JEE MAIN - Chemistry (2021 - 18th March Evening Shift - No. 17)
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Consider the above reaction where 6.1 g of Benzoic acid is used to get 7.8 g of m-bromo benzoic acid. The percentage yield of the product is __________
(Round off to the Nearest Integer).
[Given : Atomic masses : C : 12.0 u, H : 1.0 u, O : 16.0 u, Br : 80.0 u]
Answer
78
Explanation
Moles of Benzoic acid = $${{6.1} \over {122}}$$ = moles of m-bromobenzoic acid
So, weight of m-bromobenzoic acid = $${{6.1} \over {122}}$$ $$\times$$ 201 gm
= 10.05 gm
% yield = $${{Actual\,weight} \over {Theoretical\,weight}} \times 100$$
$$ = {{7.8} \over {10.05}} \times 100$$ = 77.61%
So, weight of m-bromobenzoic acid = $${{6.1} \over {122}}$$ $$\times$$ 201 gm
= 10.05 gm
% yield = $${{Actual\,weight} \over {Theoretical\,weight}} \times 100$$
$$ = {{7.8} \over {10.05}} \times 100$$ = 77.61%
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