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.

If the equivalent conductivity of 1 M Phthalic acid is $12.8\, S\, cm^2\, eq^{-1}$ and the limiting equivalent conductivities of the hydrogen and Phthalate ions are 42 and $288.42\, S\, cm^2\, eq^{-1}$, respectively, the degree of dissociation of Phthalic acid will be:

0.0387

The correct answer is Option (4) → 0.0387 **

Core Concept

For weak electrolytes:

$\alpha = \Lambda / \Lambda^\circ$

where

$\Lambda$ = equivalent conductivity at given concentration

$\Lambda^\circ$ = limiting equivalent conductivity

Step 1: Calculate $\Lambda^\circ$

Phthalic acid $\rightarrow$ H$^+$ + Phthalate$^-$

$\Lambda^\circ = \lambda^\circ(\text{H}^+) + \lambda^\circ(\text{Phthalate}^-)$

$\Lambda^\circ = 42 + 288.42$

$\Lambda^\circ = 330.42 \text{ S cm}^2 \text{ eq}^{-1}$

Step 2: Calculate degree of dissociation

$\alpha = \Lambda / \Lambda^\circ$

$\alpha = 12.8 / 330.42$

$\alpha \approx 0.0387$