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Example 2.4

Chapter 2: Electrochemistry · CHEMISTRY

Example

Example . and L m = k × cm L – molarity – – – – – . × S cm × cm L . mol L = S cm mol – The electrical resistance of a column of . mol L – NaOH solution of diameter cm and length cm is . × ohm. Calculate its resistivity, conductivity and molar conductivity. A = p r = . × . cm = . cm = . × – m l = cm = . m l R A  or      . .785cm 50cm RA l = . W cm Conductivity =   .       S cm – = .01148 S cm – Molar conductivity, × c  cm L – – – – .01148 S cm × cm L . mol L = . S cm mol – If we want to calculate the values of different quantities in terms of ‘m’ instead of ‘cm’,  = RA l – . × × . × . m  = . × – W m   = .  = . S m – and c  – – . S m mol m = . × – S m mol – .

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