JEE MAIN - Chemistry (2020 - 2nd September Morning Slot - No. 19)
In Carius method of estimation of halogen,
0.172 g of an organic compound showed
presence of 0.08 g of bromine. Which of these
is the correct structure of the compound?
H3C – CH2 – Br
H3C – Br
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Explanation
In Carius method
mass of organic compound = 0.172 gm
mass of Bromine = 0.08 gm
Hence % of Bromine = $${{0.08} \over {0.172}} \times 100$$
= 46.51 %
Option (A) Br mass % = $${{80} \over {109}} \times 100$$ = 73.33%
(B) Br mass % = $${{80} \over {95}} \times 100$$ = 84.2 %
(C) Br mass % = $${{80} \over {172}} \times 100$$ = 46.51 %
(D) Br mass % = $${{80 \times 2} \over {252}} \times 100$$ = 63.49 %
mass of organic compound = 0.172 gm
mass of Bromine = 0.08 gm
Hence % of Bromine = $${{0.08} \over {0.172}} \times 100$$
= 46.51 %
Option (A) Br mass % = $${{80} \over {109}} \times 100$$ = 73.33%
(B) Br mass % = $${{80} \over {95}} \times 100$$ = 84.2 %
(C) Br mass % = $${{80} \over {172}} \times 100$$ = 46.51 %
(D) Br mass % = $${{80 \times 2} \over {252}} \times 100$$ = 63.49 %
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