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Q. The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non-volatile and non-electrolyte solid weighing 2.175 g is added to 39.08 of benzene. If the vapour pressure of the solution is 600 mm of Hg, what is the molecular weight of solid substance ?

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Solution:

$\frac{P^{\circ}-P_{s}}{P^{\circ}}=\frac{\frac{w}{m}}{\frac{w}{m}+\frac{W}{M}} $
$\because \frac{W}{M}>\frac{w}{m} $
$\Rightarrow \frac{640-600}{640}$
$=\frac{w}{m} \times \frac{M}{W}$
$ \Rightarrow \frac{40}{640}=\frac{2.175 \times 78}{m \times 39.08}, $
$m=\frac{2.175 \times 78}{39.08} \times \frac{640}{40}$
$m =69.45 .$