What will be the mass of AgBr obtained when 1 gm of product is treated with AgNO₃ in the Carius method?
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In the Carius method, halogen atoms in the organic product react with AgNO₃ to form insoluble silver halide salts. The mass of the halide salt depends on the number of moles of halogen atoms in the product.
Step 1: Understanding the reaction.
The given reaction involves the conversion of the nitrobenzene derivative (methyl nitrobenzene) into a product through a series of reactions:
1. \(\text{Sn} + \text{HCl} \) (reduction to an amine group),
2. Ac₂O (acetylation),
3. FeBr₃ + Br₂ (bromination), and
4. H₃O⁺ (hydrolysis).
When this product is treated with AgNO₃ in the Carius method, it forms AgBr.
Step 2: Write the reaction and identify the number of moles of Br.
For this process, 1 gm of the product is treated with AgNO₃, and the product undergoes a substitution reaction, leading to the formation of AgBr. The mass of AgBr depends on the number of moles of Br atoms in the product.
For 1 gm of the product, we assume it contains a specific amount of Br, and we will calculate how much AgBr is formed by this substitution.
Step 3: Calculate the molar mass of AgBr.
The molar mass of AgBr is calculated as:
\[
\text{Molar mass of AgBr} = \text{Ag} + \text{Br} = 107.87 + 79.90 = 187.77 \, \text{g/mol}
\]
Step 4: Calculate the mass of AgBr formed.
From the stoichiometry of the reaction, 1 mole of product will form 1 mole of AgBr. Assuming the molecular weight of the product (formed in the given reaction steps) is approximately 1 g, the mass of AgBr obtained will be directly proportional to the amount of Br in the product.
Since 1 gm of product gives 1 gm of AgBr (due to the 1:1 ratio of the reactant to product), we conclude that the mass of AgBr obtained is:
\[
\boxed{1 \, \text{gm}}
\]
