Solutions Question 410

Question: 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:

Options:

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

Show Answer

Answer:

Correct Answer: D

Solution:

$ {p_{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.



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