Answer:
Freezing point = -30.0 °C
Boiling point = 108.25 °C
Explanation:
Step 1: Data given
Kf for water is 1.86 ∘C/m.
fluid is 50% antifreeze by mass
Antifreeze = ethylene glycols (C2H6O2)
Step 2: Determine mass of antifreeze
Let's suppose a total mass of 1000 grams
Since 50 % is antifreeze (ethylene glycol) it has a mass of 500 grams
The other 50 % is water
Step 3: Calculate moles antifreeze
500 grams / 62.07 g/mol = 8.055 moles C2H6O2
Step 4: Calculate molality
The molality = 8.055moles / 0.5 kg = 16.11 moles / kg = 16.11 molal
Step 5: Calculate freeing point
ΔT = Kf * molality
ΔT = 1.86 °C/m * 16.11 molal = 30.0 °C
This means the freezing point is 30 °C below the freezing point of water (0°C)
0°C - 30.0°C = -30.0°C
The freezing point is -30.0 °C
What is the boiling point of radiator fluid that is 50% antifreeze by mass?
Kb for water is 0.512 ∘C/m.
Step 1: Data given
Kb for water is 0.512 ∘C/m
The fluid is 50% antifreeze by mass
Step 2: Determine mass of antifreeze
Let's suppose a total mass of 1000 grams
Since 50 % is antifreeze (ethylene glycol) it has a mass of 500 grams
The other 50 % is water
Step 3: Calculate moles antifreeze
500 grams / 62.07 g/mol = 8.055 moles C2H6O2
Step 4: Calculate molality
The molality = 8.055moles / 0.5 kg = 16.11 moles / kg = 16.11 molal
Step 5: Calculate boiling point
ΔT = Kf * molality
ΔT =0.512 °C/m * 16.11 molal = 8.25 °C
This means the boiling point is 8.25 °C higher than the boiling point of water (100°C)
Boiling point = 100 °C + 8.25 °C = 108.25 °C
