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 4.60 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 need help with this, if anyone could help me.

Explanation / Answer

Volume of ethylene glycol = 4.6/2=2.3 gallons and volume of water =2.3 gallons

1 Gallon =3.87 L, 2.3 gallons = 2.3*3.87=8.9 L, =8.9*1000ml

mass = Volume* density

Masses : ethylene glcycol = 8900*1.11=9790 gm , water = 8900*0.998=8882.2 gm =8.882 Kg

Moles = mass/Molar mass

molar masses : ethylene glycol = 62 water =18

moles : ethylene glycol = 9790/62 =157.9

Molality = moles/kg solvent = 157.9/8.882=17.76

elevation in boiling point = kb*m*i, i= 1 for ethylene glycol

Kb= 0.512 deg.c/m for water

boiling point elevation = 17.76*0.512=9.1 deg.c

Boiling point of solution = boiling point of water+9.1= 100+9.1= 109.1 deg.c

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