At a certain temperature, the vapour pressure of pure water is 25 mm Hg and that of a very dilute aqueous solution of urea is 24.5 mm Hg. What is the molality of the solution (in mol kg−1)? (Molar mass of water = 18 g mol−1)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.11
For a non-volatile solute, the relative lowering of vapour pressure equals the mole fraction of the solute. In a very dilute solution n2 is tiny beside n1, so the mole fraction becomes n2/n1:
\( \displaystyle \frac{p_1^0-p_1}{p_1^0}=\frac{n_2}{n_1} \)
\( \displaystyle \frac{n_2}{n_1}=\frac{25-24.5}{25}=0.02 \)
Molality is the moles of solute in 1 kg (1000 g) of water. In 1000 g of water:
\( \displaystyle n_1=\frac{1000\,\mathrm{g}}{18\,\mathrm{g\,mol^{-1}}}=55.56\,\mathrm{mol} \)
\( \displaystyle m=n_2=0.02\times55.56 \)
\( \displaystyle m=1.11\,\mathrm{mol\,kg^{-1}} \)
Without the dilute-solution step, n2/n1 = 0.02/0.98 and m = 1.13 mol kg−1, which is still nearest to 1.11.
NCERT (2026-27 edition):
1.6.1 Relative Lowering of Vapour Pressure – “… Here n1 and n2 are the number of moles of solvent and solute respectively present in the solution. For dilute solutions n2 << n1, hence neglecting n2 in the denominator we have”
\( \displaystyle \frac{p_1^0-p_1}{p_1^0}=\frac{n_2}{n_1} \qquad (1.27) \)