Subjective Type Questions: Some Basic Concepts of Chemistry- 3 | JEE Advanced | 35 Years Chapter wise Previous Year Solved Papers for JEE PDF Download

Q. 29. Calculate the molality of 1 litre solution of 93% H₂SO₄ (weight/volume). The density of the solution is 1.84 g/ml. (1990 - 1 Marks) 

Ans. Sol. TIPS/Formulae :

Molality = 

Mass of H₂SO₄ in 100 ml of 93% H₂SO₄ solution = 93 g

∴   Mass of H₂SO₄ in 1000 ml of the H₂SO₄ solution = 930 g

Mass of 1000 ml H₂SO₄ solution = 1000 × 1.84 = 1840 g

Mass of water in 1000 ml of solution = 1840 – 930 g    = 910 g = 0.910 kg 

Moles of H₂SO₄ = 

∴ Moles of H₂SO₄ in 1 kg of water

 = 10.43 mol

∴    Molality of solution = 10.43m

Q. 30. A solution of 0.2 g of a compound containing Cu²⁺ and C₂^2- O₄ ions on titration with 0.02 M KMnO₄ in presence of H₂SO₄ consumes 22.6 ml. of the oxidant. The resultant solution is neutralized with Na₂CO₃, acidified with dil. acetic acid and treated with excess KI. The liberated iodine requires 11.3 ml of 0.05 M Na₂S₂O₃ solution for complete reduction.
 Find out the molar ratio of Cu²⁺ to C₂O₄^2- in the compound.
 Write down the balanced redox reactions involved in the above titrations. (1991 - 5 Marks)

Ans. Sol. In th e given problem, a solution contain in g Cu²⁺ and ^C₂O₄^2- is titrated first with KMnO₄ and then with Na₂S₂O₃ in presence of KI. In titration with KMnO₄, it is the C₂O₂^4- ions that react with the MnO^-₄ ions. The concerned balanced equation may be written as given below.
2MnO₄^- + 5 C₂O₂^4- + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

Thus according to the above reaction 2 mmol MnO^-₄ ≡ 5 mmol C₂ O₂^4-

However, No. of mmol of MnO^-₄ used in titration = Vol. in ml × M = 22.6 × 0.02 = 0.452 mmol MnO^-₄

Since 2 mmol MnO^-₄ ≡ 5 mmol C₂O₂^4-0.452 mmol MnO^-₄ ≡  × 0.452  = 1.130 mmol C₂ O₂^4-

 Titration with Na₂S₂O₃ in the presence of KI.

Here Cu²⁺ react and the reactions involved during titrationare

Thus  

No. of m mol of S₂O₃^2- used in titration        

 = 0.05 × 11.3 = 0.565 mmol S₂O₃^2-

Now since 2 mmol S₂O₃^2-  ≡ 2 mmol Cu²⁺ [From above equation]

0.565 mmol S₂O₃^2- =× 0.565 mmol Cu²⁺  

= 0.565 mmol Cu²⁺

∴ Molar ratio of Cu²⁺ to  = 1 : 2

Balanced equations in two cases Case I. Mn⁺⁷ + 5e^– → Mn⁺²

C₂⁺³ → 2C⁺⁴ + 2e^–

Case II. 2Cu⁺² + 2e^– → Cu²⁺ 2I^– → I₂ + 2e^– and I₂ + 2e^– → 2I^– 2S₂⁺² → S₄^+3/2 + 2e^–

Q. 31. A 1.0 g sample of Fe₂O₃ solid of 55.2% purity is dissolved in acid and reduced by heating the solution with zinc dust.
 The resultant solution is cooled and made upto 100.0 ml. An aliquot of 25.0 ml of this solution requires 17.0 ml of 0.0167 M solution of an oxidant for titration. Calculate the number of electrons taken up by the oxidant in the reaction of the above titration. (1991 - 4 Marks) 

Ans. Sol. Mass of Fe₂O₃ in the sample = × 1 = 0.552 g
Number of moles of Fe₂O₃ =  = 3.454 × 10^–3

