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. 




$\displaystyle \ce{x\_{solute}} = \ce{\frac{n\_{solute}}{n\_{solute} + n\_{solvent}}}$


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 $\ce{x\_{solute} + 0.974 = 1\implies x\_{solute} = 0.026}$ 


 1