1. Write the dissociation equilibrium

For the weak electrolyte $A_2B_3 \rightleftharpoons 2A^{3+} + 3B^{2-}$

Let the initial concentration of $A_2B_3$ be $c$ and degree of dissociation be $\alpha$.

Then at equilibrium:

• $[A_2B_3] = c(1-\alpha)$
• $[A^{3+}] = 2c\alpha$
• $[B^{2-}] = 3c\alpha$

2. Write the equilibrium constant expression

By law of mass action, $K_{eq} = \frac{[A^{3+}]^2[B^{2-}]^3}{[A_2B_3]}$

Substitute the equilibrium concentrations: $K_{eq} = \frac{(2c\alpha)^2(3c\alpha)^3}{c(1-\alpha)}$

3. Simplify

First compute the numerical factor: $2^2 \cdot 3^3 = 4 \cdot 27 = 108$

And powers of $c$ and $\alpha$: $K_{eq} = \frac{108\,c^5\alpha^5}{c(1-\alpha)} = \frac{108\,c^4\alpha^5}{1-\alpha}$

4. Use the condition for a concentrated solution of a weak electrolyte

Since it is a weak electrolyte, dissociation is small: $\alpha \ll 1$ So, $1-\alpha \approx 1$

Hence, $K_{eq} \approx 108\,c^4\alpha^5$

5. Solve for $\alpha$

$\alpha^5 = \frac{K_{eq}}{108c^4}$

Therefore, $\alpha = \left(\frac{K_{eq}}{108c^4}\right)^{1/5}$

6. Compare with options

This matches Option B.

Final Answer: $\boxed{\alpha = \left(\frac{K_{eq}}{108c^4}\right)^{1/5}}$ So the correct option is B.