1.

Molar conductance of `1M` solution of weak acid `HA` is `20 ohm^(-1) cm^(2) mol^(-1)`. Find `%` dissocation of `HA`: `((^^_(m)^(o)(H^(+)) =350 S cm^(2) mol^(-1)),(^^_(m)^(o)(A^(-)) =50 S cm^(2) mol^(-1)))`

Answer» Correct Answer - 5
`^^_(m)^(o) (HA) = 350 +50 = 400`
`alpha =(^^_(m)^(C))/(^^_(m)^(o)) xx 100 = (20)/(400) xx100 = 5%`


Discussion

No Comment Found

Related InterviewSolutions