Subjective Type Questions: Analytical Chemistry- 1 | JEE Advanced | 35 Years Chapter wise Previous Year Solved Papers for JEE PDF Download

Q.1. Account for the following. Limit your an swer to two sentences: The precipitation of second group sulphides in qualitative analysis is carried out with hydrogen sulphide in presence of hydrochloric acid and not nitric acid. (1979)

Ans. Sol. HNO₃ is strong oxidising agent and it oxidises H₂S to S. So HNO₃ cannot be used to precipitate second group elements.

Q.2. Compound A is a light green crystalline solid. It gives the following tests: (1980) (i) It dissolves in dilute sulphuric acid. No gas is produced. (ii) A drop of KMnO₄ 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 gr een solution from step (iv) g ives a white precipitate E with a solution of barium nitrate. (vi) Residue D from step (iii) is heated on charcoal in a reducing flame. It gives a magnetic substance.
 Name the compounds A, B, C, D and E 

Ans. Sol. (i) (A) is FeSO₄.7H₂O because it is light green crystalline solid. Which dissolves in water containing H₂SO₄
(ii) On strong heating FeSO₄ both SO₂ (B) and SO₃ (C) are evolved. The colour of KMnO₄ disappears due to the formation of MnSO₄.
(iii) SO₂ being a reducing agent turns a dichromate solution green and forms H₂SO₄ in the solution. SO₃ dissolves in water to give H₂SO₄. Therefore, white ppt of BaSO₄ is formed with a solution of Ba(NO₃)₂
(iv) The brown residue left behind (D) is Fe₂O₃ which is reduced to Fe on heating in charcoal cavity. Fe is magnetic substance.

Q.3. When 16.8 g of white solid X were heated, 4.4 g of acid gas A that turned lime water milky was driven off together with 1.8 g of a gas B which condensed to a colourless liquid. The solid that remained, Y, dissolved in water to give an alkaline solution, which with excess barium chloride solution gave a white precipitate Z. The precipitate effervesced with acid giving off carbon dioxide. Identify A, B and Y and write down the equation for the thermal decomposition of X. (1984 - 4 Marks) 

Ans. Sol. 

Representing the given facts in the form of equation.

The above equation leads to the following facts : (i) Since the gas A turned lime water milky, it must be CO₂.
(ii) NOTE : The compound Y gives alkaline solution in water which when treated with BaCl₂ forms a white precipitate of Z. Since the compound Z when treated with acid gives effervescenes of CO₂, Z and hence Y must be metal carbonate, CO₃^2- . Hence Y may be written as metal carbonate MCO₃ or M₂CO₃.
(iii) When X is heated, it yields a carbonate (Y) along with the evolution of CO₂ (A) and another gas (B), it must be a bicarbonate.
(iv) The above facts point out that B may be water vapour Thus the above reaction can be written as below.

Calculation of molecular weight of MHCO₃ 4.4.g of CO₂ is given by 16.8 g of MHCO₃  

Since two molecules of MHCO₃ are taking part in the reaction, the molecular weight of

Calculation of atomic weight of metal M MHCO₃  = 84; M + 1 + 12 + 48 = 84
M + 61  =  84; M = 84 – 61 = 23
Thus the metal must be Na and hence the given salt X is NaHCO₃. The above facts coincide with the given thermal decomposition.

Q.4. A mixture of two salts was treated as follows : (1987 - 5 Marks)
 (i) The mixture was heated with manganese dioxide and concentrated sulphuric acid when yellowish green gas was liberated.
 (ii) The mixture on heating with sodium hydroxide solution gave a gas which turned red litmus blue.
 (iii) Its solution in water gave blue precipitate with potassium ferricyanide and red colouration with ammonium thiocyanate.
 (iv) The mixture was boiled with potassium hydroxide and the liberated gas was bubbled through an alkaline solution of K₂HgI₄ to give brown precipitate.
 Identify the two salts. Give ionic equations for reactions involved in the tests (i), (ii) and (iii). 

