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Question

Ethylene glycol (molar mass=62gmol-1) is a common automobile antifreeze. calculate the freezing point of a solution containing 12.4g of this substance in 100gof water. would it be advisable to keep this substance in the car radiator during summer?

Given: kf for water =1.86Kkgmol-1, kb for water =0.512Kkgmol-1


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Solution

Step 1: Analyze the formula

The formula for depression in freezing point is
ΔTf=Kf×b×i

Where, i=1 for ethylene glycol

b=molality=molesofsolutekgofsolvent

Kf is called the molal freezing point depression constant

Step 2: Find the value of Molality

One mole of Ethylene glycol =62g

Therefore, 12.4g constitutes 12.462=0.2mole

Solvent =100g=0.1kg

Hence, b=0.20.1=2molkg-1

Step 3: Find freezing point of solution

By the formula for depression in freezing point:

ΔTf=1.86×2Kelvin=3.72K

The difference in freezing point =3.72K (depression)

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

Therefore freezing point of solution: =2733.6=269.4K=-3.72°C (since temp in K=273+temp in °C)

Step 4: Find boiling point of solution.

Boiling point elevation

ΔTb=Kb×b×i=0.512×2=1.024K

The boiling point of pure water =100°C=373K

Therefore, the boiling point of the solution =373+1.024=374.024K=101.024°C

The boiling point of aqueous ethylene glycol, on the other hand, rises as the amount of ethylene glycol increases.

Thus, the inclusion of ethylene glycol not only lowers the freezing point but also raises the boiling point, making it ideal for keeping in a radiator throughout the summer.


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