An aqueous solution contains 1.00 g/L of a derivative of the detergent laurel alcohol. The osmotic pressure of this solution is 17.8 torr, at 25°C. What is the molar mass (in g/mol) of this detergent?

mm = 1043.33 g/mol

Explanation:

osmotic pressure (π):

π = CbRT

∴ π = 17.8 torr = 0.0234 atm

∴ Cb: solute concentration

∴ T = 25°C = 298 K

∴ R = 0.082 atm. L/K. mol

⇒ Cb = π/RT

⇒ Cb = (0.0234 atm) / ((0.082 atm. L/K. mol) (298 K))

⇒ Cb = 9.585 E-4 mol/L

molar mass (mm):

⇒ mm = (1.00 g/L) (L/9.585 E-4 mol)

⇒ mm = 1043.33 g/mol