Question

Total vapour pressure of mixture of

1 mol volatile component $A\left(\right.P^\circ A=100 \, mmHg\left.\right)$ and $3 \, mol$ of volatile component $B\left(\right.P^\circ B=60 \, mmHg\left.\right)$ is $75 \, mm$ . For such case-

• A. There is positive deviation from Raoult’s law
• B. Boiling point has been lowered
• C. Force of attraction between $A \, and \, B$ is smaller than that between $A \, and \, A$ or between $B \, and \, B.$
• D. All the above statements are correct

Answer 1

Answer: (D)

$P_{T}^{o}=P_{A}^{o}X_{A}+P_{B}^{o}X_{B}$

$P^{o}_{T}=\frac{1}{4}\times 100+\frac{3}{4}\times 60=\frac{280}{4}=70$

(According to Raoult’s law)