Question

Ethylene glycol (EG), CH2(OH)CH2(OH), is a common automobile antifreeze. It is water soluble and fairly nonvolatile (b.p. 197°C). Calculate the boiling point and freezing point of a solution containing 475.5 g of ethylene glycol in 3503 g water. The molar mass of ethylene glycol is 62.07 g/mol. Freezing point: Boiling point: °C °C

Answer 1

1) Boiling point

Boiling point elevation (∆T_b) is calculated by the following formula

∆T_b = K_b × b × i

where,

K_b = boiling point elevation constant, for water it is 0.512℃/m

b = molality of solute

i = Van't Hoff factor of solute , for nonelectrolyte it is 1

given moles of ethelene glycol = 475.5g/62.07g/mol = 7.661mol

Molality is defined as number of moles of solute per kilogram of solvent

molality of ethelene glycol = (7.661mol/3503g)×1000g = 2.187m

∆T_b = 0.512℃/m × 2.187m × 1

∆T_b = 1.12℃

Boiling point of the solution = poiling point of solvent + ∆T_b

_{= 100℃ + 1.12℃}

_{​​​​​​= 101.12℃}

2) Freezing point

Freezing point of the solution(∆T_f) is calculated by the following formula

∆T_f = K_f × b × i

K_f = freezing point depression constant , for water it is 1.86℃/m

b = molality of solute

i = Van't Hoff factor of solute , 1

∆T_f = 1.86℃/m × 2.187m × 1

∆T_f = 4.07℃

Freezing point of the solution = freezing point of solvent - ∆T_f

= 0℃ - 4.07℃

  = - 4.07℃