1. IIT–JEE Syllabus

Mole concept, Calculations involving Oxidation – Reduction, neutralisation and displacement reactions

2. Introduction

Stoichiometry deals with the calculation of the quantities of various reactants and products involved in a chemical reaction. The numerals used to balance a chemical equation are known as stoichiometric coefficients. These numerals are essential for solving problems based on chemical equation, hence called stoichiometric calculation.

In stoichiometric calculations, the mole relationship between different reactants and products are required as from them the mass-mass, mass-volume and volume-volume relationship between different reactants and products can be obtained.

Stoichiometric calculations can be carried out by two methods; Gravimetric Analysis and Volumetric Analysis. In the first method the amount of chemical species is determined by measurement of mass, whereas, in second method it is determined by measurement of volume. Stoichiometric calculation can be carried in terms of two important concept: (i) Mole Concept; (ii) Equivalent concept.

3. Mole Concept

A mole (symbol mol) is defined as the amount of substance that contains as many atoms, molecules, ions electrons or any other elementary entities as there are carbon atoms in exactly 12 gm of 12C. The number of atoms in 12 gm of 12C is called Avogadro’s number (N0).

N0 = 6.023 × 1023

One atomic mass unit (amu) = = 1.66 × 10–24 gm = 1.66 × 10–27 kg

The number of moles of a substance can be calculated by various means depending on data available, as follows.

1. a) Number of moles of molecules =
2. b) Number of moles of atoms =
3. c) Number of moles of gases =

(Standard molar volume at STP = 22.4 lit)

1. d) Number of moles of particles e.g. atoms, molecules ions etc =
2. e) Number of moles of solute = molarity × volume of solution in litres
   =
3. f) for a compound AxBy, y moles of A = x moles of B
4. g) Mole fraction = fraction of the substance in the mixture expressed in terms of mol is called its mol fraction (X)

E.g. for a mixture of substance A & B

XA = XB = (n terms of denote number of moles)

XA + XB = 1 & XA= (1 – XB)

Principle of Atom Conservation (POAC)

In chemical reaction atoms are conserved, so moles of atoms shall also be conserved. This is known as principle of atomic conservation. This principle is helpful in solving problems of nearly all stoichiometric calculations e.g. 

KClO3(s) ⎯→ KCl(s) + O2(g)

Applying POAC for K atoms

Moles of K atoms in KClO3  = Moles of K atoms in KCl

Since one mole of KClO3 contains 1 mol of K atom. Similarly 1 mol of KCl contains one mole of K atoms.

∴ 1 × (Mass-mass relationship)

Applying POAC for ‘O’ atoms

Moles of O atom in KClO3 = Moles of O atom in O2

∴ 3 × = 2 ×

⇒ 3 × (Mass volume relationship of reactant and product)

In this way applying POAC we can break the chemical equation into a number arithmetic equations without balancing the chemical equation. Moreover number of reactions and their sequence from reactants to products are not required. It is important to note that POAC can be applied for the atoms which remain conserved in chemical reaction.

4. Concept of Limiting Reagent

In reactions involving more than one reactant, one has to identify first, of all the reactant, which is completely consumed (limiting reagent), one can identify the limiting reagent as follow:

N2 + 3H2 ⎯→ 2NH3

Initial mole 5 12 ––

If N2 is the limiting reactant moles of NH3 produced = 10. If H2 is the limiting reactant moles of NH3 produced = ×12 = 8. The reactant producing the least number of mole of the product is the limiting reactant, hence H2 is the limiting reactant.

The limiting reactant can also be ascertained by knowing the initial number of equivalents or milli equivalents of each reactant. The reactant with least number of equivalents or milli equivalent is the limiting reactant. The equivalent methods to identify the limiting reactant used not require balancing of chemical equation.

5. Gravimetric Analysis

 

5.1 Calculation involving in mass – mass relationship

In general, the following steps are adopted in making necessary calculations:

1. Write down balanced molecular equation for the chemical change 
2. Write down the no of moles below the formula of each of the reactant and product 
3. Write down the relative masses of the reactants and the products with the help of formula below the respective formula. These shall be the theoretical amounts of reactants and product.
4. By the applications of unitary method, mole concept or proportionality method, the unknown factor or factors are determined.

5.2 Calculation involving percent yield

In general, when a reaction is carried out in the laboratory we do not obtain actually the theoretical amount of the product. The amount of the product that is actually obtained is called the actual yield. Knowing the actual yield and theoretical yield the  percentage yield can be calculated as 

% yield = × 100

5.3 Weight –  Volume Relationship

2Mg(s) + O2(g) ⎯→ 2MgO(s)

In the above reaction, one can find out the volume of O2  at STP required to react with 10 gm of Mg. The moles of Mg is . The moles of O2 required would be ½ the moles of Mg. Therefore moles of O2 = .  Since 1 mole of a gas (ideal) occupies 22.4L at STP, therefore moles of O2 would occupy, ×22.4L= 4.67L.

5.4 Volume – Volume Relationship

Let us consider the reaction H2(g) + ½O2(g) → H2O(l). We are given 10L of H2 at a given temperature and pressure. How many liters of O2 would react with hydrogen at the same temperature and pressure? 

From the ideal gas equation [PV = nRT] it is clear that the volume of an ideal gas is directly proportional to its no. of moles. Therefore under the same conditions of P and T, Since the molar ratio is 2:1 (H2:O2),   ∴ the volume ratio would also be 2:1. 

Therefore the volume of O2 required would be 5L.

On the other hand if we need to calculate the volume of O2 at a different T and P, then 

and dividing we get , 
∴ 

Illustration 1: A sample of CaCO3 and MgCO3 weighed 2.21 gm is ignited to constant weight of 1.152 gm. What is the composition of the mixture. Also calculate the volume of CO2 evolved at 0°C and 76 cm of pressure.

Solution: CaCO3 ⎯→ CaO + CO2

x gm

MgCO3 ⎯→ MgO + CO2

y gm

∴(x + y) = 2.21 gm

 100 gm of CaCO3  gives 56 gm of CaO

∴ x gm of CaCO3 ≡ CaO

Similarly 84 gm of MgCO3  gives 40 gm of MgO

∴ y gm of MgCO3  = gm

∴ Wt. of residue = = 1.152 …(2)

Solving equations (1) and (2)

x = 1.19 gm

y = 1.02 gm

Mole of CO2 formed = Mole of CaCO3 + Mole of MgCO3

= = 0.0241

∴ Volume of CO2 at STP = 0.0421 × 22.4 litre = 539.8 ml

Illustration 2: A mixture of FeO and FeO3 when heated in air to constant weight gains 5% in weight. Find out composition of mixture.

Solution: 2FeO + O2 ⎯→ Fe2O3

2Fe3O4 + O2 ⎯→ 3Fe2O3

Let, weight of FeO = x

Weight of Fe3O4 = y

∴ x + y = 100 …(1)

 2 × 72 gm of FeO give Fe2O3= 160 gm

∴ x gm FeO gives Fe2O3 = gm

2 × 232 gm of Fe3O4 gives Fe2O3 = 3 × 160 gm

∴ y gm Fe3O4 gives Fe2O3 = gm

∴ = 105 …(2)

Solving equation (1) & (2)

x = 20.25 gm

y = 79.95 gm

Illustration 3: Calculate the weight of FeO from 2 gm VO and 5.75 gm of Fe2O3. Also report the limiting reagent.

VO + Fe2O3 ⎯→ FeO + V2O5

Solution: Balanced equation

2VO + 3Fe2O3 ⎯→ 6FeO +        V2O5

Moles before reaction  

= 0.0298 0.0359   0   0

Moles after reaction (0.0298 – 0.0359)   0  

As 2 mole of VO react with 3 mole of Fe2O3

∴ 0.0298 gm mole of VO = × 0.0298 = 0.0447 mole of Fe2O3

Moles of Fe2O3 available = 0.0359 only

Hence, Fe2O3 is the limiting reagent

Mole of FeO formed = × 0.0349

∴ Weight of FeO formed = 0.0359 × 2 × 72 = 5.17 gm

⇒

Illustration 4: A mixture of KBr, NaBr weighing 0.56 gm was treated with aqueous solution of Ag+ and the bromide ion was recovered as 0.97 gm of pure AgBr. What was the weight of KBr in the sample?

Solution: KBr + NaBr + Ag+ ⎯→ AgBr

a gm (0.56 – a) gm 0.97 gm

Applying POAC for Br atoms

Moles of Br in KBr + moles of Br in NaBr = moles of Br in AgBr

or 1 × moles of KBr + 1 × moles of NaBr = 1× moles of AgBr

⇒ (MKBr = 199, MNaBr = 103, MAgBr = 188)

∴ a = 0.1332 gm

Percentage of KBr in the sample = × 100 = 23.78

Illustration 5: In a gravimetric determination of P of an aqueous solution of dihydrogen phosphate in H2PO4– is treated with a mixture of ammonium and magnesium ions to precipitate magnesium ammonium phosphate Mg(NH4)PO4.6H2O. This is heated and decomposed to magnesium pyrophosphate. Mg2P2O7 which is weighed. A solution of H2PO4– yielded 1.054 gm of (Mg2P2O7) which is weighed. What weight of NaH2PO4 was present originally?

Solution: NaH2PO4 + Mg2+ + NH4+ ⎯→ Mg(NH4)PO4.6H2O Mg2P2O7

As P atoms are conserved, apply POAC for P atoms, moles of P in NaH2PO4 = moles of P in Mg2P2IO7

⇒ 1 × moles of NaH2PO4 = 2 × 2 moles of Mg2P2O7

∴

∴ = 1.14 gm

Illustration 6: Pyrolusite is an ore of manganese and it contains 80% by mass MnO2, 15% inert impurities and the rest is water. When MnO2 is heated it converts to Mn3O4. A sample of pyrolusite was heated till constant weight is obtained. Calculate the % of Manganese in the residue.

Solution: First of all think what is being asked. We have to calculate the % of Mn in the residue. This means that we must have the mass of manganese in the residue and the total mass of residue. Since the problem does not mention the mass of pyrolusite that was heated, then probably it was redundant. So, let us assume the mass of the sample to be x g.

Mass of MnO2 in pyrolusite = 0.8 x g.

⇒ Mass of inert impurities = 0.15 x g.

