lon 20 of 26 Sapling Learning A 50/50 blend of engine coolant and water (by volu

Question

lon 20 of 26 Sapling Learning A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If your cars cooling system holds 4.00 gallons, what is the boiling point of the solution? Make the following assumptions in your calculation: at normal fling 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 (HOCH CH2OH), 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 boilling-point elevation constant for water. Number o C Peevicus rch

Explanation / Answer

Since we have a 50:50 ratio then it is correct to say that there are 2 gallons of water and 2 gallons of solvent

2 + 2 = 4 gallons in the cooling system

calculate the mass of water

1 gallon = 3785.41 ml

2 gallons = 7570.82 ml

mass = density * volume = 0.998 * 7570.82 = 7555.68 grams

calculate mass of coolant

mass = density * volume = 1.11 * 7570.82 = 8403.61 grams

calculate moles of etylenglicol, molar mass etylenglicol is 62.07 g/gmol

moles = mass / molar mass = 8403.61 / 62.07 = 135.389 moles

kg of solvent = 7.555 kg

calculate molality = moles of solute / kg of solvent

molality = 135.389 / 7.555 = 17.92 m

Apply the next equation

DeltaT = Kb * m

Kb is the ebulloscopic constant, for water is 0.512 C / m so

Delta T = 0.512 * 17.92 = 9.175 C

The new boiling temperature is

100 + 9.175 = 109.175 C

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