4% solution of a non volatile solute is isotonic with 0.702% urea solution. If molar mass of urea is 60.What is the molar mass of the non volatile solute.

Isotonic solution means the solution have same osmotic pressure.
Osmotic pressure, $\pi =\frac{nRT}{V}$
0.702 % urea solution means 0.702 g urea is dissolved in 100 mL.
mass of urea = 0.702 g
Molar mass of urea = 60 g mol^-1
Number of moles = 0.702 / 60
Osmotic pressure of urea solution, $\pi = \frac{0.702\times 0.082\times T}{60\times 0.1}$
For a non-volatile solution
4% solution means 4 g non-volatile solute is dissolved in 100 mL or 0.1 L.
Mass of Non-volatile solute = 4 g
Volume of solution = 0.1 L
Number of moles = 4 / M
where M is molar mass of non-volatile solute.
Osmotic pressure of non-volatile solution,$\pi =\frac{4\times 0.082\times T}{M\times 0.1}$
Equating the osmotic pressures

$$\begin{gathered} \frac{0.702\times 0.082\times T}{60\times 0.1}=\frac{4\times 0.082\times T}{M\times 0.1} \\ M = \frac{4\times 60}{0.702} \\ M = 341.88 g mo{l}^{-1} \end{gathered}$$

Molar mass of non-volatile solute = 341.88 g mol^-1