The resistance of an aqueous solution containing `0.624g` of `CuSO_(4).5H_(2)O` per `100 cm^(3)` of the solution in a conductance cell of cell constant `153.7` per meter iss 520 ohms at 298K. Calculate the moalr conductivity. `(CuSO_(4).5H_(2)O = 249.5)`

Answer

`118.2 mho cm^(2) mol^(-1)`

Explanation

`K=(153.7)/(520)=2.96 xx 10^(-1) "mho m"^(-1)`
`=2.96 xx 10^(-3) "mho cm"^(-1)`
`lamda_(m)=(K xx 1000)/(c)=(2.96 xx 10^(-3) xx1000)/(2.5 xx 10^(-2))`
`1.184 xx10^(2)`
`=118.4 "mho cm"^(2) "mol"^(-1)`.