SATHEE: Solutions Question 410

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:

$p_{i d e a l} = p_{A}^{o} x_{A} + p_{B}^{o} x_{B}$

$= 100 \times \frac{1}{4} + 80 \times \frac{3}{4} = 85, m m, H g$

$p_{a c t u a l} = 90 m m, H g$

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

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Solutions Question 410

Question: The total vapor pressure of a mixture of 1 mol of volatile component A ( PAO = 100 mm Hg) and 3 mol of volatile component B ( PBO =80 mm Hg) is 90 mm Hg. For such case:

Options:

Answer:

Solution:

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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: