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Q.
Boiling point of $0.01 M AB _{2}$ which is $10 \%$ dissociated in aqueous medium $\left( K _{ b _{ H_2O }}=0.52\right)$ as $A ^{+2}$ and $B ^{-}$
Solutions
Solution:
$\Delta T _{ b }= i \left( K _{ b } \times m \right)$
$T _{ b }- T _{ b }^{\circ}= i \left( K _{ b } \times m \right)$
$i =\frac{1-\alpha+\alpha+2 \alpha}{1}=1+2 \alpha $
$T _{ b }-373=1.2[0.52 \times 0.01]$
$=1+\frac{2 \times 10}{100}$
$T_{b}-373=0.00624$
$i =\Rightarrow 1.2 $
$T _{ b }=373.006\, K$
