Question

A 0.05 M NH₄OH solution offers the resistance of 50 ohms to a conductivity cell at 298 K. If the cell constant is 0.50 cm⁻¹ and molar conductance of NH₄OH at infinite dilution is 471.4 ohm⁻¹ cm² mol⁻¹, calculate:

1. Specific conductance
2. Molar conductance
3. Degree of dissociation

Answer 1

Molarity (C) = 0.05 mol/L

Resistance (R) = 50 ohms

Cell constant, `l/a` = 0.50 cm⁻¹

`Lambda_m^infty` = 471.4 ohm⁻¹ cm² mol⁻¹

C = `1/R`

= `1/50` ohm⁻¹

a. Specific conductance:

κ = conductance × cell constant

= `1/R xx K`

= `1/50 xx 0.50`

= `1/100`

= 0.01 ohm⁻¹ cm⁻¹

b. Molar conductance:

`Lambda_m = (kappa xx 1000)/(C)`

= `(0.01 xx 1000)/(0.05)`

= `(10)/(0.05)`

= 200 ohm⁻¹ cm² mol⁻¹

c. Degree of dissociation:

`alpha = Lambda_m/Lambda_m^infty`

= `200/471.4`

= 0.424