Answer:
The correct answer is No, it would boil at 109.13°C
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 × [tex]K_{b}[/tex] ×[tex]m_{solute}[/tex]
Where [tex]K_{b}[/tex] = 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