Question

In carius method of estimation of halogen 0.172 g of an organic compound showed presence of 0.08 g of bromine. Which of these is the correct structure of the compound?

विकल्प

• CH₃–Br

• CH₃ CH₂–Br

• 

• 

Answer 1

Explanation:

No. of mole of bromine = `0.08/80` = 0.001 mole = 10⁻³ mole.

Let the molecular weight of the compound is M.

When 10⁻³ mole of bromine is present, weight of organic compound is 0.172 g.

When 1 mole of bromine is present, weight of organic compound = `0.172/10^-3` g mol⁻¹

∴ M = `0.172/10^-3` = 172 g mol⁻¹

Therefore the compound is

As its molecular weight = 80 + 72 + 14 + 6 = 172