The conductivity of a saturated solution of AgCl at 298 K is 3.45x 10^-6 S cm^-1, and that of the water used is 1.55x 10^-6 S cm^-1. Limiting molar ionic conductivities: lambda deg(Ag+) = 61.9 and lambda deg(Cl-) = 76.3 S cm^2 mol^-1. The solubility product of AgCl is K_sp = yx 10^-10. The value of y (two decimals) is

JEE Advanced levelNumerical answerDifficulty 3 of 3ec196

The conductivity of a saturated solution of $AgCl$ at 298 K is $3.45 \times 1 0^{- 6} S c m^{- 1}$ , and that of the water used is $1.55 \times 1 0^{- 6} S c m^{- 1}$ . Limiting molar ionic conductivities: $λ^{\circ} (Ag X^{+}) = 61.9$ and $λ^{\circ} (Cl X^{-}) = 76.3 S c m^{2} mo l^{- 1}$ . The solubility product of $AgCl$ is $K_{s p} = y \times 1 0^{- 10}$ . The value of $y$ (two decimals) is

Numerical answer type, tolerance 0.03.

Conductivity due to AgCl alone:

κ = 3.45 \times 1 0^{- 6} - 1.55 \times 1 0^{- 6} = 1.90 \times 1 0^{- 6} S c m^{- 1}

Λ_{m}^{\circ} (AgCl) = 61.9 + 76.3 = 138.2 S c m^{2} mo l^{- 1}

. For a sparingly soluble salt

Λ_{m} \approx Λ_{m}^{\circ}

, so solubility

s = \frac{1000 κ}{Λ _{m}^{\circ}} = \frac{1000 \times 1.90 \times 1 0 ^{- 6}}{138.2} = 1.375 \times 1 0^{- 5} mol L^{- 1}

K_{s p} = s^{2} = 1.89 \times 1 0^{- 10}

, so

y = 1.89