JEE MAIN - Chemistry (2020 - 6th September Morning Slot - No. 4)
In an estimation of bromine by Carius method,
1.6 g of an organic compound gave 1.88 g of
AgBr. The mass percentage of bromine in the
compound is _______.
(Atomic mass, Ag = 108, Br = 80 g mol–1)
(Atomic mass, Ag = 108, Br = 80 g mol–1)
Answer
50
Explanation
Mass of organic compound = 1.6 gm
Mass of AgBr = 1.88 gm
Moles of Br = Moles of AgBr = $${{1.88} \over {188}}$$ = 0.01
Mass of Br = 0.01 × 80 = 0.80 gm
% of Br = $${{0.80 \times 100} \over {1.60}}$$ = 50 %
Mass of AgBr = 1.88 gm
Moles of Br = Moles of AgBr = $${{1.88} \over {188}}$$ = 0.01
Mass of Br = 0.01 × 80 = 0.80 gm
% of Br = $${{0.80 \times 100} \over {1.60}}$$ = 50 %
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