A `0.01 m` aqueous solution of potassium ferricy freezes at `-0.062^(@)C`. What is the apparent percentage of dissociation? `(K_(3)` for `H_(2)O -1.86)`.
Strategy: Potassium is `K_(3)[Fc(CN)_(6)]`. The appearent percentage of dissociation is degree of dissociation x 100%. To calculate degree of dissociation we must determine the value of van't hoff factor for which we need to work out calculated colligative property.

Explanation

Step 1. Find calculate colligative property We are given observed `DeltaT_(f)`. We work out calculate (or normal) colligative property as follows:
`DeltaT_(f)=K_(f)m`
`=(1.86^(@)C//m)(0.01 m)`
`=0.0186^(@)C`
Now, `i=("Observed" DeltaT_(f))/("Calculated" DeltaT_(f))
=(0.062^(@)C)/(0.0186^(@)C)`
`=3.33`
Step 2. Calculate degree of dissociation let's assume that `alpha` is the degree of dissociation of `K_(3)[Fe(CN)_(6)]`. `{:(,,[Fe(CN)_(6)],hArr,3K^(+),+,[Fe(CN)_(6)]^(3-)),("Moles before dissociation",,1 mol,,0mol,,0mol),("Moles after dissociation",,(1-alpha)mol,,3alphamol,,alphamol):}`
`i=("Total moles after dissociation")/("Total moles before dissociation")`
`=((1-alpha)+3alpha+alpha)/1=(1+3alpha)/1`
or `alpha=(i-1)/3`
`=(3.03-1)/3`
`=0.78`
Therefore
Percent dissociation `=0.78xx100%`
`=78%`