In carius method of estimation of halogen, 0.45 g of an organic compound gave 0.36 g of AgBr . Find out the percentage…

Step 1: Recall the Carius Method Formula

In the Carius method for bromine: $\ce{Organic-Br ->[ {HNO3, AgNO3}] AgBr(v)}$

$\%Br = \frac{80}{188} \times \frac{m_{AgBr}}{m_{\text{compound}}} \times 100$

Step 2: Substitute Given Values

• $m_{AgBr}$ = 0.36 g
• $m_{\text{compound}}$ = 0.45 g
• Molar mass AgBr = 188 g/mol; Molar mass Br = 80 g/mol

$m_{Br} = \frac{80}{188} \times 0.36 = \frac{28.8}{188} = 0.15319 \text{ g}$

Step 3: Calculate Percentage

$\%Br = \frac{0.15319}{0.45} \times 100 = \frac{15.319}{45} \times 100 = 34.04\%$

Answer: $34.04\%$ → Option (a)

Verification: $\%Br = \frac{80 \times 0.36}{188 \times 0.45} \times 100 = \frac{28.8}{84.6} \times 100 = 34.04\% \checkmark$

Key Points to Remember:

• Bromine formula: $\%Br = \frac{80}{188} \times \frac{m_{AgBr}}{m_{\text{compound}}} \times 100$
• $\frac{80}{188} \approx 0.4255$
• AgBr: pale yellow precipitate, insoluble in dilute HNO₃, partially soluble in NH₃
• Carius tube: sealed borosilicate glass tube for high-pressure reactions