Question

0.400 g of an organic compound (X) gave 0.376 g of AgBr in Carius method for estimation of bromine. % of bromine in the compound (X) is____
(Given: Molar mass AgBr=188g mol^-1)

Answer 1

Correct Answer: 40

To determine the percentage of bromine in the compound (X), we need to follow a series of logical steps:

1. First, calculate the moles of AgBr produced:
   Mass of AgBr = 0.376 g
   Molar mass of AgBr = 188 g/mol
   Moles of AgBr = 0.376 g / 188 g/mol ≈ 0.002 mol
2. Since 1 mole of AgBr contains 1 mole of Br, the moles of Br in AgBr is also 0.002 mol.
3. Calculate the mass of Br:
   Molar mass of Br = 80 g/mol
   Mass of Br = 0.002 mol × 80 g/mol = 0.16 g
4. Determine the percentage of bromine in the compound (X):
   Mass of compound (X) = 0.400 g
   % Br = (0.16 g / 0.400 g) × 100 = 40%
5. Verification: The calculated percentage (40%) is exactly at the lower boundary of the expected range of 40 to 40.

Thus, the percentage of bromine in the compound (X) is 40%, which fits within the provided range.