The mole fraction of the solute in 1 molal aqueous solution is :

Mole fraction of solute $x_{solute} = ?$

Mole of solute $= 1$ ; since the solution is one $molal$ , the

wt. of solvent water $= 1 kg = 1000 g$

Also $g. mol.$ wt. of $H_{2} O = (2 \times 1) + 16 = 18 g . So, no. of moles of H_{2} O = \frac{1000}{18} = 55.5$

Hence, $x_{solute} = \frac{no . of moles of solute}{no . of moles of solute + no . of moles of H_{2} O}$

$= \frac{1}{1 + 55.5} = \frac{1}{56.5} = 0.0177 ≃ 0.018$

The mole fraction of the solute in $1 molal$ aqueous solution is :