Ans. Sol. (A) Test (i) of the problem indicates that the mixture contains Cl^– ion which is liberated as Cl₂ (yellowish green gas) when heated with MnO₂ and conc. H₂SO₄. (B) Test (ii) indicates the presence of NH⁴⁺ ion in the mixture which gives ammonia when heated with  NaOH solution.
Since ammonia is basic in nature, it turns red litmus blue. Presence of NH⁴⁺ in the mixture is further confirmed by the given test (iv) according to which the gas (NH₃) gives brown precipitate with Nessler’s reagent (alkaline solution of K₂[HgI₄]. (C) Test (iii) indicates Fe²⁺ ion in the mixture which gives blue precipitate with potassium ferricyanide (note that potassium ferricyanide gives brown ppt. with Fe³⁺ ions). (D) Red colouration with ammonium thiocynate indicates that the mixture also contains Fe³⁺ ions which are believed to be formed by the oxidation of Fe²⁺ ions by air.

Thus the mixture contains FeCl₂ and NH4Cl.
Ionic reactions :

Q.5. A hydrated metallic salt A, light green in colour, on careful heating gives a white anhydrous residue B. B is soluble in water and its aqueous solution reacts with NO to give a dark brown compound C. B on strong heating gives a brown residue D and a mixture of two gases E and F. The gaseous mixture when passed through acidified permanganate, discharges the pink colour and when passed through acidified BaCl₂ solution gave a white precipitate. Identify A, B, C, D, E and F. (1988 - 3 Marks) 

Ans. Sol. (i) (A) on heating loses water of crystallization and thus it is a hydrated salt.
(ii) Anhydrous salt (B) on heating gives two gases and brown residue and so (B) is FeSO₄. Thus (A) is  

(iii) (B) is soluble in water and reacts with NO to give brown compound.

(iv) Gaseous mixture decolorizes acidified KMnO₄. 5SO₂ + 2KMnO₄ + 2H₂O → K₂SO₄ + 2MnSO₄ + 2H₂SO₄

(v) Gaseous mixture on passing through BaCl₂, gives white ppt. of BaSO₄.

Q.6. When 20.02 g of a white solid X is heated 4.4 g of an acid gas A and 1.8 g of a neutral gas B are evolved, leaving behind a solid residue Y of weight 13.8 g. A turns lime water milky and B condenses into a liquid which changes anhydrous copper sulphate blue. The aqueous solution of Y is alkaline to litmus and gives 19.7 g of white precipitate Z with barium chloride solution. Z gives carbon dioxide with an acid. Identify A, B, X, Y and Z. (1989 - 5 Marks)

Ans. Sol. Representing the given facts in the form of equation, we get

The above equation leads to the following facts : (i) Since the gas A turned lime water milky, it must be CO₂.
(ii) NOTE : The compound Y is alkaline to litmus and when treated with BaCl2 forms a white precipitate of Z. Since the compound Z when treated with acid gives effervescenes of CO₂, Z and hence Y must contain carbonate, CO₃^2- . Hence Y may be written as metal carbonate MCO₃ or M₂CO₃.
(iii) When X is heated, it yields a carbonate (Y) along with the evolution of CO₂ (A) and a neutral gas (B), it must be a bicarbonate. (iv) B changes anhydrous CuSO₄ blue, which point out that B is water.
Thus the above reaction can be written as below :

Calculation of molecular weight of MHCO₃ 4.4 g of CO₂ is given by 20.02 g of MHCO₃

44 g of CO₂ is given by =

Since two molecules of MHCO₃ are taking part in the reaction, the molecular weight of

MHCO₃ (X)  =

Calculation of atomic weight of Metal M MHCO₃ = 100; M + 1 + 12 + 48 = 100
M + 61 = 100; M = 100 – 61 = 39
Thus the metal must be K and hence the given salt X is KHCO₃.

The above facts coincide with the given thermal decomposition.

Hence, we have

Q. 7. The gas liberated on heating a mixture of two salts with NaOH, gives a reddish brown precipitate with an alkaline solution of K₂[HgI₄]. The aqueous solution of the mixture on treatment with BaCl₂ gives a white precipitate which is sparingly soluble in conc. HCl. On heating the mixture with K₂Cr₂O₇ and conc. H₂SO₄, red vapours of A are produced.
 The aqueous solution of the mixture gives a deep blue colouration B with potassium ferricyanide solution. Identify the radicals in the given mixture and write the balanced equations for the formation of A and B. (1991 - 4 Marks) 

Ans. Sol. Let us summarise the given facts of the question.

