A solution containing $10 \mathrm{~g}$ of an electrolyte $\mathrm{AB}_2$ in $100 \mathrm{~g}$ of water boils at $100.52^{\circ} \mathrm{C}$. The degree of ionization of the electrolyte $(\alpha)$ is $\underline{\hspace{2cm}}$$\times 10^{-1}$. (nearest integer) [Given : Molar mass of $$\mathrm{AB}_2=200 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{~K}_{\mathrm{b}}$$(molal boiling point elevation const. of water) $=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, boiling point of water $=100^{\circ} \mathrm{C} ; \mathrm{AB}_2$ ionises as $$\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2 \mathrm{~B}^{-}]$$