JEE MAIN - Chemistry (2021 - 18th March Morning Shift - No. 20)
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]
(Given : Atomic masses : C : 12.0 u, H : 1.0 u, N : 14.0 u, Br : 80.0 u]
Answer
3
Explanation
Number of moles of benzyl trimethyl
ammonium bromide formed = $${{23} \over {230}}$$ = 0.1
$$ \therefore $$ No. of moles of bromomethane consumed
= 3 $$ \times $$ 0.1
= 3 $$ \times $$ 10ā1
ammonium bromide formed = $${{23} \over {230}}$$ = 0.1
$$ \therefore $$ No. of moles of bromomethane consumed
= 3 $$ \times $$ 0.1
= 3 $$ \times $$ 10ā1
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