JEE Advanced levelSingle correct optionDifficulty 3 of 3si193
A dibromoalkane has the general formula CXnHX2nBrX2. In a Carius estimation, 0.404 g of it gave 0.752 g of AgBr (Ag = 108, Br = 80, C = 12, H = 1). The molecular formula of the compound is
ACX3HX6BrX2
BCX2HX4BrX2
CCX4HX8BrX2
DCHX2BrX2
Answer: A
Moles of AgBr = 0.752/188 = 0.0040 mol. Each molecule has two Br, so moles of compound = 0.0020 mol and molar mass = 0.404/0.0020 = 202 g/mol. For CXnHX2nBrX2, 14n+160=202 gives n=3.