A pale green crystalline salt X (molar mass , a heptahydrate) dissolves in water to give a pale green solution. This solution gives a deep blue precipitate with , a white precipitate with that is insoluble in dilute HCl, and it decolourises acidified . On strong heating, X first loses its water of crystallisation and then decomposes to give a brown-red solid residue Y and a gaseous mixture Z of two oxides of sulphur in equal mole ratio. Z is passed through aqueous solution and gives a white precipitate. Take atomic masses Fe = 55.8, S = 32.0, O = 16.0, Ba = 137.3, H = 1.0.
Question 1
Single correct option
The number of moles of acidified decolourised by mol of X, and the identity of Y, are respectively
- A. mol;
- B. mol;
- C. mol;
- D. mol;
Show answer and solution
Answer: (A) mol;
The salt X is (molar mass 278), consistent with the blue Turnbull's blue precipitate for Fe and the barium sulphate test. In acid, . So mol of Fe needs mol of permanganate (the sulphate does not take part). On strong heating, , giving the brown-red residue Fe2O3 together with equal moles of the two gases. Hence Y is and the answer is option A. The other options are wrong: 0.50 mol ignores the five-electron change of permanganate, is black, and is not formed because Fe(II) is oxidised to Fe(III) in the SO3-forming decomposition.
Question 2
Single correct option
When g of X is strongly heated and all the gaseous mixture Z is passed through excess aqueous solution, the mass of white precipitate formed is closest to
- A. g
- B. g
- C. g
- D. g
Show answer and solution
Answer: (B) g
Moles of X mol. The decomposition means mol of gives mol each of and . In water gives and then insoluble (): the HCl formed keeps any sulphite in solution, because dissolves in acid, so gives no precipitate. Mass of g, option B. Option C (4.67 g) wrongly assumes that both and precipitate as barium sulphate, option A (1.17 g) counts only half of the , and option D (6.99 g) corresponds to three times the correct amount.
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