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Liquids in Liquids

The vapour pr...

Question

The vapour pressure of pure benzene at a certain temperature is $0.850$ bar. A non-volatile, non-electrolyte solid weighing $0.5$ $g$ when added to $39$ $g$ of benzene (molar mass $78$ $g$ ${m o l}^{- 1}$ ), vapour pressure becomes $0.845$ bar. What is the molar mass of the solid substance?

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Solution

$\frac{p^{0} - p}{p^{0}} = \frac{W_{2} M_{1}}{W_{1} M_{2}}$
$p^{0} = 0.850 b a r =$ vapour pressure of pure benzene.
$p = 0.845 b a r =$ vapour pressure of solution.
$W_{1} = 39 g =$ mass of benzene
$W_{2} = 78 g =$ mass of solute
$M_{1} = 78 g / m o l =$ Molar mass of benzene.
$M_{2} =$ Molar mass of solute.
$\frac{0.850 b a r - 0.845 b a r}{0.850 b a r} = \frac{0.5 g \times 78 g / m o l}{39 g \times M_{2}}$
$M_{2} = 170 g / m o l$

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