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Ethylene glycol (molar mass = 62 g mol¯1) is a common automobile antifreeze. Calculate the freezing point of a solution containing 12.4g of this substance in 100 g of water. Would it be advisable to keep this substance in the car radiator during summer? Given : Kf for water = 1.86K kg/mol Kb for water = 0.512K kg/mol

Ethylene glycol (molar mass = 62 g mol¯1) is a common automobile antifreeze. Calculate the freezing point of a solution containing 12.4g of this substance in 100 g of water. Would it be advisable to keep this substance in the car radiator during summer?
Given : Kf for water = 1.86K kg/mol Kb for water = 0.512K kg/mol

Grade:11

2 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
9 years ago
Freezing point depression ΔTf = Kf *b * i

i = 1 for ethylene glycol

b = moles of solute / kg of solvent

1 mole of Ethylene glycol = 62g



Therefore, 12.4g constitutes 12.4/62 = 0.2 mole

Solvent = 100g = 0.1kg

Hence, b = 0.2/0.1 = 2 moles/kg



ΔTf = 1.86 * 2 Kelvins

ΔTf = 3.72 Kelvins

Difference in freezing point = 3.72K (depression)

We know freezing point of pure water is 0 deg C = 273K

Therefore freezing point of solution =273 – 3.6 = 269.4K = -3.72 deg C

(since temp in K = 273 + temp in C)



Boiling point elevation ΔTb= Kb *b * i

= 0.512 * 2 = 1.024K

Boiling point of pure water = 100 deg C = 373K

Therefore, boiling point of solution = 373 + 1.024 = 374.024K

= 101.024 deg C

However, the boiling point for aqueous ethylene glycol increases with increasing ethylene glycol percentage. Thus, the use of ethylene glycol not only depresses the freezing point, but also increases the boiling point so it is good top keep in radiator during summer.
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem.
 
Freezing point depression ΔTf = Kf *b * i
i = 1 for ethylene glycol
b = moles of solute / kg of solvent
1 mole of Ethylene glycol = 62g
Therefore, 12.4g constitutes 12.4/62 = 0.2 mole
Solvent = 100g = 0.1kg
Hence, b = 0.2/0.1 = 2 moles/kg
ΔTf = 1.86 * 2 Kelvins
ΔTf = 3.72 Kelvins
Difference in freezing point = 3.72K (depression)
We know freezing point of pure water is 0 deg C = 273K
Therefore freezing point of solution =273 – 3.6 = 269.4K = -3.72 deg C (since temp in K = 273 + temp in C)
Boiling point elevation ΔTb= Kb *b * i = 0.512 * 2 = 1.024K
Boiling point of pure water = 100 deg C = 373K
Therefore, boiling point of solution = 373 + 1.024 = 374.024K = 101.024 deg C
However,
the boiling point for aqueous ethylene glycol increases with increasing ethylene glycol percentage. Thus, the use of ethylene glycol not only depresses the freezing point, but also increases the boiling point so it is good top keep in radiator during summer.
 
Hope it helps.
Thanks and regards,
Kushagra

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