Verified Solution • 2025–26 Session moderate

QJEEM22Ch9df0: JEE Main 2022 Chemistry - Class 11 JEEM22Ch9df0 Solution: In Carius method of estimation of halogen, 0.45   g of an…

In Carius method of estimation of halogen, 0.45 g of an organic compound gave 0.36 g of AgBr. Find out the percentage of bromine in the compound.

(Molar masses : AgBr=188 g mol−1;Br=80 g mol−1)

Multiple Choice Options

  • A 34.04%
  • B 40.04%
  • C 36.03%
  • D 38.04%

Step-by-Step Verified Solution

Mass of organic compound =0.45 gm=0.45 \,\mathrm{gm}

Mass of AgBr\mathrm{AgBr} obtained =0.36 gm=0.36 \,\mathrm{gm}

∴\therefore Moles of AgBr=0.36188\mathrm{AgBr}=\frac{0.36}{188}

∴\therefore Mass of Bromine =0.36188×80=0.1532 gm=\frac{0.36}{188} \times 80=0.1532 \,\mathrm{gm}

∴% Br\therefore \%\, \mathrm{Br} in compound =0.15320.45×100=34.04 %=\frac{0.1532}{0.45} \times 100=34.04 \,\%

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