A solution containing 8.6 g L^–1 of urea is isotonic with a 5% solution of unknown solute. The molar mass of the solute will be

$$\begin{gathered} \text{ Urea }\to 8.6\mathrm{g}/\mathrm{L}=\text{ i.e., }\left(\frac{\mathrm{w}}{\text{ volume }}\right) \\ \text{ Molar mass }={\mathrm{NH}}_2{\mathrm{CONH}}_2\Rightarrow 14\times 2+4+12+16\Rightarrow 60 \end{gathered}$$
$\begin{array}{l}\text{ isotonic }\to \left[{\mathrm{\pi }}_1={\mathrm{\pi }}_2\right]\\ {\mathrm{C}}_1\mathrm{ST}={\mathrm{C}}_2\mathrm{ST}\\ {\mathrm{C}}_1={\mathrm{C}}_2\\ \frac{\mathrm{W}}{{\mathrm{M}}_{\mathrm{w}}\times \mathrm{V}}=\frac{\mathrm{W}}{{\mathrm{M}}_{\mathrm{w}}\times \mathrm{V}}\\ \frac{8.6}{60}\Rightarrow \frac{5\times 1000}{{\mathrm{M}}_{\mathrm{w}}\times 100}\\ {\mathrm{M}}_{\mathrm{w}}=\frac{5\times 60\times 10}{8.6}=348.8\mathrm{g}/\mathrm{mol}\end{array}$
Errata : 5% solution of unknown solute but it must be written 5% weight/volume means 5 g of solute present in 100 mL of solution