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In an estimation of bromine by Carius method, 1.6 g of an organic compound gave 1.88 g of  AgBr. The mass % of Bromine in the compound is: 
 

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Moles of AgBr= \frac{1.88}{188}

Thus, the mole of  AgBr = 0.01

Thus, moles of bromine in organic compound = 0.8 g

Now, % of bromine is given as:

\mathrm{\%\: of\: bromine\: =\: \frac{0.8}{1.6}\: x\: 100\: =\: 50\%}

Thus, the correct answer is 50.

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Kuldeep Maurya

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