Q. On adding a solute to a solvent having vapour pressure 0.80 atm vapour pressure reduces to 0.60 atm. Mole fraction of solute is
A) 0.25
B) 0.75
C) 0.50
D) 0.33
Explanation
(a)
According to Raoult’s law
$\frac{p - p_{s}}{p} = x_{solute}$
Where, p= vapour pressure of pure solvent =0.80 atm
$p_{s} =$ vapour pressure of solute =0.60 atm
$X_{solute} =$ mole fraction of solute
$\frac{0.80 - 0.60}{0.80} = X_{solute}$
or $\frac{0.20}{080} = X_{solute}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{solute} = 0.25$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.