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 4.90 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.

Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: Solvent Formula Kf Norma Kb Normal boiling valuefreezing (°C/m) point (°C) (°C/m)point C°C) 1.86 0.00 value 0.512 100.00 2.53 80.1 2.92 80.7 1.22 78.4 5.03 76.8 H20 C6H65.125.49 water benzene cyclohexane C6H12 20.8 6.59 ethanol carbon tetrachloride C2H60 1.99 CCI4 29.8-22.9 camphor 10H160 37.8176

Explanation / Answer

DTb = i*kb*m

DTb = Tb-T0

i = vanthoff factor of solute = 1

T0 = boiling point of solvent(water) = 100 C

Tb = boiling point of solution = x C

kb for water = 0.512°C·kg/mol

m = molality of solution = (w/M)*1000/W

w = Amount of ethylene glycol = 10239.75 g

M = molarmass of ethylene glycol = 62.07 g/mol

W = Amount of water = 9206.55 g

DTb = i*Kb*(w/M)*(1000/W)

x-100 = 1*0.512*(10239.75/62.07)*(1000/9206.55)

x = 109.17 c

Tb = boiling point of solution = x = 109.17 c

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