InterviewSolution
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InterviewSolution
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Calculating molar conductance: The resistance of `0.01 M` solution of an electrolyte was found to be `210 ohm` at `25^(@)C`. Calculate them molar conductance of the colution at `25^(@)C`, if the cell constant is `0.88 cm^(-1)` Strategy: First conductivity in `Sm^(-1)` (SI units) and then divide it by molar concentration in mole `m^(-3)` (SI units) to get molar conductance in `Sm^(2) mol^(-1)`. |
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Answer» `G = (1)/R = (1)/(210)S` `K_("cell") = 0.88 cm(-1)` According to Equation (3.18), we have `kappa = ("conductance")("cell constant")` `((1)/(210)S) (0.88 cm^(-1))` `= 0.00419 Scm^(-1)` `= 0.419 Sm^(-1)` Molar concentration of the solution: `C = 0.01 mol L^(-1) = 0.01 mol ((1)/(1000 m^(3)))^(-1)` `= 10 mol m^(-3)` Thus, `Lambda_(m) = (kappa)/(C) = (0.419 Sm^(-1))/(10 mol m^(-3))` `= 0.0419 Sm^(2)mol^(-1)` Alternatively, use the Equation `Lambda_(m) = (kappa(Sm^(-1)))/("molarity "(molL^(-1))xx(1000 L m^(-3))` `= (0.419 Sm^(-1))/((0.01 mol L^(-1)) xx (1000 L m^(-3))` `= 0.0419 Sm^(2) mol^(-1)`. |
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