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

Solute is `K_(3) [Fe(CN)_(6)]`.

Explanation

`DeltaT_(f)` (theroretical)`=mK_(f)` ltbRgt `=1.00xx10^(-3)xx1.86=1.86xx10^(-3)K`
`DeltaT` (observed) `=7.10xx10^(-3)K`
`i=(DeltaT_(f)("observed"))/(DeltaT_(f)("theroretical"))=(7.10xx10^(-3))/(1.86xx10^(-3))=3.81`
`therefore 1+x=3.82`
`x=2.82approx3`.
`therefore` Solute is `K_(3)[Fe(CN)_(6)]`.