A 5% solution of cane sugar (mol. wt. = 342) is isotonic with 1% solution of substance X. The molecular weight of X is

Given 5% solution of Cane Sugar

5g of solute present in 100ml of solution  

For isotonic solutions , 

$$\begin{gathered} {\mathrm{\pi }}_1={\mathrm{\pi }}_2 \\ {\mathrm{C}}_1={\mathrm{C}}_2 \\ \frac{5\times 10}{342}=\frac{1\times 10}{{\mathrm{M}}_{\mathrm{x}}} \left(\because \mathrm{C}=\mathrm{M}=\frac{(\mathrm{w}/\mathrm{v})\%\times 10}{\mathrm{Mol}.\mathrm{wt}.}\right) \\ {\mathrm{M}}_{\mathrm{x}}=\frac{342}{5}=68.4\mathrm{g} \end{gathered}$$