Question

A conductivity cell when filled with 0.05 M solution of KCl records a resistance of 410.5 ohm at 25°C. When filled with CaCl₂ solution (11 g in 500 mL), it records a resistance of 990 ohms. If specific conductance of 0.05 M KCl is 0.00189 ohm⁻¹ cm⁻¹, calculate

1. cell constant; 
2. specific conductance of CaCl₂;
3. equivalent conductance of CaCl₂ solution;
4. molar conductance of CaCl₂ solution.

Answer 1

Given:

Resistance of KCl solution R₁ = 410.5 Ω

Specific conductance of KCl κ₁ = 0.00189 S cm⁻¹

Resistance of CaCl₂ solution R₂ = 990 Ω

Mass of CaCl₂ = 11 g in 500 mL = 0.5 L

Molar mass of CaCl₂ = 40.08 + 2 × 35.45 = 110.98 g/mol

i. Cell constant = κ₁​ × R₁​

= 0.00189 × 410.5

= 0.776 cm⁻¹

ii. Specific conductance of CaCl₂

`kappa_2 = 0.776/990`

= 0.0007848 ohm cm⁻¹

iii. Equivalent conductance of CaCl₂

Equivalents of CaCl₂ = mass/equivalent weight = `11/55.49` = 0.198 eq

Volume = 500 mL = 0.5 L

Normality N = `0.198/0.5` = 0.396 N

`Lambda_"eq" = (kappa_2 xx 1000)/N`

= `(0.0007848 xx 1000)/0.396`

= 1.981 ohm cm² eq⁻¹

iv. Molar conductance of CaCl₂

Moles of CaCl₂ = `11/110.98` = 0.0991

Molarity M = `0.0991/0.5` = 0.1982 

`Lambda_"eq" = (kappa_2 xx 1000)/M`

= `(0.0007848 xx 100)/0.1982`

Λ_eq = 3.96 ohm cm² mol⁻¹