JEE Advanced (Subjective Type Questions): Some Basic Concepts of Chemistry- 1 | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q. 1. What weight of AgCl will be precipitated when a solution containing 4.77 g of NaCl is added to a solution of 5.77 g of AgNO₃?
 (1978)
 Ans. Sol. TIPS/Formulae : Write the balance chemical equation and use mole concept for limiting reagent.

From the given data, we find AgNO₃ is limiting reagent as NaCl is in excess.

∵ 170.0 g of AgNO₃ precipitates AgCl = 143.5 g
∴ 5.77 g of AgNO₃ precipitates AgCl

=  4.87 g

Q. 2. One gram of an alloy of aluminium and magnesium when treated with excess of dil. HCl forms magnesium chloride, aluminium chloride and hydrogen. The evolved hydrogen, collected over mercury at 0°C has a volume of 1.20 litres at 0.92 atm. pressure. Calculate the composition of the alloy. [H = 1, Mg = 24, Al = 27] (1978) 

Ans. Sol. TIPS/Formulae : (i) Find volume of H₂ at N.T.P. (ii) Total amount of H₂ liberated = H₂ liberated by Mg & HCl + H₂ liberated by Al & HCl.
Conversion of volume of H₂ to N.T.P
Given conditions                                              N.T.P conditions
P1 = 0.92 atm.                                                 P₂ = 1 atm.
V₁ = 1.20 litres                                                 V₂ = ?
T₁ = 0 + 273 = 273 K                                       T₂ = 273 K
Applying ideal gas equation,   

 = 1.104 litres = 1104 ml

The relevant chemical equations are

(i) 

= 54 g                                  = 67200 ml at NTP

(ii) 

Wt. of alloy = 1 g Let the wt. of aluminium in alloy = x g
∴ Wt. of magnesium in alloy = (1 – x) g According to equation
(i) 54 g of Al = 67200 ml of H₂ at N.T.P ∴ x g of Al = 6 = 1244.4 x ml of H₂ at N.T.P

Similarly, from equation
(ii) 24 g of Mg = 22400 ml of H₂ at N.T.P (1 – x) g of Mg =   × (1 – x) = 933.3 (1 – x) ml of H₂

Hence total vol. of H₂ collected at N.T.P = 1244.4 x + 933.3 (1 – x) ml

But total vol. of H₂ as calculated above = 1104 ml

∴ 1244.4 x + 933.3 (1 – x) = 1104 ml
1244.4 x – 933.3 x = 1104 – 933.3
311.1 x = 170.7, x = 0.5487

Hence 1 g of alloy contains Al = 0.5487 g

∴ Percentage of Al in alloy == 54.87%
% of Mg in alloy = 100 – 54.87 = 45.13%

Q. 3. Igniting MnO₂ converts it quantitatively to Mn₃O₄. A sample of pyrolusite is of the following composition : MnO₂ 80%, SiO₂ and other inert constituents 15%, rest being water. The sample is ignited in air to constant weight. What is the percentage of Mn in the ignited sample? (1978) [O = 16, Mn = 54.9] 

Ans. Sol. 

Let the amount of pyrolusite ignited = 100.00 g
∴ Wt. of MnO₂ = 80 g (80% of 100 g = 80 g) Wt. of SiO₂ and other inert substances = 15 g Wt. of water = 100 – (80 + 15) = 5 g
According to equation, 260.7 g of MnO₂ gives = 228.7 g of Mn₃O₄ 
∴ 80 g of MnO₂ gives = × 80 = 70.2 g of Mn₃O₄

NOTE :
During ignition, H₂O present in pyrolusite is removed while silica and other inert substances remain as such.
∴ Total wt. of the residue = 70.2 + 15 = 85.2 g
Calculation of % of Mn in ignited Mn3O4
3 Mn                  =   Mn₃O₄
3 × 54.9 = 164.7 g               3 × 54.9 + 64 = 228.7g
Since, 228.7 g of Mn₃O₄ contains 164.7 g of Mn
70.2 g of Mn₃O₄ contains =  × 70.2 = 50.55 g of Mn

Weight of residue = 85.2 g

Hence, percentage of Mn is the ignited sample

x 100 = 59.33%

Q. 4. 4.215 g of a metallic carbonate was heated in a hard glass tube and the CO₂ evolved was found to measure 1336 ml at 27°C and 700 mm pressure. What is the equivalent weight of the metal? (1979) 

Ans. Sol. TIPS/Formulae : (i) Find the volume of CO₂ at NTP
(ii) Find molecular wt. of metal carbonate
(iii) Find the wt. of metal
(iv) Calculate equivalent weight of metal
Given P₁ = 700 mm,  P₂ = 760 mm, V₁ = 1336 ml, V₂ = ?
T₁ = 300 K, T₂ = 273 K

