A $1.00 \text{ m}$ aqueous solution means that $1 \text{ mole}$ of solute is dissolved in $1 \text{ kg}$ ($1000 \text{ g}$) of the solvent (water) . Number of moles of solute ($n_2$) = $1 \text{ mol}$. Molar mass of water ($\text{H}_2\text{O}$) = $18 \text{ g mol}^{-1}$ . Number of moles of solvent (water), $n_1 = \frac{\text{Mass of water}}{\text{Molar mass of water}} = \frac{1000 \text{ g}}{18 \text{ g mol}^{-1}} = 55.55 \text{ mol}$ . The mole fraction of the solute ($x_2$) is given by the formula: $x_2 = \frac{n_2}{n_1 + n_2}$ . Substituting the values: $x_2 = \frac{1}{1 + 55.55} = \frac{1}{56.55} = 0.01768 \approx 0.0177$ .