The molecular conductivity of acetic acid at infinite dilution is `390 S cm^(2) "mole"^(-1)` and for `0.1` acetic and solution is `5.8 cm^(3) "mole"^(-1)` .The hydrogen ion concentration of the solution is

Options

A. `15 xx 10^(-3)`
B. `15 xx 10^(-4)`
C. `66 xx 10^(-3)`
D. `66 xx 10^(-4)`

Answer

b

Explanation

`alpha = (^^_C)/(^^ ^(prop)) = (5.8)/(390) ~= 15 xx 10^(-3)`
`[H^(-)] = Cu = 0.1 xx 15 xx 10^(-3) = 15 xx 10^(-4)`