Question

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

Answer 1

Answer: 92

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, $w_A=28g,w_{H_{2}o}=72g,P_A^{\circ }=?$

$P_{H_{2}O}^{\circ }=155$

$M_A=56g$

$M_{H,o}=18g$, and $P_M=200mm$

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

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

$\therefore P_A^{\circ }=92mm$