The conductivity of 0.20 M solution of KCl at 298 K is 0.248S `cm^(-1)` . Calculate its molar conductivity.

The resistance of a conductivity cell when filled with 0.05M solution of an electrolyte x is 100 Omega at 40^(@)C . The same conductivity cell filled with 0.01M solution of electrolyte y has a resistance of 50 Omega . The conductivity of 0.05 M solution of electrolyte x is 1.0 xx 10^(-4) S cm^(-1) . Calculate (i) cell constant (ii) conductivity of 0.01 M solution (iii) molar conductivity of 0.01 M solution

Conductivity of 0.00241 M acetic acid solution is 7.896 xx 10^(-5) S cm^(-1) . Calculate its molar conductivity in this solution. If Lambda_(m)^(@) for acetic acid be 390.5 S cm^(2) "mol"^(-1) , what would be its dissociation constant? (i) First , find molar conductivity using the formula,, Lambda_(m) = ( K xx 1000)/( C ) (ii) Then find degree of dissociation ( alpha) and dissociation constant ( K_(a)) by using formula alpha = ( Lambda_(m))/( Lambda_(m)^(@)) and K_(a) = ( C alpha^(2))/( 1-alpha) respectively.

Conductivity of 0.00241M acetic acid is 7.896 xx 10^(-5) S cm^(-1) . Calculate its molar condcutivity. If Lambda_(m)^(@) acetic acid is 390.5 S cm^(2) "mol"^(-1) , what is its dissociation constant?

Conductivity of 2.5 xx 10^(-4) M methonoic acid is 5.25 xx 10^(-5) S cm^(-1) . Calculate its molar Condcutivity and degree of dissociation. Given : lambda^(@) ( H^(+) = 349.5 S cm^(2) "mol"^(-1) and lambda^(@) ( HCOO^(-)) = 50.5 S cm^(2) "mol"^(-1) .

At a certain temperature the specific conductance of 0.40M KCl solution is 4.96xx10^-2 ohm^-1cm^-1 . Calculate its molar conductance.