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Resistance of a conductivity cell filled with 0.1 mol L ^(-1) KCl solution is 100 Omega. If the resistance of the same cell when filled with 0.02 mol L^(-1) KCl solution is 520 Omega, calculate the conductivity and molar conductivity of 0.02 mol L^(-1)KCl solution. The conductivity of 0.1 mol L^(-1) KCl solutin is 1.29 S/m. |
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Answer» SOLUTION :The cell constant is given by the equation : Cell constant `=G^(**)=` CONDUCTIVITY `xx` RESISTANCE `=1.29 S//m m 100Omega =129m^(-1)=1.29 cm^(-1)` Conductivity of `0.02 mol L^(-1)` Kcl solution = cell constant /resistance `=(G^(**))/(R)=(129m^(-1))/(520Omega)=0.248Sm^(-1)` Concentration `=0.02 mol L^(-1)` `=1000xx0.02 mol m^(-3) =20 mol m^(-3)` Molar conductivity `=^^_(m)=k/c ` `(248xx10^(-3)Sm^(-1))/(20 mol m ^(-3))=124xx10^(-4) Sm^(2) mol ^(-1)` ALTERNATIVELY, `k=(1.29 cm^(-1))/(520Omega)=0.248xx10^(-2) S cm^(-1)` `and ^^_(m) =k xx1000 cm^(3) L-1 "molarity"^(-1)` `=(0.248 xx10^(-2) S cm^(-1)xx1000 cm ^(23) L^(-1))/(0.02 mol L^(-1))` `=124S cm^(2) mol ^(-1)` |
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