How many liters of the antifreeze ethylene glycol [CH2(OH)CH2(OH)] would you add

Question

How many liters of the antifreeze ethylene glycol [CH2(OH)CH2(OH)] would you add to a car radiator containing 6.50 L of water if the coldest winter temperature in your area is -28C? (The density of ethylene glycol is 1.11 g/mL. Assume the density of water at -28C is 1.00 g/mL.) Calculate the boiling point of this water-ethylene glycol mixture.

Explanation / Answer

water freezes at 0 Celsius. But now it is depressed by36 degrees. Use the formula: delT = Kf * m, where m is molality and Kf is freezing pointdepression constant (1.86 C*kg/mol) 36 C = 1.86 C*kg/mol * m m = 19.35 mol ethylene glycol/ kg water. You know you have 6.50 L of water, which is 6500 mLof water => 6500 g of water (6.5 kg) So, mass of ethylene glycol: 6.5 kg * 19.35 mol/kg =125.806 mol MW ethylene gylcol: 62.068 g/mol So liters: 125.806 mol * 62.068 g/mol * 1 mL/1.1g =7098.6 mL (~ 7L) Boiling Point: delT = Kb*m = 0.512 C*kg/mol * 19.35 mol/kg= 9.907 C (the 0.512 is a constant) So new boiling point = old bp + del T = 109.9 C for further interest you can visit http://www.engineeringtoolbox.com/ethylene-glycol-d_146.html

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