Answer:
109.09°C
Explanation:
Given that:
the capacity of the cooling car system = 5.6 gal
volume of solute = volume of the water; since a 50/50 blend of engine coolant and water (by volume) is used.
∴ [tex]\frac{5.60}{2}gallons = 2.80 gallons[/tex]
Afterwards, the mass of the solute and the mass of the water can be determined as shown below:
mass of solute = [tex](M__1}) = Density*Volume[/tex]
[tex]= 1.1g/mL *2.80*\frac{3785.41mL}{1gallon}[/tex]
[tex]= 11659.06grams[/tex]
On the other hand; the mass of water = [tex](M__2})= Density*Volume[/tex]
[tex]= 0.998g/mL *2.80*\frac{3785.41mL}{1gallon}[/tex]
[tex]= 10577.95 grams[/tex]
Molarity = [tex]\frac{massof solute*1000}{molarmassof solute*massofwater}[/tex]
= [tex]\frac{11659.06*1000}{62.07*10577.95}[/tex]
= 17.757 m
≅ 17.76 m
∴ the boiling point of the solution is calculated using the boiling‑point elevation constant for water and the Molarity.
[tex]\Delta T_{boiling} = k_{boiling}M[/tex]
where,
[tex]k_{boiling}[/tex] = 0.512 °C/m
[tex]\Delta T_{boiling}[/tex] = 100°C + 17.56 × 0.512
= 109.09 °C
