If the observed and theoretical molecular mass of NaCl is found to be 49.1 and 58.50, then the percent of dissociation of NaCl is

Given:

Observed molecular mass of NaCl = 49.1

Theoretical molecular mass of NaCl 58.50

n = no. of ions dissociated in NaCl = 2  (${\mathrm{Na}}^{+}, {\mathrm{Cl}}^{-}$)
 

Van't Hoff factor, $\mathrm{i}=\frac{\mathrm{Theoretical} \mathrm{molecular} \mathrm{mass}}{\mathrm{Observed} \mathrm{molecular} \mathrm{mass}}=\frac{58.50}{49.1}=1.19$

Percent of dissociation of NaCl , $\mathrm{\alpha }=\frac{\mathrm{i}-1}{\mathrm{n}-1}=\frac{1.19-1}{2-1}=0.19$

                                                           $\Rightarrow \mathrm{\alpha }=0.19 \mathrm{or} 19\%$

The percent of dissociation of NaCl is 19%.