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In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of $A g B r$ . The percentage of bromine in the compound is: $(a t . m a s s A g = 108; B r = 80)$

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Detailed Solution

$% X = \frac{A t o m i c m a s s o f B r}{M o l e c u l a r m a s s o f A g B r} \times \frac{w t . o f A g B r}{w t . o f o r g a n i c b r o m i d e} \times 100$

Thus $% B r = \frac{80}{188} \times \frac{141}{250} \times 100 = 24$

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