The molality of a solution prepared by dissolving 4.0 g NaOH in 1000 mL distilled water is _____
(Assume that the density of water is \( 1\, g\, mL^{-1} \))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(0.1\, mol\, kg^{-1}\)
Molality (m) is the number of moles of solute dissolved in 1 kg of solvent:
\( \displaystyle m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)
The molar mass of NaOH is 23 + 16 + 1 = 40 g mol−1, and 1000 mL of water at 1 g mL−1 has a mass of 1000 g = 1 kg.
\( \displaystyle \text{moles of NaOH} = \frac{4.0\ \mathrm{g}}{40\ \mathrm{g\,mol^{-1}}} = 0.1\ \mathrm{mol} \)
\( \displaystyle m = \frac{0.1\ \mathrm{mol}}{1\ \mathrm{kg}} = 0.1\ \mathrm{mol\,kg^{-1}} \)
So the molality of the solution is 0.1 mol kg−1.