Consider the following sequence of reactions:
Chlorobenzene $\xrightarrow[ii) CO_2, H_3O^{+}]{i) Mg, \text{dry ether}} \text{Benzoic acid}$ $\xrightarrow{NH_3, \Delta} \text{Benzamide (A)}$ $\xrightarrow{Br_2, NaOH} \text{Aniline (B)}$
$11.25 \ mg$ of chlorobenzene will produce $.......... \times 10^{-1} \ mg$ of product $B$.
(Consider the reactions result in complete conversion.)
[Given molar mass of $C, H, O, N$ and $Cl$ as $12, 1, 16, 14$ and $35.5 \ g \ mol^{-1}$ respectively]

• A
  $90$
• B
  $91$
• C
  $92$
• D
  $93$