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Question

(B) The dissociation reaction is $AB_2 \rightarrow A^{2+} + 2B^-$. The van't Hoff factor $i$ is given by $i = 1 + (n-1)\alpha$,where $n=3$. So,$i = 1

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) The dissociation reaction is $AB_2 ightarrow A^{2+} + 2B^-$. The van't Hoff factor $i$ is given by $i = 1 + (n-1)\alpha$,where $n=3$. So,$i = 1 + 2\alpha$.The boiling point elevation is $\Delta T_b = T_b - T_b^{\circ} = 100.52^{\circ} C - 100^{\circ} C = 0.52 \ K$.The molality $m$ is calculated as $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{10/200}{100/1000} = \frac{0.05}{0.1} = 0.5 \ mol \ kg^{-1}$.Using the formula $\Delta T_b = i \cdot K_b \cdot m$:$0.52 = (1 + 2\alpha) 	imes 0.52 	imes 0.5$$1 = (1 + 2\alpha) 	imes 0.5$$2 = 1 + 2\alpha$$2\alpha = 1$$\alpha = 0.5 = 5 	imes 10^{-1}$.Thus,the value is $5$.

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