   = 1119.78 ml  = 1.12 L at NTP

∴  1.12 L of CO₂ is given by carbonate = 4.215 g

Molecular weight of metal carbonate = 

= 84.3
Metal carbonate is MCO₃ = M + 12 + 48 = M + 60

Atomic weight of M = 84.3 – 60 = 24.3

Eq. wt. of metal =  × M. wt. x 24.3 = 12.15

Q. 5. (a) 5.5 g of a mixture of FeSO₄. 7H₂O and Fe₂(SO₄)₃. 9H₂O requires 5.4 ml of 0.1 N KMnO₄ solution for complete oxidation. Calculate the number of gram mole of hydrated ferric sulphate in the mixture.
 (b) The vapour density (hydrogen = 1) of a mixture consisting of NO₂ and N₂O₄ is 38.3 at 26.7°C. Calculate the number of moles of NO₂ in 100 g of the mixture. (1979)

Ans. Sol. (a) Equivalents of KMnO₄ = Equivalents of FeSO₄ . 7H₂O

5.4 ml 0.1 N KMnO₄ == 5.4 × 10^–4 equivalentsAmount of FeSO₄ = 5.4 × 10^–4 × Mol wt. of FeSO₄.7H₂O                      
= 5.4 × 10^–4 × 278 = 0.150 g
Total weight of mixture = 5.5 g
Amount of ferric sulphate = 5.5 – 0.150 g = 5.35 g
Hence Moles of ferric sulphate = 
= 9.5 × 10^–3 gram-mole

(b) Using the relation, Mol. wt. = 2 × vapour density, we get Mol. wt. = 2 × 38.3 = 76.6

No. of moles =   = 1.30              ....(i)

Let weight of NO₂ in mixture = x g

Then weight of N₂O₄ in mixture = 100 – x No. of moles of NO₂ =   ....(ii)
No. of moles of N₂O₄ =      ...(iii)
- According to problem

On solving the equation we find, x = 20.1

∴ weight of NO₂ = 20.1 g Moles of NO₂ =  = 0.437 moles.

Q. 6. 5 ml of a gas containing only carbon and hydrogen were mixed with an excess of oxygen (30 ml) and the mixture exploded by means of an electric spark. After the explosion, the volume of the mixed gases remaining was 25 ml. On adding a concentrated solution of potassium hydroxide, the volume further diminished to 15 ml of the residual gas being pure oxygen. All volumes have been reduced to N.T.P.
 Calculate the molecular formula of the hydrocarbon gas. (1979)
 Ans. Sol. Volume of oxygen taken = 30 ml, Volume of unused oxygen = 15 ml
Volume of O₂ used = Volume of O₂ added – Volume of O₂ left              
= 30 – 15 = 15 ml
Volume of CO₂ produced = Volume of gaseous mixture after explosion – Volume of unused oxygen or Volume of CO₂ produced = 25 – 15 = 10 ml
Volume of hydrocarbon = 5 ml General equation for combustion of a hydrocarbon is as follows -

∴  Volume of CO₂ produced = 5x, Since Volume of CO₂ = 10 ml

∴  5x = 10 ⇒ x = 2, Volume of O₂ used = 15 ml

⇒  2 +  = 3 (∵ x = 2) ⇒  8 + y = 12 ∴ y = 4

Hence Molecular formula of hydrocarbon is C₂H₄.

Q. 7. In the analysis of 0.500 g sample of feldspar, a mixture of chlorides of sodium and potassium is obtained which weighs 0.1180g. Subsequent treatment of mixed chlorides with silver nitrate gives 0.2451g of silver chloride. What is the percentage of sodium oxide and potassium oxide in feldspar.
 (1979)

Ans. Sol. TIPS/Formulae : (i) Equate given mass of AgCl against mass obtained from NaCl and KCl
(ii) 2NaCl ≡ Na₂O & 2KCl ≡ K₂O Let amount of NaCl in mixture = x gm
∴ amount of KCl in mixture = (0.118 - x) gm
NaCl + AgNO₃ —→ AgCl + NaNO₃  58.5 g                    
143.5 g

∵ 58.5 g NaCl gives AgCl = 143.5g
∴ x g NaCl gives AgCl =  × x g

Again
KCl + AgNO₃ —→ AgCl + KNO₃ 
74.5 g                             143.5 g

∵ 74.5 g KCl gives AgCl = 143.5g

∴ (0.118 – x) g KCl gives AgCl =

Total weight of AgCl = 0.2451g

= 0.2451

∴ x = 0.0338g
∴ Amount of NaCl in mixture = 0.0338g
∴ Amount of KCl in mixture = 0.118 – 0.0338 = 0.0842g
Since            
 2NaCl     ≡   Na₂O              
2 × 58.5             62
  = 117.0
∵ 117g NaCl is equivalent to = 62.0g Na₂O

