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 cars cooling system holds 6.30 galions, 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. Number o c

Explanation / Answer

Ans. Given, the volume of cooling system mixture = 6.30 gallons = 23.8481 L

So,

            Volume of water = Volume of coolant = 50% of 23.8481 L = 11.92405 L

# Step 1: Mass of water = Volume x Density = 11.92405 L x (0.998 kg/L) = 11.90 kg

# Mass of ethylene glycol = 11.92405 L x (1.11 kg/ L) = 13.2356955 kg = 13235.696 g

            Moles of EG = Mass / MW = 13235.696 g / (62.07 g/ mol) = 213.238 mol

Now,

            Molality of EG = Moles of EG / Mass of water in kg

                                    = 213.238 mol / 11.90 kg

                                    = 17.92 m

# Step 2: Elevation in boiling point of a solution is given by-

dTb = i Kb m            - equation 1

Kb = molal boiling point elevation constant of solvent (= 0.5120C / m)

dTb = Boiling point elevation = BP of solution – BP of pure solvent

Putting the values in equation 1-

            dTb= 1 x (0.5120C / m) x 17.92 = 9.20C

Now,

            BP of solution = dTb + BP of pure solvent (water) = 9.20C + 100.00C = 109.20C

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