How many grams of estrogen, C18H24O2 (272.4 g/mol), must be dissolved in 205.0 grams of benzene to reduce the freezing point by 0.400°C ?

Question

The freezing point of benzene C₆H₆ is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is estrogen (estradiol).

How many grams of estrogen, C₁₈H₂₄O₂ (272.4 g/mol), must be dissolved in 205.0 grams of benzene to reduce the freezing point by 0.400°C ?

_____________ g estrogen

Answer 1

∆T = imK

∆T = change in freezing point = 0.400

i = van't Hoff factor = 1 for estrogen since it is a non electrolyte

m = molality = mols estrogen per kg water = ?

K = freezing constant for benzene = 5.12º/m

Solving for m, we have ...

m = ∆T / (i)(K) = 0.400 / (1)(5.5)

m = 0.0727 mol/kg

0.0727 mol / kg x 0.205 kg = 0.01490 mols

grams = 0.01490 mols x 272.4 g/mol = 4.06 g estrogen