Q29 / 90 · JEE Main 2024 (09 Apr Shift 1) · Chemistry
Molarity \((\mathrm{M})\) of an aqueous solution containing \(x \mathrm{~g}\) of anhyd. \(\mathrm{CuSO}_4\) in \(500 \mathrm{~mL}\) solution at \(32{ }^{\circ} \mathrm{C}\) is \(2 \times 10^{-1} \mathrm{M}\). Its molality will be ______ \(\times 10^{-3} \mathrm{~m}\). (nearest integer). [Given density of the solution \(=1.25 \mathrm{~g} / \mathrm{mL}\)]
Correct answer: 164
Solution
$\begin{aligned} & \begin{aligned} & \mathrm{M}_{\text {sol }^{\mathrm{I}}}=\mathrm{v}_{\text {sol }^{\mathrm{n}}} \times \mathrm{d}_{\text {sol }^{\mathrm{n}}} \\ &=500 \times 1.25=625 \mathrm{~g} \\ & \text { Mass of solute }(\mathrm{x})=0.2 \times 0.5 \times 159.5 \\ & \quad=15.95\end{aligned} \\ & \mathrm{n}_{\text {solute }}=0.1, \quad \\ & \text { Mass of solvent = Mass of solution }- \text { Mass of solute } \\ & =625-15.95 \\ & =609.05 \\ & \mathrm{~m}=\frac{0.1}{\frac{609.05}{1000}} \\ & \mathrm{~m}=0.164=164 \times 10^{-3}\end{aligned}$
Hence, the correct answer is 164.