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):

∴ 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

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