Molarity (M) of an aqueous solution containing xg of anhyd. $\mathrm{CuSO}_{4}$ in 500 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 $=1.25 \mathrm{~g} / \mathrm{mL}$.]

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Some Basic Concepts of Chemistry

Molarity (M) of an aqueous solution containing xg of anhyd.

CuSO_{4}

in 500 mL solution at

32^{\circ} C

2 \times 1 0^{- 1} M

. Its molality will be ____

\times 1 0^{- 3} m

. (nearest integer).
[Given density of the solution

= 1.25 g / mL

Answer: 164

Moles of

CuSO_{4} = M \times V = 0.2 \times 0.5 = 0.1

mol, so mass of solute

x = 0.1 \times 159.5 = 15.95

g. Mass of solution

= V \times d = 500 \times 1.25 = 625

g, giving solvent mass

= 625 - 15.95 = 609.05

= 0.60905

kg. Molality

= \frac{0.1}{0.60905} = 0.164 m = 164 \times 1 0^{- 3} m

. (The correct answer is 164; NTA's key of 81 was withdrawn — the question was marked BONUS in the exam.)

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Integer answer