JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Practical and Analytical Chemistry

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In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide ( AgBr ). The percentage of Bromine in the organic compound is ______

\times 1 0^{- 1}

% (Nearest integer).
(Given : Molar mass of Ag is 108 and Br is

80 g mol^{- 1}

)

Answer: 255

Solution

In the Carius method, all the halogen in the organic compound is converted quantitatively to silver halide, so the mass of

AgBr

formed is used to back-calculate the bromine content:

% Br = \frac{molar mass of Br}{molar mass of AgBr} \times \frac{mass of AgBr}{mass of sample} \times 100

With molar mass of

AgBr = 108 + 80 = 188 g/mol

, mass of

AgBr = 0.15 g

and sample mass

= 0.25 g

% Br = \frac{80}{188} \times \frac{0.15}{0.25} \times 100 = 25.53%

Expressed in the required units of

1 0^{- 1} %

25.53% = 255.3 \times 1 0^{- 1} %

, which rounds to 255.

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Integer answer

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JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Practical and Analytical Chemistry

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