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0.157g of a certain gas collected over water occupies a volume of 135ml at $27^{0} C$ and 750mm of $H g$ . Assuming ideal behaviour, the molecular weight of the gas is (aqueous tension at $27^{0} C$ is 26.7 mm of $H g$ )

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Detailed Solution

$P_{g a s} = P_{m o i s t . g a s} - a q u e o u s t e n s i o n$

$= 750 - 26.7$

$= 723.3 m m$

$= \frac{723.3}{760} a t m$

$M = \frac{W R T}{P V}$

$= \frac{0.157 \times 0.0821 \times 300}{\frac{723.3}{760} \times \frac{135}{1000}}$

$= 30$

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