Answer : $\large\frac{K_b(K_bm -\Delta T_b)}{(2 \Delta T_b- K_bm)^2}$

$2A \rightleftharpoons A_2$

$(1- \alpha)( \alpha/2) $;

Total amount $=(1-\alpha/2)$

The Van't Hoff factor is $i =\large\frac{Total\;amount\;in\;solution}{Amount\;of\;A\;to\;start\;with}$

$\qquad= \large\frac{(1-\alpha/2)}{1}$

$\qquad= 1- \alpha/2$

$\alpha =2 (1-i)$

Hence $K_{eq} =\large\frac{m(\alpha)}{m^2(1- \alpha)^2}$

$\qquad= \large\frac{(1-i)}{m(2i -1)^2}$

Where $i = \large\frac{\Delta T_b }{(\Delta T_b)_0}$

$\qquad= \large\frac{\Delta T_b}{K_b m}$

Thus $K_{eq} =\large\frac{K_b(K_bm -\Delta T_b)}{(2 \Delta T_b- K_bm)^2}$