MolSynapsequestion bank

MolSynapse / Question bank / Structural Identification and Practical Organic Chemistry / Estimation of elements: Lassaigne, Kjeldahl, Carius

JEE Main levelSingle correct optionDifficulty 2 of 3si181

In a Carius estimation, 0.32 g of an organic compound gave 0.453 g of $AgBr$ . With Ag = 108 and Br = 80, the percentage of bromine in the compound is

A $81.32%$
B $60.24%$
C $104.86%$
D $30.12%$

Answer: B

Each mole of

AgBr

(188 g) contains 80 g of Br. Mass of Br =

\frac{80}{188} \times 0.453 = 0.1928

g, so %Br =

\frac{0.1928}{0.32} \times 100 = 60.24%

.

Practice Structural Identification and Practical Organic Chemistry More on Estimation of elements: Lassaigne, Kjeldahl, Carius

Previous questionNext question

More questions on Estimation of elements: Lassaigne, Kjeldahl, Carius