1.

Find out molar conductivity of AgCl at infinite dilution. Given, Lamda_(m)^(alpha)=133.4(AgNO_(3)),Lamda_(m)^(alpha)=149.9(KCl), Lamda_(m)^(alpha)=144.9" S "cm^(2)mol^(-1)(KNO_(3)).

Answer»

140 S `cm^(2)mol^(-1)`
138 S `cm^(2)mol^(-1)`
134 S `cm^(2)mol^(-1)`
132 S `cm^(2)mol^(-1)`

Solution :`Lamda_(m)^(0)(AgCl)=Lamda_(m)^(0)(AgNO_(3))+Lamda_(m)^(0)(KCl)-Lamda_(m)^(0)(KNO_(3))`
`=133.4+149.9-144.9`
=138 S `cm^(2)mol^(-1)`


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