JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Practical and Analytical Chemistry

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In Carius method of estimation of halogen. 0.45 g of an organic compound gave 0.36 g of AgBr . Find out the percentage of bromine in the compound.
(Molar masses :

AgBr = 188 g mol^{- 1} : Br = 80 g mol^{- 1}

)

(A) $34.04%$ ✓ Correct
(B) $40.04%$
(C) $36.03%$
(D) $38.04%$

Answer: (A)

Solution

In the Carius method, the organic compound is heated with fuming nitric acid and silver nitrate. The halogen forms a silver halide precipitate.

Given:

Mass of organic compound = 0.45 g
Mass of AgBr obtained = 0.36 g
Molar mass of AgBr = 188 g/mol
Atomic mass of Br = 80 g/mol

Calculation:

Moles of AgBr =

\frac{0.36}{188}

mol

Since 1 mole of AgBr contains 1 mole of Br:

Mass of Br =

\frac{0.36}{188} \times 80 = 0.1532

g

Percentage of Br =

\frac{0.1532}{0.45} \times 100 = 34.04%

Therefore, the answer is (A) 34.04%.

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Practical and Analytical Chemistry

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