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Calculating molar conductance: The resis...

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)`.

Text Solution

Verified by Experts

`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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