The conductivity of 0.001 M acetic acid is `5xx10^(-5)S cm^(-1)` and `^^^(@)` is 390.5 `S cm^(2) "mol"^(-1)` then the calculated value of dissociation c onstnat of acetic acid would be

The conductivity of 0.001 M acetic acid is 5xx10^(-5)S cm^(-1) and ^^^(@) is 390.5 S cm^(2) "mol"^(-1) then the calculated value of dissociation constant of acetic acid would be

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

At a given temperature, specific conductance of 0.001(M) acetic acid solution is 5xx10^(-5)"S.cm"^(-1) and equivalent conductivity at infinite dilution is "390.5 S.cm"^(2).mol"^(-1) . What is the value of dissociation constant of acetic acid at the same temperature.

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 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).

Conductivity of 0.00241 M acetic acid is 7.896 x 10^(-5) S cm^(-1). Calculate its molar conductivity. If A°_m for acetic acid is 390.5 S cm^2 mol^(-1) , what is its dissociation constant?

The equivalent conductance of 0.1 N acetic acid is 5 cm^2 ohm^-1 gm eq^-1 and at infinite dilution is 390 cm^2ohm^-1gm eq^-1 . Calculate the degree of dissociation of acetic acid.

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