Solutions Question 276
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.
BETA
