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45 g of ethylene glycol C(2)H(6)O(2) is ...

`45 g` of ethylene glycol `C_(2)H_(6)O_(2)` is mixed with `600 g` of water. Calculate (a) the freezing point depression and (b) the freezing point of solution.
Given`K_(f)=1.86 K kg mol^(-1)`.

Text Solution

Verified by Experts

`M_(B)-(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))`
Mass of ethylene glycol`(W_(B))=45 g`
Mass of water `(W_(A))=600 g=0.6 kg`
Molar mass of ethylene glycol `(M_(B))=(2xx12+xx6+2xx16)=62g mol^(-1)`
Molal depression cnstant` (K_(f))=1.86 K kg mol^(-1)`
`DeltaT_(f)=((45g)xx(1.86 kg mol^(-1)))/((62g mol^(-1))xx(0.6kg))=2.25 K`
Calculation of freezing point of solution
=(273K-2.25 K)=270.75 K
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