7. Solved Problems

 

7.1       Subjective

 

Problem -1:      0.5 g fuming sulphuric acid (oleum) is diluted with water. This solution is completely neutralised by 26.7 ml of 0.4 N NaOH. Find the percentage of free SO₃ in the sample solution.

 

Solution:        Oleum consists of SO₃ and H₂SO₄

Let the mass of SO₃ in the given sample be x gm and mass of H₂SO₄ in the given sample = (0.5 – x) gm

Eq. mass of SO₃ =  = 40

No. of g equivalents of SO₃ =

2NaOH + SO₃ ¾® Na₂SO₄ + H₂O

2NaOH + H₂SO₄ ¾® Na₂SO₄ + 2H₂O

Eq. mass of H₂SO₄ =  = 49

No. of gm equivalents of H₂SO₄ =

Total no. of gm equivalents = +

26.7 ml of 0.4 N NaOH contain no. of eq. NaOH = × 26.7

Thus,  =

x =

% of free SO₃ = = 20.72

 

Problem 2:       A solution of 0.2 g of a compound containing Cu⁺² and  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 dilute acetic acid and treated with excess KI. The liberated iodine requires 11.3 mL of 0.05 N Na₂S₂O₃ solution for complete reduction. Find out the molar ratio of Cu⁺² to  in the compound. Write down the balanced redox reactions involved in the above titration.

 

Solution:        The mixture of Cu²⁺ and  are reacting separately first with KMnO₄ solution and then with solid KI to liberate iodine. It can be seen that Cu⁺² cannot be oxidised and cannot be reduced. This is because Cu is already in its highest oxidation state +2.

\  Equivalents of KMnO₄ solution=   =   2.26 ´ 10^-3
\  moles of =      =    1.13 ´ 10^-3

This is because only  is oxidised by KMnO₄ to CO₂ (‘n’ factor 2)

Equivalents of Na₂S₂O₃ solution = =     5.65 ´ 10^-4

\ moles of Cu⁺²   =    = 5.65 ´ 10^-4

This is because only Cu⁺² is reduced by KI to Cu⁺

                        (Note: Whenever a metal ion is reduced it always goes to lower oxidation state but generally never goes to oxidation state zero).

\ molar ratio of Cu⁺² to  =   = 0.5

Reactions:       2Mn + 5C₂ + 16H⁺ ® 2Mn⁺² 10CO₂ + 8H₂O

(2Cu⁺² + 4I^– ® Cu₂I₂ + I₂ )I₂ + 2S₂  ®  2I^– + S₄.

 

Problem- 3:      An equal volume of a reducing agent is treated separately with 1 M KMnO₄ in acid, neutral and alkaline medium. The volumes of KMnO₄ required are 20 ml in acid, 33.4 ml in neutral and 100 ml in alkaline medium. Find out the oxidation state of manganese in each reaction product. Find out the volume of 1 M K₂Cr₂O₇ consumed, if the same volume of the reducing agent is treated in acid medium.

 

Solution.        Let N₁, N₂ and N₃ be the normalities of 1 M KMnO₄ solution in acidic, neutral and alkaline medium respectively.

20 ml N₁ º 33.4 ml N₂ º 100 ml N₃

In acidic medium

1 M KMnO_­4 = 5 N KMnO₄

Thus N₂ = N₁=  = 3 N

N₃ = N₁ = = 1N

The volume required for titration of the same volume of reducing agent with acidified K₂Cr₂O₇ solution

20 ml 5N KMnO₄ = V6NK₂Cr₂O₇

V =  = 16.66 ml

Problem – 4:     H₂O₂ is reduced rapidly by Sn²⁺, the products being Sn⁴⁺ and water. H₂O₂ decomposes slowly at room temperature to yield O₂ and water. Calculate the volume of O₂ produced at 20^oC and 1 atm. when 200 gm. of 10% by mass H₂O₂ in water is treated with 100 ml. of 2 M Sn²⁺ and then the mixture is allowed to stand until no further reaction occurs.

      Solution:        Equivalents of H₂O₂ intially    =    = 1.176

Equivalents of Sn²⁺           =

Equivalents of H₂O₂ left         =    1.176 – 0.4 = 0.776

Moles of H₂O₂ left                  =

Moles of O₂ produced            =     = 0.194

Volume of O₂                    =    =     4.66 L

 

Problem – 5:    Calculate the % of MnO­₂ in a sample of pyrolusite ore, 1.5 g which was made to react with 10 g of Mohr’s salt (FeSO₄.(NH₄)₂SO₄. 6H₂O) and dilute H₂SO₄. MnO₂ was converted to Mn²⁺ . After the reaction the solution was diluted to 250 ml and 50 ml of this solution, when titrated with 0.1 N K₂Cr₂O₇, required 10 ml of the dichromate solution.

 

      Solution:        Equivalents of K₂Cr₂O₇          =

Equivalents of Fe²⁺ in 50 ml   = 1 ´ 10^–3

Equivalents of Fe²⁺ in 250 ml    = 1 ´ 10^–3 ´ 5 = 5 ´ 10^–3

Initial equivalents of Fe²⁺        =

Equivalents of Fe²⁺ consumed by MnO₂ =    2.55 ´ 10^–2 – 5 ´ 10^–3

=    0.0205

\ Equivalents of MnO₂    =    0.0205

Moles of MnO₂                 =

Mass of MnO₂                  =    1.025 ´ 10^–2 ´ 87 = 0.89175g

% MnO₂                             =     =  59.45%

 

Problem – 6:    The H₂S and SO₂ concentrations of a gas were determined by passage through three absorber solutions connected in series. The first contained an ammoniacal solution of Cd²⁺ to trap the sulphide as CdS. The second contained 10.0 ml of 0.0396 N I₂ to oxidize SO₂ to SO₄^2-. The third contained 10.0 ml 0.0345 N thiosulphate solution to retain any I₂ carried over from the second absorber. A 25.0 litre gas sample was passed through the apparatus followed by an additional amount of pure N₂ to sweep the last traces of SO₂ from the first and second absorber. The solution from the first absorber was made acidic and 20.0 ml of the 0.0396 N I₂ were added. The excess I₂ was back  titrated with 7.45 ml of the thiosulphate solution. The solutions in the second and third absorbers were combined and the resultant iodine was titrated with 1.44 ml of the thiosulphate solution. Calculate the concentrations of SO₂ and H₂S in mg/L of the sample

 

      Solution:        Equivalents of I₂ that reacted with the first absorber

=    = 5.35 ´ 10^–4

Equivalents of S^2–                   =    5.35 ´ 10^–4

Moles of S^–2                            =

Mass of H₂S                           =    2.675 ´ 10^–4 ´ 34 = 9.095 ´ 10^–3 g.

Equivalents of I₂ that reacted with

SO₂

\ Equivalents of SO₂        =   1.32 ´ 10^–6

Moles of SO₂              =

Mass of SO₂         =    4.224 ´ 10^–5 gm.

\  Concentration in mg/L of H₂S

Concentration in mg/L of SO₂ = 1.6896 ´ 10^–3 mg / L

 

Problem – 7:     A 2.5 g sample containing As₂O₅, Na₂HAsO₃ and inert substance is dissolved in water and the pH is adjusted to neutral with excess NaHCO₃. The solution is titrated with 0.15 M I₂ solution, requiring 11.3 ml to just reach the end point, then the solution is acidified with HCl, KI is added and the liberated I₂  requires 41.2 ml of 0.015 M Na₂S₂O₃ under basic conditions where it converts to SO₄^2-. Calculate % composition of the mixture.

 

Solution:        Equivalents of I₂ reacting =  = 3.39 ´ 10^–3

\ Equivalents of Na₂HAsO₃ = 3.39 ´ 10^–3

Moles of Na₂HAsO₃ =  = 1.695 ´ 10^–3

Mass of Na₂HAsO₃ = 1.695 ´ 10^–3 ´ 170 = 0.289 gm

% Na₂HAsO₃ = 11.6%

Equivalents of Na₂S₂O₃ =   = 4.944 ´ 10^–3

\ Equivalents of I₂ liberated = 4.994 ´ 10^–3

\ Equivalents of As₂O₅ and Na₂HAsO₃ = 4.994 ´ 10^–3

\ Equivalents of As₂O₅ = 4.994 ´ 10^–3 – 3.39 ´ 10^–3 = 1.554 ´ 10^–3

Moles or As₂O₅ =  = 3.885 ´ 10^–4

Mass of As₂O₅ = 3.885 ´ 10^–4 ´ 230 = 0.0893 gm

% As₂O₅ =  ´ 100 = 3.57%

Problem- 8:      1.03 g mixture of sodium carbonate and calcium carbonate required 20 ml N HCl for complete neutralisation. Calculate the percentage of sodium carbonate and calcium carbonate in the given mixture.

 

Solution:        Na₂CO₃ + 2HCl ¾® 2NaCl + H₂O + CO₂

CaCO₃ + 2HCl ¾® CaCl₂ + H₂O + CO₂

Let x g CaCO₃ be present in the mixture

So, mass of Na₂CO₃ in the mixture will be (1.03 – x)g

No. of gm equivalents of CaCO₃ =

No. of gm equivalents of Na₂CO₃ =

No. of gm equivalents in 20 ml HCl =  =

But

No. of gm equivalents of CaCO₃ + No. of gm of Na₂CO₃ = No. of gm eq,. of HCl

,  x = 0.50

CaCO₃ = 0.50g %                   CaCO₃ =  = 48.54 %

Na₂CO₃ = 0.53g %                  Na₂CO₃ =  = 51.46 %

 

Problem – 9:     A compound on analysis gave 73.4% Pb and 3.2% H₂O. 0.235 gms of the substance when treated with an excess of KI solution acidified with HCl acid, liberated an amount of iodine which was equivalent to 25 ml of N/20 Na₂S₂O₃ solution. On igniting the substance a residue of PbO and Cr­₂O₃ was left behind. Compute  the percentage of chromium in the compound. The compound was insoluble in water but on digestion with Na₂SO₄ solution the latter became strongly alkaline and yellow. Give the formula and name of the compound.

 

Solution:        The substance must be a chromate since it leaves a residue of Cr₂O₃ on ignition.

+14H⁺ + 6I^– ¾® 2Cr³⁺ + 3I₂

equivalents of hypo used =  =  equivalents of  I₂ liberated

=  equivalents of  present in 0.235 g of substance. Moles of  CrO₄^-2   in 0.235

sample =

[as n factor = 3]

\ mole of Cr atoms =

\ weight of Cr in 0.235 g sample =  g

percentage of Cr =   = 9.21

Now we have

Pb              73.4%              73.4/207          = 0.354

Cr              9.2%                9.2/52              = 0.177

H₂O           3.2%                3.2/18              = 0.177

Hence Pb : Cr : H₂O                                 =  0.354 : 0.177: 0.177

= 2:1:1

Hence the probable composition is PbCrO₄. PbOH₂O or PbCrO₄, Pb(OH)₂

The fact that digestion with Na₂SO₄ gives an alkaline solution is possible only when the substance is PbCrO₄. Pb(OH)₂ as

PbCrO₄. Pb(OH)₂ + 2 Na₂SO₄ ¾® Na₂CrO₄ + 2 NaOH + 2 PbSO₄ ¯

base

Thus the compound is chrome red.

 

Problem – 10:   Sufficient amount of H₂S gas is passed through 5 ml solution of tincture of iodine to convert its all iodine into iodide ion. The sulphur precipitated is filtered off and the solution is made upto 1 litre and the solution is acidified with HCl. 250 ml of this solution requires 28 ml of 0.05 N Ce⁴⁺ for the conversion of entire I^– into ICl only. 2 ml of same sample of tincture of iodine gave 0.0313 gm of yellow precipitate in another experiment when treated with AgNO₃ solution. What weight percent of iodine is present in the form of free iodine. ( Tincture of iodine contains free I^– and I₂ both)

 

Solution:        I₂ + H₂S ¾® 2HI + S ¯

250 ml diluted  tincture of iodine solution has

equivalents of I^– of n factor 2.

                              [Note: In the first reaction n factor of I^– produced is 1 but when it is reacting with Ce⁴⁺ to give ICl its n factor is 2 \ equivalents of I^– in the second reaction with n factor 2 to be converted into equivalents of I^–  of n factor 1.].

