Question

The vapour pressure of a solution in benzene is 722 mm and that of pure solvent (benzene) is 760 mm. The molality of solution is :

• A. 0.128
• B. 0.641
• C. 1.28
• D. 6.67

Answer 1

The correct option is B 0.641

$\begin{array}{c}\text{solvent}=H_{2}O\\ \text{Benzene-}{C}_6{H}_6=78\end{array}$

$\begin{array}{c}\frac{p^0-p_s}{p^0}=\frac{n}{N}\\ \frac{760-722}{760}=\frac{n}{N}\end{array}$

$\frac{n}{N}=\frac{38}{760}=\frac{1}{20}\frac{W_b}{W_w}\times \frac{18}{78}=\frac{1}{20}$

$\text{molality of}{sol}^n=\frac{\text{mole of solute}}{\text{weight of solvent in kg}}$

$\begin{array}{c}=\frac{\text{mole of solute}}{\text{mole of solvent}}\times \frac{1}{Mw\text{. of solvent}}\\ =\frac{1}{20}\times \frac{1}{18}\times \text{1000}\\ =\frac{1}{360}\times 1000\end{array}$

$\begin{array}{cc}=\frac{W_b}{78\times W_b}& \times \frac{78}{18\times 20}=\frac{1}{360}\times \frac{1000}{10}\\ & =0.267\end{array}$

$\begin{array}{cc}\text{molality of}{sol}^n& =1-0.267\\ & =0.633\end{array}$