∴ 0.0338g NaCl is equivalent to = × 0.0338 g Na₂O

   = 0.0179g

% of Na₂O in 0.5g of feldspar = × 100 = 3.58%

2KCl         ≡       K₂O
2 × 74.5 = 149                 94
∵ 149g of KCl is equivalent to = 94g K₂O

∴ 0.0842g of KCl is equivalent to =      = 0.0531g K₂O

∴ % of K₂O in 0.5g of feldspar = × 100 = 10.62%
% of Na₂O in feldspar = 3.58%
% of K₂O in feldspar = 10.62%

Q. 8. A compound contains 28 percent of nitrogen and 72 percent of metal by weight. 3 atoms of metal combine with 2 atoms of N. Find the atomic weight of metal. (1980) 

Ans. Sol. According to problem, three atoms of M combine with 2 atoms of N
∴  Formula of compound is M₃N₂ (Where M is the metal) Equivalent wt of N =   (∵ valency of N in compound is 3)
∵ 28 g N combines with = 72g metal
∴ 14/3 N combines with =  = 12
∴ Eq. wt. of metal = 12
At wt of metal = Eq. wt × valency = 12 × 2 = 24 [Valency of metal = 2]

Q. 9. (i) A sample of MnSO₄.4H₂O is strongly heated in air. The residue is Mn₃O₄.
 (ii) The residue is dissolved in 100 ml of 0.1 N FeSO₄ containing dilute H₂SO₄.
 (iii) The solution reacts completely with 50 ml of KMnO₄ solution.
 (iv) 25 ml of the KMnO₄ solution used in step (iii) requires 30 ml of 0.1 N FeSO₄ solution for complete reaction.
 Find the amount of MnSO₄.4H₂O present in the sample. (1980)
 Ans. Sol. Following reactions take place

Mn₃O₄ + 2FeSO₄ + 4H₂SO₄ —→ Fe₂(SO₄)₃ + 3MnSO₄ + 4H₂O
Milliequivalents of FeSO₄ in 30 ml of 0.1N FeSO₄ = 30 × 0.1 = 3 m. eq.
According to problem step (iv) 25 ml of KMnO₄ reacts with = 3 m eq of FeSO₄

Thus in step (iii) of the problem, 50 ml of KMnO₄ reacts with = x m.eq. of FeSO₄                                
= 6 meq of FeSO₄
Milli  eq. of 100 ml of 0.1N FeSO₄ = 100 × 0.1 = 10 m eq.
FeSO₄ which reacted with Mn₃O₄ = (10^–6) = 4 m eq.
Milli eq of FeSO₄ = Milli eq. of Mn₃O₄ (∵ Milli eq of oxidising agent and reducing agent are equal)
∵ Mn₃O₄ ≡ 3MnSO₄ .4H₂O
∴ 1 Meq of Mn₃O₄ = 3 Meq of MnSO₄ . 4H₂O
∴ 4 Meq of Mn₃O₄ = 12 Meq of MnSO₄ . 4H₂O
Eq. wt of MnSO₄.4H₂O == 111.5

Wt of MnSO₄.4H₂O in sample = 12 × 111.5  = 1338 mg = 1.338g.

Q. 10. (a) One litre of a sample of hard water contains 1 mg of CaCl₂ and 1 mg of MgCl₂. Find the total hardness in terms of parts of CaCO₃ per 106 parts of water by weight.
 (b) A sample of hard water contains 20 mg of C^a++ ions per litre. How many milli-equivalent of Na₂CO₃ would be required to soften 1 litre of the sample?
 (c) 1 gm of Mg is burnt in a closed vessel which contains 0.5 gm of O₂.
 (i) Which reactant is left in excess?
 (ii) Find the weight of the excess reactants?
 (iii) How may milliliters of 0.5 N H₂SO₄ will dissolve the residue in the vessel. (1980) 

Ans. Sol. (a)

From this it is evident, that 111 mg CaCl₂ will give CaCO₃ = 100mg

∴ 1 mg CaCl₂ will give CaCO₃ =  mg = 0.90 mg

95 mg MgCl₂ gives CaCO₃ = 100 mg

∴ 1 mg MgCl₂ gives CaCO₃ =  mg = 1.05 mg

∴ Total CaCO₃ formed by 1 mg CaCl₂ and 1 mg MgCl₂ = 0.90 + 1.05 = 1.95 mg
∴ Amount of CaCO₃ present per litre of water = 1.95mg
∴ wt of 1 ml of water = 1g = 10³ mg
∴ wt of 1000 ml of water = 10³ × 10³ = 10⁶mg
∴ Total hardness of water in terms of parts of CaCO₃ per 10⁶ parts of water by weight = 1.95 parts.

