A `0.1 M` solution of potassium sulphate `K_(2)SO_(4)` is dissolved to the extent of `80%`. What would be its osmotic pressure at `27 (@)C`.

Explanation

Since `Na_(2)SO_(4)` dissociates as
`Na_(2)SO_(4) hArr 2Na^(+) + SO_(4)^(2-)`
If `alpha` is the degree of dissociation , then .
` alpha = (i-1)/(m-1)`
` m = 3 , alpha = 95% = 0.95`
` 0.95 = (i-1)/(3-1)`
` i-1 =0.95 xx 2 = 1.90`
` i=1.90 + 1 = 2.90`
` pi = iCRT `
` = 2.90 xx 0.1 xx 0.0821 xx 300`
` = 7.143` atm.