Chemistry - Liquid Solution Question with Solution | TestHub

Molality of solution $=\frac{\text{ Mass of solute }}{\text{ Mass of solvent }(\mathrm{kg})}$ 3 M NaCl solution means 3 mole of NaCl are present in 1 L of the solution. Mass of 1 L of the solution $=$ Volume $\times$ Density $=1000\times 1.25\mathrm{g}=1250\mathrm{g}$ Mass of 3 mole of NaCl (solute) $\begin{array}{rl}& =3\times 58.5=175.5\mathrm{g}\\ & \therefore \text{ Mass of solvent }=1250-175.5=1074.5\mathrm{kg}\\ & \therefore 1074.5\mathrm{g}=1.0745\mathrm{kg}\\ & \therefore \text{ Molality }=\frac{3\mathrm{mol}}{1.0745\mathrm{kg}}=2.7\mathrm{mol}{\mathrm{kg}}^{-1}\end{array}$