Number of moles of Fe³⁺ ions =  2 × 3.454 × 10^–3                                  
= 6.9 × 10^–3 mol = 6.90 mmolSince its only 1 electron is exchanged in the conversion of Fe³⁺ to Fe²⁺, the molecular mass is the same as equivalent mass.
∴  Amount of Fe²⁺ ion in 100 ml. of sol. = 6.90 meq Volume of oxidant used for 100 ml of Fe²⁺ sol.            
= 17 × 4 = 68 ml.
Amount of oxidant used = 68 × 0.0167 mmol                          
= 1.1356 mmolLet the number of electrons taken by the oxidant = n
∴ No. of meq. of oxidant used = 1.1356 × n
Thus 1.1356 × n = 6.90   ⇒ n = = 6

Q. 32. A 2.0 g sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of CO₂ ceases. The volume of CO₂ at 750 mm Hg pressure and at 298 K is measured to be 123.9 ml. A 1.5g of the same sample requires 150 ml. of (M/10) HCl for complete neutralisation. Calculate the % composition of the components of the mixture. (1992 - 5 Marks)

Ans. Sol. 1.5 g of sample require = 150 ml. of   HCl

∴ 2 g of sample require =ml. of   HCl

= 200 ml. of  HCl

On heating, the sample, only NaHCO₃ undergoes decomposition as given below.

 2 equ.
Neutralisation of the sample with HCl takes place as given below.

NaHCO₃ + HCl     →  NaCl + H₂O + CO₂ 
1 eq.            1 eq.

Na₂CO₃ +  2HCl   →  2NaCl + H₂O + CO₂ 
1 mole        2 mole
2 eq.            2 eq.
Hence, 2 g sample ≡ 200 ml. of M/10 HCl              
= 200 ml. of N/10 HCl = 20 meq = 0.020 eq
Number of moles of CO₂ formed, i.e.

= 0.005

Moles of NaHCO₃ in the sample (2 g) = 2× 0.005 = 0.01

Equivalent of NaHCO₃ = 0.01

Wt. of NaHCO₃ = 0.01 × 84 = 0.84 g

% of NaHCO₃ = = 42%

Equivalent of Na₂CO₃ = 0.02 – 0.01 = 0.01

Wt. of Na₂CO₃ = 0.01 × 53 = 0.53 g

∴ % of Na₂CO₃ =  = 26.5%

∴ % of Na₂SO₄ in the mixture = 100 – (42 + 26.5) = 31.5%

Q. 33. One gram of commercial AgNO₃ is dissolved in 50 ml. of water. It is treated with 50 ml. of a KI solution. The silver iodide thus precipitated is filtered off. Excess of KI in the filterate is titrated with (M/10) KIO₃ solution in presence of 6M HCl till all I^– ions are converted into ICl. It requires 50 ml. of (M/10) KIO₃ solution. 20 ml. of the same stock solution of KI requires 30 ml. of (M/10)KIO₃ under similar conditions.
 Calculate the percentage of AgNO₃ in the sample.
 (Reaction : KIO₃ + 2KI + 6HCl → 3ICl + 3KCl + 3H₂O) (1992 - 4 Marks) 

Ans. Sol. Reaction involved titration is 1 mole 2 moles KIO₃ + 2KI + 6HCl   →  3ICl + 3KCl + 3H₂O

20 ml. of stock KI solution ≡ 30 ml. of   KIO₃ solution
Molarity of KI solution = 

Millimoles in 50 ml. of KI solution = 50 ×= 15

Millimoles of KI left unreacted with AgNO₃ solution

= 2 × 50 ×  = 10

∴ Millimoles of KI reacted with AgNO₃ = 15 – 10 = 5

Millimoles of AgNO₃ present in AgNO₃ solution = 5

Molecular weight of AgNO₃ = 170

∴ Wt. of AgNO₃ in the solution = 5 × 10^–3 × 170 = 0.850 g

% AgNO₃ in the sample =× 100 = 85%

Q. 34. Upon mixing 45.0 ml. of 0.25 M lead nitrate solution with 25.0 ml of 0.10 M chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also, calculate the molar concentrations of the species left behind in the final solution. Assume that lead sulphate is completely insoluble. (1993 - 3 Marks) 

