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wedge (eq) of 0.10N solution of CaI(2) i...

`wedge _(eq)` of `0.10N` solution of `CaI_(2)` is `100.0 Scm^(2)eq^(-1)` at `298 K.G^(**)` of the cell `=0.25 cm^(-1)`. How much current will flow potential difference between the electrode is `5V` ?

Text Solution

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`wedge_(eq)=(kxx1000)/(N),k=(wedge_(eq)xxN)/(1000)`
`=(100xx0.1)/(1000)=0.01S cm^(-1)`
`G(` conductance `)=(K)/(G^(**)( cell con s t ant))`
`=(0.1)/(0.2)=0.4ohm^(-1) or S`
`R=(1)/(G)=(1)/(0.04)=25ohm`
Current in ampere `=("Potential difference ( by Ohm's law)")/("Resistance (ohm)")`
`=(5)/(25)=0.2A`
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