# How to find mole fraction of solvent and solute in a vapor pressure lowering problem?

At $25$ degrees celsius, the vapor pressure of pure benzene ($\ce{ C6H6}$) is $93.9$ torr. When a non-volatile solute is dissolved in benzene, the vapor pressure of benzene is lowered to $91.5$ torr. What is the concentration of the solute and the solvent, expressed in mole fraction?

Through Raoult's formula, I found that the mole fraction of the solvent, benzene, is:

$$P_{solution}=x_{\ce{C6H6}}\cdot P^{o}_{\ce{C6H6}}$$

$$91.5=x_{\ce{C6H6}}\cdot 93.9$$

$$0.974 = x_{\ce{C6H6}}$$

Now the only thing I am having trouble finding is the mole fraction of the solute:

$$x_{\ce{C6H6}}=\frac{n_{\ce{C6H6}}}{n_{\ce{C6H6}}+n_{\ce{solute}}}$$

I was wondering, is it even possible to calculate the mole fraction with just the given info? I think that it might not be possible, since I need the grams of solute dissolved or grams of solvent to do this.

• Hint: What does the sum of all the mole fractions within the system have to equal? – J. Ari Apr 12 '17 at 18:03

$\displaystyle \ce{x_{solute}} = \ce{\frac{n_{solute}}{n_{solute} + n_{solvent}}}$
Thus, we have that $\displaystyle\ce{x_{solute}} + \ce{x_{solvent}} = \ce{\frac{n_{solute}}{n_{solute} + n_{solvent}}} + \ce{\frac{n_{solvent}}{n_{solute} + n_{solvent}}} = \ce{\frac{n_{solute} + n_{solvent}}{n_{solute} + n_{solvent}}} = 1,$ so $\ce{x_{solute} + 0.974 = 1\implies x_{solute} = 0.026}$