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At 298K the resistance of a 0.5N NaOH so...

At 298K the resistance of a 0.5N NaOH solution is 35.0 ohm. The cell constant is 0.503 `cm^(-1)` the electrical conductivity of the solution is

A

`1.437xx10^(-2)ohm ^(-1)cm^(-1)`

B

`1.473ohm^(-1)cm^(-1)`

C

`1.06ohm^(-1)cm^(-1)`

D

`3.5ohm^(-1)cm^(-1)`.

Text Solution

Verified by Experts

The correct Answer is:
A
The electrical conductivity, `k=(1)/(R)xx(l)/(a)`
`=(1)/(35)xx0.503=1.437xx10^(-2)ohm^(-1)cm^(-1)`
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