The density of a $3 \mathrm{M}$ aqueous solution of $\mathrm{NaCl}$ is accurately measured to be $1.

Question

The density of a $3 \mathrm{M}$ aqueous solution of $\mathrm{NaCl}$ is accurately measured to be $1.25 \mathrm{~g/mL}$. The molality of this solution is (Given the molar mass of $\mathrm{NaCl} = 58.5 \mathrm{~g \cdot mol}^{-1}$)

Professor PC Thomas Classes · Answer

(a) : Molarity is $3 \mathrm{M}$, meaning $1 \mathrm{~L}$ ($1000 \mathrm{~mL}$) of the solution contains $3 \mathrm{~moles}$ of $\mathrm{NaCl}$. Mass of the solution = $	ext{Volume} 	imes 	ext{Density} = 1000 \mathrm{~mL} 	imes 1.25 \mathrm{~g/mL} = 1250 \mathrm{~g}$. The mass of the solute ($\mathrm{NaCl}$) = $3 	imes 58.5 = 175.5 \mathrm{~g}$. The mass of the solvent (water) = $	ext{Mass of solution} - 	ext{Mass of solute} = 1250 - 175.5 = 1074.5 \mathrm{~g} = 1.0745 \mathrm{~kg}$. The molality of the solution $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{3}{1.0745} \approx 2.79 \mathrm{~m}$.