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A conductivity cell has a cell constant ...

A conductivity cell has a cell constant of `"0.5 cm"^(-1)`. This cell when filled with 0.01 M NaCl solution has a resistance of `"384 ohm at "25^(@)C`. Calculate the equivalent conductance of the given solution.

A

`130.2Omega.cm^(2).g-eq^(-1)`

B

`137.4Omega.cm^(2).g-eq^(-1)`

C

`154.6Omega.cm^(2).g-eq^(-1)`

D

`169.2Omega.cm^(2).g-eq^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
Equivalent conductance, `Lambda_("eq")=(kappa xx1000)/("Normality")`
where, `kappa` (specific conductance) `=Cxx(l)/(A)`
`=(1)/(R)xx(l)/(A)=(1)/(384)xx0.5=1.302xx10^(-3)"ohm"^(-1)."cm"^(-1)`
`therefore Lambda_("eq")=(1.302xx10^(-3)xx1000)/(0.01`
`="130.2ohm"^(-1)."cm"^(2)."g-eq"^(-1)`
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