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

Given that the answer is uncertain with a glucose solution, is isotonic.
So, there are osmotic pressures. same pressure exists.

$\pi =\mathrm{CRT}=\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2\mathrm{V}}\times \mathrm{RT}$

$\begin{array}{l}{\left(\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2\mathrm{V}}\times \mathrm{RT}\right)}_{\text{unknown }}={\left(\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2\mathrm{V}}\times \mathrm{RT}\right)}_{\text{glucose }}\\ {\left(\frac{1.05}{{\mathrm{M}}_2\times 0.1}\times \mathrm{RT}\right)}_{\text{unknown }}={\left(\frac{3}{180\times 0.1}\times \mathrm{RT}\right)}_{\text{glucose }}\\ {\mathrm{M}}_2=\frac{180\times 0.1\times 1.05}{0.1\times 3}=63\end{array}$

Unknown solute's molecular weight is 63 g.