An aqueous solution of glucose is labelled as \(20\%\) \(w/w\). If the density of the solution is \(1.2\, \ g/mL\) the molality of the solution will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1.38\, \ mol/kg\)
Molality is the number of moles of solute per kilogram of solvent:
\( \displaystyle m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)
20% w/w means 20 g of glucose in 100 g of solution, that is in 80 g of water.
\( \displaystyle \text{Moles of glucose} = \frac{20\,\mathrm{g}}{180\,\mathrm{g\,mol^{-1}}} = 0.111\,\mathrm{mol} \)
\( \displaystyle m = \frac{0.111\,\mathrm{mol}}{0.080\,\mathrm{kg}} = 1.39\,\mathrm{mol\,kg^{-1}} \approx 1.38\,\mathrm{mol/kg} \)
Shortcut: molality needs only masses, so the density (1.2 g/mL) can be ignored.
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 – “1.5 A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL–1, then what shall be the molarity of the solution?”