Calculate the degree of dissociation `(alpha) of CH_(3)COOH at 298K.` ltbr Given that `^^_(CH_(3)COOH)^(oo)=11.75cm^(2)mol^(-1)` `^^_(CM_(3)COO^(-))^(oo)=40.65cm^(2) mol ^(-1)` `^^_(H^(+))^(0)=349.15 cm^(2)mol ^(-1)`

Calculate the degree of dissociation `(alpha) of CH_(3)COOH at 298K.`

1.	Calculate the degree of dissociation `(alpha) of CH_(3)COOH at 298K.` ltbr Given that `^^_(CH_(3)COOH)^(oo)=11.75cm^(2)mol^(-1)` `^^_(CM_(3)COO^(-))^(oo)=40.65cm^(2) mol ^(-1)` `^^_(H^(+))^(0)=349.15 cm^(2)mol ^(-1)`
Answer» Given that `^^_(CH_(3)COOH)^(oo) =11.75 cm ^(2) mol ^(-1)` `^^_(CH_(3)COO^(-))^(oo) =40.95 cm^(2) mol ^(-1)` `^^_(H)^(oo) =349.15 cm^(2)mol ^(-1)` `^^_(CH_(3)COOH)^(oo) =^^_(CH_(3) COO^(-))^(oo) +^^_(H^(+))^(oo)=40.95+349.15=390.1` Degree of dissociation `(alpha) =(^^(alpha))/(lamda^(oo))=(11.75)/(390.1)=0.30=3xx10^(-2)` `therefore alpha=3xx10^(-2)`

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Calculate the degree of dissociation `(alpha) of CH_(3)COOH at 298K.` ltbr Given that `^^_(CH_(3)COOH)^(oo)=11.75cm^(2)mol^(-1)` `^^_(CM_(3)COO^(-))^(oo)=40.65cm^(2) mol ^(-1)` `^^_(H^(+))^(0)=349.15 cm^(2)mol ^(-1)`

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