Question

You are given that a 163.5 grams of water and an unknown amount of ethylene glycol (C2H6O2) were mixed together. The only information given is that the mole fraction of ethylene glycol is .098. Calculate the amount of grams of ethylene glycol as well as the new boiling point and freezing point.

Answer 1

Mole fraction of water = 1 - 0.098 = 0.902

163.5 grams of water = 163.5 / 18.016 = 9.075 moles

9.075 / n = 0.902

where n is total number of moles

n = 10.06 moles

number of moles of glycol = 10.06 - 9.075 = 0.985 moles

Molar mass of ethylene glycol is 62.07 g/mol

0.985 moles = 0.985 x 62.07 = 61.14 grams

Elevation in boiling point = (Kb x 1000 x w2) / (M2 x w1)

where kb is boiling point constant = 0.52^oC kg/ mol

w2 is the amount of solute

M2 is the molar mass of solute

and w1 is the amount of solvent

So putting all the values we get ,

Delta Tb = ( 0.52 x 1000 x 61.14) / (62.07 x 163.5)

Delta Tb = 3.13^oC

so the boiling point of the solution will be 100 + 3.13=

= 103.13^oC

Depression in freezing point (Delta Tf) = (kf x 1000 x w2) / (M2 x w1)

kf for water = 1.86^oC kg/mol

Delta Tf = ( 1.86 x 1000 x 61.14) / (62.07 x 163.5)

Delta Tf = 11.2^oC

So the freezing point of the solution will be = 0 - 11.2 =

= -11.2^oC