Total vapour pressure of mixture of 1 mol of volatile Component A $\left( {P_A^{\rm{o}} = 100\,\,mm\,Hg} \right)$ and 3 mol of volatile Component B $\left( {P_B^{\rm{o}} = 80\,\,mm\,Hg} \right)$ is 90 mm Hg, for such case:

There is positive deviation from Raoult’s law

Force of attraction between A and B is smaller than that between A and A or between B and B

NEET UG CHEMISTRY MCQ: SOLUTIONS

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Total vapour pressure of mixture of 1 mol of volatile
Component A

\left( {P_A^\mathrm{o} = 100\,\,mm\,Hg} \right)

and 3 mol of volatile
Component B

\left( {P_B^\mathrm{o} = 80\,\,mm\,Hg} \right)

is 90 mm Hg, for such case:

View answer and explanation

Correct option: D

All the above statements are correct

Option D (All the above statements are correct) is correct because the total vapor pressure is less than expected from Raoult's law, indicating negative deviation, which lowers the boiling point and implies weaker A-B interactions.

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