JEE Main · 2015 · 04 Apr 2015 · Offline · Practical and Analytical Chemistry

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In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr . The percentage of bromine in the compound is: (at. Mass

Ag = 108; Br = 80)

(A)36
(B)48
(C)60
(D)24✓ Correct

Answer: (D)

Solution

In the Carius method every bromine atom of the sample finishes as

AgBr

. The precipitate, molar mass

108 + 80 = 188

, weighs

141

mg, which is

\frac{0.141}{188} = 7.5 \times 1 0^{- 4}

mol, and that equals the moles of bromine. Its mass is

7.5 \times 1 0^{- 4} \times 80 = 0.060

= 60

mg, so the bromine content is

\frac{60}{250} \times 100 = 24%

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option

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