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Practical Organic Chemistry question

2023 · 13 Apr · Shift 2 · Q14
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Practical Organic Chemistry question

2023 · 13 Apr · Shift 2 · Q14

JEE MainChemistryPractical Organic ChemistryNumerical+4 / −1
0.400 g0.400 \mathrm{~g}0.400 g of an organic compound (X)(\mathrm{X})(X) gave 0.376 g0.376 \mathrm{~g}0.376 g of AgBr\mathrm{AgBr}AgBr in Carius method for estimation of bromine. %\%% of bromine in the compound (X)(\mathrm{X})(X) is ‾\underline{\hspace{2cm}}​. (Given: Molar mass AgBr=188 g mol−1Br=80 g mol−1\mathrm{AgBr=188~g~mol^{-1}}\mathrm{Br}=80 \mathrm{~g} \mathrm{~mol}^{-1}AgBr=188 g mol−1Br=80 g mol−1)
Numerical answer
View written solutionFree

Correct answer: 40

  1. Principle of Carius method

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

    So, the mass of bromine in the sample is obtained from the mass of AgBr\mathrm{AgBr}AgBr formed.

  2. Given data

    • Mass of organic compound, mX=0.400 gm_X = 0.400\,\mathrm{g}mX​=0.400g
    • Mass of AgBr\mathrm{AgBr}AgBr formed, mAgBr=0.376 gm_{\mathrm{AgBr}} = 0.376\,\mathrm{g}mAgBr​=0.376g
    • Molar mass of AgBr=188 g mol−1\mathrm{AgBr} = 188\,\mathrm{g\,mol^{-1}}AgBr=188gmol−1
    • Atomic mass of Br=80 g mol−1\mathrm{Br} = 80\,\mathrm{g\,mol^{-1}}Br=80gmol−1

  3. Find mass of bromine present in 0.376 g0.376\,\mathrm{g}0.376g of AgBr\mathrm{AgBr}AgBr

    In 188 g188\,\mathrm{g}188g of AgBr\mathrm{AgBr}AgBr, mass of bromine =80 g= 80\,\mathrm{g}=80g.

    Therefore, in 0.376 g0.376\,\mathrm{g}0.376g of AgBr\mathrm{AgBr}AgBr,

    Mass of Br=0.376×80188\text{Mass of Br} = 0.376 \times \frac{80}{188}Mass of Br=0.376×18880​ =0.376×2047= 0.376 \times \frac{20}{47}=0.376×4720​ =0.160 g= 0.160\,\mathrm{g}=0.160g

  4. Calculate percentage of bromine in the compound

    % Br=mass of Brmass of compound×100\%\,\mathrm{Br} = \frac{\text{mass of Br}}{\text{mass of compound}} \times 100%Br=mass of compoundmass of Br​×100 % Br=0.1600.400×100=40\%\,\mathrm{Br} = \frac{0.160}{0.400} \times 100 = 40%Br=0.4000.160​×100=40

  5. Final answer

    40\boxed{40}40​

  6. Comparison with stored correct answer

    Stored correct answer = 404040

    My derived answer also = 404040, so they agree.

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