Calculate the molality of KI if the density of 20% (mass/mass) aqueous solution of KI is \( 1.202\, g\, mL^{-1} \). (Molar mass of KI is \( 166\, g\, mol^{-1} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(1.5\, mol\, kg^{-1}\)
Molality is the number of moles of solute per kilogram of solvent. Because it is based on masses, a mass percentage gives it directly; the density is needed only for molarity.
A 20% (mass/mass) solution means 20 g of KI in 100 g of solution, so the mass of water is 100 − 20 = 80 g = 0.080 kg.
\( \displaystyle n_{\ce{KI}} = \frac{20\ \mathrm{g}}{166\ \mathrm{g\,mol^{-1}}} = 0.1205\ \mathrm{mol} \)
\( \displaystyle m = \frac{0.1205\ \mathrm{mol}}{0.080\ \mathrm{kg}} = 1.506\ \mathrm{mol\,kg^{-1}} \approx 1.5\ \mathrm{mol\,kg^{-1}} \)
So the molality of KI is 1.5 mol kg−1.
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):
Intext Questions – “Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.”