Part C: Calculate the freezing point of the solution 580 g of the antifreeze pro

Question

Part C: Calculate the freezing point of the solution 580 g of the antifreeze propylene glycol, C3H8O2, a nonelectrolyte, dissolved in 1.40 kg of water. Express your answer using three significant figures.
Part D: Calculate the boiling point of the solution 580 g of the antifreeze propylene glycol, C3H8O2, a nonelectrolyte, dissolved in 1.40 kg of water. Express your answer using four significant figures.
Part C Calculate the freezing point of the solution 580 g of the antifreeze propylene glycol, CHsO, a nonelectrolyte, dssolved in 1.40 kg of water. Express your answer using three significant figures. the freezing point oC Submit My Answers Give Up Part D Calculate the bolling point of the solution 580 g of the antifreeze propylene glycol, CaHsO2. a nonelectrolyte, dissolved in 1.40 kg of water Express your answer using four significant figures. the boiling point Submit My Answers Give Up

Explanation / Answer

Part C.

moles of antifreeze (n):
n = 580 g/(76.09 g/mol)
n = 7.6226 mol

Calculate molality:

Equation:
b = n/S
b, molality
n, moles of solute
S, mass of solvent (kg)

b = 7.6226 mol/1.40 kg
b = 5.4447 mol/kg

Calculate freezing point depression:

deltaT = -Kf . b
Kf, freezing point depression constant
b, molality

deltaT = -(1.86) . (5.4447 mol/kg)
deltaT = -10.127 ºC

Calculate freezing point:
Tf = 0 ºC - 10.127 ºC
Tf = -10.127 ºC
*assuming standard conditions

Part D.

Calculate boiling point increment:

deltaT = -Kb . b
Kf, boiling point elevation constant
b, molality

deltaT = (0.512) . (5.4447 mol/kg)
deltaT = 2.7877 ºC

Calculate boiling point:
Tb = 100 ºC + 2.7877 ºC
Tb = 102.788 ºC
*assuming standard conditions

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