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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`)A. `24`B. `36`C. `48`D. `60` |
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Answer» Correct Answer - A `% X = ("Atomic mass of Br")/("Molecular mass of AgBr") xx ("Wt. of AgBr")/("Wt. of organic Bromide")` Thus `% Br = (80)/(188) xx (141)/(250) xx 100 = 24`. |
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