\ equivalents of I^– reacted of n–factor 1 in 250 ml =

Equivalents of I^– produced ion 1 litre= = equivalents of I^– produced in 5 ml.

= moles of I^– produced in 5 ml.

\ weight of entire I^– =  g

1n  5 ml                 = 0.3556 g

2 ml tincture of iodine gives 0.0313 g AgI (yellow ppt) this means that weight of I^– in 2 ml tincture of iodine

=  = 0.0169 g

\ In 5 ml tincture of iodine 0.0169 ´2.5 = 0.042 g I^–

\ % weight of Iodine in the form of free iodine  = ´100

= 88.2%

 

 

 

 

 

7.2       Objective

 

Problem 1:       If 0.5 mole of BaCl₂ is mixed with 0.2 mole of Na₃PO₄ the maximum number of moles of Ba₃(PO₄)₂ that can be formed is

                        (A)  0.7                                            (B)  0.5

                        (C)  0.3                                            (D)  0.1

 

Solution:        3 BaCl₂ + 2Na₃PO₄ ¾® Ba₃(PO₄)₂ + 6NaCl

From the molar ratio we see

BaCl₂ : Na₃PO₄ : Ba₃(PO₄)₂ : NaCl

3        :      2      :      1           :   6

                        \(D)

 

Problem 2:       A compound was found to contain nitrogen and oxygen in the ratio nitrogen 28g and 80g of oxygen. The formula of the compound is

                        (A)  NO                                           (B)  N₂O₃

                        (C)  N₂O₅                                         (D)  N₂O₄

 

Solution:        gm atom of N =   = 2

gm atom of O =  = 5

Thus, the formula is N₂O₅

                        \ (C)

 

Problem 3:       The percent loss in weight after heating a pure sample of potassium chlorate (Molecular weight = 122.5) will be

                        (A)  12.25                                        (B)  24.50

                        (C)  39.18                                        (D)  49

 

Solution:        2KClO₃ ¾® 2KCl + 3O₂­

245 gm KClO₃ on heating shows a weight loss of 96 gm

\ 100 g KClO₃ on heating shows a weight loss of  =  = 39.18%

\(C)

 

Problem 4:       10 ml of gaseous hydrocarbon on combustion gives 40 ml of CO₂(g) and 50 ml of H₂O (vapour). The hydrocarbon is

                        (A)  C₄H₅                                          (B)  C₈H₁₀

                        (C)  C₄H₈                                                                                   (D)  C₄H₁₀

 

Solution:        C_aH_b +O₂  ¾® aCO₂ + b/2 H₂O

10        excess              ––            ––

0           excess              10a          5b

Therefore

10a = 40

So, a = 4

5b = 50

b = 10

                        \ (D)

Problem 5:       How many grams of KCl would have to be dissolved in 60g H₂O to give 20% by weight of solution

                        (A)  15g                                           (B)  1.5g

                        (C)  11.5g                                        (D)  31.5g

 

Solution:        % by weight = × 100

or 20 =

w = 15 gm

                        \ (A)

 

Problem 6:       The amount of oxalic acid (hydrated) required to prepare 500 ml of its 0.01 N solution is

                        (A)  0.315g                                      (B)  6.3g

                        (C)  3.15g                                        (D)  63g

 

Solution:        Meq. of oxalic acid = 500 × 0.1 = 50

= 50

W =  = 3.15g

                        \(C)

 

Problem 7:       What weight of sodium hydroxide is required to neutralise 100 ml of 0.1 N HCl

                        (A)  4.0g                                          (B)  0.04g

                        (C)  0.4g                                          (D)  2.0g

 

Solution:        Meq. of NaOH = Meq. of HCl

100 × 0.1 = 10

× 1000 = 10

w = 0.4 g

                        \(C)

 

Problem-8:       Density of water at room temperature is 1g/ml. How many molecules are there in a drop of water, if its volume is 0.05 ml.

                        (A)  1.67 ´ 10²¹                                                                       (B)  16.7 ´ 10²¹

                        (C)  6.023 ´ 10²³                                     (D)  1.67 ´ 10²³

 

 

Solution:        Density of water at room temperature = 1 gm/ml

\ 1 ml contain  moles of H₂O

0.05 ml will contain 0.05 ´  moles

\ 1 mole = 6.022 ´ 10²³ moles

Total no. of molecules in a drop of water = 6.023 ´ 10²³ ´ 0.05 ´
= 1.67 ´ 10²¹

\  (A)

 

Problem-9:       How many molecules are present in 12 L of liquid CCl₄? The density of the liquid is 1.59 g cm^–3

                        (A)  7.44 ´ 10²⁶                                      (B)  0.744 ´ 10²⁶

                        (C)  1.59 ´ 10²⁶                                      (D)  15.9 ´ 10²⁶

 

Solution:        1 cc of CCl₄ contains 1.59 gms or  =

\ 12 L of liquid CCl₄ will contain = 12 ´ 1000 ´ 0.0103

Þ 0.744 ´ 10²⁶ molecules of it

\ (B)

 

Problem-10:     13.4g of a sample of unstable hydrated salt: Na₂SO₄×nH₂O was found to contain 6.3g of water. Determine the number of water of crystallisation.

                        (A)  6                                                    (B)  5

                        (C)  7                                                    (D)  8

 

Solution:         =

= 7

\ (C)

 

Problem-11:     25.4 g of iodine and 14.2g of chlorine are made to react completely to yield a mixture of ICl and ICl₃. Calculate the ratio of moles of ICl and ICl₃.

                        (A)  1:1                                                  (B)  1:2 

                        (C)  1:3                                                  (D)  2:3

 

Solution:        I₂ + 2Cl₂ ¾® ICl + ICl₃

25.4 gm of Iodine contains 0.1 moles of it

14.2 gm of chlorine contains 0.2 moles of it

\ Moles of ICl & ICl₃ produced will be 0.1 and 0.1. Hence Molar ratio 1: 1

\ (A)

 

Problem-12:     Calculate the weight of iron which will be converted into its oxide by the action of 18g of steam on it.

                        (A)  37.3 gm                                    (B)  3.73 gm

                        (C)  56 gm                                       (D)  5.6 gm

 

 

Solution:        2Fe + 3H₂O ¾® Fe₂O₃ + 3H₂

3 ´ 18gm of steam will react with 2 ´ 56 gm of Iron

\ 18 gm of steam will convert =  = 37.3 gm of Fe

                        \ (A)

Problem-13:     The hourly energy requirement of an astronaut can be satisfied by the energy released when 34g of sucrose are burnt in his body. How many g of oxygen would be needed to be carried in space capsule to meet his requirement for one day?

                        (A)  916.2gm                                         (B) 91.62 gm

                        (C)  8.162 gm                                        (D)  9.162 gm

 

Solution:        C₁₂H₂₂O₁₁ + 12O₂ ¾® 12CO₂ + 11H₂O

1 mole of starch requires 12 mole of oxygen

\  mole of starch requires  = 916.2 gm

\ (A)

 

Problem-14:     10 ml of a solution of H₂O₂ labelled ’10 volume’ just decolorises 100 ml of potassium permanganate solution acidified with dilute H₂SO₄. Calculate the amount of potassium permanganate in the given solution.

                        (A)  0.1563 gm                                      (B)  0.563 gm

                        (C)  5.63 gm                                          (D)  0.256 gm

 

Solution:        2H₂O₂ ¾® 2H­₂O + O₂

22400 ml of O₂ evolved from 68 gm of H₂O₂

\ 10 ml of O_­2 is evolved from gm of H₂O₂

Hence 1 ml  of H₂O₂ contain  mol = 0.00178 equivalent 10 ml of H₂O₂ therefore 0.0178 equivalents which will be  present in 100 ml of KMnO₄ solution.

Amount of KMnO­₄ in given sample =  = 0.563 gm

\ (B)

 

Problem 15:     If  0.5 mole of BaCl₂ is mixed with 0.2 mol of Na₃PO₄, the maximum amount of Ba₃(PO₄)₂ that can be formed is:

                        (a) 0.7 mol                         (B) 0.5 mol

                        (C) 0.2 mol                         (d) 0.1 mol

 

Solution:        Let us  first solve this problem by writing the complete balanced reaction.

3BaCl₂ + 2 Na₃PO₄ ¾® Ba₃(PO₄)₂ ¯+ 6NaCl

We can see that the moles of BaCl₂ used is  times the moles of Na₃PO₄.
\ to react  with 0.2 mol of Na₃PO₄, the moles of BaCl₂ required would be
0.2 ´  = 0.3. Since BaCl₂ is 0.5 mol, we can conclude  that Na₃PO₄ is the limiting reagent. Therefore, moles of Ba₃(PO₄)₂ formed is 0.2 ´

= 0.1 mol.

                        \  (d)

                        Alternatively,

                              We can use the concept of equivalents to arrive at the answer. BaCl₂ and Na₃PO₄ must be reacting without undergoing any change in oxidation state since in Ba₃(PO₄)₂ the oxidation states of Ba, P and O remain fixed at +2, +5 and –2 respectively and the other product is expected to be NaCl.

Equivalents of BaCl₂ = 0.5 ´ 2  (Calculate  ‘n’ factor as number of moles of cation in 1 mole of substance ´ oxidation state of each cation).

Equivalents of Na₃PO₄     = 0.2 ´ 3

= 0.6

Therefore, Na₃PO₄ is the limiting reagent.

Þ Equivalents of Ba₃(PO₄)₂ formed = 0.6

‘n’  factor of Ba₃(PO₄)₂ = 6 ( Q ‘n’ factor of Na₃PO₄ is 3, \ Na₃PO₄ and Ba₃(PO)₄ will be in the molar ratio of 2:1)

\ Moles of Ba₃(PO₄)₂ =  = 0.1

\ (d)

 

Problem 16:     For the reaction

                         ¾®

                        the  correct coefficients of the reactants for the balanced reaction are

                                                             H⁺
(a)        2                5                      16

                        (B)        16               5                      2

                        (C)        5                16                     2

                        (d)        2                16                     5

 

Solution:        The above reaction can be balanced by using the ion electron method as under:

Oxidation reaction :  ¾® CO₂

Reduction reaction :           ¾®

Balancing atoms other than O

¾® 2 CO₂

¾®  Mn²⁺

Since medium is acidic

¾® 2CO₂  (oxygen is already balanced)

8H⁺ +  ¾® Mn²⁺ + 4H₂O

Balancing Charge

¾® 2CO₂ + 2e^– —————— (1)

5e^– + 8H⁺ +  ¾® Mn²⁺ + 4H₂O —————– (2)

Multiplying equation (1) by 5 and equation (2) by 2, and adding, we get

+  + 16H⁺ ¾® 10CO₂ + 2Mn²⁺ + 8H₂O

                        \ (a)

                        Alternatively,

‘n’ factor of  is 2 and that  is 5.

\ they would react in the molar ratio of 5:2

                        \ (a)

 

Problem 17:     It takes 0.15 mole of ClO^– to  oxidize 12.6 g of chromium oxide of a specific formula to . ClO^– became Cl^–. The formula of the oxide is (atomic weight Cr = 52, O = 16).

                        (a) CrO₃                                                (B) CrO₂           

                        (C) CrO₄                                                (d) CrO

 

Solution:       ‘n’ factor of ClO^– is 2 ( Cl⁺¹ ® Cl^–1)

\ Equivalents of ClO^– =  0.15 ´ 2 = 0.3

If the molecular weight of the chromium oxide is M, and the oxidation state of chromium in it is x. Then equivalents of chromium oxide is

(6–x) ´  ( number of chromium atoms in the oxide is 1 Q all the choices have 1 Cr per molecule)

\ (6–x) ´  = 0.3

=  or   =

Put  each choice and find out the correct result. Choice (a) and (C) can be ignored ( because oxidation state of Cr is +6 and +8 respectively. The later is an impossible oxidation state  of Cr).

