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The freezing point depression of 0.001mK...

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

A

`3`

B

`4`

C

`2`

D

`5`

Text Solution

Verified by Experts

The correct Answer is:
A
`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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