The molar conductivity of `0.025mol L^(-1)` methanoic acid is `46.1 S cm^(2)mol ^(-1).` Calculate its degree of dissociation and dissociation constant.
Given, `lamda^(0)(H^(+))=349.6S cm^(2) mol ^(-1)`
`and lamda^(0)(HCOO^(-))=54.6S cm^(2)mol^(-1)`

The molar conductivity of 0.024 "mol" L^(-1) methanoic acid is 46.1 S cm^(-2)"mol"^(-1) . Calculate its degree of dissociation and dissociation constant. Given lambda^(0)(H^(+) = 349.6 S cm^(2) mol^(-1) and lambda^(0)(HCOO^(-)) = 54.6 S cm^(2) mol^(-1) .

The conductivity of 0.001028 mol L^(-1) acetic acid is 4.95 xx 10^(-5) S cm^(-1) . Calculate its dissociation constant if Lambda_(m)^(0) for acetic is 390.5 S cm^(2) mol^(-1) ?

The conductivity of 0.001028 mol L ^(-1) acetic acid is 4.95 xx10^(-5)S cm ^(-1). Calculate its dissociation constant if ^^(m)^(0) for acetic acid id 390.5 S cm ^(2)mol ^(-1).

The conductivity of 0.20M solution of KCl at 298 K is 0.0248S cm^(-1). Calculate molar conductance.

The conductivity of 0.001 M acetic acid is 5 xx 10^(-5) S cm^(-1) and ^^""^(0) is 390.5 S cm^(2) "mol" ^(-1) then the calculated value of dissociation constant of acetic acid would be

Conductivity of 0.00241 M acetic acid is 7.896 xx 10^(-4) S cm^(-1) . Calculate its moar conductivity and if ^^_(m)^(@) for acetic is 390.5 S cm^(2) mo^(-1) , what is its dissociation constant ?

The degree of dissociation of an acid HA in 0.1 M solution is 0.1%. Its dissociation constant is

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)

Specific conductance of 0.1 M Nitric acid is 6.3 xx 10^(-2) "ohm"^(-1) cm^(-1) "mol"^(-1) The molar conductance of the solution is