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(B) The van't Hoff factor $(i)$ is calculated as: $i = \frac{	ext{Calculated molar mass}}{	ext{Observed molar mass}} = \frac{170}{92.64} = 1.835$. F

Vedclass · Accepted Answer

$83.5$

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(B) The van't Hoff factor $(i)$ is calculated as: $i = \frac{	ext{Calculated molar mass}}{	ext{Observed molar mass}} = \frac{170}{92.64} = 1.835$. For the dissociation of silver nitrate $(AgNO_3 ightarrow Ag^+ + NO_3^-)$,the relation between $i$ and the degree of dissociation $(\alpha)$ is given by: $i = 1 + (n-1)\alpha$,where $n = 2$. Substituting the values: $1.835 = 1 + (2-1)\alpha$. $\alpha = 1.835 - 1 = 0.835$. Therefore,the degree of dissociation in percentage is $0.835 	imes 100 = 83.5 \%$.

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

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