Vapour pressure of water at 293 K is 17.535 mm Hg. The vapour pressure of solution after adding 8.28 g of urea in 450 g of water will be
Given: Molar mass of urea is 60 g mol−1 and assume the solution to be dilute.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 17.438 mm Hg
By Raoult’s law, the relative lowering of vapour pressure caused by a non-volatile solute equals the mole fraction of the solute. For a dilute solution, n2 is very small compared with n1, so:
\( \displaystyle \frac{p^{\circ}-p}{p^{\circ}}=\frac{n_2}{n_1+n_2}\approx\frac{n_2}{n_1} \)
The moles of solute (urea) and solvent (water) are:
\( \displaystyle n_2=\frac{8.28}{60}=0.138\ \mathrm{mol} \)
\( \displaystyle n_1=\frac{450}{18}=25\ \mathrm{mol} \)
Substituting, with p° = 17.535 mm Hg:
\( \displaystyle \frac{17.535-p}{17.535}=\frac{0.138}{25}=0.00552 \)
\( \displaystyle 17.535-p=17.535\times0.00552=0.0968 \)
\( \displaystyle p=17.535-0.0968=17.438\ \mathrm{mm\ Hg} \)
Shortcut: A solute always lowers the vapour pressure, which rules out Option (2). The mole fraction of urea is only about 0.55%, so the lowering is only about 0.1 mm Hg. Only 17.438 is that close to 17.535.
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Exercises – “… 1.34 Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.”