JEE Advanced (Subjective Type Questions): Some Basic Concepts of Chemistry- 2 | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q. 15. Balance the following equations.

(i) Cu₂O + H⁺ + NO₃^- → Cu²⁺ + NO + H₂O (1981 - 1 Mark)
 (ii)K₄[Fe(CN)₆] + H₂SO₄ + H₂O → K₂SO₄ + FeSO₄ + (NH₄)₂SO₄ + CO (1981 - 1 Mark)
 (iii) C₂H₅OH + I₂ + OH^– → CHI₃ + HCO ₃^- + I^– + H₂O (1981 - 1 Mark)

Ans. Sol. Balance the reactions by ion electron method.

(i) Cu₂O + 2H⁺ → 2Cu²⁺ + H₂O + 2e^–] × 3 ......(i)

(ii)K₄[Fe(CN)₆] + 6H₂SO₄ + 6H₂O          
→ 2K₂SO₄ + FeSO₄ + 3(NH₄)2SO₄ + 6CO

(iii) C₂H₅OH + 4I₂ + 8OH^–                                    
→CHI₃ + HCO₃^-  + 5I^– + 6H₂O

Q. 16. Hydroxylamine reduces iron (III) according to the equation:
 2NH₂OH + 4Fe³⁺ → N₂O(g) ↑­ + H₂O + 4 Fe²⁺ + 4H⁺
 Iron (II) thus produced is estimated by titration with a standard permanganate solution. The reaction is :
 MnO₄^- + 5 Fe²⁺ + 8H⁺ → Mn²⁺ + 5 Fe³⁺ + 4H₂O
 A 10 ml. sample of hydroxylamine solution was diluted to 1 litre. 50 ml. of this diluted solution was boiled with an excess of iron (III) solution. The resulting solution required 12 ml. of 0.02 M KMnO₄ solution for complete oxidation of iron (II). Calculate the weight of hydroxylamine in one litre of the original solution. (H = 1, N = 14, O = 16, K = 39, Mn = 55, Fe = 56) (1982 - 4 Marks)

Ans. Sol. Given 2NH₂ OH + 4Fe³⁺ → N₂O + H₂O + 4Fe²⁺+ 4H⁺ .....(i)
and  MnO₄^- + 5Fe ²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺+ 4H₂O ..(ii)

∴ 10 NH₂OH + 4MnO₄^- + 12H⁺ → 5N₂O + 21H₂O+ 4Mn²⁺ [On multiplying (i) by 5 and (ii) by 4 and then adding the resulting equations]
Molecular weight of NH₂OH = 33
Thus 4000 ml of 1M MnO₄^– would react with NH₂OH = 330g
∴ 12 ml of 0.02 M KMnO₄ would react with NH₂OH

∴ Amount of NH₂OH present in 1000 ml of diluted solution

Since 10 ml of sample of hydroxylamine is diluted to one litre

∴ Amount of hydroxyl amine in one litre of original solution

 = 39.6 g

Q. 17. The density of a 3 M sodium thiosulphate solution (Na₂S₂O₃) is 1.25 g per ml. Calculate (i) the percentage by weight of sodium thiosulphate, (ii) the mole fraction of sodium thiosulphate and (iii) the molalities of N^a+ and S₂O₃^2– ions. (1983 - 5 Marks) 

Ans. Sol. TIPS/Formulae :

(i) Mole fraction =

(ii) 1 mole of Na₂S₂O₃ gives 2 moles of Na⁺ and 1 mole of S₂O₃^2– Molecular wt. of sodium thiosulphate solution (Na₂S₂O₃)
= 23 × 2 + 32 × 2 + 16 × 3= 158
(i) The percentage by weight of Na₂S₂O₃

[Wt. of Na₂S₂O₃ = Molarity × Mol wt] (ii) Mass of 1 litre solution = 1.25 × 1000 g = 1250 g  
[∵density = 1.25g/l]

Mole fraction of Na₂S₂O₃

Moles of water  = 43.1

Mole fraction of Na₂S₂O₃  =  = 0.065

(iii) 1 mole of sodium thiosulphate (Na₂S₂O₃) yields 2 moles of Na⁺ and 1 mole of S₂ O₃^2- .
Molality of Na₂S₂O₃ = = 3.87
Molality of Na⁺  = 3.87 × 2 = 7.74m

