# Calculate the mole fraction of component B in this solution? (partial pressures)

The vapour pressure of a pure liquid solvent A is $\pu{0.80atm}$. When a non volatile substance B is dissolved in the solvent, its vapour pressure drops to $\pu{0.60atm}$. Calculate the mole fraction of component B in this solution.

All I can think of is that $P_A=x_A * P_A^*$, which is Raoult's Law. But I don't know how to apply it.

• Raoult's Law is correct. Think of it as a math problem Y=a*X. – MaxW Nov 4 '15 at 4:11

$$P_i = P_i^*x_i$$ where $P_i$ is the partial vapor pressure of the component $i$ in the gaseous mixture (above the solution), $P_i^*$ is the vapor pressure of the pure component $i$, and $x_i$ is the mole fraction of the component $i$ in the mixture (in the solution).
$$x_i = P_i/P_i^*$$ $$x_i=\pu{0.60atm}/\pu{0.80atm} = 0.75$$
Of course the value calculated above is for the volatile solvent A. Since the total mole fraction of both components combined is equal to unity, the mole fraction of the non-volatile solute B must be $1 - 0.75 = 0.25$.