The solubility product of a sparingly soluble salt $\mathrm{A}_{2} \mathrm{X}_{3}$ is $1.1 \times 10^{-23}$. If specific conductance of the solution is $3 \times 10^{-5} \mathrm{~S} \mathrm{~m}^{-1}$, the limiting molar conductivity of the solution is $x \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}$. The value of $x$ is ______.

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Redox Reactions and Electrochemistry

The solubility product of a sparingly soluble salt

A_{2} X_{3}

1.1 \times 1 0^{- 23}

. If specific conductance of the solution is

3 \times 1 0^{- 5} S m^{- 1}

, the limiting molar conductivity of the solution is

x \times 1 0^{- 3} S m^{2} mol^{- 1}

. The value of

x

is ______.

Answer: 3

For the salt

A_{2} X_{3}

dissolving as:

A_{2} X_{3} ⇌ 2 A^{3 +} + 3 X^{2 -}

If solubility is

s

M, then

K_{s p} = (2 s)^{2} (3 s)^{3} = 108 s^{5}

.

Given

K_{s p} = 1.1 \times 1 0^{- 23}

108 s^{5} = 1.1 \times 1 0^{- 23}

s^{5} \approx 1 0^{- 25}

s \approx 1 0^{- 5} M = 0.01 mol/m^{3}

For a very dilute solution,

Λ_{m} \approx Λ_{m}^{\infty} = \frac{κ}{c}

, where

κ

is specific conductance and

c

is molar concentration.

Λ_{m}^{\infty} = \frac{3 \times 1 0 ^{- 5}}{0.01} = 3 \times 1 0^{- 3} S m^{2} mol^{- 1}

Therefore,

x = 3

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Integer answer