Ans. Sol. Calculation of number of moles in 45 ml. of 0.025 M Pb(NO₃)₂

Moles of Pb(NO₃)₂ = 0.25 × = 0.01125

∴ Initial moles of Pb²⁺ = 0.01125

Moles of NO^3- = 0.01125 × 2 = 0.02250 [1 mole Pb(NO₃)₂ ≡ 2 moles of NO₃]

Calculation of number of moles in 25 ml. of 0.1 M chromic sulphate

Moles of chromic sulphate (Cr₂(SO₄)₃  

= 0.1 × = 0.0025 moles

Moles of SO₂^4- = 0.0025 × 3 = 0.0075 [1 Mole of chromic sulphate ≡ 3 moles of SO₄^2–]

Moles of PbSO₄ formed = 0.0075 [SO₄^2– is totally consumed]

Moles of Pb²⁺ left = 0.01125 – 0.0075 = 0.00375

Moles of NO₃^- left = 0.02250 [NO₃^– remain unreacted]

Moles of chromium ions = 0.0025 × 2 = 0.005

Total volume of the solution = 45 + 25 = 70 ml.

∴ Molar concentration of the species left

(i) Pb²⁺ = × 1000 = 0.05357 M

(ii) NO₃^- =× 1000 = 0.3214 M

(iii) Cr³⁺ =× 1000 = 0.0714 M

Q. 35. The composition of a sample of Wustite is Fe_0.93O_1.00 .
 What percentage of the iron is present in the form of Fe (III)? (1994 - 2 Marks)

Ans. Sol. In pure iron oxide (FeO), iron and oxygen are present in the ratio 1 : 1.
However, here number of Fe²⁺ present = 0.93 or No. of Fe²⁺ ions missing = 0.07 Since each Fe²⁺ ion has 2 positive charge, the total number of charge due to missing (0.07) Fe²⁺ ions = 0.07 × 2 = 0.14 To maintain electrical neutrality, 0.14 positive charge is compensated by the presence of Fe³⁺ ions. Now since, replacement of one Fe²⁺ ion by one Fe³⁺ ion increases one positive charge, 0.14 positive charge must be compensated by the presence of 0.14 Fe³⁺ ions.
In short, 0.93 Fe²⁺ ions have 0.14 Fe³⁺ ions 100 Fe²⁺ ions have =  × 100 = 15.05%

Q. 36. 8.0575 × 10^–2 kg of Glauber ’s salt is dissolved in water to obtain 1 dm³ of a solution of density 1077.2 kg m^–3. Calculate the molarity, molality and mole fraction of Na₂SO₄ in the solution. (1994 - 3 Marks) 

Ans. Sol. The formula of Glauber’s salt is Na₂SO₄.10H₂O Molecular mass of Na₂SO₄.10H₂O          
= [2 × 23 + 32.1 + 4 × 16] + 10 (1.01 × 2 + 16) = 322.3 g mol^–1
Weight of the Glauber’s salt taken = 80.575 gm Out of 80.575 g of salt, weight of anhydrous Na₂SO₄

 × 80.575 = 35.525 g

Number of moles of Na₂SO₄ per dm³ of the solution = 0.25

Molarity of the solution = 0.25 M

Density of solution = 1077.2 kgm^–3

gm cm^–3 = 1.0772 g cm^–3

Total weight of sol = V × d = 1 dm³ × d                          
= 1000 cm³ × 1.0772 gcm^–3  = 1077.2 g
Weight of water = 1077.2 – 35.525 = 1041.67 g

