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Some Basic Concepts of Chemistry question

2022 · 27 Jul · Shift 1 · Q12
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Some Basic Concepts of Chemistry question

2022 · 27 Jul · Shift 1 · Q12

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
In Carius method of estimation of halogen, 0.45 g0.45 \mathrm{~g}0.45 g of an organic compound gave 0.36 g0.36 \mathrm{~g}0.36 g of AgBr\mathrm{AgBr}AgBr. Find out the percentage of bromine in the compound. (Molar masses : AgBr=188 g mol−1;Br=80 g mol−1\mathrm{AgBr}=188 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{Br}=80 \mathrm{~g} \mathrm{~mol}^{-1}AgBr=188 g mol−1;Br=80 g mol−1)
  1. A
    34.04%
  2. B
    40.04%
  3. C
    36.03%
  4. D
    38.04%
View written solutionFree

Correct answer: A

  1. Use Carius method relation

    In Carius method, halogen in the organic compound is converted into silver halide.
    Here, bromine is converted into AgBr\mathrm{AgBr}AgBr.

  2. Find mass of bromine present in 0.36 g0.36\,\text{g}0.36g of AgBr\mathrm{AgBr}AgBr

    Molar mass relation: 188 g of AgBr contains 80 g of Br188\,\text{g of } \mathrm{AgBr} \text{ contains } 80\,\text{g of Br}188g of AgBr contains 80g of Br

    Therefore, 0.36 g of AgBr contains 80188×0.36 g of Br0.36\,\text{g of } \mathrm{AgBr} \text{ contains } \frac{80}{188}\times 0.36\,\text{g of Br}0.36g of AgBr contains 18880​×0.36g of Br

    mBr=80×0.36188m_{\mathrm{Br}} = \frac{80\times 0.36}{188}mBr​=18880×0.36​

    mBr=28.8188=0.1532 gm_{\mathrm{Br}} = \frac{28.8}{188} = 0.1532\,\text{g}mBr​=18828.8​=0.1532g

  3. Calculate percentage of bromine in the organic compound

    Mass of organic compound =0.45 g= 0.45\,\text{g}=0.45g

    %Br=0.15320.45×100\%\mathrm{Br} = \frac{0.1532}{0.45}\times 100%Br=0.450.1532​×100

    %Br=34.04%\%\mathrm{Br} = 34.04\%%Br=34.04%

  4. Match with options

    The correct option is: A: 34.04%\boxed{\text{A: }34.04\%}A: 34.04%​

  5. Comparison with stored answer

    Stored correct answer is A, which matches the derived result.

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