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A 5% solution of anhydrous CaCI(2) at 0^...

A `5%` solution of anhydrous `CaCI_(2)` at `0^(@)C`developed 15 atm osmotic pressure. What is the degree of dissociation of `CaCI_(2)`.

Text Solution

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`5g` of `CaCI_(2)` are present n `100ml`, so `111g` (mol. wt. of `CaCI_(2))` will be present in
`(100 xx 111)/(5 xx1000) = 2.22` lit.
Now `piV = ST ( :' n = 1)`
or `pi = (0.082 xx 273)/(2.22) = (22.47)/(2.22) = 10.09` atm
`i=("Actually no.of particles in solution")/("No.of particles taken")` and `alpha =(i-1)/(n-1)`
here `n = 3`
`alpha = ((15)/(10.09)-1)/(3-1) =(4.91)/(10.09 xx2) = 0.2433` or `24.33%`
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