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? View Available Hint(s) tert-Butyl alcohol is a solvent with a of 9.10 and a freezing point of 25.5 . When 0.807 of an unknown colorless nonelectrolyte liquid was dissolved in 11.6 of tert-butyl alcohol, the solution froze at 15.3 .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,

ΔTf is the depression in the freezing point of tert-Butyl alcohol (ΔTf = freezing point of pure solvent - freezing point in presence of unknown liquid = 25.5°C - 15.3°C = 10.2°C).

Kf is the molal freezing point constant of tert-Butyl alcohol (Kf = 9.1 °C/m).

m is the molality of unknown liquid.

∵ ΔTf = Kf.m

∴ m = ΔTf/Kf = (10.2°C)/(9.1 °C/m) = 1.121 m.

• We need to calculate the molar mass of the unknown liquid:

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

∴ m = (mass/molar mass) of unknown liquid/(mass of tert-Butyl alcohol (kg))

m = 1.121 m, mass of unknown liquid = 0.807 g, mass of tert-Butyl alcohol = 11.6 g = 0.0116 kg.

∴ molar mass of unknown liquid = (mass of unknown liquid)/(m)(mass of tert-Butyl alcohol (kg)) = (0.807 g)/(1.121 m)(0.0116 kg) = 62.06 g/mol.

• So, the unknown liquid is:

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