Vapour pressure of pure benzene at any temperature is $640 m m$ $H g$ . A non volatile, non-electrolyte solid whose weight is $2.75 g m$ , is added to $39.0 g m$ of benzene. Vapour pressure of solution is $600 m m$ $H g$ . Calculate the molecular weight of the solid?

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Solution

Let M g/mol be the molecular weight of the solid. The number of moles of solid
$= \frac{2.75 g}{M g / m o l} = \frac{2.75}{M} m o l$
The molecular weight of benzene is $78$ g/mol. The number of moles of benzene
$= \frac{39.0 g}{78 g / m o l} = 0.5 m o l$ .
The mole fraction of the solid
$= \frac{\frac{2.75}{M}}{\frac{2.75}{M} + 0.5} ≃ \frac{\frac{2.75}{M}}{0.5} = \frac{5.5}{M}$
The relative lowering of vapour pressure of solution is equal to the mole
fraction of solute.
$\frac{P^{0} - P}{P^{0}} = X$
$\frac{640 - 600}{640} = \frac{5.5}{M}$
$\frac{40}{640} = \frac{5.5}{M}$
$M = 640 \times \frac{5.5}{40}$
$M = 88$ g/mol.
The molecular weight of the solid is $88$ g/mol.

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