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

24

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Solution

The correct option is A $24$
The molar mass of $A g B r$ is $108 + 80 = 188 g / m o l$
Hence, 141 mg of $A g B r$ will contain $141 m g \times \frac{80 g}{188 g} = 60 m g$ of bromine.
Hence, the percentage of bromine in the compound is $\frac{60 m g}{250 m g} \times 100 = 24 %$ .

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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:(Atomic mass Ag=108;Br=80)

The correct option is A 24The molar mass of AgBr is 108+80=188g/molHence, 141 mg of AgBr will contain 141mg×80g188g=60mg of bromine.Hence, the percentage of bromine in the compound is 60mg250mg×100=24%.

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

The correct option is A $24$
The molar mass of $A g B r$ is $108 + 80 = 188 g / m o l$
Hence, 141 mg of $A g B r$ will contain $141 m g \times \frac{80 g}{188 g} = 60 m g$ of bromine.
Hence, the percentage of bromine in the compound is $\frac{60 m g}{250 m g} \times 100 = 24 %$ .