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Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is (100Omega). The conductivity of this solution is 1.29sm^(-1).Resistance of the same cell when filled with 0.2 M of the same solution is 520Omega . The molar conductivity of (0.02)M solution of the electrolyte will be : |
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Answer» ` 124xx10^(-4) sm^(2) MOL^(-1)` For 0.1M solution `G^ast= (1.29)/(G) = 129m^(-1)`For 0.2M solution `K= Gxx(1)/(R ) = (129)/(520) sm^(-1)` If we DILUTE 0.2 M solution by 10 times, its concentration becomes 0.02 M. Thus number of ions PER unit volume is `(1)/(10)`th of INITIAL one. So K for this solution is `(129)/(520xx10) SM^(-1)` |
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