Calculate the lowering of vapour pressure caused by the addition of 50 g of sucrose (mol mass = \( 342\, g\, mol^{-1} \)) to 750 g of water, if the vapour pressure of pure water at 25°C is 23.8 mm Hg.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.083 mm Hg
By Raoult’s law, the relative lowering of vapour pressure of a solution of a non-volatile solute equals the mole fraction of the solute:
\( \displaystyle \frac{p_1^{0} - p_1}{p_1^{0}} = x_2 = \frac{n_2}{n_1 + n_2} \)
Sucrose is a non-electrolyte, so each molecule is one particle. The amounts of solute and solvent are
\( \displaystyle n_2 = \frac{50\ \mathrm{g}}{342\ \mathrm{g\,mol^{-1}}} = 0.1462\ \mathrm{mol} \)
\( \displaystyle n_1 = \frac{750\ \mathrm{g}}{18\ \mathrm{g\,mol^{-1}}} = 41.67\ \mathrm{mol} \)
\( \displaystyle x_2 = \frac{0.1462}{41.67 + 0.1462} = \frac{0.1462}{41.81} = 3.50 \times 10^{-3} \)
The lowering of vapour pressure is then
\( \displaystyle \Delta p = p_1^{0}\,x_2 = 23.8\ \mathrm{mm\,Hg} \times 3.50 \times 10^{-3} = 0.083\ \mathrm{mm\,Hg} \)
So the vapour pressure is lowered by 0.083 mm Hg (from 23.8 to about 23.72 mm Hg).
Note: NTA picked this question from an NCERT (2026-27) intext question and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Reverse Osmosis and Water Purification – “Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.”