Molality of S₂ O₃^2- =  3.87m

Q. 18. 4.08 g of a mixture of BaO and an unknown carbonate MCO₃ was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 ml of 1 N HCl. The excess acid required 16 ml of 2.5 N NaOH solution for complete neutralization.
 Identify the metal M. (1983 - 4 Marks) (At. wt. H = 1, C = 12, O = 16, Cl = 35.5, Ba = 138)

Ans. Sol. Weight of MCO₃ and BaO = 4.08 g   (given)
Weight of residue = 3.64 g  (given)
∴ Weight of CO₂ evolved on heating = (4.08 – 3.64) g      = 0.44 g

= 0.01 mole

Volume of 1N HCl in which residue is dissolved = 100 ml

Volume of 1N HCl used for dissolution = (100 – 2.5 × 16) ml        = 60 ml

= 0.06 equivalents

The chemical equation for dissolution can be written as

[Number of moles of BaO and MO = 1 + 1 = 2]

Number of moles of BaO + Number of moles of MO = 

 = 0.03
Number of moles of BaO = (0.03 – 0.01) = 0.02 moles
Molecular weight of BaO = 138 + 16 = 154
∴               Weight of BaO = (0.02 × 154) g = 3.08 g
Weight of MCO₃ = (4.08 – 3.08) = 1.0 g
Since weight of 0.01 mole of MCO₃ = 1.0 g
∴ Mol. wt. of MCO₃ =  = 100

Hence atomic weight of unknown M = (100 – 60) = 40

The atomic weight of metal is 40 so the metal M is Ca.

Q. 19. Complete and balance the following reactions :
 (i) Zn + NO₃^- → Zn²⁺ + NH ⁺₄ (1983 - 1 Mark)
 (ii) Cr ₂ O₇^2- + C₂H₄O → C₂H₄O₂ + Cr³⁺ (1983 - 1 Mark)
 (iii) HNO₃ + HCl → NO + Cl₂ (1983 - 1 Mark)
 (iv) Ce³⁺ + S2O₈^2- → SO₂^4- + Ce⁴⁺ (1983 - 1 Mark)
 (v) Cl₂ + OH^– → Cl^– + ClO^– (1983 - 1 Mark)
 (vi) Mn²⁺ + PbO₂ → MnO ^-₄ + H₂O (1986 - 1 Mark)
 (vii) S + OH^– → S₂^– + S₂O₃^2- (1986 - 1 Mark)
 (viii) ClO ₃^- + I– + H₂SO₄ → Cl^– + HSO^-₄ (1986 - 1 Mark)
 (ix) Ag⁺+ AsH₃ → H₃AsO₃ + H⁺ (1986 - 1 Mark)

Ans. Sol. TIPS/Formulae : Balance the atoms as well as charges by ion electron/ oxidation number method.
While balancing the equations, both the charges and atoms must balance.

(i) 4Zn + NO ₃^- + 10H+ —→ 4Zn²⁺ + NH ⁺₄ + 3H₂O
(ii) Cr₂O^2-₇ + 3C₂H₄O + 8H⁺ —→ 3C₂H₄O₂ + 2Cr³⁺ + 4H₂O
(iii) 2HNO₃ + 6HCl —→ 2NO + 3Cl₂ + 4H₂O
(iv) 2Ce³⁺ + S₂O₈^2-  —→ 2 SO₄^2-+ 2Ce⁴⁺ 
(v) Cl₂ + 2OH– —→ Cl^– + ClO^– + H₂O
(vi) 2Mn²⁺ + 5PbO₂ + 4H⁺ → 2 MnO ^-₄ + 2H₂O + 5Pb²⁺
(vii) 4S + 6OH^– → 2S^2– + S₂ O₃^2- + 3H₂O
(viii) ClO ₃^- + 6I– + 6H₂SO₄ → Cl^– + 6 HSO ^-₄ + 3I₂ + 3H₂O
(ix) 6Ag⁺+ AsH₃ + 3H₂O → 6Ag + H₃AsO₃ + 6H⁺

