Question

The vapour pressure of pure liquid solvent A is 0.80 atm. When a nonvolatile substance B is added to the solvent its vapour pressure drops to 0.60 atm. What is the mole fraction of component B in the solution?( If your answer is $x$, then write $x\times 100$).

Answer 1

Using Raoults law for lowering of vapour pressure.We get $\frac{P_o-P}{P_o}=x_2$
putting,Values of actual and lowered vapour pressure we get
$\frac{0.8-0.6}{0.8}=x_2$
where $x_2$ is the mole fraction
Ans-0.25