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Compound A is a light green crystalline solid. It gives the following tests: (i) It dissolves in dilute sulphuric acid. No gas is produced. (ii) A drop of KMnO base 4 is added to the above solution. The pink colour disappears. (iii) Compound A is heated strongly. Gases B and C, with pungent smell, come out A brown residue D is left behind. (iv) The gas mixture (B) and (C) is passed into a dichromate solution. The solution turns green. (v) The green solution from step (iv) gives a white precipitate E with a solution of barium nitrate.

Compound A is a light green crystalline solid. It gives the following tests:
(i) It dissolves in dilute sulphuric acid. No gas is produced.
(ii) A drop of KMnO base 4 is added to the above solution. The pink colour disappears.
(iii) Compound A is heated strongly. Gases B and C, with pungent smell, come out A brown residue D is left behind.
(iv) The gas mixture (B) and (C) is passed into a dichromate solution. The solution turns green.
(v) The green solution from step (iv) gives a white precipitate E with a solution of barium nitrate.

Grade:12th pass

1 Answers

Kevin Nash
askIITians Faculty 332 Points
6 years ago

(i)(A) is FeSO base 4.7H base 2O because it is light green crystalline solid. Which dissolves in water containing H base 2SO base 4

(ii) On strong heating FeSO base 4 both SO base 4 (B) and SO base 3 (C) are evolved. The colour of KMnO base 4 disappears due to the formation of MnSO base 4.

(iii) SO base 2 being a reducing agent turns a dichromate solution green and forms H base 2SO base 4 in the solution. SO base 3 dissolves in water to give H base 2SO base 4. Therefore, white ppt of BaSO base 4 is formed with a solution of Ba(NO base 3) base 2

(iv) The brown residue left behind (D) is Fe base 2O base 3 which is reduced to Fe on heating in charcoal cavity. Fe is magnetic substance.

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