A solution prepared by dissolving $8.95 \mathrm{mg}$ of a gene fragment in $35.0 \mathrm{mL}$ of water has an osmotic pressure of$0.335$ torr. at $25°\mathrm{C}$. Assuming that the gene fragment is a non-electrolyte, calculate its molar mass.

We know that osmotic pressure $\left(\mathrm{\pi }\right)$ is given as:

$\mathrm{\pi }=\frac{\text{ wt. of solute (in g) }}{\text{ Molar mass of solute (m) }}\times \frac{\mathrm{RT}}{\mathrm{V}}$                     …(1)

Change 0.335 torr into atm:

 $\mathrm{\pi }=\frac{0.335}{760}\mathrm{atm}$

Substituting these values in equation(1),we get:

$$\begin{gathered} \frac{0.335}{760} \mathrm{atm}=\frac{8.95}{1000}\mathrm{g}\times \frac{1}{\mathrm{m}}\times 0.0821\times 298 \mathrm{K}\left(273+25°\mathrm{C}\right)\times \frac{1000}{35} \mathrm{L} \\ \mathrm{m}=14193 \end{gathered}$$