Molality of sol. =   × 1000 g  = 0.2399 = 0.24 m

Number of moles of water in the solution =  = 57.87

Mole fraction of Na₂SO₄

= 0.0043 = 4.3 × 10^–3

Q. 37. A 3.00 g sample containing Fe₃O₄, Fe₂O₃ and an inert impure substance, is treated with excess of KI solution in presence of dilute H₂SO₄.The entire iron is converted into Fe²⁺ along with the liberation of iodine. The resulting solution is diluted to 100 ml. A 20 ml of the diluted solution requires 11.0 ml of 0.5 M Na₂S₂O₃ solution to reduce the iodine present. A 50 ml of the diluted solution, after complete extraction of the iodine requires 12.80 ml of 0.25 M KMnO₄ solution in dilute H₂SO₄ medium for the oxidation of Fe²⁺. Calculate the percentages of Fe₂O₃ and Fe₃O₄ in the original sample. (1996 - 5 Marks) 

Ans. Sol. TIPS/Formulae : Find the milliequivalents and equate them as per data given in question.
For Fe₃O₄ → 3FeO
2e + Fe₃^(8/3)+ → 3Fe²⁺ 
Thus, valence factor for Fe₃O₄ is 2 and for FeO is 2/3.
For,   Fe₂O₃ → 2FeO; 2e + Fe₂³⁺ → 2Fe²⁺ ...(1)
Thus valence factor for Fe₂O₃ is 2 and for FeO is 1.
Let Meq.of Fe₃O₄ and Fe₂O₃ be a and b respectively.
∴ Meq. of Fe₃O₄ + Meq. Fe₂O₃ = Meq. of I₂ liberated                
= Meq. of hypo used

a + b == 27.5

Now, the Fe²⁺ ions are again oxidised to Fe³⁺ by KMnO₄.
Note that in the change Fe²⁺ → Fe³⁺ + e^–; valence factor of Fe²⁺ is l.
Thus, Meq. of Fe²⁺ (from Fe₃O₄) + Meq. of Fe²⁺ (from Fe₂O₃) = Meq. of KMnO₄ used  .... (2)
If valence factor for Fe²⁺ is 2/3 from Eq. (1), then Meq. of Fe²⁺ (from Fe₃O₄) = a
If valence factor for Fe²⁺ is 1 then Meq. of Fe²⁺ (from Fe₃O₄) = 3a/2 … (3)
Similarly, from Eq. (2), Meq. of Fe²⁺ from (Fe₂O₃) = b.
∴ 3a/2 + b = 0.25 x 5 x 12.8 x 100/50 = 32 or 3a + 2b = 64 ....(4)
From Eqs. (3) and (4)
Meq. of Fe₃O₄ = a = 9  &  Meq. of Fe₂O₃ = b = 18.5

∴ WFe₃O₄ =  = 1.044g

and  WFe₂O₃ = = 1.48g

∴  % of Fe₃O₄ == 34.8

 and   % of Fe₂O₃ == 49.33

Q. 38. An aqueous solution containing 0.10 g KIO₃ (formula weight = 214.0) was treated with an excess of KI solution. The solution was acidified with HCl. The liberated I₂ consumed 45.0 mL of thiosulphate solution to decolourise the blue starch-iodine complex. Calculate the molarity of the sodium thiosulphate solution. (1998 -  5 Marks) 

Ans. Sol. TIPS/Formulae : Write the reactions taking place, balance them and equate moles of I₂ and Na₂S₂O₃.

KIO₃ + 5KI → 3K₂O + 3I₂ i.e., 2 I⁵⁺ + 10e^– → I⁰₂
2I^– → I⁰₂ + 2e^– 
Now liberated I₂ reacts with Na₂S₂O₃ I₂ + 2e^– → 2I^–
2S₂O₃^2- → S₄O₆^2- + 2e^– 
∴ millimole ratio of  I₂ : S₂O₃ = 1 : 2
Thus, m mole of I₂ liberated = m mole of Na₂S₄O₆ used x

[M is molarity of thiosulphate]

Also m mole of KIO₃ = 

Now m mole ratio of  KIO₃ : I₂ = 1 : 3

Thus, 

∴  M = = 0.062

Q. 39. How many millilitres of 0.5 M H₂SO₄ are needed to dissolve 0.5 g of copper(II) carbonate? (1999 - 3 Marks) 

Ans. Sol. TIPS/Formulae : Use molarity equation to find volume of H₂SO₄ solution.

∴ For 123.5 gms of Cu(II) carbonate 98 g of H₂SO₄ are required.
For 0.5 gms of Cu(II) carbonate weight of H₂SO₄ reqd.

  = 0.39676 g H₂SO₄

Weight of required H₂SO₄ = 0.39676 g

Weight of solute in grams

0.39676 =  or

Volume of H₂SO₄ solution = 8.097 ml

Q. 40. A plant virus is found to consist of uniform cylindrical particles of 150 Å in diameter and 5000 Å long. The specific volume of the virus is 0.75 cm³/g. If the virus is considered to be a single particle, find its molar mass.(1999 - 3 Marks)
 Ans. Sol. TIPS/Formulae : (i) Volume of virus = πr²ℓ (Volume of cylinder)

(ii) Mass of single virus = 

(iii) Molecular mass of virus = Mass of single virus × 6.02 × 10²³

Volume of virus = πr²l

 x 5000 x 10^-8

= 0.884 x 10^–16 cm³

Weight of one virus = 

= 1.178 x 10^-16 g

∴  Mol. wt. of virus = 1.178 x 10^-16 x 6.02 x 10²³ = 7.09 x 10⁷

Q. 41. Hydrogen peroxide solution (20 ml) reacts quantitatively with a solution of KMnO₄ (20 ml) acidified with dilute H₂SO₄.
 The same volume of the KMnO₄ solution is just decolourised by 10 ml of MnSO₄ in neutral medium simultaneously forming a dark brown precipitate of hydrated MnO₂. The brown precipitate is dissolved in 10 ml of 0.2 M sodium oxalate under boiling condition in the presence of dilute H₂SO₄. Write the balanced equations involved in the reactions and calculate the molarity of H₂O₂. (2001 - 5 Marks) 

Ans. Sol. TIPS/Formulae : Write the balanced chemical reaction for change and apply mole concept.
The given reactions are MnO₂ ↓ + Na 2C₂ O₄+ 2H₂SO₄
—→ MnSO₄ + CO₂ + Na 2SO₄+ 2H₂O

∴   Meq. of MnO₂ ≡ Meq of Na₂C₂O₄ = 10 × 0.2 × 2 = 4

∴   mM of MnO₂ == 2 

Now 2KMnO₄ + 3MnSO₄+ 2H₂O
—→ 5MnO₂ ↓ + K₂SO₄+ 2H₂O
Since eq. wt. of MnO₂ is derived from KMnO₄ and MnSO₄ both, thus it is better to proceed by mole concept
mM of KMnO₄ ≡ mM of MnO₂ × (2/5) = 4/5
Also 5H₂O₂ + 2KMnO₄+ 3H₂SO₄ 
—→ 2MnSO₄ + K₂SO₄ + 8H₂O+ 5O₂
∴ mM of H₂O₂ = mM of KMnO₄ ×

2KMnO₄ + 5H₂O₂ + 3H₂SO₄ —→ K₂SO₄ + 2MnSO₄ + 8H₂O + 5O₂

2KMnO₄ + 3MnSO₄ + 2H₂O —→ 5MnO₂ + 2H₂SO₄ + K₂SO₄

MnO₂ + Na₂C₂O₄  + 2H₂SO₄ —→MnSO₄ + 2CO₂ + Na₂SO₄ + 2H₂O

Q. 42. Calculate the molarity of water if its density is 1000 kg/m³. (2003 - 2 Marks)

Ans. Sol. 1 litre water = 1 kg i.e. 1000 g water (∵ d = 1000 kg/m³)

 = 55.55 moles of water

So, molarity of water = 55.55M