Question: In the estimation of halogen, 0.18gm of an organic compound gave 0.12gm of silver bromide. What is t...

Question

In the estimation of halogen, 0.18gm of an organic compound gave 0.12gm of silver bromide. What is the percentage of bromine in the compound?
[Molar mass of AgBr=188; Atomic weight of Br=80]
(A) 35.24
(B) 34.84
(C) 28.36
(D) 30.64

Answer 1

Solution

We can find the % of bromine in the compound by following formula.
% of Bromine in the compound= $\dfrac{\text{Weight of Br in the compound}\times \text{100}}{\text{Molecular weight of the compound}}$

Complete answer:
It is evident that the compound has bromine in its structure. So, we can write the given reaction as
$\underset{0.18gm}{\mathop{Compound}}\,\to \underset{0.12gm}{\mathop{AgBr}}\,$
We are given here that from 0.18gm of the compound; 0.12gm of Silver bromide gets formed. Now, we are given the molecular weight of AgBr and atomic weight of Bromine atom as well.
So, we can write that when one mole of compound will react with a silver ion containing compound and one mole of silver bromide will be generated.
Thus, if 0.12gm of AgBr is generated from 0.18gm of the compound, then 188gm of AgBr(Molecular weight) will be generated from $\dfrac{188\times 0.18}{0.12}=282$ gm of the compound.
Thus, we obtained that from 282gm of the compound, we will obtain 188gm of silver bromide. In other words, we obtained the molecular weight of the compound which is 282 $gmmo{{l}^{-}}$ .
Now, we know that the atomic weight of bromine is 80 $gmmo{{l}^{-}}$ .
So, % of Bromine in the compound = $\dfrac{\text{Weight of Br in the compound}\times \text{100}}{\text{Molecular weight of the compound}}$
So, we can put all the available values in the given formula. We will get,
% of bromine in the compound = $\dfrac{80\times 100}{282}=28.36%$
Thus, we can conclude that the% of bromine atoms in this compound is 28.36%.

Therefore, the correct answer of the question is (C).

Note:
Remember that if a compound contains any halide, then its reaction with silver nitrate solution will give precipitates of silver halides in the solution. By weighing the precipitates, we can measure the amount of halide present in the compound.