Question

The resistance of a conductivity cell with a 0.1 M KCl solution is 200 ohm. When the same cell is filled with a 0.02 M NaCl solution, the resistance is 1100 ohm. If the conductivity of 0.1 M KCl solution is 0.0129 ohm^-1 cm^-1, calculate the cell constant and molar conductivity of 0.02 M NaCl solution.

Answer 1

Given: `"R"_((0.1  "M" "KCl"))` = 200 ohm, `"R"_((0.02  "M" "NaCl"))` = 1100 ohm

`"K"_((0.1  "M" "KCl"))`  = 0.0129 ohm^-1 cm^-1

`"G"_((0.02  "M" "NaCl"))` = ?, ∧_m = ?

Cell constant (G) = Conductivity (K) × Resistance (R)

= 200 × 0.0129

= 2.58 cm^-1

Conductivity `(lambda_(0.02  "M" "NaCl")) = "Cell constant (G)"/("Resistance" ("R"_(0.02  "M" "NaCl")))`

`= (2.58)/1100`

= 2.345 × 10^-3 s cm^-1

Molar conductivity (λ_m) `= "Conductivity (K) × 1000"/"Molarity"` 

`= (2.345 xx10^-3 xx 1000)/(0.02)`

= 117.25 s m² mol^-1

Hence, the cell constant and moral conductivity of 0.02 M NaCl will be 2.345 × 10^-3 s cm^-1 and 117.25 s m² mol^-1, respectively.