From 0.2g of compound, 0.188g of AgBr is formed by Carius method. Find % of Br?

Options

A. 0.8
B. 0.2
C. 0.4
D. 0.1

Answer

To find the percentage of bromine (Br) in the compound, we can follow these steps:

### Step 1: Understand the relationship between AgBr and Br
In the Carius method, bromine is estimated by forming silver bromide (AgBr). The molecular mass of AgBr is the sum of the atomic masses of silver (Ag) and bromine (Br).

### Step 2: Determine the molecular mass of AgBr
The atomic mass of silver (Ag) is approximately 108 g/mol, and the atomic mass of bromine (Br) is approximately 80 g/mol. Therefore, the molecular mass of AgBr can be calculated as follows:
\[ \text{Molecular mass of AgBr} = \text{Atomic mass of Ag} + \text{Atomic mass of Br} = 108 + 80 = 188 \text{ g/mol} \]

### Step 3: Calculate the moles of AgBr formed
Given that 0.188 g of AgBr is formed, we can calculate the number of moles of AgBr:
\[ \text{Moles of AgBr} = \frac{\text{mass of AgBr}}{\text{molecular mass of AgBr}} = \frac{0.188 \text{ g}}{188 \text{ g/mol}} = 0.001 \text{ mol} \]

### Step 4: Calculate the moles of Br in AgBr
Since each mole of AgBr contains one mole of Br, the moles of Br will also be 0.001 mol.

### Step 5: Calculate the mass of Br
Now, we can find the mass of bromine using its atomic mass:
\[ \text{Mass of Br} = \text{moles of Br} \times \text{atomic mass of Br} = 0.001 \text{ mol} \times 80 \text{ g/mol} = 0.08 \text{ g} \]

### Step 6: Calculate the percentage of Br in the compound
Finally, we can find the percentage of bromine in the original compound:
\[ \text{Percentage of Br} = \left( \frac{\text{mass of Br}}{\text{mass of compound}} \right) \times 100 = \left( \frac{0.08 \text{ g}}{0.2 \text{ g}} \right) \times 100 = 40\% \]

### Final Answer:
The percentage of bromine (Br) in the compound is **40%**.

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