Question

(2) (4 points) Ethylene glycol (C,HO2) is a molecular compound that is used in many commercial anti-freezes. A water solution of ethylene glycol is used in vehicle radiators to lower its freezing point and thus prevent the water in the radiator from freezing. Calculate the freezing point of a solution of 350.0 g of ethylene glycol in 900.0 g of Kxkg water. Rationalize your answer. Ky(water)= 1.86

Answer 1

Lets calculate molality first

Molar mass of C2H6O2,

MM = 2*MM(C) + 6*MM(H) + 2*MM(O)

= 2*12.01 + 6*1.008 + 2*16.0

= 62.068 g/mol

mass(C2H6O2)= 350.0 g

use:

number of mol of C2H6O2,

n = mass of C2H6O2/molar mass of C2H6O2

=(3.5*10^2 g)/(62.07 g/mol)

= 5.639 mol

m(solvent)= 900.0 g

= 0.9 kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(5.639 mol)/(0.9 Kg)

= 6.266 molal

lets now calculate ΔTf

ΔTf = Kf*m

= 1.86*6.2655

= 11.6539 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 11.6539

= -11.6539 oC

Answer: -11.65 oC