Question

Be sure to answer all parts.

How many liters of the antifreeze ethylene glycol [CH₂(OH)CH₂(OH)] would you add to a car radiator containing 6.50 L of water if the coldest winter temperature in your area is −21.0°C? Calculate the boiling point of this water-ethylene glycol mixture. (The density of ethylene glycol is 1.11 g/mL.)

What is the volume of antifreeze?
  L

What is the boiling point of the solution?
  °C

Answer 1

Using expression;.

dTf = i × Kf × m

Here, dTf = depression in freezing point = freezing point of pure solvent (water here) - freezing point of solution (we need that tha solution don't freeze upto -21°C)

So , dTf here = 0°C - (- 21°C) = 21°C

i = Vant's Hoff factor of solute , i= 1 for non-electrolytes.

Kf = molal depression in freezing point constant of solvent, kf = 1.86°C/m for water.

m = molality of solution = moles of solute/ mass of solvent in kg

Using all values to calculate (m) molality;

m = dtf/i×Kf = 21°C/1×1.86°C/m = 11.29m

From molality ; moles of solute = molality × mass of solvent in kg = 11.29m × 6.50kg = 73.385moles of ethylene glycol.

Mass of ethylene glycol required = moles × molar mass = 73.385mol × 62.07g/mol = 4555g = 4.56kg

Thus we need to add 4.56kg of ethylene glycol to 6.50L of water to make antifreeze solution which will freeze at -21°C .

Volume of ethylene glycol = mass/density = 4555g/1.11g/mL = 4103.6mL = 4.104L

Total volume of antifreeze = 6.50L water + 4.104L ethylene glycol = 10.604L

For boiling point;

dTb = i × kb × m

dTb = 1 × 0.512°C/m × 11.29m = 5.78°C

Boiling point of solution = dTb + boiling point of pure solvent = 5.78°C + 100°C = 105.78°C

Thus boiling point of this antifreeze solution is 105.78°C .