Question

In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is \(\_\_\_\_\)times 10^{-1}% (Nearest integer). (Given: Molar mass of Ag is 108 and Br is 80 g mol^{-1}
 

Hint:

In Carius method, the amount of halogen can be calculated by the formation of silver halide, and the mass of the halogen can be derived using stoichiometry. Always use the molar mass of the halide to find the exact amount of halogen present.

Answer 1

Using the Carius method, the amount of bromine can be calculated as follows: We are given that 0.25 g of the organic compound yields 0.15 g of silver bromide (AgBr). The moles of AgBr are calculated using: \[ \text{Moles of AgBr} = \frac{0.15}{108 + 80} = \frac{0.15}{188} \approx 0.000796 \, \text{mol} \] Next, we calculate the mass of bromine in the organic compound: \[ \text{Mass of Br} = 0.000796 \times 80 = 0.0637 \, \text{g} \] Now, calculate the percentage of bromine: \[ \text{Percentage of Br} = \left(\frac{0.0637}{0.25}\right) \times 100 = 25.48\% \quad \text{or} \quad 2.54 \times 10^1\% \] Rounding this to the nearest integer, the final answer is: \[ \boxed{4} \]