Q. The vapour pressure of pure liquid A is 0.80 atm. When a non-volatile B is added to A its vapour pressure drops to 0.60 atm. The mole fraction of B in the solution is
A) 0.125
B) 0.25
C) 0.5
D) 0.75
Explanation
(b)
According to the Raoult’s law the relative lowering vapour pressure which is produced by dissolving a non-volatile solute in a solvent is equal to mole fraction of the solute.
$\frac{P - P_{s}}{P} = X_{B\ }$
where, p= vapour pressure of solvent
$P_{s} = \ \ vapour\ \ pressure\ of\ solution\$
$X_{B} = mole\ fraction\ of\ B\$
Given, P=0.80 atm
$P_{s} =$ 0.60 atm
$\therefore\ \$ $X_{B} = \frac{0.80 - 0.60}{0.80} = \frac{0.20}{0.80} = 0.25\$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.