0.314 g of a certain nonelectrolytic solute is dissolved in 2.95*102 mL of H2O. The osmotic pressure of this solution is 2.49*102 torr at 21.0 °C. Calculate the molar mass of the solute. (760 Torr = 1 atm)

mm = 78.329 g/mol

Explanation:

osmotic pressure (π):

π = Cb*R*T

∴ a: water (solvent)

∴ b: solute nonelectrolytic

∴ Cb [=] mol/L

∴ π = (2.49E2 torr) * (atm/760 torr) = 0.3276 atm

∴ wb = 0.314 g

∴ T = 21.0°C = 294 K

∴ R = 0.082 atm. L/K. mol

∴ molar mass [=] g/mol

⇒ Cb = π/RT

⇒ Cb = (0.3276 atm) / [ (0.082 atm. L/K. mol) (294 K) ]

⇒ Cb = 0.0136 mol/L

⇒ moles solute (nb) = (0.0136 mol/L) * (2.95E2 mL) * (L/1000 mL)

⇒ nb = 4.01 E-3 mol

∴ molar mass (mm):

⇒ mm = (0.314 g) / (4.01 E-3 mol)

⇒ mm = 78.329 g/mol