                        \ (B)

 

Problem – 18:   It takes 2.56 ´ 10^–3 equivalents of KOH to neutralise 0.1254 g H₂XO₄. The number of neutrons in X is

                        (a) 16                                                   (B) 8                

                        (C) 7                                                     (d) 32

 

Solution:        Moles of H₂XO₄ =  [ M_x = Atomic mass of X]

‘n’ factor of H₂XO₄ = 2 [QH₂XO₄ is a dibasic acid]

\ 2.56 ´ 10^–3 =  ´ 2

M_x = 31.96 g / mol » 32 g / mol

\ X is Sulphur

Sulphur has 16 protons and 16 neutrons (for these type of problems, atomic masses and atomic numbers would be supplied)

                        \ (a)

 

Problem 19:     Equal volumes of 1 M each of KMnO₄ and K₂Cr₂O₇ are used to  oxidise Fe(II) solution in acidic medium. The amount of Fe oxidised will be

                        (a) more with KMnO₄                      (C) equal with both oxidising agents

                        (B) more with K₂Cr₂O₇                     (d) cannot be determined 

 

Solution:       The ‘n’ factor of KMnO₄ is 5 while that of K₂Cr₂O₇ is 6. So for the same number of moles, K₂Cr₂O₇ will have greater equivalence than KMnO₄.

                        \ (B)

 

Problem- 20:    In an experiment, 50 ml of 0.1M  solution of a salt reacted with 25ml  of 0.1M solution of Sodium sulphite. The half equation for the oxidation of sulphite ion is : + H₂O ¾® + 2H⁺_(aq) + 2e^–

                        If the oxidation number of the metal in the salt was 3, what would be the new oxidation number of the metal?

                        (a) 0                                                           (B) 1                

                        (C) 2                                                           (d) 4

 

Solution:         get oxidised and its ‘n’ factor is 2.

The metal must have been reduced.

Applying the law of equivalence

50 ´ 0.1 ´(3–n) = 25 ´ 0.1 ´2

n = 2

                              \ (C)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

8. Assignments (Subjective Problems)

Level –  I

 

1. Balance the following equation by Ion electron method:
2. i) H₂S + HNO₃ ® NO + S + H₂O
3. ii) Cr₂O₇^2– + H⁺ + C₂O₄^2– ® Cr³⁺ + H₂O + CO₂

 

2. The reaction Cl₂ + ¾®  + Cl^– is carried out in basic medium. If 0.15 mole Cl₂, 0.01  and 0.3 mole OH^– are present in the beginning, how many moles of hydroxides ion will be left behind after the completion of reaction.

 

3. A 1.2 g. of a mixture containing H₂C₂O₄×2H₂O and KHC₂O₄×H₂O and impurities of a neutral salt, consumed 18.9 ml of 0.5 N NaOH for complete neutralization. On titiration with KMnO₄ solution 0.4 g of the same substance needed 21.55 ml of 0.25 N KMnO₄. Calculate the percentage composition of the substance.

 

4. A polyvalent metal weighing 0.1 gm and having atomic weight of 51 reacted with dilute H₂SO₄ to give 43.90 ml of hydrogen at N.T.P. This solution containing the metal in the lower oxidation state was found to require 58.8 ml of 0.02 M KMnO₄ for complete oxidation. What are the oxidation states of the metal in the two reactions.

 

5. Certain weight of hydrated oxalic acid is added to 0.5 gm of pyrolusite sample which contains 54.96% in the form of MnO₂. After the reaction in acidic medium is complete the excess oxalic acid requires 30 ml of 0.02 M KMnO₄ for oxidation. What is the weight of oxalic acid added.

 

6. One litre of O₂ at S.T.P. is passed through an ozonizer. The volume of exit gases is reduced by 10% due to partial conversion of O₂ into O₃. What amount of As₂O₃ can be oxidised by ozonide oxygen in alkaline medium.

 

7. 10 gm mixture of KF and KCl was treated with H₂O₂ solution. Now the gas liberated was treated with iodine solution to give 0.176 gm of iodic acid (HIO₃). Find the amount of KF in the mixture.

 

8. A mixture of KMnO₄ and K₂Cr₂O₇ weighing 0.24 gms on being treated with KI in acid solution, liberated just sufficient iodine to react with 60 ml of 0.1 N Na₂S₂O₃. Find the percentage of chromium and manganese in the mixture.

 

9. Solid silver sulphate is shaken well with a 500 ml of 0.01 M sodium sulphate solution. After equilibrium is reached the solution is filtered and the filtrate is diluted upto 2 Litre. 23 ml of this diluted solution when treated with excess of Barium chloride  solution gives 25.6 mg of white precipitate. What is the solubility of silver sulphate in 0.01M Na₂SO₄ solution in gms/Litre.

 

10. 0.804 gm sample of iron ore containing only Fe and FeO was dissolved in acid. Iron oxidises into +2 state and it requires 117.20 ml of 0.112 N K₂Cr₂O₇ solution for titration. Calculate the percentage of iron is the ore.

 

 

11. A sample of sodium carbonate contains sodium sulphate also. 1.5 gm of the sample is dissolved in water and volume raised to 250 ml. 25 ml of this solution requires 20 ml of N/10 H_­2SO₄ solution for neutralisation. Calculate the percentage of sodium carbonate in the sample.

 

12. 40 ml of HCl is exactly neutralised by 20 ml of NaOH solution. The resulting neutral solution is evaporated to dryness and the residue is found to have a mass of 0.117g. Calculate the normality of HCl and NaOH.

 

13. A solution of H₂O₂, labelled as ‘20 volumes’, was left open. Due to this some H₂O₂ decomposed and the volume strength of the solution decreased. To determine the new volume strength of the H₂O₂ solution, 10 mL of the solution was taken and it was diluted to 100 mL. 10mL of this diluted solution was titrated against 25 mL of 0.0245 M KMnO₄ solution under acidic condition. Calculate the volume strength of the H₂O₂ solution.

 

14. A sample of potassium chlorate is partly decomposed to give KCl and O₂ and the remaining converts to potassium perchlorate KClO₄. A 2.45 gm sample of potassium chlorate as per the reactions leaves a residue of 1.874 g whose aqueous solution gives 2.009 g of AgCl. What percentage of KClO₃ has converted into perchlorate?

 

 

15. The following facts are known concerning a solid element

• On combination with oxygen, it gives a base
• Its specific heat is 0.119.
• 10 g of element will combine with 4.298 g of oxygen
• 8 g of element will combine with 10.159 g of chlorine

What is the atomic weight of the element. What are its valencies.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

LEVEL – II

 

1. A mixture of pure K₂Cr₂O₇ and pure KMnO₄ weighing 0.561 g. was treated with excess of KI in acidic medium. Iodine liberated required 100 ml. of 0.15 M of sodium thiosulphate solution for exact oxidation. What is the % of each in the mixture?

 

2. What weight of inorganic rubber phosphonitrile dichloride can be obtained if the synthesis begins with 1.4 g N₂ , 5.6 Lit of H₂ at S.T.P., 80 g of PCl₃ and large excess of S₂Cl₂ and HCl.

N₂ + 3H₂ ¾® 2NH₃

NH₃ + HCl ¾® NH₄Cl

3PCl₃ + S₂Cl₂ ¾® 2PSCl₃ + PCl₅

 

 

3. One gram of an alloy of aluminium and magnesium when heated with excess of  dil. HCl forms  magnesium chloride, aluminum chloride and hydrogen. The evolved hydrogen collected over mercury at 0^oC has a volume of 1.2 litres at 0.92  atm.  pressure. Calculate the composition of the alloy.

 

 

4. A 2.18g sample containing a mixture of XO and X₂O₃. It takes 0.015 mole of K₂Cr₂O₇ to oxidise the sample completely to form XO₄^– and Cr³⁺. If 0.0187 mole of XO₄^– is formed, what is the atomic mass of X?

 

5. A solution contains a mixture of sulphuric acid and oxalic acid, 25 ml. of the solution requires 35.54 ml. of 0.1 M NaOH for neutralization and 23.45 ml. of 0.02 M KMnO₄ for oxidation. Calculate the molarity of solution with respect to sulphuric acid and oxalic acid.

 

6. A 1.85 g. sample of a mixture of CuCl₂ and CuBr₂ was dissolved in water and mixed thoroughly with a 1.8 g. portion of AgCl. After the reaction the solid, a mixture of AgCl and AgBr, was filtered, washed, and dried. Its mass was found to be 2.052 g. What percent by mass of the original mixture was CuBr₂?

 

7. Calculate % of MnO₂ in a sample of pyrolusite ore, 1.5 g of which was made to react with 10 g of Mohr’s salt (FeSO₄.(NH₄)₂SO₄^.6H₂O) and dilute H₂SO₄. MnO₂ was converted to Mn²⁺. After the reaction the solution was diluted to 250 ml and 50 ml of this solution, when titrated with 0.1 N K₂Cr₂O₇ required 10 ml of the dichromate solution.

 

8. The oxides of sodium and potassium contained in a 0.5g of feldspar were converted to the respective chlorides. The weight of the chlorides thus obtained was 0.118 g. Subsequent treatment of the chlorides with AgNO₃ gave 0.2451g of AgCl. What is the percentage of Na₂O and K₂O in the sample?

 

9. A 10 gm portion of an aqueous solution containing H₃PO₄ requires 46 ml of
   0.11 N NaOH to neutralize in the presence of phenolphthalein indicator. Calculate the percentage of H₃PO₄ in the solution. What weight of solution  should be evaporated to obtain 1 millimole metaphosphoric acid?
10. 5 ml of 8N HNO₃, 4.8 ml of 5N HCl and a certain volume of 17 M H₂SO₄ are mixed together and made upto 2 litre. 30 ml of this acid mixture exactly neutralises 42.9 ml of Na₂CO₃ solution containing 1 g of Na₂CO₃.10H₂O in 100 ml of water. Calculate the amount of sulphate ions in grams present in the solution.

11.       A mixture containing Li₂CO₃, Na₂CO₃  and Na₂O was strongly heated at 300°C, the gas evolved occupies 59.2 ml at 740 mm pressure. The residue reacts completely with 15 ml seminormal HCl and further evolves 45  ml gas measured at 27°C and at 740 mm pressure. Calculate the percentage of Na₂O in the mixture.

 

12. 10 litre of air at N.T.P. were slowly bubbled through 50 cc of N/25 Barium hydroxide solution rendered red with phenolphthalein. After filtering the solution, from the precipitated barium carbonate, the filtrate required 22.5 cc of 2N/25 HCl to become just colourless. Calculate the volume percentage of CO₂ gas in
    air at NTP.
13. A solution contains sodium carbonate and sodium bicarbonate, 10 ml of this solution requires 2.5 ml of 0.1M H₂SO₄ for neutralization using phenonpthalein as indicator. Methyl orange is then added when a further 2.5 ml of 0.2 M H₂SO₄ was required. Calculate the amount of Na₂CO₃ and NaHCO₃ in one litre of the solution.

 

14. A 100 ml solution ’S’ is taken which contains NaOH and Na₂CO₃. 40 ml of ‘S’ is titrated against 0.1 N HCl using phenolphthalein as indicator. When it required 80 ml of acid. Then methyl orange was added and the titration continued when a further 10 ml of the acid was required to reach the end point. Remaining portion of solution ‘S’ i.e  60 ml is diluted to 100 ml. Excess of granulated zinc is added to it. What will be the volume of Hydrogen evolved at S.T.P. Also calculate the amount of  NaOH and sodium carbonate in the original solution.

 

15. 3.1 gm of a mixture containing NaNO₃, AgNO₃ and some impurity was heated, the evolved gases were passed through alkaline pyragallol whose weight increases by 0.208 gram and the residue reacts completely with 25 ml, N/10 HCl. Calculate the percentage composition of NaNO₃ and AgNO₃ in the mixture.

                  NaNO₃ and AgNO₃ decompose on heating as follows

2 NaNO₃ ¾¾® 2NaNO₂ + O₂

2AgNO₃   ¾¾® 2Ag + 2NO₂ + O₂

 

 

 

 

 

 

 

 

 

 

 

Level – iii

1. A well known fertiliser super phosphate of lime is a mixture of Ca(H₂PO₄)₂, CaSO₄ and H₃PO₄ in the molar ratio 4:11:2 which is manufactured in a plant following the reaction:

5Ca₃(PO₄)₂ + 11H₂SO₄ ¾® 4Ca(H₂PO₄)₂ + 2H₃PO₄ + 11CaSO₄.

