Question

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

Answer 1

$\alpha =0.5$
$\underset{(1-\alpha )}{M{X}_2}\underset{\alpha }{M^{2+}}+\underset{2\alpha }{2X^{-}}$
$i=\frac{1-\alpha +\alpha +2\alpha }{1}$
$i=1+2\alpha$
$i=1+2\times 0.5=2$