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 6.00 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.

(°C/m)

Normal freezing

point (°C)

(°C/m)

Normal boiling

point (°C)

Solvent Formula Kf value*

(°C/m)

Normal freezing

point (°C)

Kb value

(°C/m)

Normal boiling

point (°C)

water H2O 1.86 0.00 0.512 100.00 benzene C6H6 5.12 5.49 2.53 80.1 cyclohexane C6H12 20.8 6.59 2.92 80.7 ethanol C2H6O 1.99 –117.3 1.22 78.4 carbon
tetrachloride   CCl4 29.8 –22.9 5.03 76.8 camphor   C10H16O 37.8 176

Explanation / Answer

Let us consider 1 ltr of solution.

Volume of coolant=500 ml

Mass of coolant=500*1.11=555 g

Moles of coolant=Mass of coolant/molecular wt=555/62=8.952 moles

Mass of water=500*.998=499 g=0.499 kg

Molality of water= moles of solute/Mass of solvent of in kg=8.952/0.499=17.94 m

Boiling point of blend=Normal boiling point+kb*molality

                                                =100.00+0.512*17.94=109.19

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