Question

The best mixture of antifreeze and water is 50% antifreeze, 50% water. The cooling system in your car has a mixture of 6.00L water and 6.00 L ethylene glycol (antifreeze). The molality of the solution is 17.9m. The chemical formula of antifreeze is C2H6O2 and its density is 1.1132 g/cm3.

If the summer temperatures rise and the coolant reaches a temperature of 108°C, will it boil?



No, it would boil at 109.13°C


No, it would boil at 123.09°C


No, it would boil at 119.13°C


Yes, it would boil at 99.0°C


Yes, it would boil at 100.13°C

Answer 1

The coolant mixture will not boil at 108°C. The calculated boiling point elevation suggests it will boil at approximately 109.16°C, based on the molality and the boiling point elevation constant for water.

Step-by-step explanation:

To determine whether the coolant mixture in your car will boil at 108°C, we can use the concept of boiling point elevation. The boiling point of a solution increases when a solute is added to a solvent due to the colligative properties of the solution. Using the molality provided (17.9m), and knowing the boiling point elevation constant (Kb) for water is approximately 0.512 °C/m, we can calculate the boiling point elevation.

ΔTb = i * Kb * m

Where ΔTb is the boiling point elevation, i is the van't Hoff factor (i = 1 for ethylene glycol as it does not dissociate in solution), Kb is the ebullioscopic constant for water, and m is the molality of the solution.

ΔTb = 1 * 0.512 °C/m * 17.9m = 9.16 °C

The normal boiling point of water is 100°C, so adding the elevation to the normal boiling point gives us:

100°C + 9.16°C = 109.16°C

Therefore, the solution will not boil at 108°C because it would boil at approximately 109.16°C.

Answer 2

The correct answer is No, it would boil at 109.13°C

Step-by-step explanation:

This question can be solved by knowing the boiling point elevation formula and the fact that ethylene glycol dissolves in water without dissociation

The boiling point elevation formula is given by

ΔT = i ×
`K_(b)` ×
`m_(solute)`

Where
`K_(b)` = 0.51 °C/mole

i = Vant't Hoff factor

m = molality of the solution

When ethylene glycol, C2H6O2, (antifreeze) enters into solution in water it disociates into

C2H6O2 (aq) ---> 2OH(-1)(aq) + C2H4(+2)(aq)

Thus one mole of C2H6O2 dissociates into two moles of hydroxyl ions and one mole of C2H4(+2) ion

Hence the Van't Hoff factor, i, = 3

Therefore the mass of the mole

Therefore ΔT = 3 × 0.51 × 17.9 = 27.387 K

However Ethylene formula = (CH2OH)2 it dissolves in water without dissociation

Therefore i = 1

and ΔT = 1 × 0.51 × 17.9 = 9.129 ≅ 9.13

Hence at the boiling point of the water with antifreeze dissolved in it it

Boiling point of water + Boiling point elevation = 100 + 9.13 = 109.13 °C

The water will not boil until it reaches 109.13 °C