A reaction of $0.1$ mole of benzylamine with bromoethane gave $23 \mathrm{g}$ of benzyl trimethyl ammonium bromide. The number of moles of bromoethane consumed in this reaction are $\mathrm{n}\times {10}^{-1}$, where $\mathrm{n}=$

$\mathrm{Ph}-{\mathrm{NH}}_2+3{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}\to \mathrm{Ph}-{\mathrm{N}}^{+}{\left({\mathrm{CH}}_3\right)}_3{\mathrm{Br}}^{-}$

Moles of benzylamine $=0.1$ mole

Moles of benzyl trimethyl ammonium bromide $=\frac{23}{230}$

$=0.1 \mathrm{mole}$

Since, 1 mole benzylamine reacts with 3 mole of bromoethane to give 1 mole of benzyl trimethyl ammonium bromide.

$0.1$ mole of benzylamine react with moles of bromoethane $=\frac{3}{1}\times 0.1$ moles $=0.3$ moles

$\Rightarrow$ The number of moles of bromoethane consumed in this reaction are $3\times {10}^{-1}$ Hence, the value of ' n ' is 3 .