The values of observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate is

The quantities provided in the question are:

Observed molecular mass of silver nitrate = 92.64

The theoretical molecular mass of silver nitrate = 170

n = no. of ions dissociated in silver nitrate ${\mathrm{AgNO}}_3$= 2  (${\mathrm{Ag}}^{+}, {{\mathrm{NO}}_3}^{-}$)
Van't Hoff factor, $\mathrm{i}=\frac{\mathrm{Theoretical} \mathrm{molecular} \mathrm{mass}}{\mathrm{Observed} \mathrm{molecular} \mathrm{mass}}=\frac{170}{92.64}=1.835$

Percent of dissociation of silver nitrate, $\mathrm{\alpha }=\frac{\mathrm{i}-1}{\mathrm{n}-1}=\frac{1.835-1}{2-1}=0.835$

$\mathrm{\alpha }=0.835 \mathrm{or} 83.5\%$

Hence, the required answer is B) 83.5%