InterviewSolution
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InterviewSolution
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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)`.A. `5xx10^(3)`B. `5xx10^(2)`C. `5xx10^(-4)`D. `5xx10^(-3)` |
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Answer» Correct Answer - c `R=(1)/(k)xx(l)/(a) ((l)/(a)=` cell constant `)` For `0.2M` solution `50=(1)/(k)xx(l)/(a)implies50=(1)/(1.4)xx(l)/(a)` `(l)/(a)=70m^(-1)` For `0.5M` solution `280=(1)/(k) xx70` `k=(1)/(4)Sm^(-1)` `^^_(m)((kxx1000)/(M))(10^(-2)m)^(3)` `^^_(m)=(1)/(4)xx((1000)/(M))(10^(-2)m)^(3)` `=(1)/(4)xx(1000)/(0.5)xx10^(-6)` `=500xx10^(-6)=5xx10^(-4)Sm^(2) mol^(-1)` |
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