A 5% solution of can sugar (molar mass = 342) is isotonic with 1% of a solution of an unknown solute. The molar mass of unknown solute in g/mol is 

Two solutions are said to be isotonic, they have the same molar concentration. Therefore, 

Concentration of cane sugar = Concentration of an unknown solute [X] 

${\mathrm{\pi }}_1={\mathrm{\pi }}_2$

$$\begin{gathered} {\mathrm{C}}_1\mathrm{ST}={\mathrm{C}}_2\mathrm{ST} \\ {\mathrm{C}}_1={\mathrm{C}}_2 \end{gathered}$$

$$\begin{gathered} {\mathrm{m}}_2=\frac{ 342}{5} \\ {\mathrm{m}}_2=68.4\mathrm{gr} \end{gathered}$$

$\frac{{\mathrm{w}}_1}{{\mathrm{m}}_1}\times \frac{1000}{{\mathrm{V}}_{1\left(\mathrm{In} \mathrm{ml}\right)}}=\frac{{\mathrm{w}}_2}{{\mathrm{m}}_2}\times \frac{1000}{{\mathrm{V}}_{2\left(\mathrm{In} \mathrm{ml}\right)}}$

$\frac{5}{342}\times \frac{1000}{100}=\frac{1}{{\mathrm{m}}_2}\times \frac{1000}{100}$

${\mathrm{m}}_2=\frac{342}{5}$

${\mathrm{m}}_2=68.4\mathrm{gr}$