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Basics of Organic Chemistry question

2015 · Shift 0 · Q3
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Basics of Organic Chemistry question

2015 · Shift 0 · Q3

JEE MainChemistryBasics of Organic ChemistryMCQ+4 / −1
In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is: (at. Mass Ag = 108; Br = 80)
  1. A
    48
  2. B
    60
  3. C
    36
  4. D
    24
View written solutionFree

Correct answer: D

  1. Principle of Carius method

In Carius method, bromine present in the organic compound is converted into silver bromide, AgBr\mathrm{AgBr}AgBr.

So, from the mass of AgBr\mathrm{AgBr}AgBr obtained, we can find the mass of bromine.

  1. Molar mass of AgBr\mathrm{AgBr}AgBr

M(AgBr)=M(Ag)+M(Br)=108+80=188M(\mathrm{AgBr}) = M(\mathrm{Ag}) + M(\mathrm{Br}) = 108 + 80 = 188M(AgBr)=M(Ag)+M(Br)=108+80=188

Thus, 188188188 g of AgBr\mathrm{AgBr}AgBr contains 808080 g of Br.

  1. Mass of bromine in 141 mg of AgBr\mathrm{AgBr}AgBr

Mass of Br=141×80188 mg\text{Mass of Br} = 141 \times \frac{80}{188} \text{ mg}Mass of Br=141×18880​ mg

Mass of Br=60 mg\text{Mass of Br} = 60 \text{ mg}Mass of Br=60 mg

  1. Percentage of bromine in the compound

Mass of organic compound taken =250= 250=250 mg

%Br=60250×100=24%\%\text{Br} = \frac{60}{250} \times 100 = 24\%%Br=25060​×100=24%

  1. Matching with options

24%24\%24% corresponds to Option D.

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