5. i) If we have to prepare 2 kg superphosphate of lime. How much litre of 5.0 M H₂SO₄ is needed ?
6. ii) If we have initially 2 kg of each calcium phosphate and sulfuric acid. How much amount of super phosphate of lime can be obtained.

 

2. A piece of aluminium weighing 2.7 gm is heated with 75 ml of H₂SO₄ (sp. gr. 1.18 containing 24.7% H₂SO₄ by mass). After the metal is carefully dissolved, the solution is diluted to 400 ml. Calculate the molarity of the free H₂SO_­4 in the resulting solution.

 

3. A steel sample is to be analysed for Cr and Mn simultaneolusly. By suitable treatment the Cr is oxidised to and the Mn to . A 10.00 g sample of steel is used to produce 250.0 mL of a solution containing  and . A 10.00 mL portion of this solution is added to a BaCl₂ solution and by proper adjustment of the acidity, the chromium is completely precipitated as BaCrO₄; 0.0549 g is obtained. A second 10.00 mL portion of this solution requires exactly 15.95 mL of 0.0750 M standard Fe²⁺ solution for its titration (in acid solution). Calculate the % of Mn and % of Cr in the steel sample

 

4. CuFeS₂ mineral was analysed for Cu and Fe percentage. 10 g of it was boiled with dil. H₂SO₄ and diluted to 1 L. 10 mL of this solution required 2 mL of 0.01 M in acidic medium. In another titration 25 mL of the same solution required. 5 mL of 0.01 M  solution iodometrically. Calculate percentage of Cu and Fe in the mineral.

 

5. A 1.87 g sample of Chromite (FeO.Cr₂O₃) was completely oxidised by the fusion of peroxide. The excess peroxide was removed. After acidification, the sample was treated with 50 ml of 0.16 M Fe²⁺. A back titration of 2.97 ml of 0.005 M Barium permanganate was required to oxidise the excess iron. What was the % of chromite in the sample.

 

6. 1.6 g pyrolusite ore was treated with 50 cm³ of 1 N oxalic acid and some sulphuric acid. The oxalic acid left undecomposed was raised to 250 cm³ in a flask. 25 cm³ of this solution when titrated with 0.1 N KMnO₄ required 32 cm³ of the solution. Find the percentage of pure MnO₂ in the sample and also the percentage of available oxygen.

 

7. 1.575 g of oxalic acid (COOH)₂.xH₂O are dissolved in water and the volume made upto 250 ml. On titration 16.68 ml of this solution requires 25 ml of N/15 NaOH solution for complete neutralisation. Calculate x.

8.         10 gm of sample of bleaching powder( CaOCl₂) were extracted with water and the solution made up to one litre. 250 ml of this solution was added to 50 ml of N/14 Mohr’s salt solution containing enough H₂SO₄. After the reaction was completed, the whole solution required 22 ml of KMnO₄ solution containing  2.2571 gm of KMnO₄ per litre for complete oxidation. Calculate the percentage of available chlorine in the sample of bleaching powder.

 

9. 1.249 g of a sample of pure BaCO₃ and impure CaCO₃ containing some CaO was treated with dil HCl and It evolved 168 ml of CO₂ at N.T.P. From this solution, BaCrO₄ was precipitated filtered and washed. The precipitate was dissolved in dil H₂SO₄ and  diluted to 100 ml. 10 ml of this solution when treated with KI solution liberated iodine which required exactly 20 ml of 0.05 N Na₂S₂O₃. Calculate the percentage of CaO in the  sample.

 

10.       12 gm of an impure sample of arsenious oxide (As₂O₃, which is acidic in nature) was dissolved in water containing sodium bicarbonate (which is basic in nature) and the resulting solution was diluted to 250 ml. 25 ml of this solution was  completely oxidised by 22.4 ml of a solution of iodine, 25 ml of this iodine solution reacted with same volume of a solution containing 24.8 gm of hydrated sodium thiosulphate (Na₂S₂O₃.5H₂O) in one litre. Calculate the percentage of arsenious oxide in the sample.

 

11. ZSM–5 a microporous shape selective catalyst used in the synthesis of petrol. Its elemental analysis shows that it has 43.98% silicon and 0.446% aluminium. What is the minimum molecular weight of ZSM–5. What is the minimum number of Si atoms per molecule if it has one Al atom per molecule?

 

12. A 8.0g sample containing Fe₃O₄, Fe₂O₃ and inert impurities. It was treated with an excess of aq. KI solution in acidic medium, which reduced all the ion to Fe²⁺ ions. The resulting solution was diluted to 50 cm³ and 10 cm³ of it was taken. The liberated iodine in this solution required 7.2 cm³ of 1 M Na₂S₂O₃ for reduction to iodide. The iodine from another 25 cm³ sample was extracted, after which the Fe²⁺ ions was titrated against 1 M KMnO₄ in acidic medium. The volume of KMnO₄ solution used was found to be 4.2 cm³. Calculate the mass percentage of Fe₃O₄ and Fe₂O₃ in the original mixture.

 

13. For estimating ozone in the air, a certain volume of air is passed through an alkaline KI solution when O₂ is evolved and Iodide is oxidised to Iodine. When such a solution is acidified, free iodine is evolved which can be titrated with standard Na₂S₂O₃ solution. In an experiment 10 L of air at 1 atm and 27°C were passed through an alkaline KI solution, and at the end, the iodine was entrapped in a solution which on titration as above required 1.5 ml of 0.01N Na₂S₂O₃ solution . Calculate volume percentage of ozone in the sample

 

14. A sample of hard water contains 96 ppm of and 183 ppm of , with Ca⁺² as the only cation. How many moles of CaO will be required to remove HC from 1000 kg of this water? If 1000 kg of this water is treated with the amount of CaO calculated above, what will be the concentration (in ppm) of residual Ca⁺² ions? (Assume CaCO₃ to be completely insoluble in water). If the Ca⁺² ions in one litre of the treated water are completely exchanged with hydrogen ions, what will be its pH? (one ppm means one part of the substance in one million part of water, mass/mass.)

 

15. Calculate the amount of SeO₃^2– in a solution on the basis of following data. 20 ml of M/60 solution of KBrO₃ was added to a definite volume of SeO₃^2– solution. The bromine evolved was removed by boiling and excess KBrO₃ was back titrated with 5.1 ml of M/25 solution of NaAsO₂. The reactions are given below: (Atomic mass Se = 79) a) SeO₃^2– + BrO₃^– + H⁺ ¾® SeO₄^2– + Br₂ + H₂O
16.              b) BrO₃^– + AsO₂^– + H₂O ¾®  Br^– + AsO₄^3– + H⁺

 

 

9. Assignments (Objective Problems)

 

Level –  I

1. It takes 2.56 × 10^–3 equivalents of KOH to neutralise 0.1254 g H₂XO₄. The number of protons in X is

(A) 16                                                        (B) 8

(C) 7                                                          (D) 32

 

2. 8g of sulphur is burnt to from SO₂ which is oxidised by Cl₂ water. The solution is treated with BaCl₂ solution. The amount of BaSO₄ precipitated is

(A) 1 mole                                                 (B) 0.5 mole

(C) 0.24 mole                                            (D) 0.25 mole

 

3. A 10.0 g sample of a mixture of calcium chloride and sodium chloride is treated with Na₂CO₃ to precipitate the calcium as calcium carbonate. This calcium carbonate is heated to convert all the calcium to calcium oxide and the final mass of Calcium oxide is 1.62 gm. The percentage by mass of calcium chloride in the original mixture is

(A) 15.2%                                                  (B) 32.1%

(C) 21.8%                                                  (D) 11.07%

 

4. The number of moles of Cr₂O₇^2– needed to oxidise 0.136 equivalents of N₂H₅⁺ by the reaction.

N₂H₅⁺ + Cr₂O₇^2– ¾® N₂ + Cr³⁺ + H_­2O is

(A) 0.136                                                   (B) 0.272

(C) 0.816                                                   (D) 0.0227

 

5. A sample of Oleum is labelled 109%. The % of free SO₃ in the sample is

(A) 40%                                                     (B) 80%

(C) 60%                                                     (D) 9%

 

6. What volume of 0.3 N Cr₂O₇^2– / H⁺ is needed for complete oxidation of 200 ml of 0.6 M FeC₂O₄ solution

(A) 1.2cc                                                   (B) 1.2 lit

(C) 120cc                                                  (D) 800 cc

 

7. 125 ml of Na₂CO₃ solution requires 100 ml of 0.1 N HCl to reach end point with phenolphthalein as indicator. Molarity of resulting solution with respect to HCO₃^– ion

(A) 0.008 M                                               (B) 0.004 M

(C) 0.16 M                                                 (D) 0.08 M

 

8. One mole of a mixture of CO and CO₂ requires exactly 20 grams of NaOH to convert all the CO₂ into Na₂CO_­3. How many more grams of NaOH would it required for conversion into Na₂CO₃ if the mixture (one mole) is completely oxidised to CO₂.

(A) 60g                                                      (B) 80g

(C) 40g                                                      (D) 20g

 

9. 11.2 litres of SO₂ is mixed with 11.2 lit of H₂S at NTP, weight of sulphur obtained is

(A) 96g                                                      (B) 48g

(C) 24g                                                      (D) 12g

 

10. If 1 gm of HCl and 1 g of MnO₂ are heated together the maximum weight of Cl₂ gas evolved will be

(A) 2 gm                                                    (B) 0.975 gm

(C) 0.486 g                                                (D) 0.972g

 

11. The equivalent weight of HCl in the given reaction is

K₂Cr₂O₇ + 14HCl ¾® 2KCl + 2CrCl₃ + 3Cl₂ + H₂O

(A) 36.5                                                     (B) 73

(C) 85                                                        (D) 16.25

 

12. 100 ml of 0.1 N KMnO₄ reacts with 0.45 g of oxalic acid. The molecular weight of oxalic acid is

(A) 45                                                        (B) 90

(C) 180                                                      (D) 22.5

 

13. 35 ml sample of hydrogen peroxide gives off 500 ml of O₂ at 27°C and 1 atm pressure. Volume strength of H_­2O₂ sample will be

(A) 10 volumes                                         (B) 12.8 volumes

(C) 11 volumes                                         (D) 12 volumes

 

14. How many moles of electron is needed for the reduction of each mole of Cr in the reaction.

CrO₅ + H₂SO₄ ¾® Cr₂(SO₄)₃ + H₂O + O₂

(A) 4                                                          (B) 3

(C) 6                                                          (D) 7

 

15. To prepare 0.5 M KCl solution from 100 ml of 0.4 M KCl

(A) Add 7.45g KCl                                     (B) Add 200 ml water

(C) Add 0.1 mole KCl                                (D) Evaporate 20 ml of water

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Level –  II

 

1. A solution of HCl containing 0.03659 g/ml and another solution of acetic acid containing 0.04509g/ml then

(A) N_HCl is more                                        (B) is more

(C) Both have same N                              (D) None of these

 

2. How many mole of Fe²⁺ions are formed, when excess of iron is treated with 50 ml of 4 M HCl under inert atmosphere assuming no change in volume

(A) 0.4                                                       (B) 0.1

(C) 0.2                                                       (D) 0.8

 

3. The weight of sulphuric acid needed for dissolving 3 gm magnesium carbonate is

(A) 3.5 g                                                    (B) 7.0g

(C) 1.7g                                                     (D) 17.0g

 

4. 1.60 g of a metal were dissolved in HNO_­3 to prepare its nitrate. The nitrate on strong heating gives 2g oxide. The equivalent weight of metal is

(A) 16                                                        (B) 32

(C) 48                                                        (D) 12

 

5. 1.35 g of pure calcium metal was quantitatively converted into 1.88g of pure calcium oxide, the atomic weight of calcium is

(A) 40.75                                                   (B) 50

(C) 60                                                        (D) 70

 

6. The density of NH₄OH solution is 0.6 g/ml. It contains 34% by weight of NH₄OH. Calculate the normality of the solution.

