JEE Main 2025 Apr 4 Shift 2, Chemistry Q24: Ionization Of Acids And Bases

JEE Main2025Apr 4, Shift 2Chemistry

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 $x \times 1 0^{- 1}$ . The value of x is _____. (Nearest integer)

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Solution

First obtain the limiting molar conductivity of ammonium ion using $N H_{4} C l$ data:

$λ^{\circ} (N H_{4}^{+}) = Λ_{m}^{\circ} (N H_{4} C l) - λ^{\circ} (C l^{-}) = 185 - 70 = 115 S cm^{2} mol^{- 1}$ .

Apply Kohlrausch's law to $N H_{4} O H$ :

$Λ_{m}^{\circ} (N H_{4} O H) = λ^{\circ} (N H_{4}^{+}) + λ^{\circ} (O H^{-}) = 115 + 170 = 285 S cm^{2} mol^{- 1}$ .

Degree of dissociation:

$α = \frac{Λ _{m}}{Λ _{m}^{\circ}} = \frac{85.5}{285} = 0.3 = 3 \times 1 0^{- 1}$ .

$∴ x = 3$ .

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JEE Main 2025 Apr 4 Shift 2, Chemistry Q24: Ionization Of Acids And Bases

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