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c) Resistance of a conductivity cell con...

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. 

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Given
Resistance `=100Omega`
Cell constant `=1.29//cm`
To find conductivity of the cell we have the formula
`K=("Cell constant")/("Resistant")`
`K=(1.29)/(100)=1.29xx10^(-2)Scm^(-1)`
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(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 .

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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) ].

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) .

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Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1M is 100 Omega. The conductivity of this solution is 1.29S m ^(-1). Resistance of the same cell when filled with 0.2M of the same solution is 520 Omega. The molar conductivity of 0.2M solution of the electrolyte will be

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