Question

tert-Butyl alcohol is a solvent with a Kf of 9.10 ∘C/m and a freezing point of 25.5 ∘C. When 0.807 g of an unknown colorless nonelectrolyte liquid was dissolved in 11.6 g of tert-butyl alcohol, the solution froze at 15.3 ∘C. Which of the following is most likely the identity of this unknown liquid? ethylene glycol (molar mass = 62.07 g/mol) 1-octanol (molar mass = 130.22 g/mol) glycerol (molar mass = 92.09 g/mol) 2-pentanone (molar mass = 86.13 g/mol) 1-butanol (molar mass = 74.12 g/mol)

Answer 1

Answer:

ethylene glycol (molar mass = 62.07 g/mol).

Explanation:

We can solve this problem using the relation:

ΔTf = (Kf)(m),

where, ΔTf is the depression in the freezing point (ΔTf = 25.5°C - 15.3°C  = 10.2 °C).

Kf is the molal freezing point depression constant of water = 9.1 °C/m,

m is the molality of the solution (m = ??? m).

∴ m = ΔTf/(Kf) = (10.2 °C)/(9.1 °C/m) = 1.12 m.

• Also, we have that:

Molality (m) is the no. of moles of solute per 1.0 kg of the solute.

m = (mass/molar mass) of the solute x (1000/mass of the solvent).

∴ 1.12 m = (0.807 g / molar mass of the solute) x (1000 / 11.6 g)

∴ molar mass of the solute = (0.807 g)(1000)/(1.12 m)(11.6 g) = 62.11 g/mol.

so, the unknown liquid is ethylene glycol (molar mass = 62.07 g/mol).