Now, think about what is meant by ‘heated till constant weight is obtained’. Well, the only thing it can mean is that the sample is heated due to which some changes took place in its weight and the sample is being heated till all the changes that were to occur have occurred. This certainly implies that all the water in the sample must have gone away.

When MnO2 is heated it converts to Mn3O4.

∴ 3MnO2→ Mn3O4 + O2 … (1)

We can see that to produce 1 mole of Mn3O4, one would need to heat 3 moles of MnO2.

Moles of MnO2 in the sample =

Moles of Mn3O4 produced =

Mass of Mn3O4 produced =  g.

Mass of inert impurities = 0.15 x g. (unchanged)

∴ Total mass of residue = g.

Now let us find the mass of Mn in the residue.

Moles of MnO2 in pyrolusite =

Moles of Mn in = moles of MnO2=

Since Mn is neither lost nor created in the conversion to Mn3O4, its no. of moles would remain constant.

∴ mass of Mn in the residue =

% Mn in the residue =

[Note: You may have noticed that the equation (1) is not balanced. To balance it , add O2 to the right hand side of the equation.

∴ we have, 3MnO2 → Mn3O4 + O2

Some of you may feel that if MnO2 is heated, the O2 should occur on the left hand side of the equation. It is not necessary that if something is heated, it would react with oxygen. It would react with oxygen only if the product has more % of oxygen than the reactant. In this case it is the other way round (36.78% O in MnO2, 27.94% O in Mn3O4)].

Illustration 7: Sodium chlorate, NaClO3, can be prepared by the following series of reactions,

2KMnO4   +  16HCl  2KCl  +  2MnCl2  +  8H2O  +  5Cl2

6Cl2   +  6Ca(OH)2   Ca(ClO3)2 +  5CaCl2  +  6H2O

Ca(ClO3)2 + Na2SO4  CaSO4 + 2NaClO3

What mass of NaClO3 can be prepared from 100 cc of concentrated HCl solution (density = 1.18 g/cc, 36% HCl by mass). Assume all other substances are present in excess amounts.

Solution: We need to find the mass of NaClO3 and we are given the amount of HCl.

100 cc of HCl solution would have a mass= 100 × 1.18 = 118 g.

Mass of HCl in it = 118 × = 42.48 g.

Moles of HCl =  = 1.1638

∴ Moles of Cl2 produced by the first reaction
= moles 

Moles of Ca(ClO3)2 produced by the second reaction
= moles

= 2 × moles of Ca(ClO3)2

= 2 × 0.0606 = 0.1212

∴ Mass of NaClO3 = 0.1212 × 106.5 gm = 12.9 g. 

6. Oxidation and Reduction

Oxidation is

1. i) the gain of oxygen
2. ii) the loss of hydrogen

iii) the loss of electrons (de-electronation)

1. iv) the increase of O.N.

Reduction is

1. i) the loss of oxygen
2. ii) the gain of hydrogen

iii) the gain of electron

1. iv) the decrease in O.N.

Fe ⎯→ Fe2+ + 2e–

O.N. = 0 +2

⇨ Fe loses electrons

⇨ There is increase in O.N.

⇨ Hence Fe is said to be oxidised

⇨ It is a source of electrons hence it can act as a reducing agent (R.A.) 

⇨ Any species that can be oxidised is a reducing agent (R.A.) 

Cu2+ + 2e ⎯→ Cu

O.N. = + 2 0

⇨ Cu2+ gains electron

⇨ There is decrease in O.N.

⇨ Hence Cu2+ is said be reduced.

Since it can gain electrons, hence it can act as an oxidising agent (O.A.)

⇨ Any species that can be reduced is an oxidising agent (O.A.), or oxidant. 

Above examples represent half reactions.

A complete reaction showing oxidation and reduction together is called a Redox reaction.

Now that you are clear on what is oxidation – reduction, we are now in the right position to know how to balance a redox (oxidation – reduction) reaction. This is important because many of the problems of stoichiometry would be based on such redox reactions.

6.1 Oxidation State / Oxidation Number

There are several chemical reactions in which oxidation – reduction takes place. Oxidation refers to a reaction in which electrons are removed from an atom and reduction refers to a reaction in which electrons are added to an atom. To describe these changes, the concept of oxidation state becomes necessary. For ionic species, the charge on each ion is said to be the oxidation state for that atom. For example in NaCl, Na exists as Na+ and Cl exists as Cl−. Therefore the oxidation state of Na in NaCl is +1 and that of Cl− is −1. But in covalent molecules, the charge on an atom would be so small that sometimes it becomes impossible to calculate the exact charge on each atom of a molecule. Therefore, the Oxidation State (O.S.) or Oxidation Number (O.N.) is defined as the charge, an atom would have in a molecule if all the bonds associated with this atom in the molecule are considered to be completely ionic. For example in H2O there are two O–H bonds. If we assume both the O–H bonds to be completely ionic, then each H would possess a charge of +1, while O possess a charge of −2. This is because oxygen is more electronegative than hydrogen. On the other hand, in H2O2 there are two O–H bonds and one O−O bond (H−O−O−H). Considering each O–H bond to be ionic both the oxygen atoms acquire a charge of −1 and both the H, +1. This is because O–O bond can not be assumed to be ionic as both the atoms have the same electronegativity.

To calculate the oxidation state of an element in a molecule you need not always know the structure of the molecule. There are certain set of rules used to assign oxidation states in polyatomic molecules:

1. The O.S.  of all elements (in any allotropic form) is zero. 
2. O.S. of oxygen is −2 in all its compounds except peroxides like H2O2 and Na2O2 where it is −1 and superoxides like KO2 where it is –½.   
3. O.S. of hydrogen  is +1 in all of its compounds except those with the metals where
   it is −1.
4. O.S. of all alkali metals is +1 and alkaline earth metals is +2 in all their compounds.
5. O.S. of all the halogens is −1 in all their compounds except where they are combined with an element of higher electronegativity. Since fluorine is the most electronegative of all elements, its O.S. is always –1.

Illustration 8: Calculate the O.S of all the atoms in the following species: 

(i) ClO−  (ii) (iii) (iv) CCl4 (v) K2CrO4 (vi) KMnO4

Solution: i) In ClO−, the net charge on the species is −1 and therefore the sum of the oxidation states of Cl and O must be equal to −1. Oxygen will have an O.S. of −2 and if the O.S. of Cl is assumed to be ‘x’ then x − 2 should be equal to −1.

∴ x is + 1

1. ii) : (2 × −2) + x = −1 (where ‘x’ is O.S. of N)

∴ x = +3

iii) : x + (3 × −2) = −1 (where ‘x’ is O.S. of N)

∴ x = +5

1. iv) In CCl4, Cl has an O.S. of −1

∴ x + 4 × −1 = 0

∴ x = +4 (where ‘x’ is O.S. of C)

1. v) K2CrO4: K has O.S. of +1 and O has O.S. of −2 and let Cr has O.S. ‘x’ then, 2 × +1 + x + 4 × −2 = 0

∴ x = + 6

1. vi) KMnO4: +1 + x + (4 × −2) = 0

∴ x = +7 (where x is O.S. of Mn).

6.2 Balancing Redox Equations

Some examples of redox reactions are

a)
b)
c)
d)

⇨ If one of the half reactions does not take place, other half will also not take place. We can say oxidation and reduction go side by side.

In this we find that Cl2 has been oxidised as well as reduced. Such type of redox reaction is called Disproportionation reaction. Examples are

                                                                                                       

How to Balance a Redox Reaction

6.2.1 Ion Electron Method

This method involves the following steps :

1. i) Divide the complete equation into two half reactions, one representing oxidation and the other reduction.
2. ii) Balance the atoms in each half reaction separately according to the following steps:
3. a) First of all balance the atoms other than H and O.
4. b) In a reaction taking place in acidic or neutral medium, oxygen atoms are balanced by adding molecules of water to the side deficient in oxygen atoms while hydrogen atoms are balanced by adding H+ ions to the other side deficient in hydrogen atoms. On the other hand, in alkaline medium (OH–), for every excess of oxygen atom on one side is balanced by adding one H2O to the same side and 2OH– to the other side. In case hydrogen is still unbalanced, then balance by adding one OH–, for every excess of H atom on the same side and one H2O on the other side.
5. c) Equalize the charge on both sides by adding suitable number of electrons to the side deficient in negative charge.

iii) Multiply the two half reactions by suitable integers so that the total number of electrons gained in one half reaction is equal to the number of electrons lost in the other half reaction.

1. iv) Add the two balanced half equations and cancel any term common to both sides.

There have been the common practices to balance the redox reaction by different methods like O.N. method and electron balance method. In the entrance examination it is never mentioned what method is to be used. We adopt here “quick” method that will certainly be a time-saving method.

Illustration 9: a) NO3– + H2S HSO4– + NH4+

1. b) Fe + N2H4 Fe(OH)2 + NH3

Solution: a) Step 1 N5+ + 8e ⎯→ N3–

S2– ⎯→ S6+ + 8e

Step 2 N5+ + S2– ⎯→ N3– + S6+

Step 3 NO3– + H2S ⎯→ NH4++ HSO4–

Step 4 NO other atom (except H and O) is unbalanced therefore no need for this step.

Step 5: Balancing O atom. This is made by using H2O and H+ ions. Add desired molecules of H2O on the side deficient in O atom and double H+ on opposite side. Therefore

NO3– + H2S + H2O ⎯→ NH4+ + HSO4– + 2H+

Step 6 Balance charge by H+

NO3– + H2S + H2O + 3H+ ⎯→ NH4+ + HSO4– + 2H+

∴ Balanced equation is  

NO3– + H2S + H2O + H+ ⎯→ NH4+ + HSO4–

1. b) Step 1 Fe ⎯→ Fe2+ + 2e

N22– + 2e ⎯→ 2N3–

Step 2 Fe + N22– ⎯→ Fe2+ + 2N3–

Step 3 Fe + N2H4 ⎯→ Fe(OH)2 + 2NH3

Step 4 No other atom (except H and O) is unbalanced and therefore no need for this step.

