We’re being asked to calculate the mole fraction of sucrose (C₁₂H₂₂O₁₁) in a solution that is prepared by dissolving 38.6 g sucrose (C₁₂H₂₂O₁₁) in 495 g of water.

Sucrose (C₁₂H₂₂O₁₁) → lesser in amount → solute
Water (H₂O) → greater in amount → solvent

Mole Fraction (X) relates the moles of solute and solvent within a solution:

$\boxed{\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{ }\mathbf{f}\mathbf{r}\mathbf{a}\mathbf{c}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{ }\mathbf{\left(}\mathbf{X}\mathbf{\right)}\mathbf{=}\frac{\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{e}}{\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}}$

We’ll calculate the mole fraction of sucrose using the following steps: