When a solution of conductanes `1.342 mho m^(-1)` was placed in a conductivity cell with parallel electrodes the resistance was found to be `170.5` ohm. The area of the electrode is `1.86 xx 10^(-4)` sq meter. Calculate the distance between the two electrodes in meter.

When a solution having conductivity 1.342 xx 10^(-2)" ohm"^(-1) cm^(-1) was placed in a cell with parallel electrodes, the resistance was found to be 170.5 ohms. If the area of the electrodes is 1.86 sq. cm., find the cell constant and the distance apart of the electrodes.

Area of each electrode of a conductivity cell is "1.25 cm"^(2) . The cell is filled with a solution of resistance of 160 ohm. If specific conductance of the solution is "0.016 ohm"^(-1)."cm"^(-1) , calculates the distance between the electrodes, and the cell constant.

a conductivity cell, the distance between two electrodes is 0.62 cm and the area of cross section of the electrode is 1.15 cm^2 . Find the cell constant of the cell.

At 25^(@)C , a conductivity cell was filled with 0.1 M NalL solution. The conductivity of this solution is 9.2 xx 10^(3) s cm^(-1) and resistance is 176.6 ohm . The cross-sectional is 176.6 ohm . What must have been the distance between the electrodes ?

The distance between the eletrodes of a conductivity cell is 0.98 cm and the cell constant is 0.5 cm^(-1) . Calculate the cross-sectional area of electrode.

The distance between the electrodes of a conductivity cell is 0.98 cm and area of cross section of theeletrodeis 1.3cm^2 . Calculate the cell constant for the cell.