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Calculate the degree of dissociation (al...

Calculate the degree of dissociation (alpha) of acetic acid if its molar conductivity is 39.05` Scm^(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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To calculate the degree of dissociation (α) of acetic acid, we can follow these steps: ### Step 1: Understand the relationship between molar conductivity and degree of dissociation The degree of dissociation (α) can be calculated using the formula: \[ \alpha = \frac{\Lambda}{\Lambda^0} \] where: ...
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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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The resistance of 0.01 M CH_(2)COOH solution was found to be 2220 ohm in a conductivity cell having cell constant 0.366 cm^(-1) . Calculate: (i) molar conductivity (wedge_(m))" of "0.01 M CH_(3)COOH (ii) wedge_(m)^(oo) (iii) degree of dissociation, alpha and (iv) dissociation constant of the acid. [lambda^(0) (H^(+))=349.1"ohm"^(-1)cm^(2)"mol"^(-1), lambda^(0)(CH_(3)COO^(-)) =40.9 "ohm"^(-1)cm^(2) "mol"^(-1)]

The molar conductivity of 0.25 mol L^(-1) methanoic acid is 46.1 S cm^(2) mol^(-1) . Calculate the degree of dissociation constant. Given : lambda_((H^(o+)))^(@)=349.6S cm^(2)mol^(-1) and lambda_(HCOO^(-))^(@)=54.6Scm^(2)mol^(-1)

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