The properties which depend on the number of solute particles irrespective of their nature relative to the total numbers of particles in the solution are called colligative properties. These are relative lowering of vapour pressure of the solvent, depression of freezing point of the solvent, elevation of boiling point of the solvent and osmotic pressure of the solution. All of these properties help in calculating the observed molar mass of the solute. Determination of osmotic pressure is considered best for the determination of molecular mass of the solute.
A 1 litre solution of glucose containing 36 g of it has an osmotic pressure of 4.98 bar at 300 K. At the same temperature, the osmotic pressure of the solution is 1.52 bar. The concentration of the solution at 1.52 bar is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.0610 mol L−1
For a dilute solution the osmotic pressure is given by the van’t Hoff equation:
\( \displaystyle \pi = CRT \)
At the same temperature, π is proportional to the molar concentration C, so:
\( \displaystyle \frac{\pi_2}{\pi_1} = \frac{C_2}{C_1} \)
The first solution has 36 g of glucose (M = 180 g mol−1) in 1 L:
\( \displaystyle C_1 = \frac{36}{180} = 0.2\ \mathrm{mol\,L^{-1}} \)
\( \displaystyle C_2 = C_1\times\frac{\pi_2}{\pi_1} = 0.2\times\frac{1.52}{4.98} \)
\( \displaystyle C_2 = 0.0610\ \mathrm{mol\,L^{-1}} \)
Shortcut: 1.52 bar is less than one-third of 4.98 bar, so the concentration must be a little below 0.2/3 ≈ 0.067 mol L−1; only 0.0610 fits.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Exercises – “1.22 At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?”