Question

At a particular temperature, the vapour pressures of 'A' and 'B' are $250mmHg$ and $180mmHg$ respectively. If the mole fraction of 'B' in the solution mixture is 0.42, the total vapour pressure of the solution mixture is $200mmHg$.
Choose the correct statement.

Answer 1

Given,
$\text{Pure vapour pressure of A,}{p}_A^{\circ }=250mmHg\text{Pure vapour pressure of B,}{p}_A^{\circ }=180mmHg$

$\text{Mole fraction of B,}{X}_B=0.42\therefore \text{Mole fraction of A,}{X}_A=1-0.42=0.58$

By Raoult's law
Total vapour pressure of solution,
$P_T=X_A.p_A^{\circ }+X_B.p_B^{\circ }$
$P_T=250\times 0.58+180\times 0.42$
$=220.6mmHg$
For the solution to be an ideal one, the vapour pressure should be $220.6mm$ as calculated from Raoult's law but the given value of vapour pressure is $160mmHg$, so the solution is not ideal. It violates the Raoult's law. Hence, it is an non-Ideal solution.
In given solution mixture,
$P_T<X_A.p_A^{\circ }+X_B.p_B^{\circ }$
$200mmHg<220.6mmHg$
Thus, the given mixture shows negative deviation from the Raoult's law.
A-B interaction is more than the A-A and the B-B interaction.