Question

The vapour pressure of pure liquids 1 and 2 are $450mmHg$ and $700mmHg$ respectively at $350K$.
When total vapour pressureof the solution is $600mmHg$, the mole fraction of component 1 in liquid phase will be:

• A. 0.4
• B. 0.6
• C. 0.3
• D. 0.7

Answer 1

The correct option is A 0.4

By Raoult's Law :
$P_T=p_1^{\circ }{\chi }_1+p_2^{\circ }{\chi }_2$
where
$P_T$ is the total pressure of the solution.
$p_1^{\circ }$ is the vapour pressure of pure component 1
$p_2^{\circ }$ is the vapour pressure of pure component 2
${\chi }_1$ is the mole fraction of component 1 in liquid phase.
${\chi }_2$ is the mole fraction of component 2 in liquid phase.

Putting the values:
$600=450{\chi }_1+700{\chi }_{2}600=450{\chi }_1+700(1-{\chi }_1)600=700-250{\chi }_1{\chi }_1=0.4$
${\chi }_2=1-0.4=0.6$
Hence option (a) is correct