Step 5 Fe + N2H4 + 4OH– ⎯→ Fe(OH)2 + 2NH3 + 2H2O

Step 6 Balance charge by H+

∴ Fe + N2H4 + 4OH– + 4H+ ⎯→ Fe(OH)2 + 2NH3 + 2H2O

Finally Fe + N2H4 + 2H2O ⎯→ Fe(OH)2 + 2NH3

6.2.2 Oxidation Number Method 

This method is based on the principle that the number of electrons lost in oxidation must be equal to the number of electrons gained in reduction. The steps to be followed are :

1. i) Write the equation (if it is not complete, then complete it) representing the chemical changes.
2. ii) By knowing oxidation number of elements, identify which atom(s) is(are) undergoing oxidation and reduction. Write down separate equations for oxidation and reduction.

iii) Add required electrons on the right hand side of oxidation reaction and on the left hand side of reduction reaction. Care must be taken to ensure that the net charge on both the sides of the equation is same.

1. iv) Multiply the oxidation and reduction reactions by suitable numbers to make the number of electrons lost in oxidation reactions equal to the number of electrons gained in reduction reactions.
2. v) Transfer the coefficients of the oxidizing and reducing agents and their products to the main equation.

By inspection, arrive at the co-efficients of the species not undergoing oxidation or reduction.

Illustration 11: Balance  the equation K2Cr2O7 + HCl ⎯⎯→ KCl + CrCl3 + H2O + Cl2

Solution: Remember that alkali and alkaline earth metals have only one oxidation number and as long as they remain in the compound they do not undergo oxidation or reduction.

+ ⎯⎯→ KCl + + H2O +

Thus here Cr of K2Cr2O7 is reduced to CrCl3(+6→ +3) and Cl of HCl is oxidised  to Cl2(–1 → 0). In short.

Oxidation: 2Cl–1 ⎯⎯→ ↑ 2 … (a)

Reduction: ⎯⎯→ 2Cr+3 ↓ 6 … (b)

Step (iii) 2Cl–1 ⎯⎯→Cl20 + 2e–; Cr2+6 + 6e–  → 2Cr3+

Step (iv) Multiply equation (a) by 6 and (b) by 2 

12Cl–1 ⎯⎯→ + 12e–

+ 12e– ⎯⎯→ 4Cr+3

—————————————

2Cr2+6 + 12Cl–1 ⎯⎯→ 4Cr3 + 6Cl20

or Cr2+6 + 6Cl– ⎯⎯→ 2Cr+3 + 3Cl20

Step (v) K2Cr2O7 + 6HCl  ⎯⎯→ 2CrCl3 + 3Cl2

Step (vi) Making provision of KCl and H2O in the product: Since the reactant has 7 oxygen atoms in the product 7H2O must be present. For accounting 14 hydrogen atoms of water in the product, the reactants must have 12 HCl molecules (the only H containing species). For accounting the 2K atoms and 14 – 12  = 2 additional Cl atoms in the reactant, the product must have 2KCl. Hence the balanced equation is.

K2Cr2O7 + 14HCl ⎯⎯→ 2KCl + 2CrCl3 + 7H2O + 3Cl2

Exercise 1:

Exercise 2: Write the balanced equation when ferrous sulphate is treated with acidified (H2SO4) potassium permanganate.

Exercise 3: Balance the equation + + H+ → CO2 + Mn+2 + H2O

7. Volumetric Analysis

Volumetric analysis deals with the determination of the strength of a unknown solution by another solution of known strength (Titration). Titrations are of various types viz., acid base titration (neutralisation reaction), redox-titration, iodometric titrations, iodiometric titrations, precipitation reaction etc.

Method of Expressing Concentration of Solution

1. i) Molarity (M): The molarity of a solution is the number of moles of solute present  in one litre (1 dm3) of the solution 
2. ii) Normaltiy (N): The normality is the number of equivalents of solute present in one litre (1 dm3) of the solution

iii) The molality (m): The molality is the number of moles of solute present in one Kg of solvent 

1. iv) Relation between molarity and molality 

(where d = density of solution)

1. v) Relation between molarity and normaltiy

Normality = Molarity × n-factor

n-factor: Factor relating molecular weight and equivalent weight 

n-factor = . 

Equivalent weight =

The n-factors in case of acids and bases are basicity and acidity respectively, in case of salt is the charge on cation or anion, in case of ions, is the charge on the ions, in case of oxidising agent and reducing agents is the change of oxidation number per mole of the substance.

8. Law of Equivalence 

The fundamental basis of all titrations is the law of equivalence. According to which at the end point of titration, the volume of the two titrants reacted have same number of equivalents or milli equivalents.

Equivalent of solute = Normality × Volume in litre

⇒ Meq. = Normaltiy × Volume in mL

Also = Eq

⇒ Also moles of solute = Molarity × Volume in litre

⇒ Milli moles of solute = Molarity × Volume in ml

∴ = Moles

⇒ Meq. = × 1000 ⇒

⇒ For a general reaction

aA + bB ⎯→ mC + nD

Eq. of A = Eq. of B = Eq. of C = Eq. of D

Or Meq. of A = Meq. of B = Meq. of C = Meq. of D

⇒ In a compound AxBy, Eq. of AxBy = Eq. of A = Eq. of B

For the chemical change

aA + bB ⎯→ mC nE X

Meq. of A used = Meq. of B used = Meq. of C formed = Meq. of D used = Meq. of E formed = Meq. of X formed.

⇒ That is equivalent or meq. of reactants react in equal number to give the same number of equivalent or meq. of the products separately.

Mole and millimole react according to the equation. It is, therefore, advisable to solve numerical problems by equivalent or meq. rather than using mole or milli mole. For this purpose change molarity into normality.

While diluting a solution from one concentration to another

1. i) M1V1   =  M2V2

(conc.)   (dilute)

1. ii) N1V1 =  N2V2

(conc.) (dilute)

For acid-base (neutralization reaction) or redox reaction

N1V1  = N2V2 is always true

But M1V1 = M2V2 (may or may not be true)

But M1 × n1 × V1 = M2 × n2 × V2 (always true where n –terms represent n-factor).

8.1  ‘n’ Factor

1. a) In Non Redox Change: 
2. i) n-factor for element: Valency of the element 
3. ii) For Acids : Acids will be treated as species which furnish H+ ions when dissolved in a solvent. The n factor of an acid is the no. of acidic H+ ions that a molecule of the acid would give when dissolved in a solvent (Basicity). 

e.g. for HCl (n = 1), HNO3 (n =1), H2SO4 (n = 2), H3PO4 (n = 3) and H3PO3 (n=2) 

iii) For Bases: Bases will be treated as species which furnish OH− ions when dissolved in a solvent. The n factor of a base is the no. of OH− ions that a molecule of the base would give when dissolved in a solvent (Acidity).

e.g. NaOH (n = 1), Ba(OH)2 (n = 2), Al(OH)3 (n = 3), etc.

1. iv) For Salts: A salt reacting such that no atom of the salt undergoes any change in oxidation state. g.   2AgNO3 +  MgCl2   Mg(NO3)2 + 2AgCl

In this reaction it can be seen that the oxidation state of Ag, N, O, Mg and Cl remains the same even in the product. The n factor for such a salt is the total charge on cation or anion.

1. b) In Redox Change: For oxidising agent or reducing agent n-factor is the change in oxidation number per mole of the substance.

Illustration 12: Calculate the n-factor in the following chemical changes.

1. i) KMnO4 Mn2+
2. ii) KMnO4 Mn4+

iii) KMnO4 Mn6+

1. iv) K2Cr2O7 Cr3+
2. v) CO2
3. vi) FeSO4 Fe2O3

vii) Fe2O3   FeSO4

Solution: i) In this reaction, KMnO4 which is an oxidising agent, itself gets reduced to Mn2+ under acidic conditions. 

n =

1. ii) In this reaction, KMnO4 gets reduced to Mn4+ under neutral or slightly (weakly) basic conditions.

n =

iii) Here KMnO4 gets reduced to Mn6+ under basic conditions.

n =

1. iv) Here K2Cr2O7 which acts as an oxidising agent gets reduced to Cr+3 under acidic conditions. (It does not react under basic conditions).

n =

1. v) (oxalate ion) gets oxidised to CO2 when it is reacted with an oxidising agent.

n =

1. vi) Ferrous ions get oxidised to ferric ions

n =

vii) Here ferric ions are getting reduced to ferrous ions

n =

Illustration 13: Predict the molar ratio in which the following two substances would react if they are assumed to be salts of category (a)

Ba3(PO4)2 and AlCl3

Solution: n factor of Ba3 (PO4) = 3 ×(+2) = 6  = n1

While n factor of AlCl3 = 1 ×(+3) = 3 = n2        

If  

molar ratio = (inverse of equivalent ratio)

∴ molar ratio in which Ba3(PO4) and AlCl3 will react = 3:6 = 1:2

Illustration 14: Find the n- factor for the reactants in each of the following cases: 

1. i) I− ⎯→    I2
2. ii) I2 ⎯→ I−

iii)  ⎯→ 

1. iv) ⎯→ I+
2. v) CuS ⎯→ Cu2+  +  SO2

Solution: i) n factor = 1 × |0–(–1)| = 1

1. ii) n factor = 2 ×| –1–0| = 2

iii) n factor = 2×|2.5–2| = 1

1. iv) n factor = 1 ×|1–5| = 4
2. v) n factor =  1 ×|4–(–2)| = 6

Illustration 15: Calculate the n-factor in the following redox change 

FeC2O4   Fe3+  +  CO2

Solution: Here Fe2+ is getting oxidised to Fe3+ and C3+ is getting oxidised to C4+. The n factor w.r.t. Fe is +1 and w.r.t. is 2 (as calculated earlier). Therefore total n factor is 3.

Exercise 4: Calculate the n factor when FeCr2O4 gets oxidised to Fe3+ and Cr3+ and Cu2S gets oxidised to Cu2+ and SO2. 

8.2 Applications of the Law of Equivalence

8.2.1 Simple titration 

In this, we can find the concentration of a substance with the help of the conc. of another substance which can react with it.

For example: Let there be a solution of a substance A of unknown concentration. We are given another substance B whose concentration is known (N1). We take a certain known volume of A in a flask (V2) and then we add B to A slowly till all the A is consumed by B. (This can be known with the help of indicators). Let us assume that the volume of B consumed is V1. According to the Law of equivalents, the number of gm equivalents of A is equal to the number of gm equivalents of B.

∴  N1V1 = N2V2,  where N2 is the conc. of A.

From this we can calculate the value of N2.

