A 2% (by mass) aqueous solution of NaCl is prepared. What is the molality of this solution? (Molar mass of NaCl = 58.5 g mol−1)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.35 mol kg−1
Molality is the number of moles of solute per kilogram of solvent. A 2% (by mass) solution has 2 g of NaCl in every 100 g of solution.
Mass of solvent: 100 g − 2 g = 98 g of water = 0.098 kg.
Moles of NaCl: \( \frac{2\,\mathrm{g}}{58.5\,\mathrm{g\,mol^{-1}}}=0.0342\,\mathrm{mol} \)
\( \displaystyle m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)
\( \displaystyle m=\frac{0.0342\,\mathrm{mol}}{0.098\,\mathrm{kg}} \)
\( \displaystyle m=0.349\,\mathrm{mol\,kg^{-1}}\approx0.35\,\mathrm{mol\,kg^{-1}} \)