# Stoichiometry

1. 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 KMnO­4 = 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 20oC 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 ´ 103

Equivalents of Fe2+ in 250 ml    = 1 ´ 103 ´ 5 = 5 ´ 103

Initial equivalents of Fe2+        =

Equivalents of Fe2+ consumed by MnO2 =    2.55 ´ 102 – 5 ´ 103

=    0.0205

\ Equivalents of MnO2    =    0.0205

Moles of MnO2                 =

Mass of MnO2                  =    1.025 ´ 102 ´ 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 SO42-. 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 ´ 104

Equivalents of S2                   =    5.35 ´ 104

Moles of S2                            =

Mass of H2S                           =    2.675 ´ 104 ´ 34 = 9.095 ´ 103 g.

Equivalents of I2 that reacted with

SO2

\ Equivalents of SO2        =   1.32 ´ 106

Moles of SO2              =

Mass of SO2         =    4.224 ´ 105 gm.

\  Concentration in mg/L of H2S

Concentration in mg/L of SO2 = 1.6896 ´ 103 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 SO42-. 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:        CaHb +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 O­2 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 =  [ Mx = Atomic mass of X]

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

\ 2.56 ´ 10–3 =  ´ 2

Mx = 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)

1. Assignments (Subjective Problems)

Level –  I

1. Balance the following equation by Ion electron method:
2. i) H2S + HNO3 ® NO + S + H2O
3. ii) Cr2O72– + H+ + C2O42– ® Cr3+ + H2O + CO2

1. 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.

1. 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.

1. 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.

1. 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.

1. 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.

1. 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.

1. 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.

1. 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.

1. 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.

1. 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 H­2SO4 solution for neutralisation. Calculate the percentage of sodium carbonate in the sample.

1. 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.

1. 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.

1. 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?

1. 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?

1. 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

1. 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 0oC has a volume of 1.2 litres at 0.92  atm.  pressure. Calculate the composition of the alloy.

1. 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?

1. 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.

1. 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?

1. 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.

1. 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?

1. 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?
2. 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.

1. 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.
2. 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.

1. 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.

1. 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.

1. i) If we have to prepare 2 kg superphosphate of lime. How much litre of 5.0 M H2SO4 is needed ?
2. 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.

1. 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 H2SO­4 in the resulting solution.

1. 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

1. 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.

1. 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.

1. 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.

1. 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.

1. 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.

1. 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?

1. 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.

1. 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

1. 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.)

1. Calculate the amount of SeO32– in a solution on the basis of following data. 20 ml of M/60 solution of KBrO3 was added to a definite volume of SeO32– 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) SeO32– + BrO3 + H+ ¾® SeO42– + Br2 + H2O
2.              b) BrO3 + AsO2 + H2O ¾®  Br + AsO43– + H+

1. 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

1. 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

1. 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%

1. The number of moles of Cr2O72– needed to oxidise 0.136 equivalents of N2H5+ by the reaction.

N2H5+ + Cr2O72– ¾® N2 + Cr3+ + H­2O is

(A) 0.136                                                   (B) 0.272

(C) 0.816                                                   (D) 0.0227

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

(A) 40%                                                     (B) 80%

(C) 60%                                                     (D) 9%

1. What volume of 0.3 N Cr2O72– / 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

1. 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

1. One mole of a mixture of CO and CO2 requires exactly 20 grams of NaOH to convert all the CO2 into Na2CO­3. 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

1. 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

1. 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

1. 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

1. 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

1. 35 ml sample of hydrogen peroxide gives off 500 ml of O2 at 27°C and 1 atm pressure. Volume strength of H­2O2 sample will be

(A) 10 volumes                                         (B) 12.8 volumes

(C) 11 volumes                                         (D) 12 volumes

1. 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

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

(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) NHCl is more                                        (B) is more

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

1. 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

1. 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

1. 1.60 g of a metal were dissolved in HNO­3 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

1. 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

1. 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

1. 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

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

(A) 941.176                                               (B) 944

(C) 945.27                                                 (D) None

1. 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

1. 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

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

(A) 12                                                        (B) 16

(C) 24                                                        (D) 48

1. 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

1. 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

1. 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

1. 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

Level – I

1. 3H2S + 2HNO3 ¾® 3S + 2NO + 4H2O

Cr2O72– + 14H+ + 3C2O42– ¾® 2Cr3+ + 7H2O + 6CO2

1. 0.2 mole 3.   H2C2O4.2H2O = 14.7%

KHC2O4.H2O = 80.9%

1. +2, +5 5.   0.586 gms
2. 0.8839 g 7.   9.6274 gm
3. % Mn = 14.22; % Cr = 20.92 9.   2.87g/lit
4. 60.9% 11. 70.67%
5. 0.1 N 13. 17.136
6. 40.16% 15. 55.8g, +2, + 3

Level – II

1. KMnO4 = 56.33 %

K2Cr2O4 = 43.67 %                             2.         11.6 g

1. Al = 0.549g 4.         99

Mg = 0.451 g

1. H2SO4 = 0.241 M 6.         34.21%

H2C2O4 = 0.0469M

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

% of K2O = 10.6

1. 2.48 %, 3.95 gm 10.       6.528 gm

1. 14.2% 12.       0.0224%
2. NaHCO3 = 4.2 gm       14.       Na2CO3 = 0.265 gm

Na2CO3 = 5.3 gm                                           NaOH = 0.7 gm

1. % 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%

1. 15.57% 6.         % of MnO2 = 48.9
2. 2             % available oxygen = 9
3. 2.84 %
4. 14.17% 10.       9.24%
5. Mol. Weight = 6058 12.       % of Fe3O4 = 17.4 %

Si = 95 atoms                                                        % of Fe2O3 = 24 %

1. 1.847 ´ 10–3 14.       2.7
2. 0.084 gms

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

At.

