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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. |
Answer» `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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