Chemistry - PRACTICAL ORGANIC CHEMISTRY Question with Solution | TestHub

ChemistryPRACTICAL ORGANIC CHEMISTRYELEMENTAL ANALYSISMedium2 minPYQ_2023

ChemistryMediumnumerical

$0.400 g$ of an organic compound $(X)$ gave $0.376 g$ of $AgBr$ in Carius method for estimation of bromine. $%$ of bromine in the compound $(X)$ is (Given: Molar mass $AgBr = 188 g {mol}^{- 1}$ $Br = 80 g {mol}^{- 1})$

Answer:

40.00

Solution:

The percentage composition of element $= \frac{mass of element}{mass of organic compound} \times 100$

Moles of $AgBr = \frac{0.376}{188}$

Moles of $Br = \frac{0.376}{188}$

Mass of $Br = \frac{0.376}{188} \times 80$

$% of Br = \frac{0.376 \times 80}{188 \times (0.400)} \times 100$

$= 40 %$

Stream:JEESubject:ChemistryTopic:PRACTICAL ORGANIC CHEMISTRYSubtopic:ELEMENTAL ANALYSIS

⏱ 2mℹ️ Source: PYQ_2023

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