Question

The freezing point of a solution that contains 1.00 g of an unknown compound, (A), dissolved in 10.0 g of benzene is found to be 2.17 oC. The freezing point of pure benzene is 5.48 oC. The molal freezing point depression constant of benzene is 5.12 oC/molal. What is the molecular weight of the unknown compound?

Answer 1

Answer:

Depression in freezing point, ∆Tf=Kf x m

∆Tf=Tf (pure solvent) -Tf (solution)

Where Tf (pure solvent)=freezing point of pure solvent,

Tf (solution)=freezing point of solution

Kf=freezing point constant of solvent

m=molality=(mass)/(molar mass) x (mass of solvent)

Given solvent is benzene.

Tf (pure solvent)=5.48°C

Tf (solution)=2.17°C

Kf=5.12 °C/molal=5.12 °C x Kg/mol (since 1 molal=mol/Kg)

Mass of compound A=1 g and mass of solvent=10 g=0.01 Kg.

(Since 1 Kg=1000 g)

Therefore

Tf(pure solvent) - Tf(solvent)=Kf x (mass)/[(molar mass) x (mass of solvent)]

(5.48 °C - 2.17 °C)=(5.12 °C x Kg/mol) x (1 g)/[(molar mass) x 0.01 Kg)]

Molar mass=[(5.12 °C x Kg/mol) x (1 g)]/[(0.01 Kg x 3.31 °C)]

Molar mass=154.683 g/mol.

Therefore the molecular weight of unknown compound is 154.683 g/mol.

Please let me know if you have any doubt. Thanks