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Question

Let’s think that you are driving in a car in winter. In a cold climate water gets frozen causing damage to radiator of a car. Ethylene glycol is used as an antifreezing agent. Calculate the amount of ethylene glycol to be added to 4 kg of water to prevent it from freezing at 6C. (Kf for water = 1.85Kkgmol1)

A
804.32g
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B
904.32 g
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C
704.32 g
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D
1004.32 g
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Solution

The correct option is A 804.32g
So bad that you are stuck with damaged radiator of the car.
Chemistry comes to rescue here.
As given in the question ethylene glycol can be used as a anti freezing agent.
We can calculate the amount of ethylene glycol to be added by our knowledge of depression in freezing point. Let’s check the given information.
ΔT=6C,W1=4kg=4000g,Mw2=62,Kf (molal depression constant) = 1.85
We just saw how to calculate mass of ethylene glycol here.
W2=Mw2×W1×ΔT1000×Kf
ΔT = Depression in freezing point
Kf = Molal depression constant of benzene
W2 = Mass of solute
Mw2 = Molecular mass of solute
W1 = Mass of solvent
62×4000×61000×1.85=804.32g

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