A 5% solution of `CaCl_(2)` at `0^(@)C` developed 15 atmospheric pressure. Calculate the degree of dissociation. (mol. wt. (M) of `CaCl_(2) = 111` a.m.u., R = 0.0821 L atm. `K^(-1) "mol"^(-1)`)

Explanation

`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%`