At a certain temperature, a 5.12% (mass/volume) solution of cane sugar (C12H22O11, molar mass 342 g mol−1) is isotonic with a 0.9% (mass/volume) solution of an unknown solute. The molar mass of the solute is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 60 g mol−1
Isotonic solutions have the same osmotic pressure at the same temperature. Since π = CRT, at one temperature they must have the same molar concentration of solute particles. Cane sugar does not dissociate, and the unknown solute is taken to behave the same way.
Cane sugar: 5.12 g in 100 mL, so 5.12/342 mol in 100 mL.
Unknown solute: 0.9 g in 100 mL, so 0.9/M mol in 100 mL.
\( \displaystyle \frac{5.12}{342}=\frac{0.9}{M} \)
\( \displaystyle M=\frac{0.9\times 342}{5.12}\ \mathrm{g\,mol^{-1}} \)
\( \displaystyle M=60.1\ \mathrm{g\,mol^{-1}}\approx 60\ \mathrm{g\,mol^{-1}} \)
NCERT (2026-27 edition):
1.6.4 Osmosis and Osmotic Pressure – “… For dilute solutions, it has been found experimentally that osmotic pressure is proportional to the molarity, C of the solution at a given temperature T.”
“Two solutions having same osmotic pressure at a given temperature are called isotonic solutions. When such solutions are separated by semipermeable membrane no osmosis occurs between them.”