Question

In Carius method of estimation of ‘Br’, 1.53 g of an organic compound gave 1 g AgBr. The % of Br in the organic compound is: (Atomic mass of Ag, Br = 108, 80 u respectively)

• A. \(35.23\)
• B. \(43.53\)
• C. \(27.81\)
• D. \(22.71\)

Hint:

In Carius method problems:

First find the molar mass of the silver halide.
Use mass ratio to extract the halogen mass.
Apply percentage formula carefully.

Answer 1

The Correct Option is B

Concept:
In the Carius method, halogens present in an organic compound are transformed into their corresponding silver halides. The percentage of halogen is determined using the expression: \[ %\text{ of halogen} = \frac{\text{Mass of halogen obtained}}{\text{Mass of organic compound}} \times 100 \] The mass of the halogen is calculated from the mass of the silver halide formed by applying molar mass relationships. Step 1: Determine the fraction of bromine in AgBr. \[ \text{Molar mass of AgBr} = 108 + 80 = 188 \] \[ \text{Fraction of Br in AgBr} = \frac{80}{188} \]
Step 2: Find the mass of bromine produced. \[ \text{Mass of Br} = 1 \times \frac{80}{188} = 0.4255 \text{ g} \]
Step 3: Compute the percentage of bromine in the organic compound. \[ %\text{Br} = \frac{0.4255}{1.53} \times 100 = 27.81% \] \[ \Rightarrow %\text{Br} = 27.81 \] Therefore, the closest correct option is (2).