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 specific conductance:

Options

• 1.29 × 10⁻⁴ ohm⁻¹ cm⁻¹

• 12.9 × 10⁻⁴ ohm⁻¹ cm⁻¹

• 1.16 × 10⁻² ohm⁻¹ cm⁻¹

• 11.6 × 10⁻² ohm⁻¹ cm⁻¹

Answer 1

1.16 × 10⁻² ohm⁻¹ cm⁻¹

Explanation:

Resistance (R) = 31.6 Ω

Cell constant (G) = 0.347 cm⁻¹

Formula for specific conductance (κ):

`kappa = "Cell constant"/"Resistance"`

= `0.347/31.6`

= 0.01097 Ω⁻¹ cm⁻¹

= 1.10 × 10² Ω⁻¹ cm⁻¹

≈ 1.16 × 10⁻² ohm⁻¹ cm⁻¹