Sodium fusion extract of an organic compound (Y) with CHCl3 and chlorine water gives violet color to the CHCl3 layer. 0.15 g of (Y) gave 0.12 g of the silver halide precipitate in Carius method. Percentage of halogen in the compound (Y) is ____ . (Nearest integer) (Given : molar mass gmol−1 C : 12, H : 1, Cl : 35.5, Br : 80, I : 127)

Percentage of Halogen using Carius Method | JEE Main Chemistry

Q. Sodium fusion extract of an organic compound (Y) with CHCl3 and chlorine water gives violet color to the CHCl3 layer. 0.15 g of (Y) gave 0.12 g of the silver halide precipitate in Carius method. Percentage of halogen in the compound (Y) is ____ . (Nearest integer)

(Given : molar mass gmol−1 C : 12, H : 1, Cl : 35.5, Br : 80, I : 127)

Correct Answer: 43%

Complete Step-by-Step Solution

Step 1: Identification of Halogen Violet colour in CHCl3 layer confirms presence of IODINE. Therefore silver halide formed = AgI

Step 2: Reaction in Carius Method Organic compound + fuming HNO3 → Iodine converted to AgI Ag+ + I− → AgI (precipitate)

Step 3: Given Data Mass of compound = 0.15 g Mass of AgI formed = 0.12 g

Step 4: Molar Mass of AgI Ag = 108 I = 127 AgI = 108 + 127 = 235 g/mol

Step 5: Mass of Iodine in 0.12 g AgI Fraction of iodine in AgI = 127 / 235 Mass of iodine = (127 / 235) × 0.12 = 0.0648 g

Step 6: Percentage of Iodine % I = (0.0648 / 0.15) × 100 = 43.2%

Nearest integer = 43%

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