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

`118.2mhocm^(2)mol^(-1)`

Explanation

`M=(0.624xx1000)/(249.5xx100)=0.025`
`(l)/(A)=153.7m^(-1).R=520Omega`
`M=(Kxx1000)/(M),R=(l)/(KA)`
or `R=(1)/(K)xxG^(*)` or `K=(G^(*))/(R)=(153.7)/(520)`
or `K=0.29557Scm^(-1)=0.29557Scm^(-1)`
`M=(2.955xx10^(-3)xx1000)/(0.025)=118.2Scm^(2)mol^(-1)`