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

The resistance of 0.05 MCH_(3)COOH solution is found to be 100ohm. If the cell constant is 0.037 cm^(-1) , the molar conductivity of the solution is

In a conductive cell, at 298K, the resistance of 0.05M NaOH solution is 31.16. Ifthe cell constant of that cell is 0.367 cm^(-1) , then calculate molar conductance of the given solution.

In a conductive cell, at 298K, the resistance of 0.05M NaOH solution is 31.16. Ifthe cell constant of that cell is 0.367 cm^(-1) , then calculate molar conductance of the given solution.

A. 0.05 M NaOH solution offered, a.resistance of 31.6 ohm in a conductivity cell at 298 K. if the cell constant is 0.367 cm^-1.calculate molar conductivity of the sodium hydroxide soIution.

At 298 K the resistance of a 0.1M KCl solution is found to be 39.0ohm. If the conductivity (k) of this solution is 1.29 xx10^(-2)ohm^(-1)cm^(-1) at 298K, what is cell constant

At 298 K the resistance of a 0.1M KCl solution is found to be 39.0ohm. If the conductivity (k) of this solution is 1.29 xx10^(-2)ohm^(-1)cm^(-1) at 298K, what is cell contant

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

0.05M NaOH solution offered a resistance of 31.69 Omega in a conductivity cell at 298K. If the cell constant of the conductivity cell is 0.367cm^(-1) the molar conductivity of the NaOH solution is