8.2.2 Back titration 

Back titration is used to calculate % purity of a sample. Let us assume that we are given an impure solid substance C weighing w gms and we are asked to calculate the percentage of pure C in the sample. We will assume that the impurities are inert. We are provided with two solutions A and B, where the concentration of B is known (N1) and that of A is not known. This type of titration will work only if the following conditions are satisfied (a) A, B and C should be such compounds that A and B can react with each other, A and C can react with each other but the product of A and C should not react with B. 

Now we take a certain volume of A in a flask (the A taken should be such that the gm equivalents of A taken should be ≥ gm equivalents of C in the sample. This can be done by taking A in excess). Now we perform a simple titration using B. Let us assume that the volume of B used is V1. In another beaker, we again take the solution of A in the same volume as taken earlier. Now, C is added to this and after the reaction is complete, the solution is being titrated with B. Let us assume that the volume of B used up is V2.

Gram equivalents of B used in the first titration = N1V1 

∴ gm. equivalents of A initially  =  N1V1

1. equivalents of B used in the second titration is N1V2

∴ gm. equivalents of A left in excess after reacting with C = N1V2

1. equivalents of A that reacted with C = N1V1 − N1V2
2. equivalents of pure C = N1V1 − N1V2.

If the n  factor of C is x, then the moles of pure C =

 ∴ the weight of C = × Molecular weight  of C.

∴ percentage of C =

8.2.3 Double titration 

The method involves two indicator (Indicators are substances that change their colour when a reaction is complete) phenolphthalein and methyl orange. This is a titration of specific compounds. Let us consider a solid mixture of NaOH, Na2CO3 and inert impurities weighing  w g. You are asked to find out the % composition of mixture. You are also given a reagent that can react with the sample, say, HCl along with its concentration (M1).

We first dissolve the mixture in water to make a solution and then we add two indicators in it, namely phenolphthalein and methyl orange. Now, we titrate this solution with HCl.

NaOH is a strong base while Na2CO3 is a weak base. So it is safe to assume that NaOH reacts with HCl first, completely and only then does Na2CO3 react.

NaOH  +  HCl     NaCl  +  H2O

Once NaOH has reacted, it is the turn of Na2CO3 to react. It reacts with HCl in two steps:

Na2CO3  +  HCl    NaHCO3  +  NaCl and then, 

NaHCO3  +  HCl   NaCl  +  CO2  +  H2O

As can be seen, when we go on adding more and more of HCl, the pH of the solution keeps on falling. When Na2CO3 is converted to NaHCO3, completely, the solution is weakly basic due to the presence of NaHCO3 (which is a weaker base as compared to Na2CO3). At this instant phenolphthalein changes colour since it requires this weakly basic solution to change its colour. Therefore, remember that phenolphthalein changes colour only when the weakly basic NaHCO3 is present. As we keep adding HCl, the pH again falls and when all the NaHCO3 reacts to form NaCl, CO2 and H2O the solution becomes weakly acidic due to the presence of the weak acid H2CO3(CO2 + H2O). At this instance methyl orange changes colour since it requires this weakly acidic solution to do so. Therefore, remember methyl orange changes colour only when H2CO3 is present.

Now, let us assume that the volume of HCl used up for the first and the second reaction, 

i.e. NaOH  +  HCl     NaCl  +  H2O and Na2CO3  +  HCl   NaHCO3 + NaCl be V1 (this is the volume of HCl from the beginning of the titration up to the point when phenolphthalein changes colour). Let the volume of HCl required for the last reaction, i.e., NaHCO3  +  HCl  →  NaCl + CO2  +  H2O be V2 (this is the volume of HCl from the point where phenolphthalein had changed colour upto the point when methyl orange changes colour). Then,

moles of HCl used for reacting  with NaHCO3 = moles of NaHCO3 reacted = M1V2 

moles of NaHCO3  produced by the Na2CO3 = M1V2

moles of Na2CO3 that gave M1V2 moles of NaHCO3 = M1V2

Mass of Na2CO3 = M1V2 × 106

% Na2CO3  = 

moles of HCl used for the first two reactions = M1V1

moles of Na2CO3 = M1V2

moles of HCl used for reacting with Na2CO3 = M1V2

moles of HCl used for reacting with only NaOH = M1V1 − M1V2

∴ moles of NaOH = M1V1 − M1V2

Mass of NaOH =

% NaOH =

8.2.4 Iodometric and Idiometric titration

The reduction of free iodine to iodide ions and oxidation of iodide ions to free iodine occurs in these titration’s.

I2 + 2e– ⎯→ 2I– (reduction)

2I– ⎯→ I2 + 2e– (oxidation)

These are divided into two types

1. a) Iodometric titration: In iodometric titrations, an oxidising agent in allowed to react in neutral medium or in acidic medium with excess of potassium iodide to librate free iodine.

KI + oxidising agent ⎯→ I2

Free iodine is titrated against a standard reducing agent usually with sodium thiosulphate Halogen, dichromates, cupric ion, peroxides, etc can be estimated by this method.

I2 + 2Na2S2O3 ⎯→ 2NaI + Na2S4O6

2CuSO4 + 4KI ⎯→ Cu2I2 + 2K2SO4 + I2

K2Cr2O7 + 6KI + 7H2SO4 ⎯→ Cr2(SO4)3 + 4K2SO4 + 7H2O + 3I2

1. b) Iodiometric titration: These are the titrations in which free iodine is used as it is difficult to prepare the solution of iodine (volatile and less soluble in water) it is dissolved in KI solution.

KI + I2 ⎯→ KI3 (Potassium triiodide)

This solution is first standardises before use with the standard solution of I2 substance such as sulphite, thiosulphate, arsenite etc. are estimated.

The iodimetric and iodometric titrations, starch solution is used as indicator. Starch solution gives blue or violet colour with free iodine. At the end point the blue or violet colour disappears when iodine is completely changed to iodide.

9. Volume of Strength of H2O2

x  volume of H2O2 means x litre of O2 is liberated by 1 litre of H2O2 on decomposition 

2H2O2 ⎯⎯→ 2H2O + O2 

68 gm 22. 4 lit at STP

22.94lit (at STP) of O2 is given by 68 gm of H2O2

x litre of O2 is released from   gm of H2O2  =

∴ x lit of O2 is given by = (strength)

Strength, S =

Normality = = = [ x = N × 5.6]

Molarity = Normality =

Illustration 16: 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. 

Solution: 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 

Illustration 17: 1.20 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100 ml of a solution. 20 ml of this solution required 40 ml of 0.1 N HCl for complete neutralisation. Calculate the weight of Na2CO3 in mixture. If another 20 ml of this solution  is treated with excess of BaCl2 what will be the weight of the precipitate?

Solution: Let weight of Na2CO3 = x gm

Weight of K2CO3 = y gm

∴ x + y = 1.20 gm …(1)

For neutralisation reaction of 100 ml of 

Meq. of Na2CO3 + Meq. of K2CO3 = Meq. of HCl

⇒

∴ 69x + 53y = 73.14 …(2)

From equation (1) and (2)

x = 0.5962 gm

y = 0.604 gm

Solution of Na2CO3 and K2CO3 gives ppt. of BaCO3 with BaCl2

(meq. of Na2CO3 + Meq. of K2CO3) in 20 ml = meq. of BaCO3

⇒ Meq. of HCl for 20 ml mixture = Meq. of BaCO3

⇒ Meq. of BaCO3 = 40 × 0.1 = 4

= 40 × 0.1 = 4

= 4

∴ = 0.394 gm

Illustration 18: What is the weight of sodium bromate and molarity of solution to prepare 85.5 mL of 0.672 N solution when half cell reaction is

BrO3– + 6H+ + 6e ⎯→ Br– + 3H2O

Solution: Meq. of sodium bromate = 85.5 × 0.672 = 57.456

Let weight of sodium bromate be w

Eq. wt. of sodium bromate = (as n-factor = 6)

∴ × 1000 = 57/456,  = 1.446 gm

As molarity × n-factor = Normality

∴ Molarity = = 0.112M

Illustration 19: A solution 0.1 M KMnO4 is used for the reaction

S2O32– + 2MnO4–+ H2O ⎯→ MnO2 + SO42– + OH–

What volume of solution in mL will be required to react with 0.158 gm of Na2S2O3

Solution: Redox changes are: Mn7+ + 3e ⎯→ Mn+4

S22+ ⎯→ 2S6+ + 8e

Meq. of KMnO4 = Meq. of S2O32–  

Let volume of KMnO4 required = V

⇒ V × 0.1 3 = ( = 158, n-factor = 8)  

  ∴ V = 26.67 ml n-factor for MnO4– = 3)

Illustration 20: A sample of MnO4.4H2O is strongly heated in air. The residue Mn3O4 left was dissolved in 100 ml of 0.1 N FeSO4 containing dil. H2SO4. This solution was completely reacted with 50 ml of KMnO4 solution. 20 ml of this KMnO4 solution was completely reduced by 30 ml of 0.1 N FeSO4 solution. Calculate the amount of MnSO4.4H2O in sample.

Solution: MnSO4 + H2O Mn3O4

Redox change 3Mn2+ ⎯→ Mn38/3+ 2e

The residue Mn3O4 is titrated by another FeSO4. The normality of KMnO4 is determined by FeSO4.

Meq. of KMnO4 = Meq. of FeSO4

25 × N = 30 × 0.1 ∴ N =

Meq. of FeSO4 added to Mn3O4 = 100 × 0.1 = 10

Meq. of FeSO4 left after reaction with Mn3O4

= Meq. of KMnO4 used = = 6

∴ Meq. of FeSO4 used for Mn3O4 = 10 – 6 = 4

∴ Meq. of Mn3O4 = 4

∴ Meq. of MnSO4.4H2O = 4

3MnSO4.4H2O ⎯→ (n-factor = 8/3 – 2 =)

∴ Let weight of hydrated salt be w and molecular weight = M

∴ × 1000 = 4 (4s × n × 1000 = Meq.) ⇒ × 1000 = 4

∴ w = 1.338 gm

Illustration 21: 25 ml of H2O2 solution were added toe excess of acidified solution of KI. The iodine so liberated required 20 ml of 0.1 N Na2S2O3 for titration. Calculate the strength of H2O2 in terms of normality and volume strength.

