Question

A radiator of motor vehicle is filled with 8L of water containing 2L of ethylene glycol (density = 1.12 g ml^-1) .What is the lowest temperature at which the vehicle can be parked without the danger of getting water to freeze in the radiator. [K_f for water is 1.86 Km^-1]
1. -8.4°C
2. -23.2°C
3. -11.6°C
4. -4.2°C

Answer 1

Correct Answer - Option 1 : -8.4°C

Concept:

• When we add a solute to a pure solvent, the solution has a change of vapor pressure.
• The change in vapor pressure results in depression of freezing point, ie the freezing point of the solution gets lowered as compared to the pure solvent.
• The depression of the freezing point is given by:

\(T{^\circ _f} - {T_f} = {K_f} × m × i\) where T_f = Freezing Point of the solution, T_f⁰, m is the molality of the solution and i is the vant Hoff factor, Kf = molal depression constant 

The molality is given by:

\({\rm{\;}}m = {\rm{\;molality\;}} = \frac{{{w_{{\rm{solute\;}}}} × 1000}}{{{M_{{\rm{solute\;}}}} × {w_{{\rm{solvent\;}}\left( {{\rm{in\;g}}} \right)}}}}\)where M = molar mass of solute, w_solute = weight of solute added w_solvent = mass of solvent

Calculation:

Here, ethylene glycol is the solvent and water is the solvent.

The mass of 8L of water is 1000 × 8 = 8000 g = wsolvent 

Mass of 2L of ethylene glycol is 1.12 × 2000 = 2240 g = wsolute

M = molar mass of solute = 62g

Kf for water is 1.86 Km-1

\({\rm{\;}}m = {\rm{\;molality\;}} = \frac{{{w_{{\rm{solute\;}}}} × 1000}}{{{M_{{\rm{solute\;}}}} × {w_{{\rm{solvent\;}}\left( {{\rm{in\;g}}} \right)}}}}\)

\(m = \frac{{2240 \times 1000}}{{62 \times 8000}} = 4.5\)

\(\Delta {T_{f}} = {K_f} \times {m} \)

\(\Delta {T_{f}} = 4.5\times 1.86=8.4\)

Hence, on adding the solute, the solvent will have depression of freezing by 8.4^o C.

The freezing point of water is 0^oC.

The new freezing point is thus 0 - 8.4 = -8.4⁰C.

Hence, the lowest temperature at which the vehicle can be parked without the danger of getting water to freeze in the radiator is -8.40C.