Q. 20. 2.68 × 10^–3 moles of a solution containing an ion Aⁿ⁺ require 1.61 × 10^–3 moles of MnO^-₄ for the oxidation of Aⁿ⁺ to AO₃^- in acid medium. What is the value of n? (1984 - 2 Marks) 

Ans. Sol. TIPS/Formulae : Equivalents of A oxidised = Equivalents of A reduced.
Since in acidic medium, Aⁿ⁺ is oxidised to AO₃^–, the change in oxidation state from
(+5) to (+n)  =  5 – n [∵ O.S. of A in AO₃^- =+5]
∴ Total number of electrons that have been given out during oxidation of 2.68 × 10^–3 moles of Aⁿ⁺
=  2.68 × 10^–3 × (5 – n)
Thus the number of electrons added to reduce 1.61 × 10^–3 moles of MnO^-4  to Mn²⁺, i.e.  (+7) to (+2)  =1.61 × 10^–3 × 5
[Number of electrons involved = + 7 – (+2) = 5]

∴ 1.61 × 10^–3 × 5 =  2.68 × 10^–3 × (5 – n)

Q. 21. Five ml of 8N nitric acid, 4.8 ml of 5N hydrochloric acid and a certain volume of 17M sulphuric acid are mixed together and made upto 2litre. Thirty ml. of this acid mixture exactly neutralise 42.9 ml of sodium carbonate solution containing one gram of Na₂CO₃.10H₂O in 100 ml. of water. Calculate the amount in gram of the sulphate ions in solution. (1985 - 4 Marks) 

Ans. Sol. TIPS/Formulae : (i) Find normality of acid mixture and Na₂CO₃ . 10H₂O.
Equate them to find volume of H₂SO₄.

(ii) Meq. of H₂SO₄ = V × N =

(iii) Equivalent of SO₄^2– = equivalents of H₂SO₄ × Eq. wt. of SO₄^{– –}
N × V (ml.) = meq.
Acid mixture contains 5 ml of 8N, HNO₃, 4.8 ml of 5N, HCl and say, ‘V’ ml of 17 M ≡ 34 N, H₂SO₄. [1MH₂SO₄ = 2N.H₂SO₄]

N of the acid mixture =

[Total volume = 2 L = 2000 ml]

∵ Eq. of wt. of Na₂CO₃.10H₂O =

= 143

N of Na₂CO₃=

= 0.069N

N₁V₁ = N₂V₂

Hence =  0.0986

64 + 34 V = 0.0986 × 2000, 64 + 34 V = 197.2
34 V = 197.2 – 64.0 = 133.2 = 3.9 ml.

Hence meq. of H₂SO₄ = V × N of H₂SO₄                    
= 3.9 × 34 = 132.6 meq.                  
= 0.1326 eq. of H₂SO₄                  
= 0.1326 eq. of SO₄^2-                  
= 0.1326 × 48 g of SO₄^2-

= 6.3648 g of SO₄^2- are in 3.9 ml of 17M H₂SO₄

Q. 22. Arrange the following in increasing oxidation number of iodine. (1986 - 1 Mark)
 I₂, HI, HIO₄, ICl 

Ans. Sol. HI < I₂ < ICl < HIO₄; O.N. of I in I₂ = 0, HI = –1, ICl = +1, HIO₄ = +7.

Q. 23. (i) What is the weight of sodium bromate and molarity of solution necessary to prepare 85.5 ml of 0.672 N solution when the half-cell reaction is -
 BrO₃^- + 6H⁺ + 6e^– → Br^– + 3H₂O
 (ii) What would be the weight as well as molarity if the half-cell reaction is : -
 2 BrO₃^- + 12H⁺ + 10e^– → Br₂ + 6H₂O (1987 - 5 Marks) 

Ans. Sol. (i) From the given half-cell reaction, Here Eq. wt. of NaBrO₃ =  = 25.17

[∵ number of electron involved = 6]

Now we know that Meq. = Normality × Vol. in ml. = 85.5 × 0.672 = 57.456
Also Meq. =  × 1000

 × 1000

 × 1000 = 57.456 g

∴  WNaBrO₃ = 1.446 g

Molarity of NaBrO₃ =

= 0.112 M

(ii) From the given half-cell reaction, Eq. wt. of NaBrO₃ =  = 30.2

[Number of electron involved per BrO₃^– = = 5]

Thus, the amount of NaBrO₃ required for preparing 1000 ml. of 1 N NaBrO₃ = 30.2 g

∴  The amount of NaBrO₃ required for preparing 85.5 ml of 0.672 N NaBrO₃.

