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The freezing point depression of 0.001 m...

The freezing point depression of `0.001 m K_(x)` `[Fe(CN)_(6)]` is `7.10xx10^(-3) K`. Determine the value of x. Given, `K_(f)=1.86 K kg "mol"^(-1)` for water.

Text Solution

Verified by Experts

`Deltax = ixx K_(F) xx m`
`7.10 xx 10^(-3) =ixx 1.86 xx 0.001`
`i= 3.817`
`alpha = (i-1)/(n-1)`
`1 = (3.817-1)/((x+1)-1)`
`x = 2.817 ~~ 3`
`:.` Molecular formula of the compound is `K_(3)[Fe(CN)_(6)]`
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