In the Carius method of estimation of halogen. 0.25 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine in the compound is
(Molar mass of AgBr = 188 g `mol^-1` )

Explanation

Here, the mass of the substance taken = 0.15 g
Mass of AgBr formed = 0.12 g
Now 1 mole of `AgBr -= 1g` atom of Br or 188 (108 + 88) g of `AgBr -= 80g` of Br
`:.` 188 of AgBr contain bromine = 80 g `:.` 0.12 g of AgBr will contain bromine `= (80)/(188) xx 0.12 g`
But this much amount of bromine is present in 0.15 g of the organic compound.
`:.` Percentage of bromine `= (80)/(188) xx (0.12)/(0.15) xx 100 = 34.04`.