Question

In carius method of estimation of 'Br' 1.53 gm of an organic compound gave 1 gm AgBr. The % of Br in organic compound is (At. mass of Ag and Br are 108 and 80 amu respectively)

• A. 35.23
• B. 43.53
• C. 27.81
• D. 22.71

Hint:

Carius Method formula: $%X = \frac{\text{At. Mass X}}{\text{Mol. Mass AgX}} \times \frac{w_{AgX}}{w_{compound}} \times 100$.

Answer 1

The Correct Option is C

To find the percentage of Bromine (Br) in the given organic compound using the Carius method, let's go through the step-by-step calculation:

1. The reaction of the organic compound with silver nitrate (AgNO₃) yields silver bromide (AgBr). The mass of AgBr formed is given as 1 g.
2. The molar masses can be calculated as:
   • Atomic mass of Ag = 108 u
   • Atomic mass of Br = 80 u
   • Molar mass of AgBr = 108 + 80 = 188 u
3. Determine the mass of Bromine in AgBr:
   • The fraction of Br in AgBr is given by the ratio of atomic mass of Br to the molar mass of AgBr: \(\frac{80}{188}\).
   • Using this ratio, the mass of Br in 1 g of AgBr is: \(\frac{80}{188} \times 1 \text{ g} = \frac{80}{188} \text{ g}\).
4. Calculate the percentage of Bromine in the organic compound:
   • The given mass of the organic compound is 1.53 g.
   • The percentage of Br is calculated as: \(\left(\frac{\text{Mass of Br in AgBr}}{\text{Mass of organic compound}}\right) \times 100\)
   • Substitute the values: \(\left(\frac{\frac{80}{188}}{1.53}\right) \times 100 = \left(\frac{80}{188} \times \frac{1}{1.53}\right) \times 100\).
   • Calculate to get: \(\left(\frac{80 \times 100}{188 \times 1.53}\right) = 27.81\%\).

Thus, the percentage of Bromine (Br) in the organic compound is 27.81%, which confirms the given correct answer.