Solution: Redox change is

O2–1 + 2e ⎯→ 2O2–

2I– ⎯→ I2 + 2e

2S22+ ⎯→ S4+5/2 + 2e

I2 + 2e ⎯→ 2I–

Meq. of H2O2 = Meq. of I2 = Meq. of Na2S2O3 = 20 × 0.1 = 2

∴

Volume strength (x) = N × 5.6

∴ Volume strength = × 5.6 = 0.448 volume

Exercise 5: 0.2M KMnO4 solution completely reacts with 0.05 M FeSO4 solution under acidic conditions. The volume of FeSO4 used is 50 ml. What volume of KMnO4 was used?

Exercise 6: 10 g. of a sample of Ba(OH)2 is dissolved in 10 ml. of 0.5N HCl solution. The excess of HCl was titrated with 0.2N NaOH. The volume of NaOH used was 20 cc. Calculate the percentage of Ba(OH)2 in the sample.

Exercise 7: 0.2 gm  of a solution of mixture of NaOH and Na2CO3 and inert impurities was first titrated with phenolphthalein and N/10 HCl 17.5 ml of HCl was required at the end point. After this methyl organge was added and 2.5 ml of same HCl was again required for next end point. Find out percentage of NaOH and Na2CO3 in the mixture.

Exercise 8: 50 ml of an aqueous solution of H2O2 was treated with an excess of KI solution and dilute H2SO4. The liberated iodine required 20 ml 0.1 N. Na2S2O3 solution for complete intersection.

Exercise 9: 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.

10. Solution to Exercises 

Exercise 1: Effected atoms are balanced

Cross multiplication

+ 3H2O2 + 4H2O ⎯→ 2MnO2 + 3O2 Add H2O on L.H.S. (excess O-side)

+ 3H2O2 + 4H2O ⎯→2MnO2 + 3O2 + 8OH–  Add double OH– on RHS

We have proceeded in a right way, but we have not considered H due to H2O2. Hnece H-atoms as well as charge are not balanced. Hence add 6H+ on R.H.S. which will neutralise OH– forming 6H2O and thus final result will be

+ 3H2O ⎯→ 2MnO2 + 3O2 + 2OH– + 2H2O

Exercise 2: Step (i) KMnO4 + H2SO4 + FeSO4 → K2SO4 + MnSO4 + Fe2(SO4)3 + H2O

Step-(ii) + H2SO4 + → K2SO4++ + H2O

Thus here Mn+7  is reduced to Mn2+ (+ 7 → +2) and Fe+2 is oxidised to Fe+3 (+2→+3).

Thus

Oxidation: 2Fe+2 ⎯⎯→ ↑ 2

Reduction: Mn+7 ⎯⎯→ Mn+2 ↓5

Step (iii) 2Fe+2 ⎯⎯→ Fe2+3 + 2e–

Mn+7 + 5e– ⎯⎯→ Mn+2

Step (iv) Multiplying equation (a) by 5 and (b) by 2.

10Fe+2 ⎯⎯→ 5Fe2+3 + 10e–

2Mn+7 + 10e– ⎯⎯→ 2Mn+2

Adding, 10Fe+3 + 2Mn+7  ⎯⎯→ 5Fe2+3 + 2Mn+2 + H2O

Thus the required equation may be written as 

10FeSO4 +2KMnO4+H2SO4 → 5Fe2(SO4)3 +2MnSO4+K2SO4+ H2O

Step (vi)  to balance ions; multiply H2SO4 by 8

10 FeSO4 + 2KMnO4 + 8H2SO4 → 5Fe2(SO4)3 + 2MnSO4 + K2SO4 + H2O

Exercise 3: + + H+ ⎯⎯→ +  + H2O

Thus there Mn+2 is reduced to Mn+2 (+7 → +2) and carbon in   is oxidised to CO2 

Thus here Mn+2 is reduced to Mn+2 (+7 → +2) and carbon in C2O4–2 is oxidised to CO2.

Thus,

Oxidation: ⎯⎯→ + 2e– ↑ 2

Reduction: Mn+7 + 5e– ⎯⎯→ Mn+2 ↓ 5

⎯⎯→ 10 CO2 + 10e–

2Mn+7+ 10e– ⎯⎯→ 2Mn+2

——————————————

Adding + 2Mn+7 ⎯⎯→ 10CO2 + 2Mn+2

+ ⎯⎯→ 2Mn+2 + 10CO2

Making provision of H+ on the L.H.S since these are given in the required reaction. It must be 16H+ because R.H.S has 8H2O. Thus the balanced equation is 

+ + 16H+ → 2Mn+2 + 10CO2 + 8H2O

Exercise 4: FeCr2O4 ⎯→ Fe3+ + Cr+3   

n factor = 1 (only Fe++ oxidises )

Cu2S ⎯→ Cu2+ + SO2

n factor = 2 (due to Cu+) + 6 (due to S– – )= 8

Exercise 5: 0.25V = 0.0550

V = 2.5 ml.

Exercise 6: Milli eq. of HCl initially = 10 × 0.5 = 5

Milli eq. of NaOH consumed = Milli eq. of HCl in excess = 10 × 0.2 = 2

∴ Milli eq. of HCl consumed =  Milli eq. of Ba(OH)2 = 5-2=3

∴ eq. of Ba(OH)2 = 3/1000 = 310-3

Mass of Ba(OH)2 = 310-3(171/2) = 0.2565 g.

% Ba(OH)2 = (0.2565/20) 100 = 1.28%

Exercise 7: Let W1  gm NaOH and W2 gm Na2CO3 was present in mixture.
At phenophthalein end point.

= × 17.5 × 10–3  ———— (1)

At second end point following reaction takes place

Eq. of NaHCO3  = eq. of HCl used ( in second titration) 

=   eq. of Na2CO3 

× = 25 × × 10–3 

W2 = 0.0265 gm 

Putting the value of W2 in equation (1), we get  

W1 = 0.06 gm 

Percentage of NaOH = × 100  = 30%

Percentage of Na2CO3 = × 100 = 13.25%

Exercise 8: H2O2 + 2KI + H2SO4 →K2SO4 + 2H2O + I2

2Na2S2O3 + I2 → Na2S4O6 + 2NaI

Equation mass H2O2 = = 17

20 ml 0.1 N Na2S2O3 = 20 ml 0.1 N I2 solution.

= 20 ml 0.1 N H2O2 solution

Amount of H2O2 in 50 ml.

Exercise 9: 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

 

11. Solved Problems 

 

11.1 Subjective 

Problem 1: 0.05 gm of commercial sample of KClO3 on decomposition liberated just sufficient oxygen for complete oxidation of 20 ml CO at 27°C and 750 mm pressure. Calculate % of KClO3 in sample

Solution: Applying PV = nRT for CO

Moles of CO (nCO) = = 8.01 × 10–4

CO + O2 ⎯→ CO2,  = 2

∴ moles of O2 required =

2KClO3 ⎯→ 2KCl + 3O2

As 3 moles of O2 are given by 2 mole of KClO3

∴ mol of O2 are given by = mol of KClO3

= 2.66 × 10–4 mol of KClO3

∴ Weight of KClO3 = 2.66 ×10–4 × 122.5 gm = 3.27 × 10–2 gm

∴ % of KClO3 in the mixture = × 100 = 65.4%

Problem 2: Upon mixing 45.0 ml of 0.25 M lead nitrate solution with 25 mL of 0.10 M chromic sulphate solution precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also calculate the concentrations of the species left behind in final solution. Assume that lead sulphate is completely insoluble.

Solution: Pb(NO3)2 + Cr2(SO4)3 ⎯→ PbSO4↓ + Cr(NO3)2

Meq. (initial) 45 × 0.25 × 2 25 × 0.1 × 6 0 0

= 22.5 = 15 (limiting reagent)

Meq. (final) 7.5 0 15 15

Meq. of PbSO4 precipitated = 15

Moles of PbSO4 = (n-factor=2) (Cr2(SO4)3 is the limiting reagent)

= 7.5 × 10–3

[Pb2+] = = 5.36 × 10–2 M

[NO3–] = = 0.32 M

[Cr3+] = = 7.14 × 10–2 M

Problem 3: A solution contains mixture of H2SO4 and H2C2O4. 25 ml of this solution requires 35.5 ml of N/10 for neutralization and 23.45 ml of N/10 KMnO4 for oxidation, calculate.

1. i) Normality of H2C2O4 and H2SO4
2. ii) Strength of H2C2O4 and H2SO4 molecular weight of H2C2O4 = 126

Solution: Meq. of H2SO4 + Meq. of H2C2O4 = Meq. of NaOH

= 3.55 × = 3.55 …(1)

Meq. of H2C2O4 = Meq. of KMnO4

= 23.45 × = 2.345

∴ Meq. of H2SO4  = 3.55 – 2.345 = 1.205

∴ = 0.0482

Strength = N × Eq. wt. = 0.0482 × = 2.362 gm/litre

Meq. of H2C2O4 = 2.345 (in 25 ml) (n-factor = 2)

∴ = 0.0938 H2C2O4 ⎯→ 2CO2

∴ Strength = N × Equivalent weight

= 0.0938 × (n-factor = 2) = 5.909 gm

Problem 4: A definite amount of BaCl2 was dissolved in HCl solution of unknown normality 20 ml of this solution was treated with 21.4 ml of N/10 NaOH for complete neutralization. Further 20 mL of solution was added to 50 ml of N/10 Na2CO3 and precipitate was filtered off. The filtrate reacted with 10.5 ml of 0.8 N/10 H2SO4 in the presence of phenolphthalein as indicator. Calculate strength of BaCl2 and HCl in the mixture.

Solution: Meq. of HCl in 20 ml solution = Meq. of NaOH

⇒ 20 × N = 21.4 × = 2.14

∴ NHCl = = 0.107

Strength of HCl = 0.107 × 36.5 = 3.9055 gm/litre

Meq. of Na2CO3 added to 20 ml solution = 50 × = 5

Meq. of Na2CO3 left after reaction in between BaCl2  +NaCO3 and (HCl + Na2CO3)

= 2 × Meq. of H2SO4 using phenolphthalein as indicator

( Meq. of Na2CO3 left = Meq. of H2SO4)

∴ Meq. of Na2CO3 left = 2 × Meq. of H2SO4)

= = 1.6

∴Meq. of Na2CO3 usedf for (HCl + BaCl2)

= Meq. of Na2CO3 added – Meq. of Na2CO3 left

= (5 – 1.68) = 3.32

∴ Meq. of HCl + Meq. of BaCl2 = 3.32

∴ Meq. of BaCl2 = (3.32 – Meq. of HCl)

= 3.32 – 2.14 (meq. of of HCl in 20 ml) = 3.32 – 2.14 = 1.18

× 2 × 1000 = 1.18 (n-factor for BaCl2 =2)

∴ in 20 ml = 0.122 of gm

∴ in 1 litre = × 1000 = 6.135 gm/lit

Problem 5: 1.6 gm of pyrolusite was treated with 50 mL of 0.5 M oxalic acid and some sulphuric acid. The oxalic acid left undecomposed was raised to 250 ml in a flask. 25 ml this solution when treated with 0.02 M KMnO4 required 32 mL of the solution. Find the % of MnO2 in the sample and also the percentage of available oxygen.

