A reaction of $0.1$ mole of Benzylamine with bromomethane gave $23\, g$ of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are $n \times 10^{-1},$ when $n =\ldots...$ (Round off to the Nearest Integer).
(Given : Atomic masses: $C$ : $12.0\, u$, $H : 1.0\, u , N : 14.0\, u , Br : 80.0\, u ]$
JEE MAIN 2021, Medium
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