In an estimation of bromine by Carius method, $1.6 g$ of an organic compound gave $1.88 g$ of $AgBr$ . The mass percentage of bromine in the compound is $\dots . .$
$($ Atomic mass, $Ag = 108, Br = 80 {gmol}^{- 1}$

Question

In an estimation of bromine by Carius method,

1.6 g

of an organic compound gave

1.88 g

of

AgBr

. The mass percentage of bromine in the compound is

\dots . .

(

Atomic mass,

Ag = 108, Br = 80 {gmol}^{- 1}

Aicharya · Accepted Answer

if molar mass of compound is M 1 . 6 &#160; / &#160; M = 1 . 88 &#160; / &#160; 188 = 0 . 01 so m = 160 gram/mole Now % of bromine = 80 / 160 &#215; 100 % = 50 %

In an estimation of bromine by Carius method, 1 . 6 g of an organic compound gave 1 . 88 g of AgBr . The…

Solution