An organic compound weighing 0.15 g gave on Carius estimation, 0.12 g of AgBr. The percentage of Br in the compound will be close to (At. mass of Ag = 108, Br = 80)

Question

An organic compound weighing 0.15 g gave on Carius estimation, 0.12 g of AgBr. The percentage of Br in the compound will be close to (At. mass of Ag = 108, Br = 80) (A) 46 % (B) 34.1 % (C) 3.41 % (D) 4.6 %

Tardigrade Admin · Accepted Answer

Percentage of bromine in the compound = (80/188) × ( textMass of AgBr formed/ textMass of substance taken) × 100 = (80/188) × (0.12/0.15) × 100 = 34.04 %

Q. An organic compound weighing $0.15 g$ gave on Carius estimation, $0.12 g$ of $A g B r$ . The percentage of Br in the compound will be close to (At. mass of $A g = 108, B r = 80$ )

$46%$

$34.1%$

$3.41%$

$4.6%$

Percentage of bromine in the compound
$= \frac{80}{188} \times \frac{Mass of AgBr formed}{Mass of substance taken} \times 100$
$= \frac{80}{188} \times \frac{0.12}{0.15} \times 100 = 34.04%$

Q. An organic compound weighing 0.15g gave on Carius estimation, 0.12g of AgBr. The percentage of Br in the compound will be close to (At. mass of Ag=108,Br=80)

46%

34.1%

3.41%

4.6%