Question

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\,g $ of benzene. If the vapour pressure of the solution is $ 600\,mm $ of $ Hg $ , what is the molecular weight of solid substance ?

• A. $ 49.50 $
• B. $ 59.60 $
• C. $ 69.60 $
• D. $ 79.82 $

Answer 1

Answer: (C)

Given, vapour pressure of benzene,

$p^{\circ}=640 mm Hg$

Vapour pressure of solution,

$p =600 mm Hg$

Weight of solute, $w=2.175 g$

Weight of benzene, $W=39.08 g$

Molecular weight of benzene,

$M=78 g$

Molecular weight of solute, $m=$?

According to Raoult's law,

$\frac{p^{\circ}-p}{p^{\circ}} =\frac{w \times M}{m \times W}$

$\frac{640-600}{640} =\frac{2.175 \times 78}{m \times 39.08}$

$\frac{40}{640} =\frac{2.175 \times 78}{m \times 39.08}$

$m =\frac{16 \times 2.175 \times 78}{39.08}$

$m =69.60$