JEE Main 2024 — 9 April, Shift 1 — Question 80

Molarity formula, $\mathrm{M = \frac{n}{V}}$

Given: $\mathrm{M = 0.2,\; V = 0.5\ L}$

Moles of $\mathrm{CuSO_4}$ $= \mathrm{n = 0.2 \times 0.5 = 0.1\ mol}$

Mass formula, $\mathrm{mass = n \times molar\ mass}$

Molar mass of $\mathrm{CuSO_4} = 160\ g\ mol^{-1}$

$$\mathrm{mass\ of\ solute = 0.1 \times 160 = 16\ g}$$

Density relation, $\mathrm{density = \frac{mass}{volume}}$

Mass of solution $= \mathrm{1.25 \times 500 = 625\ g}$

Mass of solvent $= \mathrm{625 - 16 = 609\ g = 0.609\ kg}$

Molality formula, $\mathrm{m = \frac{n}{mass\ of\ solvent\ (kg)}}$

$$\mathrm{m = \frac{0.1}{0.609} \approx 0.164}$$ $$\mathrm{m = 164 \times 10^{-3}}$$