$\Rightarrow 100 x = 5$

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