The molar mass of a solute ‘X’ in g $mol{e}^{-1}$  , if its 1% solution is isotonic with a 5% solution of cane sugar ( molar mass of sugar = 342g $mol{e}^{-1}$  ) is 

Isotonic solutions have same osmotic pressure 
 $$\begin{gathered} {\text{π}}_{\mathrm{X}}={\mathrm{\pi }}_{{\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}} \\ {\left[{\mathrm{Con}}^{\mathrm{n}}\right]}_{\mathrm{X}}\mathrm{RT}={\left[{\mathrm{Conc}}^{\mathrm{n}}\right]}_{{\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}}\mathrm{RT} \\ [\frac{\mathrm{W}}{\mathrm{GMW}}\times \frac{1000}{{\mathrm{V}}_{\left(\mathrm{ml}\right)}}]=[\frac{\mathrm{W}}{\mathrm{GMW}}\times \frac{1000}{{\mathrm{V}}_{\left(\mathrm{ml}\right)}}] \\ {[\frac{1}{\mathrm{GMW}}\times \frac{1000}{100}]}_{\mathrm{X}}=\frac{5}{342}\times \frac{1000}{100} \\ {\mathrm{GMW}}_{\mathrm{x}}=\frac{1000}{100}\times \frac{342}{5}\times \frac{100}{1000} =68.4\mathrm{g}. {\mathrm{mole}}^{-1} \end{gathered}$$