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(A) The van't Hoff factor $i$ is given by the ratio of normal molar mass to observed molar mass:$i = \frac{	ext{Normal molar mass}}{	ext{Observed mo

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(A) The van't Hoff factor $i$ is given by the ratio of normal molar mass to observed molar mass:$i = \frac{	ext{Normal molar mass}}{	ext{Observed molar mass}} = \frac{164}{65.6} = 2.5$For the dissociation of $MX_2 ightarrow M^{2+} + 2X^-$,the number of ions produced per formula unit is $n = 3$.The relationship between the van't Hoff factor $i$ and the degree of dissociation $\alpha$ is:$i = 1 + \alpha(n - 1)$Substituting the values:$2.5 = 1 + \alpha(3 - 1)$$2.5 = 1 + 2\alpha$$1.5 = 2\alpha$$\alpha = 0.75$Percentage of ionization $= \alpha 	imes 100 = 0.75 	imes 100 = 75 \%$.Thus,the correct option is $A$.

$A$ compound $MX_2$ has observed and normal molecular masses $65.6$ and $164$ respectively. Calculate the percentage of ionization of $MX_2$ ............ $\%$

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