A 50/50 blend of engine coolant and water (by volume) is usually used in an auto

Question

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If your car's cooling system holds 5.70 gallons, what is the boiling point of the solution? Make the following assumptions in your calculation: at normal filling conditions, the densities of engine coolant and water are 1.11 g/mL and 0.998 g/mL respectively. Assume that the engine coolant is pure ethylene glycol (HOCH2CH2OH), which is non-ionizing and non-volatile, and that the pressure remains constant at 1.00 atm. Also, you'll need to look up the boiling-point elevation constant for water.

Explanation / Answer

In this solution first, we to convert from 50/50 coolant and water (by volume) to molal means (moles solute/kg solvent)

Density = mass/ volume

Mass = density * volumes
For 1 liter; 50/50 coolant and water measn 500 ml each present

500ml x 1.11g/ml = 555g ethylene glycol
500ml x 0.998g/ml = 499g water

Now, find moles ethylene glycol in 555g
molar mass of ethylene glycol= 62.08g/mole, molar mass of water =

555g/62.08g/mole = 8.94 moles
So, there are 8.94 moles in 499g water but, according to definition of molality;

we need the moles in 1000g of water:

8.94mole x (1000g/499g) = 17.9 moles/kg water, so it is 17.9 molal

delt T =Kfb m, where m = molality (moles of solute / kg. of solvent)

T solution – T pure = Kfb m

T solution = T pure + Kfb m

The boiling point should be 100 C+ (17.9 molal x 0.512 C/molal)

= 109.2C

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