Question

Ethylene glycol is a common automobile antifreeze. Calculate the freezing point and boiling point of a solution containing 343 g of ethylene glycol and 1055 g of water. (kb and ki of water are 0.52 C/m and 1.86 C/m respectively)

Answer 1

1)

ethylene glycol is C2H6O2

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)= 343 g

number of mol of C2H6O2,

n = mass of C2H6O2/molar mass of C2H6O2

=(343.0 g)/(62.068 g/mol)

= 5.526 mol

m(solvent)= 1055 g

= 1.055 Kg

Molality,

m = number of mol / mass of solvent in Kg

=(5.526 mol)/(1.055 Kg)

= 5.238 molal

lets now calculate ΔTf

ΔTf = Kf*m

= 1.86*5.238101

= 9.74 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 9.74

= -9.74 oC

2)

lets now calculate ΔTb

ΔTb = Kb*m

= 0.52*5.238101

= 2.72 oC

This is increase in boiling point

boiling point of pure liquid = 100.0 oC

So, new boiling point = 100 + 2.72

= 102.72 oC

boiling point is 102.7 oC

melting point is -9.74 oC