Question

1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40K. The freezing point depression constant of benzene is 5.12 K kg/mol. Find the molar mass of the solute

Answer 1

∆Tf= Kf x weight of solute / ( molar mass of solute x weight of solvent in kg)
o.4 = 5.12 x 1x 1000/(Mx 50)
M = 256

Answer 2

Answer:

25600g

Explanation:

Freezing point depression constant of benzene = Kf = 5.12 K kg/mol (Given)

W1 = 50g (Given)

W2 = 100g (Given)

Freezing point of benzene = ΔTf = 0.40K (Given)

Thus,

ΔTf × Kf × W2 × 1000/ W1 × M2

M2 = Kf × W2 × 1000/ W1 × Tf

M2 = 5.12 × 100 × 1000/ 50 × 0.4

M2 = 25600g

Therefore, the molar mass of the solute is 25600g