Question

The normal freezing point of a certain organic solvent is 5.50 degrees C. When 0.3003 g of naphthalene (C10H8) is dissolved in 9.9876 g of this solvent, the solution has a freezing point of 4.32 Degrees C. What is the freezing point depression constant (kf) in Degree C/m, for this solvent?

Answer 1

Given that

mass of naphthalene, w = 0.3003 g

molar mass of naphthalene M = 128 g/mol

mass of solvent W = 9.9876 g = 0.0099876 kg

Kf = ? ^oC/m

depression in freezing point dT =  5.50 ^oC - 4.32 ^oC = 1.18 ^oC

We know that

   depression in freezing point ΔT = Kf x molality

molality = (w/M) x (1/W in kg)

Then, depression in freezing point ΔT = Kf x { (w/M) x (1/W) }

1.18 ^oC = Kf [ (0.3003 g / 128 g/mol) x (1 x 0.0099876 kg)

kf = 5.02 ^oC/m

Therefore, freezing point depression constant (kf) =  5.02 ^oC/m