1. Dissolution and solubility expression

For the salt $A_2X_3$:

$A_2X_3(s) \rightleftharpoons 2A^{3+} + 3X^{2-}$

If its molar solubility is $s$ mol L$^{-1}$, then

$[A^{3+}] = 2s, \qquad [X^{2-}] = 3s$

So,

$K_{sp} = [A^{3+}]^2 [X^{2-}]^3 = (2s)^2(3s)^3$

$K_{sp} = 4s^2 \cdot 27s^3 = 108s^5$

Given:

$K_{sp} = 1.1 \times 10^{-23}$

Hence,

$108s^5 = 1.1 \times 10^{-23}$

$s^5 = \frac{1.1 \times 10^{-23}}{108} \approx 1.02 \times 10^{-25}$

Now,

$s = (1.02 \times 10^{-25})^{1/5}$

Since $10^{-25/5} = 10^{-5}$ and $1.02^{1/5} \approx 1$, we get

$s \approx 1.0 \times 10^{-5}\ \text{mol L}^{-1}$

2. Use conductivity relation

For a very dilute solution,

$\Lambda_m^\circ \approx \Lambda_m = \frac{\kappa}{C}$

where:

• $\kappa = 3 \times 10^{-5}\ \text{S m}^{-1}$
• $C$ is concentration in mol m$^{-3}$

Convert solubility into mol m$^{-3}$:

$1.0 \times 10^{-5}\ \text{mol L}^{-1} = 1.0 \times 10^{-2}\ \text{mol m}^{-3}$

Thus,

$\Lambda_m^\circ = \frac{3 \times 10^{-5}}{1.0 \times 10^{-2}}$

$\Lambda_m^\circ = 3 \times 10^{-3}\ \text{S m}^2\text{ mol}^{-1}$

3. Find $x$

Given

$\Lambda_m^\circ = x \times 10^{-3}\ \text{S m}^2\text{ mol}^{-1}$

Comparing,

$x = 3$

4. Comparison with stored answer

Stored correct answer = $3$

This matches the derived answer.