JEE MAIN - Chemistry (2021 - 20th July Evening Shift - No. 21)
When 0.15 g of an organic compound was analyzed using Carius method for estimation of bromine, 0.2397 g of AgBr was obtained. The percentage of bromine in the organic compound is ______________. (Nearest integer)
[Atomic mass : Silver = 108, Bromine = 80]
[Atomic mass : Silver = 108, Bromine = 80]
Answer
68
Explanation
Moles of Br = Moles of AgBr obtained
$$\Rightarrow$$ Mass of Br = $${{0.2397} \over {188}}$$ $$\times$$ 80 g
therefore % Br in the organic compound
$$ = {{{W_{Br}}} \over {{W_T}}}$$ $$\times$$ 100
$${{0.2397 \times 80} \over {188 \times 0.15}}$$ $$\times$$ 100 = 0.85 $$\times$$ 80 = 68
$$\Rightarrow$$ Nearest integer is '68'.
$$\Rightarrow$$ Mass of Br = $${{0.2397} \over {188}}$$ $$\times$$ 80 g
therefore % Br in the organic compound
$$ = {{{W_{Br}}} \over {{W_T}}}$$ $$\times$$ 100
$${{0.2397 \times 80} \over {188 \times 0.15}}$$ $$\times$$ 100 = 0.85 $$\times$$ 80 = 68
$$\Rightarrow$$ Nearest integer is '68'.
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