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Question

A reaction of 0.1 mole of Benzyl amine with bromomethane gave 23 g of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are n×101, when n =_____
. (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]

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Solution

Answer: (3)
The given reaction is,
PhCH2NH20.1 mol+3CH3BrPhCH2N+(Me)3Br23 g
Moles of benzyltrimethylammonium bromide=23/230=0.1 mol
moles of CH3Br= 0.3 mol=3×101 mol

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