(A) 4.8 N                                                   (B) 10 N

(C) 0.5 N                                                   (D) 5.8 N

 

7. 100 ml of 0.3 N HCl solution were mixed with 200 ml of 0.6 N H₂SO₄ solution. The final normality of mixture is

(A) 0.9 N                                                   (B) 0.6N

(C) 0.5N                                                    (D) 0.4 N

 

8. Insulin contains 3.4% sulphur. The minimum molecular weight of insulin is

(A) 941.176                                               (B) 944

(C) 945.27                                                 (D) None

 

9. The minimum quantity of H₂S needed to precipitate 6.35 g of Cu²⁺will be nearly

(A) 63.5 g                                                  (B) 31.75 g

(C) 3.4 g                                                    (D) 20 g

 

10. 2.76g of silver carbonate on being strongly heated yields a residue weighing

(A) 2.16g                                                   (B) 2.48g

(C) 2.32g                                                   (D) 2.64g

 

11. A metal oxide has 40% oxygen. The equivalent weight of the metal is

(A) 12                                                        (B) 16

(C) 24                                                        (D) 48

12. A metal oxide has the formula Z₂O₃. It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is

(A) 27.90                                                   (B) 159.60

(C) 79.80                                                   (D) 55.80

 

13. 0.7g of Na₂CO₃.xH₂O were dissolved in water and the volume was made to 100 ml, 20 ml of this solution required 19.8 ml of N/10 HCl for complete neutralisation. The value of x is

(A) 7                                                          (B) 3

(C) 2                                                          (D) 5

 

14. The solution A and B are 0.1 and 0.2 molar in a substance. If 100 ml of A are mixed with 25 ml of B and there is no change in volume, then the final molarity of the solution is

(A) 0.15 M                                                 (B) 0.18 M

(C) 0.12 M                                                 (D) 0.30 M

 

15. 0.84g of a metal carbonate reacts with 40 ml of N/2 H₂SO₄. The equivalent weight of metal carbonate is

(A) 84g                                                      (B) 64g

(C) 42g                                                      (D) 38g

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

10. Answers to Subjective Assignments

 

Level – I

1. 3H₂S + 2HNO₃ ¾® 3S + 2NO + 4H₂O

Cr₂O₇^2– + 14H⁺ + 3C₂O₄^2– ¾® 2Cr³⁺ + 7H₂O + 6CO₂

 

2. 0.2 mole 3.   H₂C₂O₄.2H₂O = 14.7%

KHC₂O₄.H₂O = 80.9%

4. +2, +5 5.   0.586 gms
5. 0.8839 g 7.   9.6274 gm
6. % Mn = 14.22; % Cr = 20.92 9.   2.87g/lit
7. 60.9% 11. 70.67%
8. 0.1 N 13. 17.136
9. 40.16% 15. 55.8g, +2, + 3

 

Level – II

 

1. KMnO₄ = 56.33 %

K₂Cr₂O₄ = 43.67 %                             2.         11.6 g

3. Al = 0.549g 4.         99

Mg = 0.451 g

5. H₂SO₄ = 0.241 M 6.         34.21%

H₂C₂O₄ = 0.0469M

7. % of MnO­₂ = 72.5 %                         8.         % of Na₂O = 3.58

% of K₂O = 10.6

9. 2.48 %, 3.95 gm 10.       6.528 gm

 

11. 14.2% 12.       0.0224%
12. NaHCO₃ = 4.2 gm       14.       Na₂CO₃ = 0.265 gm

Na₂CO₃ = 5.3 gm                                           NaOH = 0.7 gm

15. % of AgNO₃ = 57.58%

% of NaNO₃ = 6.85%

 

Level – III

 

1. i) 1.68 litre 2.         0.1826 M
2. ii) 1390.62 gm
3. % of Cr = 2.821% 4.         % of Fe = 5.6%

% of Mn = 1.498                                                   % of Cu = 1.27%

5. 15.57% 6.         % of MnO₂ = 48.9
6. 2             % available oxygen = 9
7. 2.84 %
8. 14.17% 10.       9.24%
9. Mol. Weight = 6058 12.       % of Fe₃O₄ = 17.4 %

Si = 95 atoms                                                        % of Fe₂O₃ = 24 %

13. 1.847 ´ 10^–3 14.       2.7
14. 0.084 gms

 

11. Answers to Objective Assignments

 

Level – I

 

1. A                   2.         D                                 3.         B
2. D                   5.         A                                 6.         B
3. D                         8.         A                                 9.         C
4. c                   11.       C                                 12.       B
5. B                   14.       B                                 15.       D

 

Level – II

 

1. A                   2.         B                                 3.         A
2. B                   5.         A                                 6.         D
3. C                   8.         a                                 9.         C
4. A                   11.       a                                 12.       D
5. C                   14.       C                                 15.       c

 

 

Table  of Relative Atomic Weights

 

At. No.	Name of Element	Symbol of Element	Atomic Weight		At. No.	Name of Element	Symbol of Element	Atomic Weight
89.	Actinium	Ac	227.0278		80.	Mercury	Hg	200.5
13.	Aluminium	Al	26.981		42.	Molybdenum	Mo	95.9
95.	Americium	Am	234.0614		60.	Neodymium	Nd	144.2
51.	Antimony	Sb	121.7		10.	Neon	Ne	20.17
18.	Argon	Ar	39.94		93.	Neptunium	Np	237.048
33.	Arsenic	As	74.922		28.	Nickel	Ni	58.7
85.	Astatine	At	209.9871		41.	Niobium	Nb	92.906
56.	Barium	Ba	137.3		7.	Nitrogen	N	14.007
97.	Berkelium	Bk	247.07		102.	Nobelium	No	259.1
4.	Beryllium	Be	9.012		76.	Osmium	Os	190.2
83.	Bismuth	Bi	208.981		8.	Oxygen	O	15.999
5.	Boron	B	10.81		46.	Palladium	Pd	106.4
35.	Bromine	Br	79.904		15.	Phosphorus	P	30.974
48.	Cadmium	Cd	112.40		78.	Platinum	Pt	195.0
55.	Cesium	Cs	132.905		94.	Plutonium	Pu	244.06
20.	Calcium	Ca	40.08		84.	Polonium	Po	208.98
98.	Californium	Cf	251.079		19.	Potassium	K	39.10
6.	Carbon	C	12.011		59.	Praseodymium	Pr	140.908
58.	Cerium	Ce	140.12		61.	Promethium	Pm	144.91
17.	Chlorine	Cl	35.453		91.	Protactinium	Pa	231.036
24.	Chromium	Cr	51.996		88.	Radium	Ra	226.025
27.	Cobalt	Co	58.933		86.	Radon	Rn	222.017
29.	Copper	Cu	63.54		75.	Rhenium	Re	186.2
96.	Curium	Cm	247.07		45.	Rhodium	Rh	102.905
66.	Dysprosium	Dy	162.5		37.	Rubidum	Rb	85.467
99.	Einsteinium	Es	252.08		44.	Ruthenium	Ru	101.07
68.	Erbium	Er	167.27		104.	Rutherfordium	Rf	261.11
63.	Europium	Eu	151.96		62.	Samarium	Sm	150.4
100.	Fermium	Fm	257.09		21.	Scandium	Sc	44.956
9.	Fluorine	F	18.998		34.	Selenium	Se	78.9
87.	Francium	Fr	223.019		14.	Silicon	Si	28.08
64.	Gadolinium	Gd	157.2		47.	Silver	Ag	107.868
31.	Gallium	Ga	69.72		11.	Sodium	Na	22.990
32.	Germanium	Ge	72.61		38.	Strontium	Sr	87.62
79.	Gold	Au	196.966		16.	Sulphur	S	32.06
72.	Hafnium	Hf	178.49		73.	Tantalum	Ta	180.947
2.	Helium	He	4.003		43.	Technetium	Tc	98.906
67.	Holmium	Ho	164.930		52.	Tellurium	Te	127.6
1.	Hydrogen	H	1.008		65.	Terbium	Tb	158.925
49.	Indium	In	114.82		81.	Thallium	Tl	204.3
53.	Iodine	I	126.904		90.	Thorium	Th	232.038
77.	Iridium	Ir	192.2		69.	Thulium	Tm	168.934
26.	Iron	Fe	55.84		50.	Tin	Sn	118.6
36.	Krypton	Kr	83.80		22.	Titanium	Ti	47.9
57.	Lanthanum	La	138.905		74.	Tungsten	W	183.8
103.	Lawrencium	Lr	262.11		92.	Uranium	U	238.036
82.	Lead	Pb	207.2		23.	Vanadium	V	50.941
3.	Lithium	Li	6.94		54.	Xenon	Xe	131.30
71.	Lutetium	Lu	174.97		70.	Ytterbium	Yb	173.04
12.	Magnesium	Mg	24.305		39.	Yttrium	Y	88.906
25.	Manganese	Mn	54.938		30.	Zinc	Zn	65.3
101.	Mendelevium	Md	258.1		40.	Zirconium	Zr	91.22

Stoichiometry

Hints to Subjective Problems

Level – I

5. Oxalic acid will react with MnO₂ and will get oxidised.
6. Partially formed ozone will reduce the volume.
7. Chlorine gas will formed from KCl will react with HIO₃
8. Both the compounds K₂Cr₂O₇ and KMnO₄ will react with Na₂S₂O₃
9. Fe²⁺ will react with K₂Cr₂O₇
10. Find the wt. of KClO₄ formed from KClO₃ and then the % of KClO₄ formed.
11. At. wt. =

valency =

Level – II

1. Both the compounds K₂Cr₂O₇ and KMnO₄ will react with Na₂S₂O₃
2. Al and Mg in the alloy will react with dilute HCl to give H₂. Then relate the volume of H₂ formed with the weights.
3. 1 mole of XO gives 1 mole and 1 mole X₂O₃ gives 2 moles of
4. NaOH will react with both the acids while KMnO₄ will react with oxalic acid only since H₂SO₄ cannot be oxidised.
5. Chlorides of Na and K gives AgCl on reacting with AgNO₃
6. Find the moles of H₂SO₄ and Na₂CO₃ and solve for the mass of NaHCO₃
7. Find the moles of NaNO₃ and AgNO₃ and then find their percentages.

Level – III

1. i) Find the mass of H₂SO₄ needed and then the volume of H₂SO₄ .
2. ii) Find the moles of calcium phosphate and then the mass of superphosphate of lime.
3. Titration I :  oxidises Fe²⁺ to Fe³⁺

Titration II:       Cu²⁺ is reduced to Cu⁺ on treatment with I^–

6. Find the mass of MnO₂ from the given normality and then its percentage.
7. Find the moles of barium carbonate and then its mass. Similarly, solve for calcium carbonate
8. Ozone formed will oxidise KI to I₂. Find the volume of ozone.

Solutions To Subjective Problems

Level – I

1. i) Oxidation : H₂S ® S   Reduction : HNO₃ ® NO

(i)   Balancing oxidation reaction :

H₂S® S+2H⁺  Balancing the charges H₂S ® S+2H⁺ + 2e^– ——— (1)  (oxidation half-reaction)

(ii)  Balancing reduction reaction :

3e^– + 3H⁺ + HNO₃ ®NO + 2H₂O                         …(2)
(reduction half-reaction)

Multiply (1) by 3 and (2) by 2 and adding
(To cancel out the electrons)

3H₂S ® 3S + 6H⁺ + 6e^–

6e⁺ + 6H⁺ + 2HNO₃ ® 2NO + 4H₂O

2HNO₃ + 3H₂S ® 3S + 2NO + 4H₂O

1. ii) Reduction : Cr₂O₇^2- ® Cr⁺³    Oxidation : C₂O₄^2- ® CO₂

Balancing reduction reaction:

Balancing atoms other than O and H : Cr₂O₇^2- ® 2Cr⁺³

Reaction is taking place in acidic medium.

Balancing O:     Cr₂O₇^2- ® 2Cr⁺³ + 7H₂O

Balancing H:      14H⁺ + Cr₂O₇^2- ® 2Cr⁺³ + 7H₂O

Balancing charges, we get reduction half-reaction.

6e^– + 14H⁺ + Cr₂O₇^2- ® 2Cr⁺³ + 7H₂O                      … (1)

Balancing oxidation reaction :

Balancing C:   C₂O₄^2- ® 2CO₂;

By balancing C, oxygen is automatically balanced

Balancing charges, we get oxidation half-reaction

C₂O₄^2- ® 2CO₂ + 2e^–                                           … (2)

Multiplying (2) by 3 and adding to (1) (To cancel out the electrons)

                  Cr₂O₇^2- + 14H⁺ + 3C₂O₄^2- ® 2Cr⁺³ + 7H₂O + 6CO₂

 

2.

