JEE Main 2024 — Electrochemistry Question with Solution

The molar conductivity (${\Lambda }_m$) of a strong electrolyte depends on the concentration ($c$) according to Kohlrausch's law, which can be mathematically expressed as ${\Lambda }_m={\Lambda }_m^0-k\sqrt{c}$, where ${\Lambda }_m^0$ is the molar conductivity at infinite dilution, and $k$ is a constant. The graph of molar conductivity (${\Lambda }_m$) against the square root of concentration ($c^{1/2}$) is a straight line with a negative slope.

The unit of molar conductivity (${\Lambda }_m$) is Siemens meter squared per mole (${\text{S cm}}^2{\text{mol}}^{-1}$). Concentration ($c$) has the unit moles per liter (${\text{mol L}}^{-1}$), and thus its square root ($c^{1/2}$) has the unit ${\text{mol}}^{1/2}{\text{L}}^{-1/2}$.

The slope of the plot of ${\Lambda }_m$ against $c^{1/2}$ is derived from the expression $k$ in $k\sqrt{c}$. Therefore, to find the correct unit of the slope, we look at the division of the unit of ${\Lambda }_m$ by the unit of $c^{1/2}$:

$\frac{{\text{S cm}}^2{\text{mol}}^{-1}}{{\text{mol}}^{1/2}{\text{L}}^{-1/2}}={\text{S cm}}^2{\text{mol}}^{-1-1/2}{\text{L}}^{1/2}={\text{S cm}}^2{\text{mol}}^{-3/2}{\text{L}}^{1/2}$

Thus, the correct unit for the slope of a plot of molar conductivity against the square root of concentration for a strong electrolyte is ${\text{S cm}}^2{\text{mol}}^{-3/2}{\text{L}}^{1/2}$, which matches with Option D.