JEE Main 2025 — Electrochemistry Question with Solution

Determine the conductivity from the resistivity.

The resistivity is given as

$$\rho =870.0\text{m}\Omega \cdot \text{m}.$$

Convert this into ohm·meters:

$$870.0\text{m}\Omega \cdot \text{m}=870.0\times 10^{-3}\Omega \cdot \text{m}=0.87\Omega \cdot \text{m}.$$

Conductivity is the reciprocal of resistivity:

$$\kappa =\frac{1}{\rho }=\frac{1}{0.87}\approx 1.15\text{S/m}.$$

Find the concentration of NaOH in mol/m³.

A $0.2\%(\mathrm{w}/\mathrm{v})$ solution means there are 0.2 grams of NaOH per 100 mL of solution.

In 1 liter (1000 mL) there are:

$$\frac{0.2\text{g}}{100\text{mL}}\times 1000\text{mL}=2\text{g/L}.$$

The molar mass of NaOH is approximately $40\text{g/mol}$. Thus, the molarity is:

$$\text{Molarity}=\frac{2\text{g/L}}{40\text{g/mol}}=0.05\text{mol/L}.$$

Since $1\text{L}=0.001{\text{m}}^3$, converting to SI units (mol/m³):

$$c=0.05\text{mol/L}\times \frac{1}{0.001{\text{m}}^3/\text{L}}=50{\text{mol/m}}^3.$$

Calculate the molar conductivity in SI units.

Molar conductivity ${\Lambda }_m$ is given by:

$${\Lambda }_m=\frac{\kappa }{c}.$$

Substituting the values:

$${\Lambda }_m=\frac{1.15\text{S/m}}{50{\text{mol/m}}^3}=0.023\text{S}\cdot {\text{m}}^2/\text{mol}.$$

Convert the molar conductivity to the desired units (mS dm² mol⁻¹).

First, convert the conductivity:

$$0.023\text{S}\cdot {\text{m}}^2/\text{mol}\times 1000\frac{\text{mS}}{\text{S}}=23\text{mS}\cdot {\text{m}}^2/\text{mol}.$$

Now, convert the area units. Recall that:

$$1{\text{m}}^2=100{\text{dm}}^2.$$

So,

$$23\text{mS}\cdot {\text{m}}^2/\text{mol}=23\times 100\text{mS}\cdot {\text{dm}}^2/\text{mol}=2300\text{mS}\cdot {\text{dm}}^2/\text{mol}.$$

The problem asks for the answer in the form

$$\text{[blank]}\times 10^2\text{mS}\cdot {\text{dm}}^2/\text{mol}.$$

Expressing 2300 mS dm²/mol in this form:

$$2300\text{mS}\cdot {\text{dm}}^2/\text{mol}=23\times 10^2\text{mS}\cdot {\text{dm}}^2/\text{mol}.$$

Thus, the molar conductivity is

$$\boxed{23\times 10^2\text{mS}\cdot {\text{dm}}^2/\text{mol}}.$$