No.

Name 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
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

# Hints to Subjective Problems

##### Level – I
1. Oxalic acid will react with MnO2 and will get oxidised.
2. Partially formed ozone will reduce the volume.
3. Chlorine gas will formed from KCl will react with HIO3
4. Both the compounds K2Cr2O7 and KMnO4 will react with Na2S2O3
5. Fe2+ will react with K2Cr2O7
6. Find the wt. of KClO4 formed from KClO3 and then the % of KClO4 formed.
7. 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

1. Find the mass of MnO2 from the given normality and then its percentage.
2. Find the moles of barium carbonate and then its mass. Similarly, solve for calcium carbonate
3. 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 : Cr2O72- ® Cr+3    Oxidation : C2O42- ® CO2

Balancing reduction reaction:

Balancing atoms other than O and H : Cr2O72- ® 2Cr+3

Reaction is taking place in acidic medium.

Balancing O:     Cr2O72- ® 2Cr+3 + 7H2O

Balancing H:      14H+ + Cr2O72- ® 2Cr+3 + 7H2O

Balancing charges, we get reduction half-reaction.

6e + 14H+ + Cr2O72- ® 2Cr+3 + 7H2O                      … (1)

Balancing oxidation reaction :

Balancing C:   C2O42- ® 2CO2;

By balancing C, oxygen is automatically balanced

Balancing charges, we get oxidation half-reaction

C2O42- ® 2CO2 + 2e                                           … (2)

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

Cr2O72- + 14H+ + 3C2O42- ® 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

1. 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.

% H2C2O4 .2H2O =  ´ 100 = 14.7%

% KHC2O4.H2O = 80.9%

1. 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

1. 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

1. 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.

1.  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

1. 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 %

1. + 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

1. 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

1. 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

1. 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

1. 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 ´ 103

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

1. 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 %

1. 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%

1. 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

1. 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

1. 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

1. 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

1. 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 %

1. Equivalents of K2Cr2O7          =

Equivalents of Fe2+ in 50 ml   =    1 ´ 103

Equivalents of Fe2+ in 250 ml       =    1 ´ 103 ´ 5 = 5 ´ 103

Initial equivalents of Fe2+        =

Equivalents of Fe2+ consumed by MnO2 =    2.55 ´ 102 – 5 ´ 103  = 0.0205

\ Equivalents of MnO2          =    0.0205

Moles of MnO2                 =

Mass of MnO2                  =    1.025 ´ 102 ´ 87 = 1.0875g

% MnO2                            =     =  72.5%

1. 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 %

1. 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 CO=

= 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%

1. 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

1. 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 Na2CO­in 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.

1. 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

1. [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

1. Mass of H2SO4 = = 21.8595 gm

2Al + 3H2SO4 ¾® Al2(SO4)3 + 3H2

H­2SO4 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

1. 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%

1. CuFeS2 + 2H2SO4 ® CuSO4 + FeSO4 + 2H2

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%

1. 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%

1. 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 %

1. 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

1. Bleaching powder is Ca2+, OCland 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%

1. 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

#### 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%

1. 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]

1. This problem can be done by two methods.

In the first method, we break up Fe34 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 ´ 102  –  6 ´ 103

=    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%.

1. 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

1. 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

1. 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 NaAsO2,, 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 SeO32– = 1.66 ´10–3 – 3.4´10–4
=  1.32 ´10–3

\ equivalents of SeO32– = 1.32 ´10–3

\ moles of SeO32– =  = 6.6´10–4 (as n factor of SeO32– = 2)

\ amount of SeO32–  = 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 SeO32– = 3.33´10–4 – 6.8 ´10–5  = 2.65 ´10–4

\ moles of SeO32– =  = 6.625 ´10–4

\ amount of SeO32– = 0.084 g

# Solutions to Objective Problems

#### Level – I

1. 2.56 ×10-3 =

Mx = 32

Thus, x = sulphur and has 16 protons

1. 8 g sulphur = moles =  moles

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

1. Moles of CaO =

moles of CaCl2 =

mass of CaCl2 = × 111 = 3.21 g

% of CaCl2 = × 100 = 32.1 %

1. Equivalents of needed = equivalents of  = 0.136

moles of  =  = 0.0227

1. Moles of H2O =

Moles of SO3 =

Mass of SO3 = × 80 = 40

1. Fe2+ ¾® Fe3+

¾® CO2

V=   = 1200 ml = 1.2 L

1. 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

1. 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

1. 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

1. 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

1. 100 × 0.1 =

\ M = 90

1. 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 =

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

1. 100 × 0.4 = 0.5 V

V = 80 ml

## level – II

1. NHCl is more

w = 5.6 g = 0.1 mole Fe2+

w = 3.5 g

1. 1.6 g metal combines with 0.4 g O2

\ 8 g O2 will combine with  = 32

1. 1.88 oxide is formed from 1.35 g Ca

\56 g oxide will be formed from  = 40.75

1. 1000 ml contains = 204 g

normality =  = 5.8 N

1. 100 × 0.3 + 200 × 0.6 = 300 N

N = 0.5

1. = 32                         (where x is the molecular weight of insulin)

x = 941.176

1. 63.5 g Cu2+ needs 34 g H2S

\ 6.35 g Cu2+ will need  = 3.4 g

1. 276 g Ag2CO3 gives 216 g Ag

\ 2.76 g Ag2CO3 give  = 2.16 g

1. Equivalent wt. of metal = =12

1. 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

1. 19.8 × 0.1 = 20 N

N = 0.1

Now,  = 100 × 0.1

M = 140

Now, 106 + 18x = 140

X = 2

1. 0.1 × 100 + 0.2 × 25 = 125 M

M = 0.12

1. =

E = 42