Verified Solution • 2025–26 Session moderate

QJEEM21Chbecd: JEE Main 2021 Chemistry - Class 11 JEEM21Chbecd Solution: A reaction of 0.1 mole of Benzylamine with bromomethane gave 23 …

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 = __________. (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]

Step-by-Step Verified Solution

Number of moles of benzyl trimethyl

ammonium bromide formed = 23230{{23} \over {230}} = 0.1

∴\therefore No. of moles of bromomethane consumed

= 3 ×\times 0.1

= 3 ×\times 10–1

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