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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 the correct approach to solving this problem. The definition of Raoult's law is explained well in this Wikipedia page as:
a law of thermodynamics established by French chemist François-Marie Raoult in 1887. It states that the partial vapor pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. Mathematically, Raoult's law for a single component in an ideal solution is stated as:
$$P_i = P_i^*x_i$$ 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).
Because you will be solving for mole fraction, you can rearrange the equation to:
$$x_i = P_i/P_i^*$$ $$x_i=\pu{0.60atm}/\pu{0.80atm} = 0.75$$
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 $1 - 0.75 = 0.25$.
