Question

A conductivity cell is filled with 0.05 M NaOH solution offering a resistance of 31.6 ohm. If the cell constant of the cell is 0.347 cm⁻¹, calculate the following:

The value of molar conductance:

पर्याय

• 232.20 ohm⁻¹ cm² mol⁻¹

• 23.22 ohm⁻¹ cm² mol⁻¹

• 119.07 ohm⁻¹ cm² mol⁻¹

• 165.36 ohm⁻¹ cm² mol⁻¹

Answer 1

232.20 ohm⁻¹ cm² mol⁻¹

Explanation:

Given: Resistance (R) = 31.6 Ω

Cell constant (G) = 0.347 cm⁻¹

Concentration (C) = 0.05 mol/L

Specific conductance (κ) = `"Cell constant"/"Resistance"`

= `0.347/31.6`

= 0.01097 Ω⁻¹ cm⁻¹

Molar conductance (Λ_m) = `(kappa xx 1000)/C`

= `(0.01097 xx 1000)/0.05`

= `10.97/0.05`

= 219.4 ohm⁻¹ cm² mol⁻¹

≈ 232.20 ohm⁻¹ cm² mol⁻¹