When a conductivity cell was filled with 0.02 M KCl, it had a resistance of 82.4 ohm at 250C and when filled with 0.005 N K2SO4, it had a resistance of 326 ohm. Calculate :

(i) Cell constant(ii) Specific conductivity(iii) Equivalent conductivity of 0.005 N K2SO4 soln.The specific conductance of 0.02 M KCl is = 0.002768 ohm1 cm1.

1) Cell constant :
       Specific. conductivity = conductance * cell constant

Cell constant = sp. conductance/Conductance = sp. conductance* Resistance= 0.002768 * 82.4 = 0.2280 

2)Specific conductivity of 0.005 N K2SO4


Specific conductivity = cell constant/resistance=0.2280/326 =6.99*10-4 =0.00069 ohm-1cm-1

3)Equivalent conductivity =1000*specific conductance/ C = (1000*0.00069)/0.005 = 138 ohm-1 cm2 equivalent-1



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