Question

Total vapour pressure of mixture of $1mole$ of volatile component $A(P_A^O=100mmHg)$ and $3mole$ of volatile component $B(P_B^O=80mmHg)$ is $90mmHg$. 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 weaker than that between $A$ and $A$ or between $B$ and $B$
• D. All the above statement is correct.

Answer 1

The correct option is A there is positive deviation from Raoult's law

$P_{ideal}=P_A^{\circ }{x}_A+P_B^{\circ }{x}_B$;$=100\times \frac{1}{4}+80\times \frac{3}{4}\Rightarrow 85$ mm Hg$P_{actual}=90$ mm Hg;

$\because$ Actual v. pr. is greater than ideal solution v.pr. so +ve deviation from Raoult's law