Solution: Meq. of MnO2 = Meq. of oxalic acid taken – Meq. of oxalic acid left

= 50 × 0.5 × 2 – 32 × 0.02 × 5 × 10 (in 250 ml)  = 18

Redox changes are C2O42– ⎯→ 2CO2 (n-factor = 2)

MnO4– ⎯→ Mn2+ (n-factor = 5)

MnO2 ⎯→ Mn2+ (n-factor = 2)

⇒ , ∴ = 0.7821 gm

∴ % of MnO2 = × 100 = 48.88%

Meq. of MnO2 = Meq. of O2 

× 2 × 1000 = 18, ∴ = 0.144 g

% of available O2 = ×100 = 9

Problem 6: 2.80 gm of KClO3 are dissolved in conc. HCl and solution was boiled chlorine gas evolved in the reaction was them passed through a solution of KI and I2 liberated was titrated with 100 ml of hypo. 12.3 ml of same hypo solution required 24.6 ml of 0.5 N iodine for complete neutralisation calculate % purity of KClO3 sample.

Solution: 2KClO3 + 12HCl ⎯→ 2KCl + 6H2O + 6Cl2

Cl2 + 2KI ⎯→ 2KCl + I2

Meq. of hypo solution can be calculated as follows

NHypo × 12.3 = 24.6 × 0.5

∴ NHypo = 1

Meq. of KClO3 = Meq. of Cl2 produced = Meq. of I2 produced  = Meq. of hypo reacted

= 100 × 1 = 100

× 6 × 1000 = 100 (n-factor for KClO3 = 6)

∴ = = 2.042

∴ % of KClO3 = × 100 = 82.32%

Problem 7: A mixture of Na2C2O4 and KHC2O4.H2C2O4 required equal volumes of 0.1 M KMnO4 and 0.1 M NaOH separately. What is the molar ratio of Na2C2O4 and KHC2O4.H2C2O4 in the mixture.

Solution: Let moles of Na2C2O4 and KHC2O4.H2C2O4 be x and y respectively

NaOH reacts with H+ and KMnO4 reacts with C2O42–

(2x + 2y + 2y) × 1000 = V × 0.1 × 5 …(1)

(y + 2y) × 1000 = V × 0.1 …(2)

C2O42– ⎯→ 2CO2 (n-factor = 2)

For KHC2O4, n-factor = 1 in neutralization reacts

For H2C2O4, n-factor = 2  in redox reacts

From equation (1) and (2)

= 5

∴ 2x + 4y = 15y

∴ 2x = 11y

∴ = 5.5

∴ x : y : : 1 : 5.5

Problem  8: 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 of KBrO3 was back titrated with 5.1 ml of M/25 solution of NaAsO2. The reactions are given below: (Atomic mass of Se = 79).

1. a) SeO3−2 + BrO3− + H+ → SeO4−2 + Br2 + H2O
2. b) BrO3− + AsO2− + H2O → Br− + AsO4−3 + H+

Solution: 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 [   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 

Problem 9: 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 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 10: A 2.18 g sample containing a mixture of XO and X2O3. It takes 0.015 mole of K2Cr2O7 to oxidize the sample completely to form and Cr3+. If  0.0187 mole of  is formed, what is the atomic mass of X?

Solution: XO + K2Cr2O7 ⎯→ Cr+3 +

X2O3 + K2Cr2O7 ⎯→ 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

Problem 11: An aqueous solution containing 0.10 g KIO3 (formula wt = 214.0) was treated with an excess of KI solution. The solution was acidified with HCl. The liberated I2 consumed 45 ml of thiosulphate solution to decolourise the blue-starch-iodine complex. Calculate the molarity of the sodium thiosulphate solution.

Solution: Moles of KIO3 = = 0.00047

∴ Moles of I2 liberated from KIO3 = = 0.000235

Moles of KI reacting = 0.00047 × 5 = 0.00235 ( ‘n’ factor for KIO3 and KI are 5 ad 1 respectively)

Moles of I2 produced from KI = = 0.001175

Total moles of I2 produced and reacted = 0.000235 + 0.001175 = 0.00141 equivalents of I2  reacted = 0.00141 × 2 = 0.00282 = equivalents of thiosulphate solution.

Molarity = = 0.063 M (For thiosulphate ‘n’ factor = 1)

Problem 12: A 10 g sample of only CuS and Cu2S was treated with 100 ml of 1.25 M K2Cr2O7. The products obtained were Cr3+, Cu2+ and SO2. The excess oxidant was reacted with 50 ml of Fe2+ solution. 25 ml of the same Fe2+ solution required 0.875 M KMnO4 under acidic condition, the volume of which used was 20 ml. Find the % of CuS and Cu2S in the sample.

Solution: Equivalents of dichromate initially = = 0.75

Equivalents of Fe2+ in 25 ml = = 0.0875

Equivalents of Fe2+ in 50 ml = 0.0875 × 2 = 0.175

Equivalents of excess dichromate  = 0.175

∴ Equivalents of dichromate consumed by (CuS and Cu2S) 

= 0.75 – 0.175 = 0.575

If x g is the mass of CuS, the mass of Cu2S is (10 – x)g

× 8 = 0.575

∴ x = 5.74 gm

% CuS = × 100 = 57.4%

% Cu2S = 42.6%

Problem 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 

Solution: The equations required  are  H2O + KI + O3  ⎯⎯→ 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   [ ‘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

Problem 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.)

Solution: 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

Problem 15: 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.

Solution: 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  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%

 

11.2 Objective 

Problem 1: 0.5 gm of fuming H2SO4 (oleum) is diluted with water. This solution is completely neutralised by 26.7 ml of 0.4 N NaOH. The percentage of free SO3 in the sample is

(A) 30.6% (B) 40.6%

(C) 20.6% (D) 50%

Solution: Meq. of H2SO4 + Meq. of SO3 = Meq. of NaOH

∴ = 26.7 × 0.4 ∴ x = 0.103

∴ % of SO3 = = 20.6%

∴ (C)

Problem 2: The minimum quantity of H2S needed to precipitate 64.5 gm of Cu2+will be nearly.

(A) 63.5 gm (B) 31.75 gm

(C) 34 gm (D) 2.0 gm

Solution: Meq. of H2S = Meq. of Cu2+

∴ × 1000 = × 1000 ∴ = 34 gm

∴ (C)

Problem 3: 34 gm of hydrogen peroxide is present in 1120 ml of solution. This solution is called

(A) 10 volume solution (B) 20 volume

(C) 30 volume (D) 32 volume

Solution: Amount of H2O2 in one ml = gm

H2O2  ⎯→ H2O + O2

 34 gm of H2O2 gives 16 gm or 11200 ml of O2

∴ gm H2O2 = = 10 ml O2

∴ Volume strength of H2O2 = 10 volume

∴ (A)

Problem 4: 1.82 gm of a metal required 32.5 mL of N – HCl to dissolve it what is the equivalent weight of metal?

(A) 54 (B) 56

(C) 28 (D) 86

Solution: Meq. of metal = Meq. of HCl

× 1000 = 32.5 × 1

∴ EM = 56

∴ (B)

Problem 5: What volume at NTP of ammonia gas will be required to be passed into 30 mL of NH2SO4 to bring down the acid normality to 0.2N?

(A) 556.5 mL (B) 480.5 mL

(C) 537.6 mL (D) 438.4 mL

Solution: Meq. of H2SO4 (original) = 30 ×1 = 30

Meq. of H2SO4 after passing NH3 = 30 × 0.2 = 6

Meq. of H2SO4 reacted = Meq. of NH3

= 30 – 6 = 24

∴ × 1000 = 24, = 0.408 gm

∴at STP = 22.4 = 0.5376 lt

= 537.6 mL

∴ (C)

Problem 6: How many grams of NaHSO3 will be required to react with one litre of NaIO3 solution containing 5.8 gm of NaIO3  according to the reaction.

IO3– + 3HSO3– ⎯→ I– + 3H+ + 3SO42–

(A) 10.2 gm (B) 9.8 gm

(C) 9.14 gm (D) 8.2 gm

Solution: Meq. of NaHSO3 = Meq. of NaIO3

= N × V = × 6 × 1000 (n-factor for NaIO3 = 6) = 175.76

× 2 × 1000 = 175.76

∴ = = 9.14 gm

∴ (C)

Problem 7: A 5 mL solution of H2O2 liberates 0.508 gm of Iodine from an acidified KI solution. What is the volume strength of the H2O2 solution.

(A) 2.2% (B) 3.8%

(C) 4.48% (D) None of these

Solution: Meq. of H2O2 = Meq. of I2

⇒ ∴ = 0.06 gm

H2O2 ⎯→ H2O2 + O2

 34 gm H2O2 gives 11.2 litre of O2 at STP

∴ 0.068 gm gives = × 0.068 = 22.4 ml O2

∴ Volume Strength of H2O2 = = 4.48 volume

∴ (C)

Problem 8: 1.5 litre of a solution of normality N and 2.5 litre of 2M HCl are mixed together. The resultant solution had a normality 5. The value of N is

(A) 6 (B) 10

(C) 8 (D) 4

Solution: Eq. of 1.5 litre solution + Eq. of 2.5 litre solution = Eq. of resultant  of solution

⇒ 1.5 ×N + 2.5 × 2 = 4 × 5

∴ N = = 10

∴ (B) 

Problem 9: The volume of 0.25 M H3PO3 required to neutralise 25 ml of 0.03 M Ca(OH)2 is

(A) 1.32 mL (B) 3 mL

(C) 26.4 mL (D) 2.0 mL

Solution: Meq. of H3PO4 = Meq. of Ca(OH)2

⇒ V × 0.25 × 2 = 25 × 0.03 × 2 (H3PO3 is dibasic acid)

∴ V = = 3mL 

∴(B)

Problem 10: A 0.1097 gm sample of As2O3 required 26.10 mL of KMnO4 solution for its titration. The molarity of KMnO4 solution is

(A) 0.02 (B) 0.04

(C) 0.018 (D) 0.3

Solution:

Let molarity of KMnO4 solution be M

∴ Eq. of As2O3 = Eq. of KMnO4 solution

= (Equivalent weight As2O3 = )

molarity = 0.017M ≈ 0.018

∴ (C)

Problem 11: Borax in water gives

B4O72– + 7H2O ⎯→ 4H3BO3 + 2OH–

How many grams of Borax (Na2B4O7.10H2O) are required to prepare 50 ml of 0.2 M solution.

