Question

A 0.05 M NH₄OH solution offers the resistance of 30.8 ohms to a conductivity cell at 298K. If the cell constant is 0.343 cm⁻¹ and the molar conductance of NH₄OH at infinite dilution is 471.4 S cm² mol⁻¹, calculate the following:

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

Answer 1

Given:

Molarity (M) = 0.05 M

Resistance = 30.8 Ω

Cell constant = 0.343 cm⁻¹

Molar conductance `π_(m)^∞` = 471.4 s cm² mol⁻¹

(1) Specific conductance (K)

= `1/"R" xx "cell constant"`

= `1/30.8 xx 0.343`

= 0.011Ω⁻¹ cm⁻¹

(2) Molar conductance (π_m)

`π = ("K" xx 1000)/"M"`

= `(0.011 xx 1000)/0.05`

= 220 Ω⁻¹ cm²/mol

(3) Degree of dissociation (α)

`α = (π_m)/(π_(m)^∞)`

= `220/471.4`

= 0.47