10 OH^– +         4Cl₂  +  ¾®        +     8Cl^– + 5H₂O

before reaction      0.3 mole          0.15 mole 0.01 mole           0              0         0.

0.2 mole          0.11 mole  0 mole             0.02 mole 0.08 mole

                 = 0.2 mole

 

3. Let the mass of H₂C₂O₄.2H₂O and KHC₂O₄.H₂O be x and y gm. Respectively.

\  ´ 2 + ´ 1 =

´ 2 +  ´ 2 =

\ x = 0.1764 gm, y = 0.9709 gm.

% H₂C₂O_{4 .}2H₂O =  ´ 100 = 14.7%

% KHC₂O₄.H₂O = 80.9%

 

4. Let lower oxidation state of metal be n

equivalents of metal =  =  equivalents of H₂ evolved  =

\n = 2

Let final oxidation state =  n¢

Then equivalents of metal =  equivalents of oxidant

\ n¢ = 5

 

5. Meq. of oxalic acid reacting with MnO₂ = Meq. of MnO₂

Meq. of MnO₂ = ×1000 =  = 6.3172

Meq. of excess oxalic acid = 30 × 0.02 ×5 = 3

= 6.3172 +3

w = 0.586 gm

 

6. 3O₂          2O₃

1000 – x c.c.   x c.c.

\Contraction = ´ = 100 c.c

\ x = 300 c.c volume of O₂  converted

into 200 c.c. of ozone at S.T.P.

\ mole of ozone  =

\ g eq.   of O₃ =  = gm eq.  of As₂O₃

\ wt of As₂O₃ =  g = 0.8839 g Ans.

 

7.  Cl₂ HIO₃ + Cl^–

gm eq. Of HIO₃ =  = 0.005

º 0.005 gm eq. Cl₂ º 0.005 gm eq. KCl  = 0.005 ´74.5 g KCl = 0.3725 g KCl

\ weight of KF = 10–0.3725 = 9.6275 g

 

8. K₂Cr₂O₇+ KMnO₄ =0.24g

If K₂Cr₂O₇ =x g

Then KMnO₄ = (0.24 – x) gm

Meq. of K₂Cr₂O₇+ Meq. of KMnO₄ = Meq. of Na₂S₂O₃

 

× 1000 = 60 × 0.1

x = 0.1419

K₂Cr₂O₇ = 0.1419 g

KMnO₄ = 0.0981 gms

Mass of Cr = = 0.0502 gms

% of Cr =  = 20.92 %

Mass of Mn = ×0.0981 = 0.0341 gms

% of Mn =  = 14.22 %

9. + BaCl₂ ¾® BaSO₄ ¯ + 2Cl^–

25.6 mg

mole

Moles of  in 2 litre solution =  = 9.55 ´ 10^–3 mole

from Ag₂SO₄ in       500 ml of 0.1 M Na₂SO₄ = 0.0096 – 0.005 = 0.0046 mole

\ 0.0046 mol Ag₂SO₄ dissolves in 500 c.c. of 0.01 M N­a₂SO₄.

\ required solubility = 0.0046 ´312 ´2 g/ Lt = 2.87 g / litre

 

10. Let x gm Fe is present in 0.804 g of sample

Total gm eq. of Fe⁺⁺ = gm eq. of K₂Cr₂O₇

 

=

16x = 7.9, x = 0.49 g

% Fe =  = 60.9

 

11. Only Na₂CO₃ reacts with H₂SO₄

N₁V₁ = N₂V₂

N₁25 = 20 × ; N₁ =  = 0.08

Mass of Na₂CO₃ in 250 ml 0.08 N solution =  =  = 1.06

% of Na₂CO₃ =  = 70.67

 

12. HCl + NaOH ¾® NaCl + H₂O

Mass of NaCl obtained = 0.117g

No. of gm eq. of NaCl =  = 0.002

Thus, 0.002 gm eq. of HCl will react with 0.002 gm eq. of NaOH to form 0.002 g eq. NaCl.

Normality of HCl =  = 0.05N

Normality of NaOH = × 1000 = 0.10 N

13. Volume strength of a H₂O₂ solution : If a solution of H₂O₂ is labelled as ‘x volumes’ it means that 1 mL of the H₂O₂ solution on complete decomposition would release O₂ measuring x mL at STP.

Normality  of KMnO₄ solution =0.0245 ´ 5 = 0.1225 N

Equivalents of KMnO₄ used   ==3.0625 ´ 10^-3

\ Equivalents of H₂O₂ in the 10 mL that

is titrated    =    3.0625 ´ 10^-3

 

Equivalents of H₂O₂ in 100 mL=3.0625 ´ 10^-3 ´ 10= 3.0625 ´ 10^-2

Equivalents of H₂O₂ in the original 10 mL =   3.0625 ´ 10^-2

[ on adding water equivalents of a substance does not change]

Moles of H₂O₂ in the original 10 mL         =    =   1.53 ´ 10^-2

[‘n’ factor of H₂O₂ is 2 because as reacting with KMnO₄ O^-1 becomes O^-2]

Moles of H₂O₂ in 1mL of the original 10 mL =     =  1.53 ´ 10^–3

Moles of O₂ that it would give on decomposing  =                 =             7.65 ´ 10^-4

Volume of O₂ at STP in mL    =    7.65 ´ 10^-4 ´ 22400=   17.136

\ Volume strength     =    17.136

            Alternatively,

If a solution of H₂O₂ has normality = N, it means that 1 litre of the solution contains N equivalents.

\ 1 mL of it would contain equivalents.

\ moles of H₂O₂ in 1 mL =

Moles of O₂ it gives =

Volume of O₂ at STP in mL    = 22400 ´    =    5.6 ´ N

\ Volume strength         =    5.6 ´ Normality

\ Normality of H₂O₂ solution       =    =  0.306

Normality of the 100mL H₂O₂ solution=   0.306

(that has been diluted)

Normality of the original 10 mL H₂O₂ solution  = 0.306 ´ 10 = 3.06

\ Volume strength     =  5.6 ´ 3.06 =  17.136

 

14. KClO₃ ¾® KCl + 3/2 O₂

2.45–x g

4KClO₃      ¾® 3 KClO₄ + KCl

number  of moles of AgCl =  = number of moles of KCl

\ wt of KCl = ´ 74.5 = 1.0 43 g

\ wt  of KClO₄ = (1.874 – 1.043) g  = 0.831 g

\ wt of KClO₃ which has converted into KClO₄  =  ´

% of KClO₃ converted into KClO₄ =  = 40.16 %

15. We get the informations

(a) It is a metal

(b) At wt =  =  = 53.78 (approx).

(c) eq. weight of metal =

(d) valency =  =  = 2.88 » 3

(d) similarly valency =  = 2

(e) correct atomic weight = Eq. weight ´ valency

\ correct  at wt. = 18.613 ´ 3 = 55.8

 

 

 

 

 

 

 

Level – II

1. 1. K₂Cr₂O₇+ KMnO₄ =0.561g

If K₂Cr₂O₇ =x g

Then KMnO₄ = (0.561 – x) gm

Meq. of K₂Cr₂O₇+ Meq. of KMnO₄ = Meq. of Na₂S₂O₃

 

× 1000 = 100 × 0.15

x = 0.245

K₂Cr₂O₇ = 0.245 g

KMnO₄ = 0.316 gms

% of K₂Cr₂O₇ =  = 43.67 %

% of KMnO₄  =  = 56.33%

 

2. N₂ +          3H₂     ¾®     2NH₃

Before reaction      mole         mole

After reaction        0.5 mole          0.25 mole        1 mole

NH₃     +    HCl     ¾®     NH₄Cl

0.1 mole    0.1 mole

3PCl₃ + S₂Cl₂ ¾® 2PSCl₃ + PCl₅

 

0.58 mole  = 0.193 mole

nPCl₅   +    nNH₄Cl ¾¾® (PNCl₂)_n + nHCl

0.198 mole      0.1 mole    0.1 mole

0.1 mole (PNCl₂)_n = 11.6 g

3. Mg + 2HCl ¾® MgCl₂ + H₂

2Al + 6 HCl ¾® 2AlCl₃ + 3 H₂

Suppose Mg = x g

And Al = (1-x )g

P₁V₁ = P₂V₂

0.92 ×1.2 = 1×V₂

V₂ = 1.104 litre

= 1.104

x = 0.4514

Mg = 0.4514 g

Al = 0.5486 g

 

4. XO + K₂Cr₂O₇ ¾® Cr⁺³ +

X₂O₃ + K₂Cr₂O₇ ¾® Cr⁺³ +

Let wt of XO in the mixture be x g

Equivalent of K₂Cr₂O₇ consumed by the mixture = 0.015 ´ 6

Equivalents of XO =

Equivalents of X₂O₃ =

\  = 0.015 ´ 6

Since 1 mole of XO gives 1 mole  and 1 moles of X₂O₃ gives 2 moles of ,

\  = 0.0187

Solving this x = 99

 

5. Meq. of H₂SO₄ + Meq. of H₂C₂O₄ = Meq. of NaOH = 35.54 ×0.1

Meq. of H₂C₂O₄ = Meq. of KMnO₄

×1000 = 23.45 ×0.02 ×5

w = 0.1477 g

Meq. of H₂SO₄ = (35.54×0.1) – (23.45 ×0.02 ×5)

Thus, w = 0.0592 g

Molarity of H₂SO₄ =  = 0.0241 M

Molarity of H₂C₂O₄ =  = 0.0469 M

6. Suppose CuCl₂ = x g

and CuBr₂ = (1.85 – x) g

Wt. of AgCl consumed =

Wt. of AgCl unused = 1.8 –

1.8 – 1.284 (1.85-x) + 1.6823 (1.85-x) = 2.045

x = 1.216

CuBr₂ = 0.632

% of CuBr₂ =  = 34.21 %

 

7. Equivalents of K₂Cr₂O₇          =

Equivalents of Fe²⁺ in 50 ml   =    1 ´ 10^–3

Equivalents of Fe²⁺ in 250 ml       =    1 ´ 10^–3 ´ 5 = 5 ´ 10^–3

Initial equivalents of Fe²⁺        =

Equivalents of Fe²⁺ consumed by MnO₂ =    2.55 ´ 10^–2 – 5 ´ 10^–3  = 0.0205

\ Equivalents of MnO₂          =    0.0205

Moles of MnO₂                 =

Mass of MnO₂                  =    1.025 ´ 10^–2 ´ 87 = 1.0875g

% MnO₂                            =     =  72.5%

8. Let the mass of K₂O = x g

mass of Na₂O = y g

KCl       +      AgNO₃ ¾® KNO₃ +  AgCl

 

NaCl           +  AgNO₃ ¾® NaNO₃ + AgCl

 

+ = 0.118                                          ….(1)

+ = 0.2451                                     ….(2)

By solving equation (1) and (2) we get

x = 0.053 g and y = 0.0179 g

% of K₂O =  = 10.6 %

% of Na₂O = × 100 = 3.58 %

 

9. In the presence of phenolphthalein indicator

H₃PO₄ + 2NaOH ¾¾® Na₂HPO₄ + H₂O

\ eq. wt of H₃PO₄ =  = 49

\ % of  H₃PO₄ =  = 2.48 %

H₃PO₄ HPO₃ + H₂O

10^–3 mole 1´10^–3 mole = 10^–3 ´98 g

\ wt. of solution =  = 3.95 g

 

10.       30 N₁ = 42.9 ×0.0699

N₁ = 0.1

5×8 + 4.8 ×5 +17×2×V = 2000 ×0.1

V = 4 ml = volume of H₂SO₄

Mass of H₂SO₄ =  = 6.664 g

Mass of Sulphate ion =  = 6.528 g

11.       Out of all alkali metal carbonates, its only Li₂CO₃ which decomposes to
give oxide and CO₂, rest all alkali metal carbonates are stable
to thermal decomposition.

