For a concentrated solution of a weak electrolyte ($\mathrm{K}_{\text {eq }}=$ equilibrium constant) $\mathrm{A}_{2} \mathrm{B}_{3}$ of concentration '$c$', the degree of dissociation '$\alpha$' is :

Question

$\left(\frac{K_{e q}}{25 c^{2}}\right)^{\frac{1}{5}}$

$\left(\frac{K_{e q}}{108 c^{4}}\right)^{\frac{1}{5}}$

$\left(\frac{K_{e q}}{5 c^{4}}\right)^{\frac{1}{5}}$

$\left(\frac{K_{e q}}{6 c^{5}}\right)^{\frac{1}{5}}$

Solution

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