Question

Butylated hydroxytoluene (BHT) is used as an antioxidant in processed foods. A solution of 2.500 g of BHT in 100.0 g of benzene had a freezing point of Tf=4.880 Celcius. What is the molar mass of BHT?

The normal freezing point of benzene is 5.5 Celcius and Kf (benzene)=5.12 C/m

Answer 1

Step 1: Calculate the freezing point depression of the solution.

T_f = (Freezing point of pure solvent) - (Freezing point of solution)
(5.5 ^oC) - (4.88 ^oC) = 0.62 ^oC

Step 2 : Calculate the molal concentration of the solution using the freezing point depression.

T_f = (K_f) (m)
m = (0.62^oC) / (5.12 ^oC/molal)
m = 0.12109 molal

Step 3: Calculate the molecular weight of the unknown using the molal concentration.

m = 0.12109 molal = 0.12109 mol BHT / kg benzene

moles BHT = (0.12109 mol A / kg benzene) (0.100 kg benzene) = 1.2109 X 10^-2mol BHT

molecular weight of BHT = (2.5 g BHT) / 1.2109 X 10^-2mol A)

molecular weigh of BHT = 206.45 g/mol