Some Basic Concepts of chemistry

LAWS OF CHEMICAL COMBINATION

(i)    Law of conservation of mass: In all physical and chemical changes, the total mass of the reactants is equal to that of the products. In other words, matter can neither be created nor be destroyed. 
     

(ii)    Law of constant composition or definite proportion: A chemical compound is always found to be made up of the same elements combined together in the same fixed proportion by weight.
        For example, pure water obtained from whatever source or any country will always be made up of only hydrogen and oxygen elements combined together in the same fixed ratio of 1 : 8 by weight.
(iii)   Law of multiple proportions: When two elements combine to form two or more chemical compounds then the weights of one of elements which combine with a fixed weight of the other bear a simple ratio to one another.
        e.g. the weights of nitrogen and oxygen forming compound N₂O, NO, N₂O₃, N₂O₄, N²O₅, if we fix the weight of nitrogen as 14 then weight of oxygen bears a simple ratio 1 : 2 : 3 : 4 : 5 to one another.
(iv)   Law of reciprocal proportions: The ratio of the weights of two elements A and B which combine separately with a fixed weight of the third element C is either the same or some simple multiple of the ratio of the weights in which A and B combine directly with each other. For example, CO₂, SO₂ and CS₂, H₂S, H₂O and SO₂.
(v)    Gay Lussac’s law of gaseous volumes: When gases react together they always do so in volumes which bears a simple ratio to one another and to the volumes of the products, if these are also gases, provided all measurements of volumes are done under similar conditions of temperature and pressure. For example
        N₂ + 3H₂  →   2NH₃
       1vol. 3vol.        2vol.

AVOGADRO’S HYPOTHESIS

Equal volumes of all gases under similar conditions of temperature and pressure contains equal number of molecules.

Applications of Avogadro’s law

(i)    In the calculation of atomicity of elementary gases.
(ii)    To find the relationship between molecular weight and vapour density of a gas.
     Molecular weight = 2 x Vapour density
(iii)    To find the relationship between weight and volume of a gas.
    Molecular weight = Weight of 22.4 L of the gas at STP

MASS CONCEPTS

(i)    Atomic Mass: The atomic mass of an element is a number which indicates how many times an atom of natural isotopic composition of an element is heavier as compared with 1/12 of mass of a atom of carbon-12.
(ii)    Gram atomic mass: The atomic mass of an element expressed in grams is called Gram atomic mass e.g., 
        Atomic mass of oxygen = 16 a.m.u.
    Gram atomic mass of oxygen = 16 a.m.u
    Note: (1 a.m.u. = 1.6 x 10^-24 g) 
(iii)    Molecular mass: The molecular mass of a substance (element or compound) is the number of times, the molecule of substance is heavier than  of the mass of an atom of Carbon–12 isotope.
(iv)    Gram molecular mass: The molecular mass of a substance expressed in grams is called its Gram molecular mass.

1 gram molecule of H₂SO₄ = 98.0 g

(v)    Atomic Mass Unit: The quantity 1/12th mass of and atom of carbon-12 is known as atomic mass unit.
    
    A mole is defined as number of atoms in 12g of carbo-12, which is experimentally found to be 6.023 x 10²³.

Example 1 : In naturally occurring neon, isotopes and their relative abundance are as follows:

                  Isotope    Fractional Abundance
                    ^2oNe    0.9051
                    ²¹Ne    0.0027
                    ²²Ne    0.0922
               Average atomic mass of Ne is
              (A)20.179    (B)    21.179
              (C)22.179    (D)    20

Solution:    (A)
        Average atomic mass of Ne = 20 x 0.9051 + 21 x 0.0027 + 22 x 0.0922 = 20.179
        Aliter
       From the fraction abundance of various isotopes, clearly shows that the average atomic mass should be near to 20 (²⁰Ne). Hence, (A) is correct

MOLE CONCEPT

For counting of articles, units like dozen or score is commonly used. Similarly in chemistry ‘mole’ is the counting unit for chemical entities.
The amount of substance containing Avogadro’s number (6.023 x 10²³) atom, molecule, ions, electrons and protons, i.e.

(i)    Number of moles of molecule  
(ii)    Number of moles of atoms  
(iii)    Number of moles of gases  
(iv)    Number of milli moles  = molarity x volume in ml.
(v)    For a compound P_xQ_y
    x moles of Q = y moles of P
(vi)     

Example 2 :   

                   

Solution:    (D).
        Moles of SO2  
        Number of atoms  

Example 3 : The volume (in L) of CO₂ liberated at STP when 10 g of 90% pure lime stone is heated completely is
                     (A) 20.16    (B)    2.016
                     (C) 2.24    (D)    22.4
Solution:    (B).   CaCo₃ ⟶ CaO + CO₂
                    Moles of CaCO3  
                    Moles of CO2 = 0.09  
                    Volume at STP = 2.016 L

