JEE MAIN - Chemistry (2021 - 27th August Morning Shift - No. 13)
In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine in the compound is ______________. (Nearest integer)
[Atomic mass : Ag = 108, Br = 80]
[Atomic mass : Ag = 108, Br = 80]
Answer
40
Explanation
nAgBr = $${{0.188g} \over {188g/mol}}$$ = 10$$-$$3 mol
$$\Rightarrow$$ nBr = nAgBr = 0.001 mol
$$\Rightarrow$$ massBr = (0.001 $$\times$$ 80) gm = 0.08 gm
$$\Rightarrow$$ mass% = $${{0.08 \times 100} \over {0.2}} = 40\% $$
$$\Rightarrow$$ nBr = nAgBr = 0.001 mol
$$\Rightarrow$$ massBr = (0.001 $$\times$$ 80) gm = 0.08 gm
$$\Rightarrow$$ mass% = $${{0.08 \times 100} \over {0.2}} = 40\% $$
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