(b) Eq wt of Ca⁺⁺ == 20

Ca²⁺ + Na₂CO₃ —→ CaCO₃ + 2Na ⁺ 
1 milliequivalent of Ca²⁺ = 20 mg
1 milliequivalent of Na₂CO₃ is required to soften 1 litre of hard water.

(c)  

 ∵ 32g of O₂ reacts with = 48g Mg

∴ 0.5g of O₂ reacts with = x 0.5 = 0.75g

Weight of unreacted Mg = 1.00 – 0.75 = 0.25g

Thus Mg is left in excess.
Weight of MgO formed =  x 0.75 = 1.25g

MgO + H₂SO₄ —→ MgSO₄ + H₂O

(40g) According to reaction

∵ 40g MgO is dissolved it gives 1000 ml of 1 N. H₂SO₄
∴ 40 g MgO is dissolved it gives 2000 ml 0.5 N H₂SO₄
∴ 1.25 MgO is dissolved it gives

ml of 0.5 N H₂SO₄

= 62.5ml of 0.5N H₂SO₄

Q. 11. A hydrocarbon contains 10.5g of carbon per gram of hydrogen. 1 litre of the vapour of the hydrocarbon at 127°C and 1 atmosphere pressure weighs 2.8g. Find the molecular formula. (1980)

Ans. Sol. 

Given P = 1 atm V = 1L, T = 127°C = 127 + 273 = 400 K

PV = nRT (Ideal gas equation)

 = 0.0304

Mol. wt == 92.10

∴ Emperical formula = C₇H₈ 

Emperical formula, wt = 12 × 7 + 1 × 8 = 92

Molecular formula = n × empirical formula              
= 1(C₇H₈) = C₇H₈

Q. 12. Find (1980) (i) The total number of neutrons and (ii) The total mass of neutron in 7 mg of 14C. (Assume that mass of neutron = mass of hydrogen atom) 

Ans. Sol. 

(i) No. of C atoms in 14g of ¹⁴C = 6.02 × 10²³

∴ No. of C atom in 7 mg (7/1000g) of ¹⁴C

= 3.01 × 10²⁰

No. of neutrons in 1 carbon atom = 7

∴ Total no. of neutrons in 7 mg of ¹⁴C = 3.01× 10²⁰ ×7      
= 21.07 × 10²⁰
Wt of 1 neutron = wt of 1 hydrogen atom

∴ Wt of 3.01 × 10²⁰ × 7 neutrons

= 3.5 × 10^–3g

Q. 13. A mixture contains NaCl and unknown chloride MCl.
 (i) 1 g of this is dissolved in water. Excess of acidified AgNO₃ solution is added to it. 2.567 g of white ppt. is formed.
 (ii) 1 g of original mixture is heated to 300°C. Some vapours come out which are absorbed in acidified AgNO₃ solution,  1.341 g of white precipitate was obtained.
 Find the molecular weight of unknown chloride. (1980)

Ans. Sol. Weight of AgCl formed = 2.567 g
Amount of AgCl formed due to MCl = 1.341 g (∵ NaCl does not decompose on heating to 300°C)
∴ Weight of AgCl formed due to NaCl = 2.567 – 1.341= 1.226g

∵ 143.5g of AgCl is obtained from NaCl = 58.5g

∴ 1.226 g of AgCl is obtained from NaCl

x 1.226 = 0.4997 g

∴ Wt of MCl in 1 g of mixture = 1.000 – 0.4997 = 0.5003g

∵ 1.341 g of AgCl is obtained from MCl = 0.5003g

∴ 143.5g of AgCl is obtained from MCl

x 143.5 = 53.53 g

∴ Molecular weight of MCl = 53.53

Q. 14. A 1.00 gm sample of H₂O₂ solution containing X per cent H₂O₂ by weight requires X ml of a KMnO₄ solution for complete oxidation under acidic conditions. Calculate the normality of the KMnO₄ solution. (1981 - 3 Marks) 

Ans. Sol. The complete oxidation under acidic conditions can be represented as follows:
5H₂O₂ + 2MnO₄^- + 6H⁺ → 5O₂ + 2Mn²⁺+ 8H₂O
Since 34 g of H₂O₂ = 2000 ml of 1N . H₂O₂

∴ 34 g of H₂O₂ = 2000 ml of 1N KMnO₄ [∵ N₁V₁ = N ₂V₂]

orH₂O₂ =   of 1N KMnO₄

Therefore the unknown normality = 

or 0.588 N