Question

Determine the freezing point depression of a solution that contains 30.7 g glycerin

Answer 1

 ∆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.