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Grade: 12th pass
        
A is a binary compound of a univalent metal, 1.422 g of A reacts completely with 0.321 g of sulphur in an evacuated and sealed tube to give 1.743 g of a white crystalline solid B, that forms a hydrated double salt, C with Al base 2(SO base 4) base 3. Identify A, B and C
4 years ago

Answers : (1)

Kevin Nash
askIITians Faculty
332 Points
							

As the solid B forms a hydrated salt C with Al base 2(SO base 4) base 3; B should be sulphate of a monovalent cation, i.e. M base 2SO base 4. Now since sulphate of a monovalent cation contains one sulphur atom per mol, weight of metal sulphate obtained by 32.1 g (at. Wt. of S) should be the molecular weight of the metal sulphate. Thus, - 0.321 g of sulphur is present in 1.743 g of B

? 32.1 g of sulphur is present in = 1.743/0.321 *32.1 = 174.3 g

Thus mol. wt. of B (M base 2SO base 4) = 174.3 g mol^-1

2x + 32.1 +64 = 174.3 (at wt. of M = x]

2x = 78.2 ? x = 39.1

Atomic weight 39.1 corresponds to metal potassium, K. Thus B is K base 2SO base 4, and C is K base 2SO base 4. Al base 2(SO base 4) base 3 24H base 2O

Nature of compound A : since A is a binary compound of potassium and it reacts with sulphur to form K base 2SO base 4, it must be oxide of potassium, probably potassium superoxide (KO base 2) which is supported by the given data.

2KO base 2 +S ? K base 2SO base 4

(A) (B)

2(39.1 +32) = 142.2

32.1 g of S reacts with 142.2 g of KO base 2

0.321 g of S reacts with = 142.2/32.1 * 0.321 = 1.422 g

Similarly,

32.1 g of S gives 1174.3 g of K base 2SO base 4

0.321 g of S gives = 174.3/32.1 *0.321 = 1.743 g

Both these datas are also given in the problem. Thus A is KO base 2.

4 years ago
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