Question

the freezing point of camphor is 178.4 degree celsius and its modal freesing point depression constant is 37.7. how many grams of naphthalene , a solute , not the solvent, in this question should be mixed with 12g of camphor to lower the freezing point to 170

Answer 1

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent.

T_f = T_f(solvent) - T_f(solution) = K_f.m

Where, m = molaility of solute

Kf = Molal freezing point depression constant

T_f = Freezing point depression

m = T_f / K_f

m = (178.4 - 170) / 37.7 = 0.2228 moles / Kg

For 12g of camphor, moles of naphthalene required = (12/1000) x 0.2228 = 0.00267 moles

Mass of naphthalene = 0.00267 x 128.17 = 0.3422 g

(Molar mass of naphthalene = 128.17 g/mole)

Hence 0.3422 g of naphthalene is needed to be added to camphor to decrease its freezing point to 170^oC