Question

The vapour pressure of pure liquid A is 0.80 atm. On mixing a non-volatile B to A, its vapour pressure becomes 0.6 atm. The mole fraction of B in the solution is.

• A. 0.150
• B. 0.25
• C. 0.50
• D. 0.75

Answer 1

Answer: (B)

0.25

Answer 2

According to Raoult’s Law $\frac { P ^ { 0 } - P _ { s } } { P ^ { 0 } } = x _ { B }$

(Mole fraction of solute) $x _ { B } = \frac { 0.8 - 0.6 } { 0.8 } = 0.25$ .