= 1.7532 g

Hence, Molarity   == 0.1344 M

Q. 24. A sugar syrup of weight 214.2 g contains 34.2 g of sugar (C₁₂H₂₂O₁₁). Calculate : (i) molal concentration and (ii) mole fraction of sugar in the syrup. (1988 - 2 Marks) 

Ans. Sol. (i) Weight of sugar syrup = 214.2 g Weight of sugar in the syrup = 34.2 g

∴  Weight of water in the syrup = 214.2 – 34.2 = 180.0 g Mol. wt. of sugar, C₁₂H₂₂O₁₁ = 342

∴  Molal concentration =  = 0.56

(ii) Mol. wt. of water, H₂O = 18

∴  Mole fraction of sugar = 

= 0.0099

Q. 25. A sample of hydrazine sulphate (N₂H₆SO₄) was dissolved in 100 ml. of water, 10 ml of this solution was reacted with excess of ferric chloride solution and warmed to complete the reaction. Ferrous ion formed was estimated and it required 20 ml. of M/50 potassium permanganate solution.
 Estimate the amount of hydrazine sulphate in one litre of the solution. (1988 - 3 Marks)

Reaction : 4Fe³⁺ + N₂H₄ → N₂ + 4Fe²⁺ + 4H⁺  
 MnO₄^- + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. 

Ans. Sol. TIPS/Formulae : No. of equivalents of KMnO₄ = No. of equivatents of hydrazine sulphate.

Change in oxidation state for each N₂H₄ = 2 × 2 = 4

Equivalent weight of N₂H₆SO₄ = = 32.5

Normality of KMnO₄ = 5 × 450     (∵ valence factor = 5)

Number of equivalents of KMnO₄ = 20 ×and if weight of hydrazin sulphate be x gm then equivalents of hydrazine sulphate = 

= 0.065 g

Hence wt. of N₂H₆SO₄ in 10 ml solution = 0.065 g

∴   Wt. of N₂H₆SO₄ in 1000 ml solution = 6.5 g

Q. 26. An equal volume of a reducing agent is titrated separately with 1M KMnO₄ in acid neutral and alkaline media. The volumes of KMnO₄ required are 20 ml. in acid, 33.4 ml. neutral and 100 ml. in alkaline media. Find out the oxidation state of manganese in each reduction product. Give the balanced equations for all the three half reactions. Find out the volume of 1M K₂Cr₂O₇ consumed; if the same volume of the reducing agent is titrated in acid medium.(1989 - 5 Marks) 

Ans. Sol. TIPS/Formulae : No. of equivalents of KMnO₄  in neutral medium = No. of equivalents of reducing agent.
Assuming that KMnO₄ shows the following changes during its oxidising nature.
Acidic medium Mn⁷⁺ + n₁e^– → Mn^a+ ∴  n₁ = 7 – a
Neutral medium Mn⁷⁺ + n₂e– → Mn^b+ ∴  n₂ = 7 – b
Alkaline medium Mn⁷⁺ + n₃e– → Mn^c+ ∴  n₃ = 7 – c
Let V ml. of reducing agent be used for KMnO₄ in different medium.
∴  Meq. of reducing agent                      
= Meq. of KMnO₄ in acid mediumMeq. of KMnO₄ in neutral medium                    
= Meq. of KMnO₄ in alkaline medium= 1 × n₁ × 20 = 1 × n₂ × 33.4 = 1 × n₃ × 100 = n₁ = 1.667 n₂ = 5 n₃
Since n₁, n₂ and n₃ are integers and n₁ is not greater than 7
∴  n₃ = 1 Hence  n₁ = 5  and  n₂ = 3
∴  Different oxidation states of Mn in Acidic medium Mn⁷⁺ + 5e^– → Mn^a+ or   a = + 2
Neutral medium Mn⁷⁺ + 3e^– → Mn^b+ or   b = + 4
Alkaline medium Mn⁷⁺ + 1e^– → Mn^c+ or   c = + 6
Further, same volume of reducing agent is treated with K₂Cr₂O₇, and therefore Meq. of reducing agent = Meq. of K₂Cr₂O₇          
1 × 5 × 20= 1 × 6 × V [Q Cr⁺⁶ + 6e^– → 2Cr⁺³]
V = 16.66 mL ∴  1M = 6 × 1N

