You add 1.00kg of the antifreeze ethylenglycol(C2H6O2) to your car radiator, which contains 4450 g of water. The constants are:Kb=o.512 celsius m-1 and Kf(little f)=1.86 celsius m-1. What are the boiling and freezing points in celsius of the solution?
Can some please work it out and show me step by step how to do it.
Answer:
Delta T = k x m
m = molality = moles solute / Kg solvent
Molar Mass C2H6O2 = 62 g/mol
1.00 Kg = 1000 g
1000 / 62 = 16.1 moles
4450 g = 4.450 Kg
m = 16.1 / 4.45 = 3.62
Delta T = 0.512 x 3.62 = 1.85 °C ( the boiling point of the solution will be 100 + 1.85 = 101.85 °C)
delta T = 1.86 x 3.62 = 6.73 °C ( the freezing point will be
0 - 6.73 = - 6.73 °C )
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