Boiling point depression 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 4.20 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.

I got the answer 133 C but it keep ssaying im wrong.

It would help if you showed your work.

First, you need to convert from 50/50 volume to molal (moles solute/kg solvent)

For 1 liter;

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

Note: when you have a situation like this it is somewhat open to debate which is the solvent, but they want the boiling point for water so we will say water is the solvent.

So: find moles ethylene glycol in 555g

555g/[(2 x 12.01) + (2 x 16.00) + (6 x 1.01)]g/mole = 555g/62.08g/mole = 8.94 moles
So, there are 8.94 moles in 499g water but, we need the moles in 1000g of water:

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

The boiling point should be 100 C+ (17.9 molal x 0.512 C/molal) = 109.2C

Not absolutely sure this is correct (I haven't done these in long time).