How to find mole fraction of solvent and solute in a vapor pressure lowering problem?
$\displaystyle \ce{x_{solute}} = \ce{\frac{n_{solute}}{n_{solute} + n_{solvent}}}$

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