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