JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Practical and Analytical Chemistry

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In an estimation of bromine by Carius method, 1.6 g of an organic compound gave 1.88 g of AgBr . The mass percentage of bromine in the compound is ____
(Atomic mass,

Ag = 108, Br = 80 g mol^{- 1}

)

Answer: 50

Solution

All the bromine in the compound ends up as silver bromide. With molar mass

AgBr = 108 + 80 = 188 g mol^{- 1}

, the

1.88 g

precipitate is

\frac{1.88}{188} = 0.01 mol

, and since each

AgBr

holds one bromine, that is

0.01 mol

0.8 g

of bromine. As a fraction of the

1.6 g

sample, the bromine content is

\frac{0.8}{1.6} \times 100 = 50%

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Integer answer

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