Li₂CO₃  Li₂O + CO₂

1. of moles of Li₂CO₃  =  no.of moles  of CO₂  =

= 1.226 ´ 10^–3

In the residue Li₂O, Na₂CO₃ and Na₂O will be present when they react with HCl, CO₂ will be released by Na₂CO₃ only

\moles  of Na₂CO₃ = moles  of CO₂ evolved

=  = 1.781 ´ 10^–3

And equivalents of Na₂CO₃+equivalents of Na₂O + equivalents of Li₂O = equivalents of HCl used. (Na₂O  and Li₂O react with HCl to form the respective chlorides and water, because they are basic).

1.781 ´ 10^–3 ´ 2  +  equivalents of Na₂O +  1.226 ´ 10^–3 ´ 2  = 15 ´ ´ 10^–3

Equivalent of Na₂O = 7.5 ´ 10^–3 – 6.01 ´ 10³ = 1.49 ´ 10^–3

Equivalent  weight of Li₂CO₃ = 73.8/2 = 36.94

Equivalent weight of Na₂CO₃ = 106/2 = 53

Equivalent  weight of Na₂O = 62/2 = 31

Weight of Li₂ CO₃ = 2.45 ´ 10^–3 ´ 36.94 = 0.0905 gm

Weight of Na₂CO₃ = 3.56 ´ 10^–3 ´ 53 = 0.1886 gm

Weight  of Na₂O = 1.49 ´ 10^–3 ´ 31 = 0.0462

% of Na₂O in the mixture =       =  = 14.2%

 

12. CO₂ gas is acidic in nature, it neutralizes the Ba(OH)₂ and precipitates BaCO₃.

Equations of Ba(OH)₂ consumed = equivalents of CO₂

equivalents of CO₂ = –  =

\ moles of CO₂ =

\% of CO₂ in air at NTP =  = 0.0224

Ba(OH)₂ + CO₂ ¾® BaCO₃ + H₂O

It is a non redox process and Ba(OH)₂ has n factor of 2. It reacts with CO₂ in molar ratio of 1:1

\ n factor of CO₂ is 2

 

13. Na₂CO₃ + H₂SO­₄ ® NaHCO₃ + Na₂SO₄

phenolphthalein would change the colour after this reaction.

Moles of H₂SO₄ which give phenolphthalein end point =   = 5 ´ 10^-4

\ moles of Na₂CO­₃  in 20 cc of the solution        =  2 ´ 5 ´ 10^-4  =  1 ´ 10^-3         

moles in 1 L = 1 ´ 10^-3 ´ 50          =   0.05

mass in 1 L   =  0.05 ´ 106           =   5.3 g

NaHCO₃ + H₂SO₄ ® Na₂SO₄ + CO₂ + H₂O

moles of H₂SO₄ which gives methylorange end point =  = 1 ´ 10^-3

\ moles of NaHCO₃ in 20cc       = 2 ´ 10^-3

moles of NaHCO₃ in 1 L   = 2 ´ 10^-3 ´ 50 =  0.1

moles of NaHCO₃ produced from Na₂CO₃ = 0.05

\ moles of NaHCO₃ initially present        = 0.1 — 0.05   =   0.05

                        \ mass of NaHCO₃ = 0.05 ´ 84 = 4.2 g.

 

14. moles of Na₂CO₃ in 40 ml  of ‘s’ =

moles . of  NaOH in  40 ml is =  =  mole

\ In 60 c.c. of ‘S’ =  moles of NaOH

Zn + 2 NaOH ¾® Na₂ZnO₂ + H₂

0.0105 moles of NaOH gives    moles of H₂

\Volume of H₂ at S.T.P. =  Lt = 0.1176 Lt   or 117 .6 ml

Mass of Na₂CO₃ in 100 ml =  = 0.265 g

Mass of NaOH in 100 ml = 0. 7 g

 

15. [AgNO₃ decomposes completely to Ag as it is covalent but NaNO₃ decomposes to NaNO₂ and O₂ as it is strongly ionic in nature]

No. of moles of gas (oxygen)  = (no of moles of NaNO₃ + no. of moles of AgNO₃)

=  +

=  = 0.013 ——— (1)

When the residue will be treated with HCl, NaNO₂ will react

NaNO₂ + HCl  ¾¾® NaCl + HNO₂

No. of moles of HCl used = no. of moles of NaNO₂ present

= no. of moles of NaNO₃ present in original sample

 

= 2.5 ´10^–3       ————–(2)

From equation  (1) and (2)

= 0.013 – 2.5 ´10^–3 = 0.0105

weight of AgNO₃ = 0.0105 ´ 170 = 1.785 g

weight of NaNO₃ = 2.5 ´ 10^–3 ´ 85 = 0.2125 gm

% of AgNO₃ =  = 57.58%

% of  NaNO₃ =  = 6.85%

 

 

 

 

 

LEVEL – III

1. (i) 2kg mass  of Ca₃(PO₄)₂  and H₂SO₄ mixture  in the molar ratio of 5:11 is needed to  get 2 kg. Super phosphate of lime. Let w g of H₂SO₄ is needed

\  =

2000–w = 1.43 w

\w =  = 823 g

\ V =  = 1.68 Lt

(ii) number of mole of Ca₃(PO₄)₂ =  = 6.45

number of mole of H₂SO₄ =  = 20.4

\ total moles of Ca₃(PO₄)₂ will react.

Weight of superphosphate of lime formed

= 6.45 ´ ´98 g = 1390.62 g

2. Mass of H₂SO₄ = = 21.8595 gm

2Al + 3H₂SO₄ ¾® Al₂(SO₄)₃ + 3H₂

H_­2SO₄ required for dissolving 2.7 gm Al   = 14.7 gm

H₂SO₄ left unreacted = (21.895 – 14.7) = 7.1595 gm

7.1595 g H₂SO₄ is present in 400 ml

Amount of H₂SO₄ present in one litre =  × 1000 = 17.898 gm

No. of gm moles of H₂SO₄ ⁼  =- 0.1826

Thus, molarity = 0.1826 M

3. Cr (in steel) ¾®

52 g                      108g                                         253g

0.0549 g

From the weight of given BaCrO₄, we can determine weight of Cr present in 10 mL solution.

Weight of chromium =  ´ 0.0549 g

= 0.0113 g in 10 mL portion

= 0.0113 ´ 25 in 250 mL solution

= 0.2821 g

This amount is in 10.00 g steel sample. Hence % of Cr in steel sample

=  = 2.821%

Fe²⁺ reduces  as well as  in acidic medium

5Fe²⁺ +  + 8H⁺ ® Mn²⁺ + 5Fe³⁺ + 4H₂O

6Fe²⁺ +  + 14H⁺ ® 2Cr³⁺ + 6Fe³⁺ + H₂O

This titration will give total normality of  and

10 mL of this mixture solution = 15.95 mL of 0.075 M = 15.95 mL of 0.075 N

\ N( + ) =  = 0.1196 N

From the weight of BaCrO₄, we can also deduce weight of  and hence is normality.

Weight of  in 0.0549 g BaCrO₄ =  = 0.0234 g in 10 mL

equivalent weight of  =

\ N( =  =            = 0.065 N

thus N () = 0.0545 N by difference

equivalent weight of =

in250 mL solution =   = 0.3243 g

Mn in 250 mL solution = 0.1498 g

This is present in 10.00 g steel sample

Hence % of Mn = = 1.498%

 

4. CuFeS₂ + 2H₂SO₄ ® CuSO₄ + FeSO₄ + 2H₂S­

Thus mineral is converted into CuSO₄ and FeSO₄

            Titration – I:    oxidises Fe²⁺ into Fe³⁺

+ 8H⁺ + 5Fe²⁺ ® Mn²⁺ + 5Fe³⁺ + 4H₂O

10 mL of Fe²⁺ (in mineral) º 2 mL of 0.01 M

º 2 mL of 0.05 N

10 ´ N (Fe²⁺) º 2 ´ 0.05 N ()

\ N (Fe²⁺) º 0.01

\ Fe²⁺ in mineral = 0.01 ´ 56 = 0.56 gL^–1 present in 10 gL^–1 solution

\ % of Fe in mineral =  = 5.6%

                  Titration II:     Cu²⁺ is reduced to Cu⁺ (as white ppt. Cu₂I₂) on treatment with I^–. I₂ formed is titrated using

2Cu²⁺ + 4I^– ® Cu₂I₂ + I₂

I₂ +  ®  + 2I^–

This titration will give amount of Cu²⁺

25 mL of Cu²⁺ solution º 5 mL of 0.01 M

º 5 mL of 0.01 N

\ 25 ´ N (Cu²⁺) = 5 ´ 0.01 N

N (Cu²⁺) = 0.002

\ Cu²⁺ in mineral = 0.002 ´ 63.5 gL^–1

                                                = 0.127 gL^–1 present in 10 gL^–1 solution

% of Cu =  = 1.27%

thus Fe content in mineral = 5.6%

and Cu content is = 1.27%

 

5. Equivalents of Barium permanganate = = 1.485 ´10^–4

Equivalents of Fe²⁺ in excess = 1.485 ´ 10^–4

Initial equivalents of Fe²⁺ =  = 8 ´ 10^–3

Equivalents of Fe²⁺ consumed by Cr⁶⁺ = 8 ´ 10^–3 – 1.485 ´ 10^–4 = 7.85 ´ 10^–3

\Equivalents of Cr⁶⁺ = 7.85 ´ 10^–3

Equivalents of FeO. Cr₂O₃ = 7.85 ´ 10^–3

Moles of FeO. Cr₂O₃ =  = 1.3 ´ 10^–3

Mass of FeO. Cr₂O₃ = 1.3 ´ 10^–3 ´ 224 = 0.2912 gm.

% FeO.Cr₂O₃ =  ´ 100 = 15. 57%

 

6. 25 cm³ of oxalic acid requires 32cm³ 0.1 N KMnO₄ solution

Thus, 250 cm³ oxalic acid requires = 320 cm³ 0.1N KMnO₄ solution

= 32 cm³ 1 N KMnO₄ = 32 cm³ 1 N oxalic acid solution

oxalic acid used by pyrolusite = (50 – 32)cm³ 1 N solution

= 18 cm³ 1 N solution = 18 cm³ 1 N MnO₂ solution

Mass of MnO₂ =  = 0.783 g

% MnO₂ = × 100 = 48.9 %

MnO₂ ¾® MnO + O

Oxygen given by 0.783 g MnO₂ = × 0.783 = 0.144g

% available oxygen = × 100 = 9 %

7. Mass of oxalic acid present in 25 ml oxalic acid solution

=  = gm

Now,

16.68 ml oxalic acid solution contains =  gm

\ 250 ml oxalic acid solution will contain  = 1.575

\ x =2

 

8. Bleaching powder is Ca²⁺, OCl^– and Cl^– (CaOCl₂). Here one Cl is in –1 oxidation state while the other is in +1 oxidation state and the net oxidation state is zero.

When bleaching powder is added to Fe²⁺  solution, Cl gets reduced to –1 while Fe²⁺ is  oxidised to Fe³⁺. The excess of  Fe²⁺ is back titrated  with KMnO₄ solution.

                               Number of eq. of Mohr’s salt used = Initial eq. – eq. of KMnO₄ used

= – ´ 5      = 3.57 ´ 10^–3 – 1.57 ´ 10^–3 = 2 ´ 10^–3

number of equivalents of 2Cl in 1000  ml = 2 ´ 10^–3 ´ 4 = 8 ´ 10^–3

Moles of Cl =

[ Q there are 2Cl with average oxidation state zero and both of them became –1 in the product]

W = 8 ´ 10^–3 ´  = 0.284 g

% of available Cl₂ = ´ 100 = 2.84%

9. 10 gm equivalent of Na₂S₂O₃ = = gm eq of I₂ = gm eq of KI in 10 ml

\ total gm eq of Cr₂O₇^{– –} in 100 ml  =   = gm eq of  BaCrO₄

\ number of mole of BaCrO₄ =  = mole of BaCO₃

\ weight of BaCO₃ =  = 0.656 g

number of mole of CO₂ =

number of mole of CaCO₃ =

weight of CaCO₃ = ´ 100 = 0.416 g

                        \ weight percentage of CaO =  = 14.17 %

10.       In this reaction, As₂O₃ acts as acidic oxide and NaHCO₃ as a base, giving acid base neutralization reaction which is non – redox process. Here As₂O₃ does not act as basic oxide Q in that case, it will form As₂(CO₃)₃ which does not exist as non-metals do not form carbonates.

n-factor of As₂O₃ is 6 and that of NaHCO₃ is 1. After the reaction As³⁺ is oxidised by I₂ to As⁺⁵ while I₂ is reduced to I^–.

