In Carius method of estimation of halogen. $0.45 g$ of an organic compound gave $0.36 g$ of $AgBr$ . Find out the percentage of bromine in the compound.
(Molar masses : $AgBr = 188 g {mol}^{- 1} : Br = 80 g {mol}^{- 1}$ )

Question

In Carius method of estimation of halogen.

0.45 g

of an organic compound gave

0.36 g

of

AgBr

. Find out the percentage of bromine in the compound.
(Molar masses :

AgBr = 188 g {mol}^{- 1} : Br = 80 g {mol}^{- 1}

)

Aicharya · Accepted Answer

The Carius halogen method in analytical chemistry is a method for the quantitative determination of halogens in chemical substances. A known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as Carius tube, in a furnace.
Mass of organic compound $= 0.45 gm$
Mass of $AgBr$ obtained $= 0.36 gm$
$∴$ Moles of $AgBr = \frac{0.36}{188}$
$∴$ Mass of Bromine $= \frac{0.36}{188} \times 80 = 0.1532 gm$
$∴ % Br$ in compound $= \frac{0.1532}{0.45} \times 100 = 34.04 %$

In Carius method of estimation of halogen. 0 . 45   g of an organic compound gave 0 . 36   g of…

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In Carius method of estimation of halogen. 0 . 45 &#160; g of an organic compound gave 0 . 36 &#160; g of…

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In Carius method of estimation of halogen. 0 . 45 g of an organic compound gave 0 . 36 g of…