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0.5 g KCl was dissolved in 100 g water, ...

`0.5 g` `KCl` was dissolved in `100 g` water, and the solution, originally at `20^(@)C` froze at `-0.24^(@)C`. Calculate the percentage ionization of salt. `K_(f)` per `1000 g` of water =`1.86^(@)C`.

Text Solution

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Calculate the (theoretical) molecular mass of `KCl `
`=39+35.5=75.5`
`DeltaT_(f)=iK_(f).m`
`0.24=(i)xx(1.86 xx 0.5)/(74.5 xx 100)xx1000`
`:. i=1.92`

"Total number of moles after dissociation"
`=1 - alpha + alpha + alpha=1+alpha`
`:.i=(1+alpha)/(1)` or `alpha=i-1=1.92-1=0.92`
Percentage ionization=`0.92 xx 100 = 92%`
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