Question

An ethylene glycol solution contains 27.6 g of ethylene glycol (C2H6O2) in 90.4 mL of water. (Assume a density of 1.00 g/mL for water.) Kb of water=0.51C/m and Kf of water=-1.86C/m

Part A: Determine the freezing point of the solution. Express your answer in degrees Celsius.

Part B: Determine the boiling point of the solution. Express your answer in degrees Celsius.

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)= 27.6 g

use:

number of mol of C2H6O2,

n = mass of C2H6O2/molar mass of C2H6O2

=(27.6 g)/(62.07 g/mol)

= 0.4447 mol

m(solvent)= 90.4 g

= 9.04*10^-2 kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(0.4447 mol)/(9.04*10^-2 Kg)

= 4.919 molal

A)

lets now calculate ΔTf

ΔTf = Kf*m

= 1.86*4.919

= 9.1493 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 9.1493

= -9.1493 oC

Answer: -9.15 oC

B)

lets now calculate ΔTb

ΔTb = Kb*m

= 0.512*4.919

= 2.5185 oC

This is increase in boiling point

boiling point of pure liquid = 100.0 oC

So, new boiling point = 100 + 2.5185

= 102.5185 oC

Answer: 102.5 oC