Question

Sufficient amount of H2S gas is passed through 5 ml solution of tincture of iodine to

convert its all iodine into iodide ion. The sulphur precipitated is filtered off and the

solution is made upto 1 litre and the solution is acidified with HCl. 250 ml of this

solution requires 28 ml of 0.05 N Ce4+ for the conversion of entire I–

into ICl only. 2

ml of same sample of tincture of iodine gave 0.0313 gm of yellow precipitate in

another experiment when treated with AgNO3 solution. What weight percent of iodine

is present in the form of free iodine. ( Tincture of iodine contains free I– and I2 both)

Please answer with explanation. I will mark it as brainliest.​

Answer 1

Sufficient amount of H2S gas is passed through 5 ml solution of tincture of iodine to

convert its all iodine into iodide ion. The sulphur precipitated is filtered off and the

solution is made upto 1 litre and the solution is acidified with HCl. 250 ml of this

solution requires 28 ml of 0.05 N Ce4+ for the conversion of entire I–

into ICl only. 2

ml of same sample of tincture of iodine gave 0.0313 gm of yellow precipitate in

another experiment when treated with AgNO3 solutionSufficient amount of H2S gas is passed through 5 ml solution of tincture of iodine to

convert its all iodine into iodide ion. The sulphur precipitated is filtered off and the

solution is made upto 1 litre and the solution is acidified with HCl. 250 ml of this

solution requires 28 ml of 0.05 N Ce4+ for the conversion of entire I–

into ICl only. 2

ml of same sample of tincture of iodine gave 0.0313 gm of yellow precipitate in

another experiment when treated with AgNO3 solution .