Question

The vapour pressure of water at $293K$ is $0.0231$ bar and the vapour pressure of a solution of $108.24g$ of a compound in $1000g$ of water at the same temperature is $0.0228bar$. Calculate the molar mass of the solute.

Answer 1

vapour pressure of pure water

$P_{\omega }^o=0.0231$ bar

Let molar mass of compound M

Number of mole = $\frac{108.24}{M}$

Mole fraction = $\frac{\frac{108.24}{M}}{\frac{108.24}{M}+\frac{1000}{18}}$

$x=\frac{108.24\times 18}{108.24\times 18+1000M}$

$P_{\omega }^{\circ }(1-x)=0.0228$

$P_{\omega }^{\circ }x=0.0231-0.0228$

$x=\frac{0.0003}{0.0231}$

$\frac{1948.32}{1948.32+1000M}=\frac{3}{231}$

$450061.92=5844.96+3000M$

$M=148.07$