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QJEEM15Ch02a0: JEE Main 2015 Chemistry - Class 11 JEEM15Ch02a0 Solution: In Carius method of estimation of halogens, 250 mg of an…

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)

Multiple Choice Options

  • A 48
  • B 60
  • C 36
  • D 24

Step-by-Step Verified Solution

Mass of substance =250=250 mgmg =0.250=0.250 gg

Mass of AgBrAgBr =141=141 mg=0.141mg=0.141 gg

11 mole of AgBrAgBr =1=1 gg atom of BrBr

188188 gg of AgBrAgBr =80=80 gg of BrBr

188188 gg of AgBrAgBr contain bromine =80=80 gg

0.1410.141 gg of AgBrAgBr contain bromine =80188×0.141= {{80} \over {188}} \times 0.141

This much amount of bromine present in 0.2500.250 gg of organic compound

∴\therefore %\% of bromine =80188×0.4140.250×100=24%= {{80} \over {188}} \times {{0.414} \over {0.250}} \times 100 = 24\%

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