#### Question

Resistance of conductivity cell filled with 0.1 M KCl solution is 100 ohms. If the resistance of the same cell when filled with 0.02 M KCl solution is 520 ohms, calculate the conductivity and molar conductivity of 0.02 M KCl solution. [Given: Conductivity of 0.1 M KCl solution is 1.29 S m^{-1} .]

#### Solution

Given: For 0.1 M KCl, k = 1.29 S m^{-1} = 0.0129 S cm^{-1}, R = 100 ohms or 0.02 M KCl, R = 520 ohms

To find: Conductivity and molar conductivity of 0.02 M KCl solution.

Formulae: a. Cell constant, b = k x R

b. Conductivity, k_{(KCl, 0.02 M)} = Cellconstant /Resistance

c.Molar conductivity = 1000 k/C

Calculation: From formula (a),

b = k x R

= 0.0129 x 100 = 1.29 cm^{-1}

From formula (b),

Conductivity, k_{(KCl, 0.02 M)} = Cellconstant/Resistance

= 1.29/520

= 2.48 x 10^{-3} ohm^{-1} cm^{-1}

From formula (c),

Molar conductivity of 0.02 M KCl solution = 1000k/C

=(1000 x 2.48 x 10 ^{-3})/0.02

=124 ohm^{-1} cm^{2} mol^{-1}