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Conductivity of 2xx10^(-3)M methanoic acid is 8xx10^(-5)" S cm"^(-1). Calculate its molar conductivity and degree of dissociation if Lambda_(m)^(@) for methanoic acid is 404" S cm"^(2)" mol"^(-1). |
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Answer» Solution :MOLAR conductivity `Lamda_(m)=(kxx1000)/(C)=(8xx10^(-5)" S CM"^(-1)xx1000)/(2XX10^(-3)" mol L"^(-1))` `=(8xx10^(-2))/(2xx10^(-3))40" S cm"^(2)" mol"^(-1)` Degree of dissociation `(Lamda_(m))/(Lamda_(m)^(@))=(40)/(404)=0.099` |
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