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A solution containing 0.5 g of KCI disso...

A solution containing 0.5 g of KCI dissolved in 100 g of water and freezes at `-0.24^(@)C`. Calculate the degree of dissociation of the salt. (`K_(f)` for water =`1.86^(@)C`. [Atomic weight K = 39, Cl = 35.5]

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Given : Wt. of KCl (w) = 0.5g
Wt. of water (W) = 100g
`K_(f)` for water = `1.86^(@)C`
Molecular wt(m) of KCl = `39+35.5=74.5`
Apply the relation of freezing point depression `(DeltaT_(f))` with molecular weight of the solute :
`DeltaT_(f_(cal))=(K_(f)xx1000xxw)/(Wxxm)`
`=(1.86xx1000xx0.5)/(100xx74.5)`
`=0.124^(@)C`
`DeltaT_(f_(obs))=0^(@)C+0.24^(@)C`
`=0.24^(@)C`
van.t Hoff factor (i) = `(DeltaT_(fobs))/(DeltaT_(f_(caI)))`
`=(0.24)/(0.124)=1.935`
The degree of dissociation `(alpha)` of KCl is :
`alpha=(i-1)/(n-1)` where, n = number of ions
Here `n=2` for `K^(+)andCl^(-)`
`=(1.93-1)/(2-1)=(0.93)/(1)`
`therefore alpha=0.93or93%`
Thus, the degree of dissociation of the salt is 93% or 0.93.
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