MX_{2} dissociates into M^{2+} and X^{-} ions | vedclass

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(B) The dissociation reaction is: $MX_{2} \rightleftharpoons M^{2+} + 2X^{-}$.Initially,let the concentration be $C$. At equilibrium,the concentration

Vedclass · Accepted Answer

$2$

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(B) The dissociation reaction is: $MX_{2} \rightleftharpoons M^{2+} + 2X^{-}$.Initially,let the concentration be $C$. At equilibrium,the concentrations are: $[MX_{2}] = C(1-\alpha)$,$[M^{2+}] = C\alpha$,and $[X^{-}] = 2C\alpha$.The total number of particles (van't Hoff factor $i$) is given by: $i = 1 - \alpha + n\alpha$,where $n$ is the number of ions produced per formula unit.For $MX_{2}$,$n = 3$ ($1$ $M^{2+}$ and $2$ $X^{-}$ ions).Given $\alpha = 0.5$,we have: $i = 1 + (3-1)\alpha = 1 + 2(0.5) = 1 + 1 = 2$.The ratio of the observed depression of freezing point to the theoretical depression (in the absence of dissociation) is equal to the van't Hoff factor $i$.Therefore,the ratio is $2$.

$MX_{2}$ dissociates into $M^{2+}$ 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

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