During electrolysis of an aqueous solution of sodium sulphate, 2.4 L of oxygen at STP was liberated at anode. The volume of hydrogen at STP, liberated at cathode would be

4.8 L

The correct answer is option 4. 4.8 L.

Given:

\(2.4 L\) of oxygen gas \((O_2)\) is liberated at the anode during the electrolysis of an aqueous solution of sodium sulfate.

According to the stoichiometry of water electrolysis:

The reaction at the anode produces oxygen gas \((O_2)\):

\(4 OH^- \longrightarrow 2H_2O + O_2 + 4e^-\)

For every mole of oxygen gas produced, two moles of hydrogen gas \((H_2)\) are produced at the cathode:

\(2 H_2O + 2 e^- \longrightarrow H_2 + 2 OH^-\)

Now, to find the volume of hydrogen gas produced at STP, we use the ratio of their molar volumes (22.4 L/mol at STP):

Calculate the number of moles of oxygen gas produced:

2.4 L of O₂ at STP corresponds to \( \frac{2.4}{22.4} \) moles of O₂.

\( \frac{2.4}{22.4} \) moles of O₂ is approximately 0.1071 moles of O₂.

According to the stoichiometry, 1 mole of O₂ produces 2 moles of H₂:

So, \( 0.1071 \) moles of O₂ will produce \( 2 \times 0.1071 = 0.2142 \) moles of H₂.

Convert moles of H₂ to volume at STP:

\( 0.2142 \) moles of H₂ at STP corresponds to \( 0.2142 \times 22.4 = 4.794 \approx 4.8\) L of H₂.

Therefore, the volume of hydrogen gas liberated at STP during the electrolysis of the aqueous solution of sodium sulfate is approximately \( \mathbf{4.8 \, L} \), which corresponds to option 4.