What mass of glucose should be dissolved in 50 g of water to produce the same lowering of vapour pressure as is produced by dissolving 1 g of urea in the same quantity of water? (Molar masses: urea = 60 g mol−1, glucose = 180 g mol−1)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3 g
By Raoult’s law, the relative lowering of vapour pressure equals the mole fraction of the solute, so it depends only on the number of solute particles, not on their identity:
\( \displaystyle \frac{p_1^0-p_1}{p_1^0}=x_2=\frac{n_2}{n_1+n_2} \)
Both solutions have the same 50 g of water (same n1). The same lowering therefore needs the same number of moles of solute.
\( \displaystyle n(\text{urea})=\frac{1}{60}\,\mathrm{mol} \)
\( \displaystyle \frac{w}{180}=\frac{1}{60} \)
\( \displaystyle w=\frac{180}{60}=3\,\mathrm{g} \)
NCERT (2026-27 edition):
1.6.1 Relative Lowering of Vapour Pressure – “… Raoult established that the lowering of vapour pressure depends only on the concentration of the solute particles and it is independent of their identity. …
\( \displaystyle \frac{\Delta p_1}{p_1^0}=\frac{p_1^0-p_1}{p_1^0}=x_2 \qquad (1.25) \)
The expression on the left hand side of the equation as mentioned earlier is called relative lowering of vapour pressure and is equal to the mole fraction of the solute.”