Question

2. a. What is the molality used of 2.00 g ethylene glycol in 20.0 g TBA?

b. Predict the freezing point of this solution

Answer 1

Answer – a) We are given, mass of ethylene glycol = 2.0 g ,

Mass of t-butyl alcohol(TBA) = 20.0 g

First we need to calculate the moles of ethylene glycol

moles of ethylene glycol = 2.0 g / 62.07 g.mol^-1

                                        = 0.0322 moles

We know molality = moles / kg of solvent

Molality of ethylene glycol = 0.0322 moles / 0.020 kg

                                             = 1.61 m

b) Freezing point of the solution –

Freezing point of the pure solvent TBA = 25^oC, Kf for TBA = 9.1^cC/m

We know the formula,

∆Tf = Kf *m

      = 9.1^cC/m * 1.61 m

      = 14.66^oC

So, freezing point of solution = Freezing point of the pure solvent - ∆Tf

                                                = 25^oC – 14.66^oC

                                                = 10.34^oC