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The resistance of a 0.01 N solution of a...

The resistance of a 0.01 N solution of an electrolyte was found to 210 ohm at `25^(@)C` using a conductance cell with a cell constant `0.88 cm^(-1)`. Calculate the specific conductance and equivalent conductance of the solution.

Text Solution

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`R=210 ohm, (l)/(a)=0.88cm^(-1)`
`k=(1)/(R)xx(1)/(a)=(1)/(210)xx0.88=4.19ohm^(-1)cm^(-1)` or `S cm^(-1)`
`wedge_(eq)=(kxx1000)/(N)=(4.19xx10^(-3)xx1000)/(0.01)`
`=419.05 S ch^(2)eq^(-1)`
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