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IN A CONDUCTIVITY CELL THE TWO PLATINUM ELECTRODES EACH OF AREA 10 SQ CM ARE FIXED 1.5CM APART.THE CELL CONTAINED 0.05M SOLUTION OF A SALT . IF TWO ELECTRODES ARE JUST HALF DIPPED IN A SOLUTION WHICH HAS  A RESISTANCE OF 50 OHMS FIND MOLAR CONDUCTANCE OF THE SALT SOLUTION.
 
 
 
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Asked by sudheerkapoor67 9th October 2017, 6:13 PM
Answered by Expert
Answer:
 Area of each electrode = 10 sq.cm, half dipped so, a = 10 sq.cm
distance = 1.5 cm
concentration = 0.05 M solution
resistance = 50 ohms
Molar conductivity = ?
 
r e s i s t i i t y comma space rho space equals space R space A over l space equals space fraction numerator 50 cross times 10 over denominator 1.5 end fraction space equals space fraction numerator 500 over denominator 1.5 end fraction space equals space 333.33
K space left parenthesis c o n d u c t i v i t y space K space equals space 1 over rho space equals space fraction numerator 1 over denominator 333.33 end fraction space equals space 0.0030 space s space c m to the power of negative 1 end exponent
M o l a r space c o n d u c t i v i t y space space equals space fraction numerator k x 1000 over denominator m o l a r i t y end fraction space equals space fraction numerator 0.0030 x 1000 over denominator 0.05 end fraction space equals space fraction numerator 3 over denominator 0.05 end fraction space equals space 60 space S space c m squared space m o l to the power of negative 1 end exponent
 
Answered by Expert 11th October 2017, 12:12 PM
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