Discussion Forum

 MX₂  dissociates into M²⁺ and X^-  ions in an aqueous solution, with a degree of dissociation ( $\alpha$) 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:

Explanation:

$MX_{2}\rightarrow M^{2+}+2X^{-}$

 van't Hoff factor for any salt can be calculated using equation

               $i=1+\alpha(n-1)$

 where ,n= number of constituent ions

 $\therefore$        $i(MX_{2})=1+\alpha(3-1)$

                                  $= 1+2\alpha$

$\frac{(\triangle T_{f})observed}{(\triangle T_{f})teoretical}=i=1+2\alpha$

   $\therefore$        i=1+2 x0.5

                       i=2