The resistance of 0.5 M solution of and electrolyte enclosed enclosed between two platinuim electrodes 1.56 cm apart and having an are of `2.0cm^(3)` was found to be 30 cm. Calculate the molar conductivity of the electrolyte/.

The resistance of 0.5 M solution of and electrolyte enclosed enclosed

1.	The resistance of 0.5 M solution of and electrolyte enclosed enclosed between two platinuim electrodes 1.56 cm apart and having an are of `2.0cm^(3)` was found to be 30 cm. Calculate the molar conductivity of the electrolyte/.
Answer» Correct Answer - `0.05 ohm^(-1)cm^(2)mol^(-1)` Step I. Calculation of specific conductance. Specific conductance (k) `=(1)/(R )xx"cell constant" =(1)/((30 ohm))xx((1.5 cm))/((2.0 cm^(2)))=0.025 ohm^(-1)cm^(-1)` Step II. Calculation of molar conductance. Molar conductance `(Lambda_(m))=(k)/(C )=((0.025" ohm"^(-1)"cm"^(-1)))/((0.5" mol cm"^(-3)))=0.05" mol"^(-1)ohm^(-1)cm^(2)`

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The resistance of 0.5 M solution of and electrolyte enclosed enclosed between two platinuim electrodes 1.56 cm apart and having an are of `2.0cm^(3)` was found to be 30 cm. Calculate the molar conductivity of the electrolyte/.

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