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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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Answer» SOLUTION :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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