Read the passage carefully and answer the Questions.

Molar conductivity ($Λ_m$) of a solution at a given concentration (c) is the conductance of volume, V of solution, containing one mole of electrolyte kept between the two electrodes with area of cross-section A and at a distance of unit length. It increases with the decrease in concentration and when the concentration approaches zero, the molar conductivity is called limiting molar conductivity ($Λ^0_m$). For a strong electrolyte, $Λ_m$ increases linearly with dilution and is given by $Λ_m = Λ^0_m -Ac^{1/2}$. The value of the constant $Λ_m$ for a given solvent depends on the type of electrolyte along with temperature. According to Kohlrausch law, the value of $Λ^0_m$ for an electrolyte is $Λ^0_m=v_+λ^0_+v_-λ^0_-$, where $v_+$ and $v_-$ are the number of cations and anions, respectively, per molecule of the electrolyte and $λ^0_+$ and $λ^0_-$ are limiting molar conductivities of cation and anion, respectively. Kohlrausch law finds many applications, like determining the solubility of a sparingly soluble salt, determining the degree of dissociation ($Λ^0_m/$Λ_m$) and the dissociation constant of a weak electrolyte.

The unit of constant A is:

$S\, cm^2\, mol^{-3/2}\,L^{1/2}$

The correct answer is Option (3) → $S\, cm^2\, mol^{-3/2}\,L^{1/2}$ **

Core Concept:

Step 1: Analyze the Equation The Debye-Hückel-Onsager equation for strong electrolytes is given by:

$\Lambda_m = \Lambda_m^0 - Ac^{1/2}$

In this equation:

• $\Lambda_m$ is the molar conductivity at concentration $c$.
• $\Lambda_m^0$ is the limiting molar conductivity (at infinite dilution).
• $c$ is the concentration of the electrolyte.
• $A$ is a constant that depends on the type of electrolyte, solvent nature, and temperature.

Step 2: Determine Units for Each Term

According to the principle of dimensional homogeneity, all terms in an equation must have the same units.

• Molar Conductivity ($\Lambda_m$): The standard units are $\text{S cm}^2 \text{ mol}^{-1}$.
• Concentration ($c$): The standard units for concentration in this context are moles per liter ($\text{mol L}^{-1}$).

Step 3: Perform Dimensional Analysis

Since $\Lambda_m$ and the term $Ac^{1/2}$ must have identical units, we can set up the following relationship:

$\text{Unit of } \Lambda_m = (\text{Unit of } A) \times (\text{Unit of } c)^{1/2} \quad \text{}$

$\text{S cm}^2 \text{ mol}^{-1} = (\text{Unit of } A) \times (\text{mol L}^{-1})^{1/2} \quad \text{}$

$\text{S cm}^2 \text{ mol}^{-1} = (\text{Unit of } A) \times \text{mol}^{1/2} \text{ L}^{-1/2} \quad \text{}$

Step 4: Solve for the Unit of A

Rearrange the equation to isolate the unit of $A$:

$\text{Unit of } A = \frac{\text{S cm}^2 \text{ mol}^{-1}}{\text{mol}^{1/2} \text{ L}^{-1/2}} \quad \text{}$

$\text{Unit of } A = \text{S cm}^2 \text{ mol}^{-1} \times \text{mol}^{-1/2} \times \text{L}^{1/2} \quad \text{}$

$\text{Unit of } A = \text{S cm}^2 \text{ mol}^{-3/2} \text{ L}^{1/2} \quad \text{}$