`17.4% K_(2)SO_(4)` solition at `27^(@)C` is isotonic with 4% NaOH solution at the same temperature. If NaOH is 100% ionised, What is the degree of innisation of `K_(2)SO_(4)` in aqueous solution ?

Explanation

Calculation of molar concerntration of both the solution.
`"Molar concentration of NaOH solution"("Mass of NaOH / Molar mass")/("Volume of solution solution in litres")`
`=((4g)//(40g mol^(-1)))/((0.1 L))=1.0 mol^(-1)=1.0 M`
molar concentration of `K_(2)SO_(4)` solution = `(Mass of K_(2)SO_(4)//Molar mass)/("Colume of solution in liters")`
`=((17.4g)//(174h mol^(-1)))/((0.1L))=10 molL^(-1)=1.0 M`
Calculation of Van't Hoff factor (i) for `K_(2)SO_(4)`
Since the two solutions are isotonic, they have same osmotic pressure.
`alpha=1, n=2`
`alpha=(i-1)/(n-1)or1=(i-1)/(2-1)ori=2,c=1 M`
`pi_(NaOH)=iCRT=2xx(1M)xxRxxT`
C=1 M, i=? ( to be calcutaed)
`pi_(K_(2)SO_(4))=oCRT=i(1)xxRxxT`
As the twpsp,itopm are oisptpmec. `pi_(NaOH) or 2xxRxxT=ixxRxxT or o=2`
Calculation of dgree of ionisation of `K_(2)SO_(4)`
`K_(2)SO_(4): alpha=(i-1)/(n-1)=(2-1)/(3-1)=1/2=0.5`
The degree of ionisation of `K_(2)SO_(4)=0.5` or its percentage ionisation=o.5xx100=50%