100 mL of an aqueous solution contains 10 g of a dibasic acid (Mol. wt = 100). If the density of the solution is 1.5 g /cc, then the mole fraction of the solute is :

Moles of dibasic acid $\frac{10}{100}=0.10\mathrm{mol}$
wt. of sol = vol x density = 100 × 1.5 = 150 g
wt. of water = 150 – 10 = 140 g
moles of water $=\frac{140}{18}=7.8 \mathrm{mole}$
${\mathrm{x}}_1=\frac{0.10}{0.10+7.8}=\frac{0.10}{7.9}=0.013$