 White ppt.sparingly soluble in conc.
HCl.
The given reactions lead to the following conclusions.
(i) Formation of reddish brown precipitate on treatment with alk. K₂[HgI₄] indicates the evolution of NH₃ gas and hence the presence of NH⁴⁺ in the mixture of salts.
(ii) Heating of mixture with K₂Cr₂O₇ and conc. H₂SO₄ to give red vapours (of chromyl chloride) indicates the presence of Cl^– ion in the mixture.
(iii) Reaction of aqueous solution of the mixture with barium chloride solution to give white ppt. (of BaSO₄) sparingly soluble in conc. HCl indicates the presence of SO₄^2-ions in the mixture.
(iv) NOTE : Reaction of aqueous solution of the mixture with potassium ferricyanide solution to give deep blue colour indicates the presence of Fe²⁺ ions in the mixture.
Hence the mixture contains following four ions :

Equations for the formation of A and B.

Q.8. A light bluish green crystalline compound responds to the following tests :
 (i) Its aqueous solution gives a br own pr ecipitate or colour with alkaline K₂[HgI₄] solution.
 (ii) Its aqueous solution gives a blue colour with K₃[Fe(CN)₆] solution.
 (iii) Its solution in hydrochloric acid gives a white precipitate with BaCl₂ solution.
 Identify the ions present and suggest the formula of the compound. (1992 - 4 Marks)

Ans. Sol. Compound gives brown ppt. with alkaline K₂[HgI₄] and so contain NH⁺₄ ions.’ ‘Compound gives blue colour with K₃[Fe(CN)₆] and so contains Fe²⁺ ions.’ ‘Solution of compound in HCl gives white ppt. with BaCl₂ and so it contains SO₄^2- ions.’ ‘Bluish green compound with NH⁺₄ , Fe²⁺ and ^2-
SO₄ suggests that it is Mohr's salt i.e.’ FeSO₄(NH₄)₂SO₄.6H₂O Reactions :

Q.9. An  orange solid (A) on heating gave a green residuce (B), a colourless gas (C) and water vapour. The  dry gas (C) on passing over heated Mg gave a white solid (D). (D) on reaction with water gave a gas (E) which formed dense white fumes with HCl. Identify (A) to (E) and give reactions involved. (1993 - 3 Marks) 

Ans. Sol. Let us summarise the given facts

(i) Formation of white dense fumes by gas (E) with HCl indicates that the gas (E) is ammonia (NH₃).
(ii) Formation of ammonia (E) by the hydrolysis of white solid (D) indicates that (D) should be magnesium nitride, Mg₃N₂.
(iii) Since compound (D) is formed by reaction of gas (C) with magnesium, the colourless gas (C) must be nitrogen.
(iv) Orange colour of the original compound (A) and green colour of the residue (B) indicates that compound (A) is ammonium dichromate, (NH₄)₂Cr₂O₇.
Reactions :

Q.10. A is a binary compound of a univalent metal, 1.422 g of A reacts completely with 0.321 g of sulphur in an evaccuated and sealed tube to give 1.743 g of a white crystalline solid B, that forms a hydrated double salt, C with Al₂(SO₄)₃. Identify A, B and C (1994 - 5 Marks) 

Ans. Sol. As the solid B forms a hydrated salt C with Al₂(SO₄)₃; B should be sulphate of a monovalent cation, i.e. M₂SO₄.
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 = = 174.3 g

Thus mol. wt. of B (M₂SO₄) = 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 K₂SO₄, and C is K₂SO₄ . Al₂(SO₄)₃ . 24H₂O
Nature of compound A : Since A is a binary compound of potassium and it reacts with sulphur to form K₂SO₄, it must be oxide of potassium, probably potassium superoxide (KO₂) which is supported by the given data.

2 (39.1 + 32) = 142.2
32.1g of Sreacts with 142.2g of KO₂ 0.321g of Sreacts with =  = 1.422 g

Similarly, 32.1 g of S gives 174.3 g of K2SO4
0.321 g of S gives = =  1.743 g

Both these datas are also given in the problem. Thus A is KO₂.

Q.11. A scarlet compound A is treated with conc. HNO₃ to give a chocolate brown precipitate B. The precipitate is filtered and the filtrate is neutralised with NaOH. Addition of KI to the resulting solution gives a yellow precipitate C. The precipitate B on warming with conc. HNO₃ in the precence of Mn(NO₃)₂ produces a pink-coloured solution due to the formation of D. Identify A, B, C and D. Write the reaction sequence. (1995 - 4 Marks) 

Ans. Sol. Summary of the given facts.

From the colour of the known compound and reaction involved, it is clear  that (A) is red lead (Pb₃O₄) and its various reactions can be represented as below.