Question

An ethylene glycol solution contains 16.2 g of ethylene glycol (C2H6O2) in 87.4 mL of water.

Part A Calculate the freezing point of the solution. (Assume a density of 1.00 g/mL for water.) Part B Calculate the boiling point of the solution.

Answer 1

volume of water is 87.4 mL

since density is water 1 g/mL

So, mass of water = 87.4 g

A)

Lets calculate molality first

Molar mass of C2H6O2 = 2*MM(C) + 6*MM(H) + 2*MM(O)

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

= 62.068 g/mol

mass of C2H6O2 = 16.2 g

we have below equation to be used:

number of mol of C2H6O2,

n = mass of C2H6O2/molar mass of C2H6O2

=(16.2 g)/(62.068 g/mol)

= 0.261 mol

mass of solvent = 87.4 g

= 8.74*10^-2 kg [using conversion 1 Kg = 1000 g]

we have below equation to be used:

Molality,

m = number of mol / mass of solvent in Kg

=(0.261 mol)/(0.0874 Kg)

= 2.986 molal

lets now calculate deltaTf

deltaTf = Kf*m

= 1.86*2.9863

= 5.5545 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 5.5545

= -5.55 oC

Answer: -5.55 oC

B)

lets now calculate deltaTb

deltaTb = Kb*m

= 0.512*2.9863

= 1.529 oC

This is increase in boiling point

boiling point of pure liquid = 100.0 oC

So, new boiling point = 100 + 1.529

= 101.5 oC

Answer: 101.5 oC