Q12 / 110 · VITEEE 2024 · Chemistry
The density of 3 M solution of NaCl is $1.25 \mathrm{~g} / \mathrm{mL}$. What is the molality of solution?
- A3.5 m
- Correct answerB2.7 m
- C5.0 m
- D6.7 m
Solution
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)
$\begin{aligned} & \qquad=3 \times 58.5=175.5 \mathrm{~g} \\ & \therefore \text { Mass of solvent }=1250-175.5=1074.5 \mathrm{~kg} \\ & \therefore 1074.5 \mathrm{~g}=1.0745 \mathrm{~kg} \\ & \therefore \text { Molality }=\frac{3 \mathrm{~mol}}{1.0745 \mathrm{~kg}}=2.7 \mathrm{~mol} \mathrm{~kg}^{-1}\end{aligned}$
Hence, the correct answer is option: 'B'.