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

The molar concentration of the gene fragment

Here, W_B = 8.95 mg = 8.95 x 10^-3 g, R = 0.0821 L atm Mol^-1 K^-1

T = 25⁰C = (25 + 273) K = 298 K. = 0.355torr =0.355/760 atm

V = 35 mL = 35 x 10^-3 L

Substituting these values in the above equation, we get