∆T = i * Kf * m Kf(cryoscopic constant of water) = -1.858°C·kg/mol i = 1 (for glycerin) m = molality of solute(glycerin) = nos of moles of glycerin / mass of solvent(water)[n/M] n = 30.7g/92.09g = 0.33337moles M is deduced from the density of water which is 1Kg/1000ml(1Kg/L) M = 376ml * (1Kg/1000ml) = 0.376Kg m is thus; 0.33337moles/0.376Kg = 0.8866mol/Kg => ∆T = 0.8866mol/Kg * 1 * -1.858°C·kg/mol ∆T = -1.647°C ~ -1.65°C.