(A) 0.32 gm (B) 3.82 gm

(C) 0.28gm (D) None of these

Solution: The n-factor or valency factor of borax in the above reaction is 2

Meq. of Borax in 50 ml solution = 50 × 0.2 × 2 = 20

Let weight of borax required = W

× 2 × 1000 = 20

∴ = 3.82

∴(B)

Problem 12: 50 mL of a solution containing 1 gm each of Na2CO3, NaHCO3 and NaOH was titrated with N–HCl. What will be the titre reading when only phenolphthalein is used as indicator.

(A) 35 mL (B) 32.5 mL

(C) 24.5 mL (D) 34.4 mL

Solution: At the end point using phenolphthalein as indicator uses complete NaOH and Meq. of Na2CO3.

∴ Meq. of NaOH + meq. of Na2CO3 = Meq. of HCl

∴ V = 34.4 ml

∴ (D)

Problem 13: 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/mol

∴ 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 14: How many molecule 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 of it =

∴ 12 L of liquid CCl4 will contain = 12 × 1000 × 0.0103⇒ 0.744 × 1026 moles of it

∴ (B)

Problem 15: 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: =

x = 7

∴ (C)

12. Assignments (Subjective Problems) 

 

LEVEL –  I

1. A definite amount of NH4Cl was boiled with 100 mL of 0.8 N NaOH for complete reaction of NH4Cl. After the reaction, the reactant mixture containing excess of NaOH was neutralized with 12.5 mL of 0.75 NH2SO4. Calculate the amount of NH4Cl taken.
2. In a quality control analysis for sulphur impurity 5.6 gm steel sample was burnt in a stream of oxygen and sulphur was converted to sulphate by using H2O2 solution to which had been added 30 mL of 0.04 M NaOH. The equation for reaction is

SO2(g) + H2O2(g) + 2OH–(aq) ⎯→ SO42– + 2H2O(l)

22.48 ml of 0.624 M HCl was required to neutralize the base remaining after oxidation reaction. Calculate % of sulphur in given sample.

3. 2.480 gm KClO3 are dissolved in conc. HCl solution and chlorine gas evolved in the reaction was then passed through a solution of KI and liberated iodine was treated with 100 mL of hypo solution 12.3 mL of same hypo solution required 24.6 mL of 0.5 N iodine for complete neutralization. Calcualte % purity of KClO3 sample.
4. To a 25 ml H2O2 solution, excess of acidified solution of KI was added. The iodine liberated required 20 ml 0.3 N Na2S2O3 solution. Calculate the volume strength of H2O2.
5. 1 gm sample of KClO3 was heated under such conditions that a part of it decomposed according to the equation
6. i) 2KClO3 ⎯→ 2KCl + 3O2
7. ii) and the remaining underwent change according to the equation

4KClO3 → 3KClO4 + KCl

If the amount of O2 evolved was 146.8 ml at STP, calculate percentage by weight of KClO4 in the residue.

6. A solid mixture (5 gm) consisting of lead nitrate and sodium nitrate was heated below 600°C until the weight of the residue is constant. If the loss in weight is 28% find the amount of lead nitrate and sodium nitrate in the mixture.
7. A 8.0 g sample contained Fe3O4, Fe2O3 and inert materials. It was treated with an excess of aqueous KI solution in acidic medium, which reduced all the iron to Fe+2 ions. The resulting solution was diluted to 50.0 cm3 and a 10.0 cm3 of it was taken. The liberated iodine in this solution required 7.2 cm3 of 1.0 M Na2S2O3 for reduction to iodide. The iodine from another 25.0 cm3 sample was extracted, after which the Fe+2 ions was titrated against 1.0 M KMnO4 in acidic medium. The volume of KMnO4 solution used was found to be 4.2 cm3. Calculate the mass percentages of Fe3O4 and of Fe2O3 in the original mixture.
8. 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.
9. 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 2 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.
10. 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.

 

LEVEL – II

1. A sample of ferrous sulphate and ferrous oxalate was dissolved in dil. H2SO4. The complete oxidation of reaction mixture required 40 mL of KMnO4. After the oxidation, the reaction mixture was reduced by Zn and H2SO4. On again oxidation by same KMnO4, 25mL were required. Calculate the ratio of Fe in ferrous sulphate and oxalate.
2. A sample of hydrazine sulphate (N2H6SO4) was dissolved in 500 mL water. 10 mL of this solution was reacted with excess of FeCl3 solution and warmed to complete the reaction. Ferrous ions formed were estimated and it required 20 mL of KMnO4 solution. Estimate the amount of hydrazine sulphate in one litre solution.

Given that 4Fe3+ + N2H4 ⎯→ N2 + 4Fe2+ + 4H+

MnO4– + 5Fe2+ + 8H+ ⎯→ Mn2++ 5Fe3+ + 4H2O

3. In an experiment 10 litre of air at 1 atmosphere pressure and 27°C were passed through an alkaline KI solution. At the end the iodine entrapped in a solution reacted with 1.5 mL of 0.01 N Na2S2O3 solution. Calculate the % volume of O3 in the sample
4. 1 gm mixture containing equal number of moles of carbonates of two alkali metals required 44.4 ml of 0.5 N HCl for complete neutralisation. The atomic weight of one metal is 7 find the atomic weight of other metal. Also calculate the amount of sulphate formed on quantitative conversion of 1.0 gm of the mixture into sulphates.
5. 1 gm of mixture of Na2CO3 and K2CO3 was made upto 250 ml in aqueous solution was neutralised by 20 ml of HCl of unknown concentration. The neutralized solution required 16.24 ml of 0.1 N AgNO3 for precipitation. Calculate.
6. a) The % of K2CO3 in mixture
7. b) of HCl in gm/litre
8. c) Molarity of HCl
9. n-Butane is produced by the monobromination of ethane, followed by the Wurtz reaction. Calculate the volume of ethane at STP that would be required to produce 55 g of n-Butane if the monobromination occurs with 85% (by mole) yield and the Wurtz reaction occurs with 90% (by mole) yield.

The reactions are :

C2H6      +  Br2  C2H5Br  +  HBr (monobromination)

(Ethane)

2C2H5Br  +  2Na  C4H10 + 2NaBr (Wurtz reaction)

(Butane)

7. A sample weighing 0.3 g is suspected to contain either pure NaCl, or pure KCl or pure NH4Cl or a mixture of any of these two mixed in any proportion by weight or a mixture of all the three mixed in any proportion by weight. What minimum volume of 5% (by weight) of AgNO3 solution (specific gravity = 1.04) must be added to the sample to ensure complete precipitation of chloride  as AgCl in every possible case.
8. 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. 
9. 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)

10. 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.

 

LEVEL – III

1. Chile salt peter a source of NaNO3 also contains NaIO3. The NaIO3 can be used as a source of Iodine produced in the following reactions:

IO3– + 3HSO3– ⎯→ I– + 3H+ + 3SO42–

5I– + IO3– + 6H+ ⎯→ 3I2 + 3H2O

One litre of chile salt peter solution containing 5.80 gm NaIO3, is treated with stoichiometric quantity of NaHSO3. Now additional amount of same solution is added to the reaction mixture to bring about the second reaction. How many grams of NaHSO3 are required in step 1 and what additional volume of chile salt peter must be added in step II to bring in complete conversion of I– to I2?

2. 1 gm sample of AgNO3 is dissolved in 50 mL of water. It is titrated with 50 mL of KI solution. The AgI precipitated is filtered off. Excess of KI in filtrate is titrated with M/10 KIO3 in presence of 6 M HCl till all I– converted into ICl. It requires 50 mL of M/10 KIO3 solution. 20 mL of the same stock solution of KI requires 30 mL of M/10 KIO3under similar condition. Calculate % of AgNO3 in sample. The reaction is

KIO3 + 2KI + 6HCl ⎯→ 3ICl + 3KCl + 3H2O

3. CuSO4 reacts with KI in acidic medium to liberate I2

2CuSO4 + 4KI ⎯→ Cu2I2 + 2K2SO4 + I2

Mercuric periodate Hg5(IO6)2 reacts with a mixture of KI and HCl according to the following equation.

Hg5(IO6)2 + 4KI + 24HCl ⎯→ 5K2HgI4 + 8I2 + 24KCl + 12H2O

The liberated iodine is titrated against Na2S2O3 solution 1 ml of which is equivalent to 0.0499 gm of CuSO4.5H2O. What volume in mL of Na2S2O3 solution will be required to react with I2 liberated from 0.7245 gm of Hg5(IO6)2?

Molecular weight of Hg5(IO6)2 = 1448.5

Molecular weight of CuSO4.5H2O = 299.5

4. 30 mL of a solution containing 9.15 gm/litre of an oxalate KxHy (C2O4)z.nH2O are required for titrating 27 mL of 0.12 N NaOH and 36 mL of 0.12 N KMnO4 separately calculate x, y, z and n. Assume all H atom are replaceable and x, y, z are in the sample ratio of gm atoms.
5. A sample of Mg was burnt in air to give a mixture of MgO and Mg3N2. The ash was dissolved in 60 meq of HCl and the resulting solution was back titrated with NaOH. 12 Meq. of NaOH were required to reach the end point. An excess of NaOH was then added and the solution distilled. The ammonia released was then trapped in 10 meq. of acid. Back titration of this solution required 6 meq of base. Calculate the % of Mg burnt to the nitride.
6. 1.0 gm of moist sample of mixture of potassium chlorate (KClO3) and potassium chloride (KCl) was dissolved in water and solution made upto 250 ml of this solution was treated with SO2 to produce all ClO3– to Cl– and excess of SO2+ was removed by boiling. The total chloride was precipitated as silver chloride. The weight of precipitate was found to be 0.1435 gm. In another experiment 25 ml of the original solution was heated with 30 ml 0.2 N FeSO4 and unused FeSO4 required 37.5 ml of 0.08 N KMnO4 solution. Calculate the molar ratio of the ClO3– to the Cl– in the given mixture. 

