6 gram of non volatile, non electrolyte solute X is present in 1 litre and this solution is isotonic with 0.2 925 weight by volume percentage NaCl solution. Molar mass of X is

Dear Student,
Isotonic solution means the solution have the same osmotic pressure.
Osmotic pressure, 
.
mass of   NaCl = 0.2925 g
Molar mass of NaCl = 58 g mol^-1
Number of moles =0.2925/58
Osmotic pressure of NaCl solution,  $\mathrm{\pi }=\frac{0.2925\times 0.082\times \mathrm{T}}{58\times 0.1}$
For a non-volatile solution

Mass of Non-volatile solute = 6 g
Volume of solution = 1 L
Number of moles = 6 / M
where M is molar mass of non-volatile solute.
Osmotic pressure of non-volatile solution,$\mathrm{\pi }=\frac{6\times 0.082\times \mathrm{T}}{\mathrm{M}\times 1}$
Equating the osmotic pressures

  $$\begin{gathered} \frac{ 0.2925\times 0.082\times \mathrm{T}}{58\times 0.1}=\frac{6\times 0.082\times \mathrm{T}}{\mathrm{M}\times 1} \\ => M =\frac{6\times 58\times 0.1}{0.2925}=118.97 \end{gathered}$$

Molar mass of non-volatile solute, X = 118.97 g mol-1

Regards,