JEE Main 2025 Apr 2 Shift 2, Chemistry Q22: Colligative Properties

Step 1: Molarity. A 0.2% (w/v) solution contains 0.2 g of NaOH in 100 mL, i.e. 2 g L${}^{-1}$.

$M=\frac{2}{40}=0.05{\text{mol L}}^{-1}=0.05{\text{mol dm}}^{-3}$.

Step 2: Conductivity from resistivity. $\rho =870\text{m}\Omega \text{m}=0.870\Omega \text{m}$.

$\kappa =\frac{1}{\rho }=\frac{1}{0.870}{\text{S m}}^{-1}\approx 1.149{\text{S m}}^{-1}$.

Converting to mS dm${}^{-1}$: $1{\text{S m}}^{-1}=100{\text{mS dm}}^{-1}$, so $\kappa \approx 114.9{\text{mS dm}}^{-1}$.

Step 3: Molar conductivity.

${\Lambda }_m=\frac{\kappa }{c}=\frac{114.9{\text{mS dm}}^{-1}}{0.05{\text{mol dm}}^{-3}}\approx 2298{\text{mS dm}}^2{\text{mol}}^{-1}$.

Expressing in the requested form: ${\Lambda }_m\approx 23\times 10^2{\text{mS dm}}^2{\text{mol}}^{-1}$.

$\therefore$ Nearest integer $=23$.