The ionization constant of a weak acid is 1.6 xx 10^(-5) and the molar conductivity at infinite dilution is 380 xx 10^(-4)sm^(2) mol^(-1). If the cell constant is 0.01m^(-1), then conductance of 0.01M acid solution is

The ionization constant of a weak acid is 1.6 xx 10^(-5) and the molar

1.	The ionization constant of a weak acid is 1.6 xx 10^(-5) and the molar conductivity at infinite dilution is 380 xx 10^(-4)sm^(2) mol^(-1). If the cell constant is 0.01m^(-1), then conductance of 0.01M acid solution is
Answer» `1.52 XX 10^(-5) s ` `1.52 s ` `1.52 xx 10^(-3)s` `1.52 xx 10^(-4)` s Solution :`K_a = C alpha^(2) , alpha = sqrt((K_a)/(c)) = sqrt((1.6 xx 10^(-5))/(10^(-2))) = sqrt(1.6 xx 10^(-3)) = sqrt(16 xx 10^(-4))` ` alpha = 4 xx 10^(-2) = 0.04 , alpha = (^^_M)/(^^_(M)^(0)) implies ^^_M = alpha = ^^_(M)^(0) = 0.04 xx 380 xx 10^(-4) implies ^^_M(K xx 1000)/(M)` ` K=(^^_M xx M)/(1000) = (0.04 xx 380 xx 10^(-4) xx 0.01)/(1000) , K=0.152 xx 10^(-7) = 1.52 xx 10^(-8)` `C = (K)/(G) = (1.52 xx 10^(-8))/(0.01) = 1.52 xx 10^(-6) , C = 1.52 xx 10^(-6) S - m^(3) =1.52S CM^(3)`