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

Options

A. `1.52 xx 10^(-5) s `
B. `1.52 s `
C. `1.52 xx 10^(-3) s`
D. `1.52 xx 10^(-4)` s

Answer

B

Explanation

`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)`