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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.

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`Deltax=ixxK_(f)xxm`
`7.10xx10^(-3)=ixx1.86xx0.001`
`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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