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