When 9.65 A current was passed for 1.0 h in | vedclass

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(C) The total charge passed is $Q = I 	imes t = 9.65 \ A 	imes 3600 \ s = 34740 \ C$.Number of moles of electrons passed $= \frac{Q}{F} = \frac{3474

Vedclass · Accepted Answer

$9.81$

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(C) The total charge passed is $Q = I 	imes t = 9.65 \ A 	imes 3600 \ s = 34740 \ C$.Number of moles of electrons passed $= \frac{Q}{F} = \frac{34740 \ C}{96500 \ C/mol} = 0.36 \ mol$.The reduction of nitrobenzene to $p$-aminophenol involves $4$ electrons:$C_6H_5NO_2 + 4H^+ + 4e^- 	o HOC_6H_4NH_2 + H_2O$.Since $4 \ mol$ of electrons produce $1 \ mol$ of $p$-aminophenol,$0.36 \ mol$ of electrons will produce $\frac{0.36}{4} = 0.09 \ mol$ of $p$-aminophenol.The molar mass of $p$-aminophenol $(C_6H_7NO)$ is $109.13 \ g/mol$.Mass of $p$-aminophenol $= 0.09 \ mol 	imes 109.13 \ g/mol = 9.82 \ g \approx 9.81 \ g$.

When $9.65 \ A$ current was passed for $1.0 \ h$ into nitrobenzene in an acidic medium,the amount of $p$-aminophenol produced is .............. $g$.

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