An ethylene glycol solution contains 30.0g of ethylene glycol (C2H6O2) in 95.6mL of water. (Assume a density of 1.00 g/mL for water.)
1.Determine the freezing point of the solution.
Express you answer in degrees Celsius
2.Determine the boiling point of the solution.
Express you answer in degrees Celsius.
first calculate the molality of the solution
molality =(W/MW ) (1000 / mass of solvent in g)
molality = (30.0 / 62.07) (1000 / 95.6)
molality = 5.05 m
now
Tf = Kf x m
Tf = 1.86 x 5.05
Tf = 9.40
freezing point = 0 - 9.40 = -9.40
freezing point = - 9.40 0C
2) Tb = Kb x m
Tb = 0.512 x 5.05
Tb = 2.58
boiling point = 100 + 2.58
boiling point = 102.58 0C
freezing point will decrease so we substract.
boiling point will increase so we are adding.
An ethylene glycol solution contains 30.0g of ethylene glycol (C2H6O2) in 95.6mL of water. (Assume a...
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