The ionic molar conductivities of `C_(2)O_(4)^(2-),K^(o+)`, and `Na^(o+)` ions are `x^('),y^('),` and `z^(')S cm^(2) mol ^(-1)` , respectively. Calculate `wedge_(m)^(@)` and `wedge_(eq)^(@)` of `(NaOOC-COO)`.

Explanation

`wedge_(m)^(@)=lambda_(m(C_(2)O_(4)^(2-))^(@))+lambda_(m(Na^(o+)))^(@)+lambda_(m(K^(o+)))^(@)`
`=(x^(')+y^(')+z^('))S cm^(2) mol^(-1)`
`wedge_(eq)^(@)=(wedge_(m)^(@))/(Tot a l charg e)=((x^(')+y^(')+z^(')))/(2)S cm^(2)Eq^(-1)`
Comparing the solution with that of illustration , we get
`(x+(y)/(2)+(z)/(2))=((x^(')+y^(')+z^('))/(2))`