Answer:
108.25 ºC
Explanation:
The boiling point elevation for a given solute in water is given by the expression:
ΔTb = i Kbm
where ΔTb is the boiling point elevation
i is the van´t Hoff factor
Kb the boiling constant which for water is 0.512 ºC/molal
m is the molality of the solution
The molality of the solution is the number of moles per kilogram of solvent. Here we run into a problem since we are not given the identity of the coolant, but a search in the literature tells you that the most typical are ethylene glycol and propylene glycol. The most common is ethylene glycol and this it the one we will using in this question.
Now the i factor in the equation above is 1 for non ionizable compounds such as ethylene glycol.
Our equation is then:
ΔTb = Kbm
So lets calculate the molality and then ΔTb:
m = moles ethylene glycol / Kg solvent
Converting gal to L
4.90 g x 3.785 L/gal = 18.55 L
in a 50/50 blend by volume we have 9.27 L of ethylene glycol, and 9.27 L of water.
We need to convert this 9.27 l of ethylene glycol to grams assuming a solution density of 1 g/cm³
9.27 L x 1000 cm³ / L = 9273.25 cm³
mass ethylene glycol = 9273.25 cm³ x 1 g/cm³ = 9273.25 g
mol ethylene glycol = 9273.25 g/ M.W ethylene glycol
= 9273.25 g / 62.07 g/mol =149.4mol
molality solution = 149.4 mol / 9.27 Kg H₂O = 16.12 m
( density of water 1 kg/L )
Finally we can calculate ΔTb:
ΔTb = Kbm = 0.512 ºC/molal x 16.12 molal = 8.25 ºC
boiling point = 100 º C +8.25 ºC = 108.25 ºC
( You could try to solve for propylene glycol the other popular coolant which should give around 106.7 ºC )