Given that ClO3– + 6Fe2+ + 6H+ ⎯→ Cl– + 6Fe3+ + 3H2O

3SO2 + ClO3– + 3H2O ⎯→ Cl– + 3SO42– + 6H+

7. 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 
8. 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.
9. 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.
10. 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.
11. 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.
12. 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.

13. 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.

14. 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?
15. 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.

 

10. Assignments  (Objective Problems)

LEVEL –  I

1. 22.7 ml of N/10 Na2CO3 solution neutralises 10.2 ml of a dilute H2SO4. Then the volume of  water that must be added to 400 ml of same H2SO4 to make
   it exactly N/10 is 

(A) 245 ml (B) 484.6 ml

(C) 480 ml (D) 490.2 ml

2. The sulphate of a metal M contains 20% of M. This sulphate is isomorphous with ZnSO4.7H2O. The atomic weight of M is

(A) 12 (B) 24

(C) 36 (D) 48

3. 1.12 ml of a gas is produced at STP by the action of 41.2 mg of alcohol, ROH with methyl magnesium iodide. The molecular mass of alcohol is

(A) 16.0 (B) 41.2

(C) 84.2 (D) 156.0

4. Rearrange the following (I to IV) in the order of increasing maasses and choose the correct answer from (A), (B), (C) and (D) (Atomic mass of N = 14, O = 16, Cu = 63).

(I) 1 molecule of oxygen

(II) 1 atom of nitrogen

(III) 1 × 10–10 gm molecular weight of oxygen

(IV) 1 × 10–10 gm atomic weight of copper

(A) (II) <  (I) < (III) < (IV) (B) (IV) < (III) < (II) < (I) 

(C) (II) < (III) < (I) < (IV) (D) (III) < (IV) < (I) < (II)

5. One mole of calcium phosphide on reaction with excess of water gives

(A) One mole of phosphine (B) Two moles of phosphoric acid

(C) Two moles of phosphine (D) One mole of phosphorus pentaoxide

6. An aqueous solution of 6.3 gm oxalic acid dihydrate is made upto 250 ml. The volume of 0.1 N NaOH required to completely neutralise 10 ml of this solution is

(A) 40 ml (B) 20 ml

(C) 10 ml (D) 4 ml

7. If 0.5 mol of BaCl2 is mixed with 0.2 mol of Na3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed is

(A) 0.7 (B) 0.5

(C) 0.30 (D) 0.10

8. 0.5 mol of H2SO4 is mixed with 0.2 mole of Ca(OH)2. The maximum number of mole of CaSO4 formed is

(A) 5 (B) 0.5

(C) 0.4 (D) 1.5

9. The hydrated salt Na2SO4.xH2O undergoes 55% loss in weight on heating and becomes anhydrous. The value of x will be

(A) 5 (B) 3

(C) 7 (D) 10

10. The percent loss in weight after heating a pure sample of KClO3 (molecular
    weight = 122.5) will be

(A) 12.25 (B) 24.50

(C) 39.18 (D) 49.0

11. The normality of 0.3 M phosphorus acid (H3PO3) is

(A) 0.1 (B) 0.9

(C) 0.3 (D) 0.6

12. The number of moles of KMnO4 that will  be needed to react completely with one mole of ferrous oxalate in acidic medium is

(A) 3/5 (B) 2/5

(C) 4/5 (D) 1

13. The number of moles of KMnO4 that will be needed to react with one mole of sulphite ion in acidic solution is

(A) 2/5 (B) 3/5

(C) 4/5 (D) 1

14. In basic medium CrO42– oxidises S2O32– to form SO42– and itself change to Cr(OH)4–. How many ml of 0.154 M CrO42– are required to react with 400 ml of 0.246 M S2O32–?

(A) 200 ml (B) 156.4 ml

(C) 170.4 ml (D) 190.4 ml

15. 20 ml of 0.2 M Al2(SO4)3 is mixed with 20 ml of 0.6 M BaCl2. Concentration of Al3+ ion in the solution will be

(A) 0.2 M (B) 10.3 M

(C) 0.1 M (D) 0.25 M

16. The sodium salt of an acid dye contains 7% of sodium. What is the minimum molar mass of the dye.

(A) 336.5 (B) 286.5

(C) 300.6 (D) 306.5

17. The oxidation states of the most electronegative element in the products of the reaction, BaO2 with dil. H2SO4 are

(A) 0 and −1 (B) −1 and −2

(C) −2 and 0 (D) −2 and +1

18. If 5 litres of H2O2 produces 50 litres of O2 at NTP, H2O2 is 

(A) 50V (B) 10V

(C) 5V (D) 250V

19. 2.76 gm of Ag2CO3 on being strongly heated yields a residue weighing.

(A) 2.16 gm (B) 2.48 gm

(C) 2.32 gm (D) 2.64 gm

20. A solution of 0.1 M KMnO4 is used for the reaction

S2O32– + 2KMnO4– + M2O ⎯→ MnO2 + SO42– + OH–

What volume of solution in ml will be required to react with 0.158 gm of Na2S2O3

(A) 24.6 ml (B) 26.67 ml

(C) 23.5 ml (D) 31.5 ml

LEVEL –  II

1. The weight of sodium bromate required to prepare 85.5 ml of 0.672 N solution for cell reaction BrO3– + 6H++ 6e ⎯→ Br– + 3H2O is

(A) 1.56 gm (B) 1.45 gm

(C) 1.23 gm (D) 1.32 gm

2. NaIO3 reacts with NaHSO3 according to equation 

IO3– + 3HSO3– ⎯→ I– + 3H++ 3SO42–

The weight of NaHSO3 required to react with 100 ml of solution containing 0.58 gm of NaIO3 is

(A) 5.2 gm (B) 4.57 gm

(C) 2.3 gm (D) None of these

3. 25 ml of H2O2 solution were added to excess of acidified solution of KI and iodine so liberated required 20 ml of 0.1 N Na2S2O3 for titration. The normality H2O2 is?

(A) 0.02 (B) 0.04

(C) 0.08 (D) 0.03

4. A mixture of KBr and NaBr weighing 0.560 gm was treated with aqueous Ag+ and all the bromide ion was recovered as 0.970 gm of pure AgBr. The fraction by weight of KBr in the sample is

(A) 0.25 gm (B) 0.2378

(C) 0.36 (D) 0.285

5. The number of moles of Fe(OH)3 produced by allowing 1 mol of Fe2S3 2 moles of H2O and 3 moles of O2 to react according to the equation

2Fe2S3(s) + 6H2O(l) + 3O2(g) ⎯→ 4Fe(OH)3(s) + 6S(s) are

(A) 1.26 (B) 2.4

(C) 1.34 (D) 1.43

6. The volume in ml of 0.1 N HCl required to react completely with 1.0 gm mixture of Na2CO3 and NaHCO3 containing equimolar amounts of the two compounds is

(A) 157.9 ml (B) 152.6 ml

(C) 200 ml (D) 98.5 ml

7. 25.4 g of iodine and 14.2g of chlorine are made to react completely to yield a mixture of ICl and ICl3. Calcualte the ratio of moles of ICl and ICl3.

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

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

8. Calculate the weight of ion 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

9 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

10 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

11. 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

12. For the reaction 

⎯→

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

13. 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

14. 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

15. 8 g of sulphur is burnt to form 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

16 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 CaCO3 is heated to convert all the calcium to CaO  and the final mass of CaO is 1.62 gms. The  %  by mass of CaCl2 in the original mixture is 

(A) 15.2% (B) 32.1%

(C) 21.8% (D) 11.07%

17. 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  

18. The number of moles of  needed to oxidise 0.136 equivalents of N2H5+ by the reaction : N2H5+ + ⎯→ N2 + Cr3+ + H2O is 

(A) 0.136 (B) 0.272

(C) 0.816 (D) 0.0227

19. A sample of oleum is labelled 109%. The % of free SO3 in the sample is 

(A) 40% (B) 80%

(C) 60% (D) 9%

Oleum is SO3 + H2SO4

20. What volume of 0.3 N / H+ is needed for complete oxidation of 200 ml of 0.6 M FeC2O4 solution.

(A) 1.2 cc (B) 1.2 litre 

(C) 120 cc (D) 800 cc

 

11. Answers to Subjective Assignments

 

LEVEL – I

1. 3.78 gm 2. 6.1875%
2. 82.32% 4. 1.344
3. 49.8% 6. Pb(NO3)2 = 3.3246gm
4. FeO4 = 17.4% NaNO3 = 1.6754gm

Fe2O3 = 23.7% 8. 4.54 L

9. 72.5% 10. 14.7%, 80.9%

 

LEVEL – II

1. 7/3 2. 6.5 gm/lit
2. 1.847 × 10–3% 4. 0.609 gm, 0.7920gm
3. a) 60%; (b) 2.96 gm lit; (c) 0.0812 6. 55.5 lit
4. 18.33 ml 8. 11.6%, 3.57%
5. 88.2% 10. 14.2%

LEVEL – III

1. 9.14gm, 20 mL 2. 2.85%
2. 40 mL 4. x:p:2::1:3:2, n = 2
3. 27.27% 6. 1:1
4. 2.82%, 1.498% 8. 5.6%, 1.27%
5. 15.57% 10. 14.09%
6. 0.0224, 2 12. 0.265, 0.7g
7. 9.24% 14. 6058
8. 15.57%

 

7

12. Answers to Objective Assignments

 

LEVEL – I

 

1. D 2. C
2. C 4. A
3. C 6. A
4. D 8. A
5. D 10. D
6. D 12. A
7. A 14. C
8. A 16. D
9. B 18. B
10. A 20. B

 

LEVEL – II

1. B 2. B
2. C 4. B
3. C 6. A
4. A 8. A
5. A 10. B
6. D 12. A
7. B 14. A
8. D 16. B
9. B 18. D
10. A 20. B

 

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

 

6