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 aren×10−1 , 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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answer is 3.

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Detailed Solution

PhCH2NH2+3CH3Br→PhCH2N+Me3 Br-0.1 mole 23 / 230 = 0.1 moleHence moles of CH3Br = 0.3 = 3*10-1mole

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