The conductance of a salt solution (AB) measured by two parallel electrodes of area 100 cm^(2) separated by 10 cm was found to be 0.0001 Omega^(-1). If volume enclosed between two electrode contains 0.1 mole of salt, and the molar conductivity (S cm^(2) mol^(-1)) of salt at same concentration is 1.0 xx 10^(-x), x is

The conductance of a salt solution (AB) measured by two parallel elect

1.	The conductance of a salt solution (AB) measured by two parallel electrodes of area 100 cm^(2) separated by 10 cm was found to be 0.0001 Omega^(-1). If volume enclosed between two electrode contains 0.1 mole of salt, and the molar conductivity (S cm^(2) mol^(-1)) of salt at same concentration is 1.0 xx 10^(-x), x is
Answer» Solution :`C = 10^(-4)Omega , a = 100 cm^(2) , L =10 cm , R= P. (1)/(a) implies (1)/(C) = (1)/(K) xx (1)/(a)implies K = C xx (1)/(a) = 10^(-4) xx (10)/(100)= 10^(-5)` ` M = (N)/(V) = (0.1)/( 100 xx 100) xx 1000 = 0.1M , ^^_M = (K xx 1000)/(M) = (10^(-5) xx 1000)/(0.1) = 0.1 = 1 xx 10^(-x) = 1 xx 10^(-1)` x=1