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Resistance of a conductivity cell filled with 0.02 M KCl solution is 520 Omega. Calculate the conductivity and molarconductivity of that solution. [Cell constant of the cell = 1.29 cm^(-1)]. |
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Answer» SOLUTION :Cell CONSTANT =`1.29 cm^(-1)," Resistance" =520Ohm` Molar concentration of KCl=0.02 M `"Conductivity "=("Cell constant")/("Resistance")=(1.29)/(520)=0.248xx 10^(-2)Scm^(-1)` `wedge_(m) (KCl)=(k xx 1000)/(C)=(0.248 xx 1000)/(0.02) =124 Scm^(2)mol^(-1)`. |
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