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Calculate the degree of dissociation (al...

Calculate the degree of dissociation `(alpha)` of acetic acid if its molar conductivity `(^^_(m))` is 39.05 `S cm^(2) mol^(-1)`
Given `lamda^(@) (H^(+)) = 349.6 cm^(2) mol^(-1) and lamda^(2) (CH_(3)COO^(-)) = 40.9 S cm^(2) mol^(-1)`

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The correct Answer is:
0.1
`Lambda^(@) (CH_(3) COOH) = lambda^(@) (CH_(3) COO^(-)) + lambda^(@) (H^(+))`
`=40.9 + 349.6`
`= 390.5 S cm^(2) mol^(-1)`
`Lambda = 39.05 S cm^(2) mol^(-1)`
`alpha = (Lambda^(c))/(Lambda_(0)) = (39.05)/(390.5)`
= 0.10 or 10 %
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