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.

Which one of the following has a different value of constant A in water?

$CaCl_2$

The correct answer is Option (1) → $CaCl_2$ **

Core Concept

In the equation:

Λₘ = Λₘ° − Ac¹ᐟ²

Constant A depends on:

• Nature of electrolyte
• Ionic charge type (1:1, 2:1 etc.)
• Solvent and temperature

Electrolytes producing different types of ions (in terms of charge and number) will have different

A values.

Detailed Option-wise Explanation

Option 1: CaCl₂

Dissociates as:

CaCl₂ → Ca²⁺ + 2Cl⁻

This is a 2:1 electrolyte.

Higher ionic charge leads to stronger interionic interactions.

Hence the value of constant A will be different from 1:1 electrolytes.

Option 2: KCl

Dissociates as:

KCl → K⁺ + Cl⁻

This is a uni-univalent (1:1) electrolyte.

A depends on ionic type.

Since both ions are singly charged, it follows the same A value as other 1:1 electrolytes.

Option 3: NaCl

Dissociates as:

NaCl → Na⁺ + Cl⁻

Also a 1:1 electrolyte.

Same charge type as KCl and LiCl.

Hence A remains similar.

Option 4: LiCl

Dissociates as:

LiCl → Li⁺ + Cl⁻

Again a 1:1 electrolyte.

Same ionic nature as NaCl and KCl.

Therefore A is not different.

Only CaCl₂ produces ions of different charge type (2:1), so it has a different value of constant A.