The resistance and conductivity of `0.02 M KCl` solution are `82.4 ohm` and `0.002768 S ch^(-1)` respectively . When filled with `0.005 N K_(2)SO_(4)`, the solution had a resistance of `324ohm`. Calculate `:`
`a.` Cell constant
`b.` Conductance of `K_(2)SO_(4)` solution
`c.` Conductivity of `K_(2)SO_(4)` solution
`d.` Equivalent conductivity of `K_(2)SO_(4)` solution
`e.` Molar conductivity of `K_(20SO_(4)` solution.

Explanation

`a.` From `KCl` data `: k =(1)/(R)xx(l)/(a)`
`:. (l)/(a)(` cell constant `) =kxxR=0.002768xx82.4`
`=0.2281cm^(-1)`
From `K_(2)SO_(4)` data `:`
`b.` `R=324 ohm,G=(1)/(R)=(1)/(324)=3.086xx10^(-3)S`
`c.` `k=Gxx(l)/(a)=3.086xx10^(-3)xx0.2281`
`=7.04xx10^(-4)Scm^(-1)`
`d.` `wedge _(eq)=(kxx1000)/(N)=(7.04xx10^(-4)xx1000)/(0.05)`
`=140.8 S cm^(2)eq^(-1)`
`e.` `wedge _(m)=(kxx1000)/(M)=(7.04xx10^(-4)xx1000)/(0.005//2)` ltbr. `=281.6S cm^(2)mol ^(-1)`
`[n` factor for `K_(2)SO_(4)=2,M=N//2]`