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Ethylene glycol `("molar mass"=62 g mol^(-1))` is a common automobile antyfreeze. Calculate the freezing point of a solution containing 12.4 g of this substance in 100 g of water. (Given` K_(f) for water = 1.86 K kg mol^(-1))`

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`M_(B)-(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))`
Mass of ethylene glycol`(W_(B))=12.4 g`
Mass of solvent i.e., water `(W_(A))=100g=0.1 kg`
Molar mass of glycol `(M_(B))=62 g mol^(-1)`
Molal depression constant `(K_(f))=1.86 K kg mol^(-1))`
`DeltaT_(f)=((12.4g)xx(1.86 K kgmol^(-1)))/((62 g mol^(-1))xx(0.1 kg))=3.72 K`
Freezing point of the solution =(273.0K-3.72 K)= 269.28 K.

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