Normality of Na₂S₂O₃.5H₂O =  = 0.1

Normality of I₂ = Normality of Na₂S₂O₃.5H₂O =

\ Equivalents of I₂ = 0.1 ´22.4´10^–3=Equivalent of  As³⁺ reacted in 25 ml=2.24 ´ 10^–3

\ Equivalents of As³⁺ reacted in 250 ml =  2.24 ´ 10^–2

Moles of As³⁺ in 250 ml =  = 1.12 ´ 10^–2

Moles of As₂O₃ reacted =  = 5.6 ´10^–3

            % Percentage of As₂O₃ =  = 9.24%

11. Let the mol wt be M

If each ZSM–5 has at least one Si atom

= 28 \ M = 63.66 amu

If each ZSM–5 has at least one  Al atom

= 27 \ M = 6058

\ Minimum mol. wt of ZSM–5 = 6053. In which 1 Al atom and
= 95 Si atoms will be present.

[Note: ZSM–5 has the formula the H[AlO₂(SiO₂)₉₅]16H₂O]

 

12. This problem can be done by two methods.

In the first method, we break up Fe₃O­₄ as an equimolar mixture of FeO and Fe₂O₃.

Method (1)

Fe₃O₄ is FeO. Fe₂O₃

Equivalents of Na₂S₂O₃   =          =   7.2 ´ 10^-3

Equivalents of I₂ in 10 cc =    7.2 ´ 10^-3

Equivalents of I₂ in 50 cc =    7.2 ´ 10^-3 ´ 5   =   3.6 ´ 10^-2

Since equivalents of I₂ is equal to that of KI which in turn is equal to the total equivalents of Fe₂O₃ (Fe₂O₃ in the free state and Fe₂O₃ combined with FeO).

\ equivalents of total Fe₂O₃      =      3.6 ´ 10^-2

Equivalents of KMnO₄ solution   =      = 2.1 ´ 10^-2

Since KMnO₄ reacts with the total Fe²⁺ (Fe²⁺ in FeO and Fe²⁺ that was produced by the action of KI on Fe₂O₃)

\ equivalents of total Fe⁺² in 50 mL  =    2.1 ´ 10^-2 ´ 2 = 4.2 ´ 10^-2

Since equivalents of Fe²⁺ produced from Fe₂O₃ is equal to the equivalents of Fe₂O₃

\ Equivalents of FeO       =    4.2 ´ 10^-2 – 3.6 ´ 10^-2

=    6 ´ 10^-3

\ moles of FeO                =    6 ´ 10^-3

moles of Fe₂O₃ combined with FeO =    6 ´ 10^-3

total moles of Fe₂O₃         =
(because when Fe₂O₃ ® Fe²⁺ ‘n’ factor is 2).

=    1.8 ´ 10^-2

moles of Fe₂O₃ in the free state  =    1.8 ´ 10^–2  –  6 ´ 10^–3

=    1.2 ´ 10^-2

mass of Fe₃O₄                              =    6 ´ 10^-3 ´ 232  = 1.392 g

mass of Fe₂O₃                                    =    1.2 ´ 10^-2 ´ 160 = 1.92 g

percentage Fe₃O₄                         =    17.4%

percentage of Fe₂O₃                    =    24%

            Method 2:

Here we take Fe₃O₄ as a single entity.

Equivalents of Na₂S₂O₃         =          = 7.2 ´ 10^-3

Equivalents of I₂ in 50 cc        =    7.2 ´ 10^-3 ´ 5   = 3.6 ´ 10^-2

\ Equivalents of Fe₃O₄ + Fe₂O₃ =    3.6 ´ 10^-2

Let us assume that the moles of Fe₃O₄ is x and that of Fe₂O₃ is y. Since on reacting with KI both Fe₃O₄ and Fe₂O₃ give Fe²⁺ ‘n’ factor for both is two.

\ 2x + 2y = 3.6 ´ 10^-2      ——–   (1)

Equivalents of KMnO₄      =     = 2.1´10^-2

Equivalents of Fe²⁺ in 50 mL =    2.1 ´ 10^-2 ´ 2 = 4.2 ´ 10^-2

Moles of Fe²⁺ in 50 mL     =    4.2 ´ 10^-2

Since the moles of Fe₃O₄ are x, moles of Fe²⁺ produced from Fe₃O₄ will be 3x and that produced from Fe₂O₃ will be 2y.

\ 3x + 2y = 4.2 ´ 10^-2  ——— (2)

(2) – (1) gives x = 6 ´ 10^-3                  \ y = 1.2 ´ 10^-2

Solving this, percentage of Fe₃O₄ is 17.4% and Fe₂O₃ is 23.75%.

 

13. The equations required are  H₂O + KI + O₃  ¾¾® I₂ + O₂ + KOH

Milliequivalents of Iodine  = Milliequivalents of KI  = Milli equivalents  of O₃ reacted

Milliequivalents  of Na₂S₂O₃ = 1.5 ´ 0.01 = 1.5 ´ 10^–2

Millimoles of Iodine  =  = 7.5 ´ 10^–3   [Q ‘n’ factor for Iodine = 2]

Millimoles of ozone  = 7.5 ´ 10^–3

Volume of ozone  = = = 184.725 ´ 10^–6 litre

Volume percentage of ozone =  ´ 100 = 1.847 ´ 10^–3

14. In 1000 kg of water the mass of HC= 183 g

moles of HC = = 3

moles of  =  = 1

\ total moles of Ca²⁺ in the solution = 1 + 1.5 = 2.5

Ca(HCO₃)₂ + CaO ® 2CaCO₃ + H₂O

moles of CaO required to be added to remove all HC = 1.5.

Now the Ca²⁺ in the solution will be only associated with . Therefore moles of Ca⁺² left in the solution = 1.

ppm of Ca²⁺ = 1 ´ 40 = 40 ppm

moles of Ca²⁺ ions in 1 L of H₂O = 10^-3

moles of H⁺ ions that Ca⁺² will exchange with = 2 ´ 10^-3

\ pH = -log(2´10^-3) = 2.7

 

15. If you observe both the reactions carefully, you will come to know that in the first reaction when KBrO₃ is reacting with SeO₃^–2, its n-factor is 5 but when excess of KBrO₃ is titrated with NaAsO_2,, its n factor is 6. Equivalents of excess of KBrO₃ can be subtracted from the initial equivalents of KBrO₃ only when both the reactions have same n factor for KBrO₃, which in this case cannot be done. We present the solution by two methods.

            Equivalent Method:

Initial moles of KBrO₃ =

Initial equivalents of KBrO₃ (n=5) = = 1.66 ´10^–3

Equivalents of AsO₂^–  reacted with excess of KBrO₃ =
= 4.08 ´10^–4    [ Q  n factor of AsO₂^– is 2]

\ equivalents of excess of KBrO₃ (n=6) = 4.08 ´ 10^–4

            [Note: Here the equivalents of excess of  KBrO₃ of n = 6 has to be converted to equivalents of excess of KBrO₃ of n factor 5 because equivalents of a substance with same n factor in two different reactions can only be added or subtracted.]

Moles of excess KBrO₃ =  = 6.8 ´10^–5

Eq. of excess of KBrO₃ (based on n factor 5) = 5 ´ 6.8 ´10^–5 = 3.4 ´ 10^–4

\ Equivalents of KBrO₃ reacted with SeO₃^2– = 1.66 ´10^–3 – 3.4´10^–4
=  1.32 ´10^–3

\ equivalents of SeO₃^2– = 1.32 ´10^–3

\ moles of SeO₃^2– =  = 6.6´10^–4 (as n factor of SeO₃^2– = 2)

\ amount of SeO₃^2–  = 6.6´10^–4 ´127 = 0.084 g

            Mole method:           

Initial moles of KBrO₃ =  = 3.33´10^–4

moles of AsO₂^– reacted with excess of KBrO₃ =

\ moles of excess of KBrO₃ =  = 6.8 ´10^–5

      [Note: As the n–factor of BrO₃^– and AsO₂ in second reaction is 6 and 2 respectively i.e., in the ratio of 3:1 \ their molar ratio will be 1:3]

\ moles of KBrO₃ reacted  with SeO₃^2– = 3.33´10^–4 – 6.8 ´10^–5  = 2.65 ´10^–4

\ moles of SeO₃^2– =  = 6.625 ´10^–4

\ amount of SeO₃^2– = 0.084 g

 

 

 

Solutions to Objective Problems

 

Level – I

1. 2.56 ×10^-3 =

M_x = 32

Thus, x = sulphur and has 16 protons

 

2. 8 g sulphur = moles =  moles

moles of BaSO₄ is also , as the mass of sulphur is conserved.

 

3. Moles of CaO =

moles of CaCl₂ =

mass of CaCl₂ = × 111 = 3.21 g

% of CaCl₂ = × 100 = 32.1 %

 

4. Equivalents of needed = equivalents of  = 0.136

moles of  =  = 0.0227

 

5. Moles of H₂O =

Moles of SO₃ =

Mass of SO₃ = × 80 = 40

 

6. Fe²⁺ ¾® Fe³⁺

¾® CO₂

 

V=   = 1200 ml = 1.2 L

 

7. Equivalent of HCl used = 100 × = 0.01 = equivalents of Na₂CO₃ in 125 ml solution

equivalents of NaHCO₃ formed = 0.01 in 125 ml solution

[NaHCO₃] =  = 0.08 M

8. Moles of NaOH =

moles of CO₂ =

moles of CO = 1 –  =

moles of CO₂ produced = from CO

moles of excess NaOH = × 2 =

mass of excess NaOH =   × 40 = 60

 

9. 2 SO₂ + 4 H₂S ¾® 6 S + 4 H₂O

4 × 22.4 litre gives 6 × 32 g sulphur

\ 11.2 litre will give  = 24 g

 

10. 2 MnO₂ +  8 HCl    ¾® 2 MnCl₂ + 4 H₂O  +  2 Cl₂

2 × 87        8 × 36.5                                        2 × 71

 

292 g HCl gives 142 g Cl₂

\ 1 g will give  = 0.486 g

 

x =             (where x is the n factor of HCl)

equivalent weight of HCl =  = 85.16

 

12. 100 × 0.1 =

\ M = 90

 

13. Moles of O₂ liberated from 35 ml H₂O₂ = = 0.02

volume of O₂ liberated from 35 ml H₂O₂ at STP = 0.02 × 22400 = 448 ml

volume strength of H₂O₂ =

 

14. Change in oxidation state in chromium = (+6) – (+3) = 3

 

15. 100 × 0.4 = 0.5 V

V = 80 ml

 

level – II

 

1. N_HCl is more

 

w = 5.6 g = 0.1 mole Fe²⁺

 

w = 3.5 g

 

4. 1.6 g metal combines with 0.4 g O₂

\ 8 g O₂ will combine with  = 32

 

5. 1.88 oxide is formed from 1.35 g Ca

\56 g oxide will be formed from  = 40.75

 

6. 1000 ml contains = 204 g

normality =  = 5.8 N

 

7. 100 × 0.3 + 200 × 0.6 = 300 N

N = 0.5

 

8. = 32                         (where x is the molecular weight of insulin)

x = 941.176

 

9. 63.5 g Cu²⁺ needs 34 g H₂S

\ 6.35 g Cu²⁺ will need  = 3.4 g

 

10. 276 g Ag₂CO₃ gives 216 g Ag

\ 2.76 g Ag₂CO₃ give  = 2.16 g

 

11. Equivalent wt. of metal = =12

 

12. Equivalent wt. of metal oxide = = 26.6

Equivalent wt. of metal + Equivalent wt. of O₂ = 26.6

\ Equivalent wt. of metal = 26.6 – 8 = 18.6

Atomic wt. of metal = 18.6 × 3 = 55.8

13. 19.8 × 0.1 = 20 N

N = 0.1

Now,  = 100 × 0.1

M = 140

Now, 106 + 18x = 140

X = 2

 

14. 0.1 × 100 + 0.2 × 25 = 125 M

M = 0.12

 

15. =

E = 42