Calculate molality of 2.5 g of ethanoic acid (CH₃COOH) in 75g of benzene.

Molar mass of ${\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{O}}_2:$
$12\times 2+1\times 4+16\times 2=60\mathrm{g}{\mathrm{mol}}^{-1}$
Moles of ${\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{O}}_2=\frac{2.5\mathrm{g}}{60\mathrm{g}{\mathrm{mol}}^{-1}}$ = 0.0417mol
Mass of banzene in Kg = $75\mathrm{g}/1000\mathrm{g}{\mathrm{kg}}^{-1}$
$=75\times {10}^{-3}\mathrm{kg}$
$\text{ Molality of }{\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{O}}_2\frac{\text{ molesof }{\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{O}}_2}{\mathrm{kg}\text{ of benzene }}$
$=\frac{0.0417\mathrm{mol}\times 1000\mathrm{g}{\mathrm{kg}}^{-1}}{75\mathrm{g}}=0.556\mathrm{mol}{\mathrm{kg}}^{-1}$