Question

At ${30}^{o}C$,pure benzene (molecular weight 78.1) has a vapour pressure of 121.8 mm. Dissolving 15.0 g of a non-volatile solute in 250 g of benzene produced a solution having vapour pressure of 120.2 mm.Determine the approximate molecular weight of the solute.

• A. $152.3$
• B. $355.8$
• C. $392.4$
• D. $409.2$

Answer 1

Answer: (B)

$355.8$

This is the case of relative lowering of vapour pressure when solute is added to pure solution.

$\frac{\Delta P}{P^o}=\frac{121.8-120.2}{121.8}=\frac{1.6}{121.8}=0.013$

$\frac{\Delta P}{P^o}=\frac{n_2}{n_2+n_1}=\frac{15/M{w}_2}{\frac{15}{M{w}_2}+\frac{250}{78.1}}$$=\frac{15}{M{w}_2(\frac{15}{M{w}_2}+3.2)}$

$0.013=\frac{15}{15+3.2M{w}_2}$

$0.013\times 15+3.2\times 0.013M{w}_2=15$

$0.195\times 0.0146M{w}_2=15$

$0.0416M{w}_2=15-0.195=14.805$

$\therefore M{w}_2=14.805/0.0416=355.8$

Hence, option B is correct.