Question

A mixture of an organic liquid $A$ and water distilled under one atmospheric pressure at $99.2^{o}C$. How many grams of steam will be condensed to obtained $1.0g$ of liquid $A$ in the distillate? (Vapour pressure of water at ${99.2}^{o}C$ is $739mmHg$. Molecular weight of $A=123$)

Answer 1

Total pressure of the solution(A+Water)$=1$ atm$=760$mm Hg

Vapour pressure of water$=739$mm Hg

Vapour pressure of organic liquid A is$=760-739=21$mm Hg

Moleucular mass of A$=123$

$\frac{W_A}{W_{water}}=\frac{P_A\times M{M}_A}{P_B\times M{M}_{w}ater}$

Where $W_A$ &$W_{water}$ represents the mass of organic liquid A and mass of water respectively.

$M{M}_A$ &$M{M}_{w}ater$ represents the molar mass of A and molar mass of water respectively

$\frac{1}{W_{water}}=\frac{21\times 123}{739\times 18}=5.19$ g