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3 years ago

What is the freezing point of radiator fluid that is 50% antifreeze by mass?

Chemistry

1 answer:

crimeas [40]3 years ago

7 0

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

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