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

Answer

B

Explanation

Specific conductance
=Conductance xx Cell constant
0.012`ohm^(-1) cm^(-1)=(1)/(55)ohmxx` Cell constant
ohm xx Cell constant
Cell constant `=53xx0.012cm^(-1)=0.66cm^(-1)`