Question

Consider the following reaction

On estimation of bromine in $100 \,g$ of $R$ using Carius method, the amount of $AgBr$ formed (in $g$ ) is __ [Given : Atomic mass of $H =1, C =12, O =16, P =31, Br =80, Ag =108$ ]

Answer 1

Molecular weight of (R) = 7 x 12 + 1 x 6 + 80 x 2= 250 g/mol
Mole of AgBr formed = 2 x mole of (R)
= 2 x $\frac{1}{250}$ x 188
= 8 x 10-3 x 188
= 1.504 g
The Amount of $AgBr$ formed is $\underline{1.50}.$