JEE Main & Advanced Chemistry Solutions / विलयन Sample Paper Topic Test - Solutions

The total vapor pressure of a mixture of 1 mol of volatile component A (\[P_{A}^{O}\]= 100 mm Hg) and 3 mol of volatile component B (\[P_{B}^{O}\]=80 mm Hg) is 90 mm Hg. For such case:

A) there is a 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

Correct Answer: D

Solution :

[d] \[{{p}_{\text{ideal}}}=p_{A}^{o}{{x}_{A}}+p_{B}^{o}{{x}_{B}}\]

\[=100\times \frac{1}{4}+80\times \frac{3}{4}=85\,mm\,Hg\]

\[{{p}_{actual}}=90mm\,Hg\]

Actual vapor pressure is greater than the vapor pressure of ideal solution. Hence, a positive deviation from Raoult's Jaw.