Question

A 0.5 M NaOH solution offers a resistance of 31.6 ohm in a
conductivity cell at room temperature. What shall be the
approximate molar conductance of this NaOH solution if cell
constant of the cell is 0.367 cm⁻¹.
(a) 234 S cm² mole⁻¹ (b) 23.2 S cm² mole⁻¹
(c) 4645 S cm² mole⁻¹ (d) 5464 S cm² mole⁻¹

Answer 1

The molar conductance of this NaOH solution is 23.2 S cm² mole⁻¹

Option (B) is correct.

Explanation:

Given data:

We know that

Conductivity K = 1 / R x cell constant

Cell constant = 0.357 cm^-1

Resistance = 31.6 ohm

Therefore

Conductivity = 1 / 31.6 x  0.367  

Conductivity =  0.0316 x 0.367

Conductivity = 0.0115 ohm-1 cm-1

Molar conductivity = Conductivity / Molarity

Molar conductivity = 0.0113 / 0.5

Molar conductivity = 23.2 S cm² mole⁻¹

Thus the molar conductance of this NaOH solution is 23.2 S cm² mole⁻¹

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