Q. 27. A mixture of H₂C₂O₄ (oxalic acid) and NaHC₂O₄ weighing 2.02 g was dissolved in water and solution made upto one litre. Ten millilitres of the solution required 3.0 ml. of 0.1 N sodium hydroxide solution for complete neutralization. In another experiment, 10.0 ml. of the same solution, in hot dilute sulphuric acid medium. require 4.0 ml. of 0.1 N potassium permanganate solution for complete reaction.
 Calculate the amount of H₂C₂O₄ and NaHC₂O₄ in the mixture. (1990 - 5 Marks) 

Ans. Sol. TIPS/Formulae : No. of equivalents of KMnO₄            
= No. of equivatents of reducing agents.
Case I. Reaction of NaOH with H2C₂O₄ and NaHC₂O₄.
(i) H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O
(ii) NaHC₂O₄ + NaOH → Na₂C₂O₄ + H₂O
Number of milliequivalents of NaOH = N × V = 3.0 × 0.1 = 0.3
∴  Combined normality of the mixture titrated with NaOH

 = 0.03

Case II. Reaction of C₂O₄^– ion and KMnO₄
(iii) 5C₂O₄^– + MnO₄^– + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O KMnO₄ will react in same manner with both NaHC₂O₄ and H₂C₂O₄ as it can be seen from the above reaction.
Number of milliequivalents of KMnO₄ = 4.0 × 0.1 = 0.4
∴  Combined normality of the mixture titrated with KMnO₄

 = 0.04

The difference (0.04 N – 0.03 N = 0.01 N) is due to NaHC₂O₄ The total normality of NaHC₂O₄ will be = 0.01 + 0.01 = 0.02 N From equation (ii) in case I.
Eq. wt. of NaHC₂O₄ = 112 Amount of NaHC₂O₄ in one litre of solution formed  = 0.01 × 112 = 1.12 g and amount of H₂C₂O₄ = 2.02 – Wt. of NaHC₂O₄ = 2.02 – 1.12 = 0.90 g

Q. 28. A solid mixture (5.0 g) consisting of lead nitrate and sodium nitrate was heated below 600ºC until the weight of the residue was constant. If the loss in weight is 28.0 per cent, find the amount of lead nitrate and sodium nitrate in the mixture. (1990 - 4 Marks) 

Ans. Sol. TIPS/Formulae : Let the amount of NaNO₃ in the mixture = x g
∴  The amount of Pb(NO₃)₂ in the mixture  = (5 – x) g
Heating effect of sodium nitrate and lead nitrate

Now since, 170 g of NaNO₃ gives = 32 g of O₂ 

∴  xg of NaNO₃ gives = × x g of O₂

Similarly, 662 g of Pb(NO₃)₂ gives = 216 g of gases (NO₂ + O₂)
(5 – x) g of Pb(NO₃)₂ gives = × (5 – x) g of gases(NO₂ + O₂)
Actual loss, on heating, is 28% of 5 g of mixture

= 1.4 g

× (5 – x) = 1.4

32 x × 662 + 216(5 – x) × 170 = 1.4 × 170 × 662

21184 x + 183600 – 36720 x = 157556 – 15536 x = – 26044, x = 1.676 g

Wt. of NaNO₃ = 1.676 g and

Wt. of Pb(NO₃)₂ = 5 – 1.676 g = 3.324 g