Ionic conductance at infinite dilution of `Al^(3+) " and " SO_(4)^(2-)` are 189 and 160 mho `cm^(2)" equiv"^(-1)`. Calculate the equivalent and molar conductance of the electrolyte `Al_(2)(SO_(4))_(3)` at infinite dilution.

Explanation

The electrolytes is `Al_(2)(SO_(4))_(3)`
Equivalent conductance at infinite dilution
`lambda_(infty)Al_(2)(SO_(4))_(3)=1/3lambda_(infty)Al^(3)+1/2lambda_(infty)SO_(4)^(2-)`
`lambda_(infty)Al_(2)(SO_(4))_(3)=189/3+160/2=63+80`
`" "=143 " mho "cm^(2)" gm "equ^(-1).`
Molar conductance at infinite dilution
`mu_(infty)Al_(2)(SO_(4))_(3)=2 times 189+3 times 160`
`" "=858" mho "cm^(2)mol^(-1)`