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(B) The degree of dissociation $(\alpha)$ is related to the van't Hoff factor $(i)$ by the formula: $\alpha = \frac{i - 1}{n - 1}$.For $AgNO_3$,the di

Vedclass · Accepted Answer

$83.5$

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(B) The degree of dissociation $(\alpha)$ is related to the van't Hoff factor $(i)$ by the formula: $\alpha = \frac{i - 1}{n - 1}$.For $AgNO_3$,the dissociation reaction is $AgNO_3 ightarrow Ag^+ + NO_3^-$,so $n = 2$.The van't Hoff factor $(i)$ is the ratio of the calculated molecular mass to the observed molecular mass: $i = \frac{	ext{Calculated mass}}{	ext{Observed mass}} = \frac{170}{92.64} \approx 1.835$.Substituting the values into the formula: $\alpha = \frac{1.835 - 1}{2 - 1} = 0.835$.To express this as a percentage: $\alpha 	imes 100 = 0.835 	imes 100 = 83.5 \%$.

The observed and calculated molecular mass of silver nitrate $(AgNO_3)$ are $92.64$ and $170$ respectively. The degree of dissociation of silver nitrate is ........ $\%$.

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