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The molar conductivity of 0.05 M of solu...

The molar conductivity of `0.05 M` of solution of an electrolyte is `200 Omega^(-1) cm^2`. The resistance offered by a conductivity cell with cell constant `(1//3) cm^(-1)` would be about .

A

` 11. 11 Omega`

B

` 22.22 Omega`

C

` 33.33 Omega`

D

` 44.444 Omega`

Text Solution

Verified by Experts

The correct Answer is:
C

`k=Lambda_C = (200 s cm2 "mol"^(-1)) (.0.5 xx 10^(-3) "mol" cm^(-1))`
` = 0.01 S cm^(-1)`
` R=1/k (e/A) = 1/((0.01 S cm^(-1)) (1/3 cm^(-1)) = 33. 33 Omega`.
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