1. Identify the electrochemical reduction involved

In acidic medium, nitrobenzene is electrolytically reduced to $p$-aminophenol.

The overall reduction requires 6 electrons per mole of nitrobenzene (and hence per mole of $p$-aminophenol formed):

$\text{C}_6\text{H}_5\text{NO}_2 + 6e^- + 6H^+ \rightarrow \text{C}_6\text{H}_5\text{NHOH} + H_2O$

followed by rearrangement to $p$-aminophenol. Thus,

$1\text{ mol } p\text{-aminophenol} \leftrightarrow 6\text{ mol } e^-$

2. Calculate total charge passed

Given:

• Current, $I = 9.65\ \text{A}$
• Time, $t = 1.0\ \text{hour} = 3600\ \text{s}$

So total charge passed is:

$Q = It = 9.65 \times 3600 = 34740\ \text{C}$

3. Calculate moles of electrons passed

Using Faraday constant:

$1\text{ F} = 96500\ \text{C mol}^{-1}$

Hence moles of electrons:

$n(e^-) = \frac{34740}{96500} = 0.36\ \text{mol}$

4. Calculate moles of $p$-aminophenol formed

Since 6 moles of electrons produce 1 mole of $p$-aminophenol,

$n(p\text{-aminophenol}) = \frac{0.36}{6} = 0.06\ \text{mol}$

5. Calculate molar mass of $p$-aminophenol

$p$-Aminophenol has formula $\text{C}_6\text{H}_7\text{NO}$.

Molar mass:

$M = 6(12) + 7(1) + 14 + 16 = 72 + 7 + 14 + 16 = 109\ \text{g mol}^{-1}$

6. Calculate mass formed

$m = nM = 0.06 \times 109 = 6.54\ \text{g}$

So the amount of $p$-aminophenol produced is:

$\boxed{6.54\ \text{g}}$

7. Compare with given options

Options are:

• A: $9.81\ \text{g}$
• B: $10.9\ \text{g}$
• C: $98.1\ \text{g}$
• D: $109.0\ \text{g}$

Our calculated answer $6.54\ \text{g}$ does not match any option.

A common mistake would be to assume a 4-electron process:

$\frac{0.36}{4} = 0.09\ \text{mol}, \quad 0.09 \times 109 = 9.81\ \text{g}$

This gives option A, but formation of $p$-aminophenol from nitrobenzene in acidic medium proceeds via reduction to phenylhydroxylamine, which is a 6-electron reduction from the nitro group. Therefore, option A appears inconsistent with the proper electron balance.