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)`

Calculate the degree of dissociation `(alpha)` of acetic acid if its m

1.	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)`
Answer» Given: (i) `^^_(m)^(c) = 39.05 S cm^(2)//mol` (ii) `lamda^(@) (H^(+)) = 349.6 S cm^(2)//mol and lamda^(@) (CH_(3)COO^(-)) + lamda^(@) (H^(+))` `= 40.9 + 349.6 = 390.5 S cm^(2)//mol` Degree of dissociation `(alpha) = (^^_(m)^(c))/(^^_(m)^(@)) = (39.05)/(390.5) = alpha = 0.1`

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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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