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.