If the radiator of an automobile contains 12L of water, how much would the freezing point be lowered by the addition of 5kg of prestone (glycol C2H4(OH)2)? How many kg of zerone (methyl alcohol, CH3OH) would be required to produce the same result? Assume 100% purity.
Risep
1) Mr C₂H₄(OH)₂ = 62 12 L water = 12 Kg water
Mr C₂H₄(OH)₂ = 62
12 L water = 12 Kg water
2)
Mr CH₃OH = 32