Question

The vapour pressure of pure water at ${30}^{\circ }C$ is $31.50\text{mm Hg}$. When 3.0 g of a non-volatile solute was dissolved in 54 g of water, the vapour pressure of the solution was found to be $31.30mmHg$. Calculate the molar mass of the solute.

Answer 1

Given : $P_A^o=31.50mmHg,P_A=31.30mmHg,$
$W_B=3.0g,W_A=54g,M_B=?$
$M_A=2\times 1+16=18g/mol$
$\frac{P_A^o-P_A}{P_A^o}=X_B=\frac{\frac{W_B}{M_B}}{\frac{W_A}{M_A}+\frac{W_B}{M_B}}$
$\Rightarrow \frac{31.50-31.30}{31.50}=\frac{3/M_B}{\frac{54}{18}+\frac{3}{M_B}}$
$\Rightarrow \frac{0.20}{31.50}=\frac{\left(3/M_B\right)\times \left(18M_B\right)}{54M_B+18\times 3}$
$\frac{54}{54M_B+54}=\frac{1}{M_B+1}=\frac{0.20}{31.50}$
$0.20(M_B+1)=31.50\times 1$
$0.20M_B+0.20=31.50$
$0.20M_B=31.50-0.20=31.30$
$M_B=\frac{31.30}{0.20}=156.5\text{g/mol}$