Question

$M{X}_2$ dissociates into $M^{2+}$ and $X^{-}$ ions in an aqueous solution, having a degree of dissociation $\left(\alpha \right)$ of 0.5. The ratio of the observed depression in freezing point of the aqueous solution to the value of the depression in freezing point in the absence of ionic dissociation will be

Answer 1

$\alpha$ = 0.5
$M{X}_2\rightleftharpoons M^{+2}+2X^{-}$

i = $\frac{observedcolligativeproperty}{normalcolligativeproperty}$

$\alpha$ = $\frac{i-1}{n-1}$

0.5 = $\frac{i-1}{3-1}$
i = 2
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