Chemistry Q72 – JEE-MAIN 2025

Home Chemistry ElectrochemistryQuestion #403

The molar conductance of an infinitely dilute solution of ammonium chloride was found to be

185 {S cm}^{2} {mol}^{- 1}

and the ionic conductance of hydroxyl and chloride ions are

170

and

70 {S cm}^{2} {mol}^{- 1}

, respectively. If molar conductance of

0.02 M

solution of ammonium hydroxide is

85.5 {S cm}^{2} {mol}^{- 1}

, its degree of dissociation is given by

x \times 10^{- 1}

. The value of

x

is _______. (Nearest integer)

Enter Calculated Numerical Value:

Clue 1 of 3: Kohlrausch's Law

This law helps determine the limiting molar conductivity of weak electrolytes from strong electrolytes.

🧠 Full Solution Path

Step 1: Calculate limiting ionic conductivity of

N H_{4}^{+}

ions✦ Active

Use Kohlrausch's Law for ammonium chloride ( $N H_{4} C l$ ) to find the limiting ionic conductivity of ammonium ions ( $λ_{N H_{4}^{+}}^{\circ}$ ).

Λ_{N H_{4} C l}^{\circ} = λ_{N H_{4}^{+}}^{\circ} + λ_{C l^{-}}^{\circ}

185 = λ_{N H_{4}^{+}}^{\circ} + 70

λ_{N H_{4}^{+}}^{\circ} = 185 - 70 = 115 {S cm}^{2} {mol}^{- 1}

💡 Teacher's Secret Hint

Remember that limiting molar conductivity of a strong electrolyte is the sum of limiting ionic conductivities of its constituent ions.

Step 2: Calculate limiting molar conductivity of

N H_{4} O H

○ Expand

Now, calculate the limiting molar conductivity of ammonium hydroxide ( $Λ_{N H_{4} O H}^{\circ}$ ) using the limiting ionic conductivities of $N H_{4}^{+}$ and $O H^{-}$ ions.

Λ_{N H_{4} O H}^{\circ} = λ_{N H_{4}^{+}}^{\circ} + λ_{O H^{-}}^{\circ}

Λ_{N H_{4} O H}^{\circ} = 115 + 170 = 285 {S cm}^{2} {mol}^{- 1}

💡 Teacher's Secret Hint

For weak electrolytes, $Λ^{\circ}$ cannot be determined by extrapolation but can be calculated using Kohlrausch's Law from strong electrolytes.

Step 3: Determine the degree of dissociation and the value of

x

○ Expand

Calculate the degree of dissociation ( $α$ ) using the given molar conductivity at $0.02 M$ concentration ( $Λ_{c}^{N H_{4} O H}$ ) and the calculated limiting molar conductivity ( $Λ_{N H_{4} O H}^{\circ}$ ). Then, find the value of $x$ from the given expression for $α$ .

α = \frac{Λ_{c}^{N H_{4} O H}}{Λ_{N H_{4} O H}^{\circ}} = \frac{85.5}{285} = 0.3

Given α = x \times 10^{- 1}

0.3 = x \times 0.1

x = \frac{0.3}{0.1} = 3

💡 Teacher's Secret Hint

The degree of dissociation is a dimensionless quantity, representing the fraction of electrolyte molecules that dissociate into ions.

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Chemistry Question 72 – JEE-MAIN 2025

Clue 1 of 3: Kohlrausch's Law

🧠 Full Solution Path