QJEEM24Ch5b88: JEE Main 2024 Chemistry - Class 12 JEEM24Ch5b88 Solution: For a strong electrolyte, a plot of molar conductivity against…

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.