1.

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.

Answer»

Solution :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)`


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