Conductivity of 0.00241 M acetic acid is `7.896xx10^-5 S cm^-1`.Calculate its molar conductivity and what is its dissociation constant?

Conductivity of 2.5xx10^(-4)M methanoic acid is 5.25xx10^(-5)Scm^(-1) . Calculate its molar conductivity and degree of dissociation (Given lambda^@(H^+)=349.5Scm^2mol^(-1) and lambda^@(HCOO)=50.5 Scm^2mol^(-1) )

The conductivity of 0.20M solution of KCl at 298K is 0.0248 Scm-1 . Calculate its molar conductivity.

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

The conductivity of 0.20M solution of KCl at 298K is 0.0248 S//cm .Calculate its molar conductivity.

The conductivity of 0.20M solution of KCL at 298K is 0.0248Scm^-1 , Calculate its molar conductivity.

The conductivity of 0.001M acetic acid is 4X10^(-2)S//m . Calculate the dissociation constant of acetic acid if lambda_m^@ for acetic acid is 390Scm^2mol^(-1) .

Give reasons for the following Resistance of a conductivity cell filled with 0.1M KCl solution is 100 Omega . If the resistance of the same cell when filled with 0.02M KCl solution is 520 Omega . Calculate the conductivity and molar conductivity of 0.02M KCl solution. Conductivity of 0.1M KCl solution in 1.29 S//m .

The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 O. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146xx10^-3 S//cm .

A monoprotic acid in a 0.1M solution ionizes to 0.001%. Its ionization constant is

The limiting molar conductances of sodium chloride, hydrochloric acid and sodium acetate are 126.45, 426.16 and 91.0 Scm^2mol^-1 respectively at 298K. Calculate the limiting molar conductance of acetic acid at 298K.