Question

An aqueous solution of liquid '$X$' (mol. weight $56$) $28 \%$ by weight has a vapour pressure $148 \,mm$. Then the vapour pressure of ' $X$ ' if vapour pressure of water is $155 \,mm$ of $Hg$ is _______.

Answer 1

According to Raoult's law for liquid mixtures,

$P_{M}=P_{A}+P_{B}$

$\therefore P_{M}=P_{A}^{\circ} \times\left\{\frac{\frac{w_{A}}{M_{A}}}{\frac{w_{A}}{M_{A}}+\frac{w_{B}}{M_{B}}}\right\}+P_{B}^{\circ} \times\left\{\frac{\frac{w_{B}}{M_{B}}}{\frac{w_{A}}{M_{A}}+\frac{w_{B}}{M_{B}}}\right\}$

Given that, $w_{A}=28 \,g, w_{H_{2} o}=72\, g, P_{A}^{\circ}=?$

$P _{ H _{2} O }^{\circ}=155$

$M_{A}=56 \,g$

$M_{ H , o }=18\, g$, and $P _{ M }=200 \,mm$

$\therefore 148= P _{ A }^{\circ} \times\left\{\frac{\frac{28}{56}}{\frac{28}{56}+\frac{72}{18}}\right\}+155 \times\left\{\frac{\frac{72}{18}}{\frac{28}{56}+\frac{72}{18}}\right\}$

$148= P _{ A }^{\circ} \times \frac{1}{2} \times \frac{2}{9}+155 \times 4 \times \frac{2}{9}$

$\therefore P_{A}^{\circ}=92\, mm$