1.

The resistance of 0.1 N solution of formic acid is 200 ohm and cell constant is 2.0cm^(-2). The equivalent conductivity (in Scm^(2)eg^(-1)) of 0.1 N formic acid is 1.0 xx 10^(x) , x is

Answer»


Solution :`R = S . (1)/(A) IMPLIES K =(1)/(S) = (2)/(200) = 10^(-2) , ^^_(e) = K xx (1000)/(N) = 10^(-2) xx (1000)/(10^(-1)) = 10^(2)`


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