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A conductivity cell when filled with `"0.1 N KCl"` at `25^(@)C` the resistance becomes 307.62 ohm at the same temperature. At `25^(@)C`, the specific conductance of 0.1 N KCl solution is `"0.01286 ohm"^(-1)."cm"^(-1)`. From these data calculate (1) cell constant (2) equivalent conductance of `"0.1 N AgNO"_(3)` solution and ressistance 362.65 ohm.

Text Solution

Verified by Experts

`"We know, specific conductance "=("cell constant")/("resistance").`
`kappa_(KCl)=(l//A)/(R_(KCl))`
or, `(l)/(A)=kappa_(KCl)xxR_(KCl)=0.01286xx307.62`
`="3.956 cm"^(-1)`
`therefore" cell constant of the cell = 3.956 cm"^(-1)`
`therefore" "kappa_(AgNO_(3))=(3.956)/(362.65)="0.0109 ohm"^(-1)."cm"^(-1)`
`therefore" Equivalent conductance "(Lambda_(eq))" of 0.1 N AgNO"_(3)`
Solution be -
`Lambda_(eq)(AgNO_(3))=kappa_(AgNO_(3))xx(1000)/(0.1)=0.0109xx(1000)/(0.1)`
`="109 ohm"^(-1)."cm"^(2)."eq"^(-1)`
Therefore, equivalent conductance of 0.1N `AgNO_(3)` solution is `"109 ohm"^(-1)."cm"^(2)."eq"^(-1)`.
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