Practical Organic Chemistry and PurificationhardNUMERICAL

In the Carius method for estimation of halogens, of an organic compound gave of . Find the percentagPractical Organic Chemistry and Purification Chemistry Question

Question

In the Carius method for estimation of halogens, $0.25\ \text{g}$ of an organic compound gave $0.141\ \text{g}$ of $AgBr$. Find the percentage of bromine in the compound. (Atomic masses: $Ag = 108$, $Br = 80$)

Answer: 24

💡 Solution & Explanation

The percentage of bromine is determined by the formula: $\%Br = \frac{80}{188} \times \frac{\text{Mass of } AgBr}{\text{Mass of compound}} \times 100$. Using the values: $\%Br = \frac{80}{188} \times \frac{0.141}{0.25} \times 100 = \frac{80 \times 0.75 \times 10^{-3} \times 10^3}{0.25} \times 100$ (simplifying fractions). Molar mass $AgBr = 188$. $0.141 / 188 = 0.00075\ \text{moles}$. Mass of Br $= 0.00075 \times 80 = 0.06\ \text{g}$. $\%Br = \frac{0.06}{0.25} \times 100 = 24\%$.

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