When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of p-aminophenol produced is:

First we have to calculate the total charge passed through the electrolytic cell. The relation between current $$I$$, time $$t$$ and charge $$Q$$ is

$$Q = I \times t$$

The current is $$I = 9.65\ \text{A}$$ and the time is $$t = 1.0\ \text{hour} = 1.0 \times 60 \times 60\ \text{s} = 3600\ \text{s}$$. Substituting, we obtain

$$Q = 9.65\ \text{A} \times 3600\ \text{s} = 34\,740\ \text{C}$$

Faraday’s first law of electrolysis states that the mass $$m$$ of a substance produced is related to the charge by

$$m = \dfrac{Q\,M}{nF}$$

where

$$M = \text{molar mass of the substance},$$ $$n$$ = number of electrons required per mole of product, $$F = 96\,500\ \text{C mol}^{-1}\ (\text{Faraday’s constant}).$$

In an acidic medium nitrobenzene $$\text{C}_6\text{H}_5\text{NO}_2$$ is electro-reduced to p-aminophenol $$\text{H}_2N\!\!-\!\!\text{C}_6\text{H}_4\!\!-\!\!\text{OH}$$. For every mole of p-aminophenol formed, four electrons are taken up, so

$$n = 4$$

The molar mass of p-aminophenol is

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

Now we substitute every quantity into the mass formula:

$$m = \dfrac{34\,740\ \text{C} \times 109\ \text{g mol}^{-1}}{4 \times 96\,500\ \text{C mol}^{-1}}$$

Simplifying step by step, first divide the charge by Faraday’s constant to get the moles of electrons:

$$\dfrac{34\,740}{96\,500} = 0.36\ \text{mol of e}^-$$

Then divide by the electron requirement $$n = 4$$ to get the moles of product:

$$\dfrac{0.36}{4} = 0.09\ \text{mol of product}$$

Finally multiply by the molar mass to get the mass formed:

$$m = 0.09\ \text{mol} \times 109\ \text{g mol}^{-1} = 9.81\ \text{g}$$

Hence, the correct answer is Option C.