A solution is prepared by dissolving 6 g of urea (molar mass = 60 g mol−1) in one litre of solution. Another solution is prepared by dissolving 34.2 g of cane sugar (molar mass = 342 g mol−1) in one litre of solution at the same temperature. The lowering of vapour pressure in the first solution is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → The same as that of the second solution
Lowering of vapour pressure is a colligative property: for a non-volatile solute it depends on the number of solute particles, not on what the solute is. Urea and cane sugar are both non-electrolytes, so each molecule is one particle.
Urea: \( \displaystyle n=\frac{6\,\mathrm{g}}{60\,\mathrm{g\,mol^{-1}}}=0.1\,\mathrm{mol} \) in 1 L of solution.
Cane sugar: \( \displaystyle n=\frac{34.2\,\mathrm{g}}{342\,\mathrm{g\,mol^{-1}}}=0.1\,\mathrm{mol} \) in 1 L of solution.
Both solutions hold 0.1 mol of solute particles per litre at the same temperature, so the mole fraction of solute, and with it the lowering of vapour pressure, is the same in both.
NCERT (2026-27 edition):
1.6 Colligative Properties and Determination of Molar Mass – “… These are: (1) relative lowering of vapour pressure of the solvent (2) depression of freezing point of the solvent (3) elevation of boiling point of the solvent and (4) osmotic pressure of the solution. All these properties depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution.”