∆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.
