Question

In carius method of estimation of 'Br' 1.53 gm of an organic compound gave 1 gm AgBr. The % of Br in organic compound is (At. mass of Ag and Br are 108 and 80 amu respectively)

• A. 35.23
• B. 43.53
• C. 27.81
• D. 22.71

Hint:

Carius Method formula: $%X = \frac{\text{At. Mass X}}{\text{Mol. Mass AgX}} \times \frac{w_{AgX}}{w_{compound}} \times 100$.

Answer 1

The Correct Option is C

Let's solve the problem using the Carius method for halogen estimation in an organic compound. We need to determine the percentage of bromine (Br) in the compound.

First, let's outline the given data and understand the process:

• Mass of the organic compound = \(1.53 \, \text{g}\)
• Mass of AgBr formed = \(1 \, \text{g}\)
• Atomic masses: \(\text{Ag} = 108 \, \text{amu}\), \(\text{Br} = 80 \, \text{amu}\)

The Carius method is used for the quantitative estimation of halogens in an organic compound by converting them into their corresponding silver halides.

We use the molecular weight of AgBr to find out how much bromine was originally present:

• Molecular weight of AgBr: \(= 108 + 80 = 188 \, \text{amu}\)

Now, calculate the amount of bromine in the AgBr:

• Mass of bromine in AgBr: \(\frac{80}{188} \times 1 \, \text{g} \approx 0.4255 \, \text{g}\)

Now find the percentage of bromine in the organic compound:

• Percentage of bromine: \(= \left(\frac{0.4255}{1.53}\right) \times 100 \approx 27.81\%\)

Thus, the percentage of bromine in the organic compound is 27.81%.

This confirms the correct answer is 27.81.