InterviewSolution
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InterviewSolution
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Resistance of `0.2 M` solution of an electrolyte is `50 Omega`. The specific conductance of the solution is ` 1.3 S m^(-1)`. If resistance of the `0.4 M` solution of the same electrolyte is `260 Omega`, its molar conductivity is .A. `6250 S m^(2) mol^(-1)`B. `6.25 xx 10^(-4) S m^(2) mol^(-1)`C. `625 xx 10^(-4) S m^(2) mol^(-1)`D. `62.5 S m^(2) mol^(-1)` |
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Answer» Correct Answer - B `k = (1)/(R) xx(1)/(a)` `1.3 = (1)/(50) xx (1)/(a)` `:. (l)/(a) = 65 m^(-1)` k of `0.4m` solution is `k = (1)/(R) xx(1)/(a)` `= (1)/(260) xx 65 = (1)/(4) ohm^(-1) m^(-1) = (1)/(4) Sm^(-1)` `^^_(m) = (k)/("molarity") = (1)/(4 xx 0.4) S m^(-1) mol^(-1) sm^(3)` `= (1)/(4 xx 0.4 xx 1000) Sm^(-) mol^(-1) m^(3)` `=(1)/(1.6) xx 10^(-3)` `=6.25 xx 10^(-4) S m^(2) mol^(-1)` |
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