The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is `1500 Omega`. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is `0.146 X 10^(-3) Scm^(-1)`?

(a) Explain why electrolysis of aqueous solution of NaCl gives H_2 at cathode and Cl_2 at anode. Write overall reaction. (b) The resistancee of a conductivity cell containing 0.000M KCl solution at 298K is 1500 Omega . Calculate the cell constant if conductivity of 0.001M KCl solution at 298K is 0.146 times 10^-3 S cm^-1 .

c) Resistance of a conductivity cell containing 0.1 M KCl solution is 100Omega . Cell constant of the cell is 1.29/cm. Calculate the conductivity of the solution at the same temperature.

The pH of 0.001 M NaOH solution is

Resistance of a conductivity cell filled with 0.02 M KCl solution is 520 Omega . Calculate the conductivity and molar conductivity of that solution. [Cell constant of the cell = 1.29 cm^(-1) ].

Calculate the pH of "0.001 M HCl" solution

The conductivity of 0.20M solution of KCL at 298K is 0.025 S cm^-1 . Calculate its molar conductivity.

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

The boiling point of 0.1 M KCl solution is ………..than 100^(@)C .