The specific conductance of a `0.1 N KCl` solution at `23^(@)C` is `0.012 ohm^(-1) cm^(-1)`. The resistance of cell containing the solution at the same tempreature was found to be `55 ohm`. The cell constant will be

Options

A. `0.142 cm^(-1)`
B. `0.66 cm^(-1)`
C. `0.918 cm^(-1)`
D. `1.12 cm^(-1)`

Answer

B

Explanation

(b) Specific conductivity, `k=0.012Omega^(-1)cm^(-1)` ,
Resistance `=55Omega`
`k=(1)/("resistance")xx(1)/(a)[G=(1)/( R)]`
`(1)/(a)`= cell constant
`(1)/(a)=55xx0.012=0.66 cm^(-1)`