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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooling system holds 5.60 gal, what is the boiling point of the solution? For the calculation, assume that at normal filling conditions, the densities of engine coolant and water are 1.11 g/mL and 0.998 g/mL respectively. Also, assume that the engine coolant is pure ethylene glycol (HOCH,CH,OH), which is non-ionizing and non-volatile, and that the pressure remains constant at 1.00 atm. The boiling-point elevation constant for water will also be needed.

Answer 1

Answer i volume of cooling system: 5.60 gal = 21198.3 ML Ag given blond is so 150 .:. udame of ethylene glycol = 21198-3/2 ML (cocant) = 10599. ISML mars of ethylene glyer = volemex density = 10599. ISMLX TillgmL) = 11765.09 Males of ethylene glyed = mars melon mess 119765.09 62 gmail E 189.76 mes volume of water = 10s99. ISML mass of water – udeme X danty = 10899 15 ML X 0.998 g MLY - 105789 = 10.58 kg molality of solutona Meleg of ethylene glycol mars of water I Solvent) - 189.76 mol 10.58kg 17.94 m Now OT: Kb of wrath x molality, - 0.512 °C 1m x 17.94m = 9.18°C > Ts - To = 9.18°C umere To = Boiling point of water in Tg = 9.18 C + To = 1000 C = 9:18°C + 100°C Tg = Balling point of solchen. - 1 og 18°C larwer)

Answer 2

3.1 coolant= 11.7335 lit
weight=12906.85 gm
moles=weight/62 =208.175
weight of water=3.1*3.875*0.998
molality= moles/weight of water in kg =17.3646
kb=0.512 c/m
boiling point =100+17.3646*0.512=108.89 c