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]
Answer (integer)
3
Solution
Number of moles of benzyl trimethyl<br><br> ammonium
bromide formed = ${{23} \over {230}}$ = 0.1
<br><br>$\therefore$ No. of moles of bromomethane consumed
<br><br>= 3 $\times$ 0.1
<br><br>= 3 $\times$ 10<sup>–1</sup>
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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