Calculate the degree of dissociation `(alpha)` of acetic acid if its molar conductivity is 39.05 S `cm^(2)mol^(-1)`. Given `lamda^(@)(H^(+))=349.6S" "cm^(2)mol^(-1) and lamda^(@)(CH_(3)CO O^(-))=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 is 39.05 S `cm^(2)mol^(-1)`. Given `lamda^(@)(H^(+))=349.6S" "cm^(2)mol^(-1) and lamda^(@)(CH_(3)CO O^(-))=40.9" S "cm^(2)mol^(-1)`.
Answer» `alpha=(wedge_(m))/(wedge_(m)^(@))` `wedge_(m)^(@)(CH_(3)COOH)=lamda^(@)(CH_(3)COO^(-))+lamda^(@)(H^(+))=40.9+349.6=390.5" S "cm^(2)mol^(-1)` `wedge_(m)=39.05" S "cm^(2)mol^(-1)` `thereforealpha=(39.05)/(390.5)=0.1`

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

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