A solution of a substance containing $1.05 \mathrm{g}$ per $100 \mathrm{mL}$ was found to be isotonic with $3\%$ glucose solution. The molecular mass of the substance is:

The degree of dissociation ($\mathrm{\alpha }$) can be calculated using the formula, $\mathrm{\alpha }=\frac{d_{\mathrm{t}}-{\mathrm{d}}_0}{(\mathrm{n}-1){\mathrm{d}}_0}$

This formula is applicable at

$0.004 \mathrm{M} {\mathrm{Na}}_2{\mathrm{SO}}_4$  is isotonic with $0.01 \mathrm{M}$ glucose. Degree of dissociation of ${\mathrm{Na}}_2{\mathrm{SO}}_4$ is-