Total vapour pressure of mixture of 1 mol volatile component A(.P° A=100 mmHg.) and 3 mol of volatile component B(.P° B=60 mmHg.) is 75 mm . For such case-

Question

Total vapour pressure of mixture of 1 mol volatile component A(.P° A=100 mmHg.) and 3 mol of volatile component B(.P° B=60 mmHg.) 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

Tardigrade Admin · Accepted Answer

PTo=PAoXA+PBoXB PoT=(1/4)× 100+(3/4)× 60=(280/4)=70 (According to Raoultâs law)

Q. Total vapour pressure of mixture of

1 mol volatile component $A (P^{\circ} A = 100 mm H g)$ and $3 m o l$ of volatile component $B (P^{\circ} B = 60 mm H g)$ is $75 mm$ . For such case-

There is positive deviation from Raoult’s law

Boiling point has been lowered

Force of attraction between $A an d B$ is smaller than that between $A an d A$ or between $B an d B .$

All the above statements are correct

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

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

(According to Raoult’s law)

Q. Total vapour pressure of mixture of 1 mol volatile component A(P∘A=100mmHg) and 3mol of volatile component B(P∘B=60mmHg) is 75mm . For such case-