The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is `1500 Omega`. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is `0.146 xx 10^(-2) S cm^(-1)` .

The resistance of a conductivity cell containing 0.001M KCl solution at 298K is 1500Omega .What is the cell constant (in mm^(-1) ). If the conductivity of 0.001 M KCl solution is 2 xx 10^(-3) S mm^(-1)

The conductivity of 0.20M solution of KCl at 298 K is 0.0248S cm^(-1). Calculate molar conductance.

The resistance of 0.05M KCl solution at 25^@C is 100 ohm in a cell whose cell constant is 0.3765cm^(-1) . The specific conductivity of KCl solution is

Resistance of a conductivity cell filled with 0.1 mol L^(-1) KCl solution is 100Omega . If the resistacne of the same cell when filled with 0.02 mol L^(-1) KCl solution is 520 Omega , calculate the conductivity and molar conductivity of 0.02 mol L^(-1) KCl solution . The conductivity of 0.1 mol L^(-1) KCl solution is 1.29 S/m.

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1M is 100Omega . The conductivity of this solution is 1.29 S m^(-1) . Resistance of the same cell when filled with 0.02M of the same solution is 520 Omega . The molar conductivity of 0.02M solution of the electrolyte will be

The resistance of 0.01N solution of an electrolyte AB at 328K is 100 ohm. The specific conductance of solution is ( cell constant =1 cm^(-1) )

The resistance of 0.1 N solution of formic acid is 200 ohm and cell constant is 2.0cm^(-2) . The equivalent conductivity (in Scm^(2)eg^(-1) ) of 0.1 N formic acid is 1.0 xx 10^(x) , x is