The percentage dissociation of 0.1M AgI solution having a resistance…

The correct answer is Option (1) → 5.7%

The degree of dissociation of an electrolyte at a concentration c is the ratio of its molar conductivity at that concentration to its limiting molar conductivity: α = Λ_m/Λ°_m. Λ°_m of AgI is obtained from those of strong electrolytes by Kohlrausch’s law of independent migration of ions.

Conductivity from the cell constant and the resistance:

    \( \displaystyle \kappa = \frac{G^*}{R} = \frac{4\ \text{cm}^{-1}}{100\ \Omega} = 0.04\ \Omega^{-1}\,\text{cm}^{-1} \)

Molar conductivity at 0.1 M:

    \( \displaystyle \Lambda_m = \frac{\kappa \times 1000}{c} = \frac{0.04 \times 1000}{0.1} = 400\ \Omega^{-1}\,\text{cm}^2\,\text{mol}^{-1} \)

Limiting molar conductivity of AgI by Kohlrausch’s law:

    \( \Lambda^\circ_{AgI} = \Lambda^\circ_{AgNO_3} + \Lambda^\circ_{KI} - \Lambda^\circ_{KNO_3} = 8000 + 9500 - 10500 = 7000\ \Omega^{-1}\,\text{cm}^2\,\text{mol}^{-1} \)

Degree of dissociation:

    \( \displaystyle \alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{400}{7000} = 0.0571 = \mathbf{5.7\%} \)

Shortcut: pick the combination so that the unwanted ions cancel: \( \ce{Ag^{+}} \) from \( \ce{AgNO3} \), \( \ce{I^{-}} \) from KI, and subtract \( \ce{KNO3} \) to remove \( \ce{K^{+}} \) and \( \ce{NO3^{-}} \).

NCERT (2026-27 edition):
Variation of Conductivity and Molar Conductivity with Concentration – …