(Mass of BHT=8500g) (Mass of cetyl alcohol=0.724g) (Freezing point of BHT=74.01 °C) (Freezing point of BHT/cetyl alcohol solution=71.61 °C)

Question

Use this data to determine 𝐾f for BHT (butylated hydroxytoluene).

𝐾f=°C/𝑚?

Answer 1

Solute = cetyl alcohol (present in smaller amount)

Solvent = BHT (present in larger amount)

molar mass ceyl alcohol (C₁₆H₃₄O) = 242 g / mole

∆T = imK

∆T = change in freezing point = 74.01º - 71.61º = 2.4º

i = van't Hoff factor = 1 for cetyl alcohol because it is a non electrolyte

m = molality = mols cetyl alcohol / kg BHT (see below for calculation of m)

K = freezing constant for BHT

m = mols cetyl alcohol / kg BHT

mols cetyl alcohol = 0.724 g x 1 mol / 242 g = 0.00299 mols

kg BHT = 8500 g BHT x 1 kg / 1000 g = 8.5 kg

m = 0.00299 mols / 8.5 kg

m = 0.000352

∆T = imK

K = ∆T / (i)(m)

K = 2.4º / (1)(0.000352 m)

K = 6818ºC/m