Example  4 :   44.8 litre of CO₂ at STP is obtained by heating x gm of pure CaCO₃. x is
                       (A)100 g    (B)    200 g
                       (C)50 g    (D)    44.8 g
Solution:     (B). 
         CaCO₃  CO₂ + CaO
        1 mol of CO₂ is obtained from 1 mole of CaCO₃
        22.4 litre of gas at STP = 1 mol of gas
        44.8 litre of gas at STP = 2 mol of gas
        Hence,
        2 mole of CO₂ is obtained from 2 mole of CaCO₃
        i.e., 2 x 100 = 200 gm of CaCO₃

DETERMINATION OF MOLECULAR FORMULAE

The percentage composition of a compound leads directly to its empirical formula. An empirical formula or simplest formula for a compound is the formula of a substance written with the smallest integer subscripts. The molecular formua of a compound is a multiple of its empirical formula.
(a)    Law of isomorphism: Isomorphism substances form crystal which have same structure. Thus both the cation and anion of isomorphous substance have the same valency.
(b)    Chemical formulae of a compound can be found out by the following steps:
    (i)    The percentage composition of compound is determined.
    (ii)    Then the percentage of each element is divided by the atomic mass and then a molar ratio of each element is obtained.
    (iii)    This molar ratio is divided by a minimum value to get the simplest ratio and an empirical formula is obtained.
    (iv)    Finally by comparing the molecular mass or vapour density of this empirical formulae to that of actual molecule its molecular formulae is obtained.

Example 5 :  An organic compound contains 49.3% carbon, 6.84% hydrogen 
                         and its vapour density is 73. Molecular formula of the 
                         compound is
                     (A) C₃H₈O₂    (B)    C₃H₁₀O₂
                     (C) C₆H₉O    (D)    C₆H₁₀O₄

Solution:    (D)    
        Step (i) 
            C = 49.3%, H = 6.84%
            O = 43.86%
        Step (ii)
            C : H : O  
        or    C : H : O = 4.11 : 6.84 : 2.74
        Step (iii)
            Dividing through out by 2.74
            C : H : O = 1.5 : 2.54 : 1
        or    C : H : O = 3 : 5 : 2
            Empirical formulae = C₃H₅O₂
        Empirical formula mass = 73
        Molecular mass of compound = 73 x 2 = 146
            Molecular formula 

Example 6 : A compound K₂XO₄ is isomorphous to potassium sulphate (K₂SO₄) and is found to contain 26.78% ‘X’. Calculate the mass of ‘X’?
                     (A) 22        (B)    26
                     (C) 52        (D)    44

Solution:    (C).
        Let the mass of ‘X’ can be m.

BALNCING REDOX REACTIONS

Quite often one has to write a balanced chemical equation while dealing with the problem on stoichiometry. After writing reactant and products, the balancing can be carried out by hit and trial method. However, systematic methods are available for balancing redox reactions. Before describing these methods, the rules governing the computation of oxidation state of an element are described in the following.

Rules to Compute Oxidation Number

The oxidation number of an element is the number assigned to it by following the arbitrary rules given below:
(i)    A free element (regardless of whether it exists in monatomic or polyatomic form, e.g. Hg, H₂, P₄ and S₈) is assigned an oxidation number of zero.
(ii)    A free monatomic ion is assigned an oxidation number equal to the charge it carries. For example, the oxidation number of Al³⁺, S^2- and Cl^- are +3, -2 and -1 respectively.
(iii)    In their compounds, the alkali and alkaline earth metals are assigned oxidation numbers of +1 and +2 respectively.
(iv)    The oxidation number of hydrogen in its compounds is generally +1 except the ionic hydrides such as LiH, LiAlH₄ where its oxidation number is -1.

(v)    The oxidation number of fluorine in all its compounds is -1. The oxidation number of all other halogens is -1 in all compound except those with oxygen (e.g.CIO^-₄) and halogens having a lower atomic number (e.g. ICI^-₃ ). The oxidation number of the latter is determined via oxygen and halogen of lower atomic number.
(vi)    The oxidation numbers of both oxygen and sulphur in their normal oxides (e.g. Na₂O) and sulphides (e.g. CS₂) is -2. The exception are the peroxides (e.g. H₂O₂ and Na₂O₂), superoxides (e.g. KO₂) and the compound OF2. Their oxidation numbers are determined by the rules 3, 4 and 5.
(vii)    The algebraic sum of oxidation numbers of atoms in a chemical species (compound or ion) is equal to the net charge on the species.
A few examples of the computation of oxidation number of atoms N in various compounds are given as follows.
If x is the oxidation number of N, we have
NH₃          x + 3(+1) = 0                    which gives x = -3
HN₃          +1 + 3(x) = 0                    which gives x = -1/3
N₂H₄        2x + 4(+1) = 0                  which gives x = -2
NO₂          x + 2(-2) = 0                    which gives x = 4
N₂O₄        2x + 4(-2) = 0                  which gives x = 4
NO^-₂         x + 2(-2) = -1                   which gives x = 3
NH₂OH    x + 2(+1) + 1(-2) + 1 = 0            which gives x = -1
NO           x + 1(-2) = 0                     which gives x = 2
HNO₃      +1 + x + 3(-2) = 0             which gives x = 5
H₂O         2x + 1(-2) = 0                   which gives x = 1
HCN        +1 + 4 + x = 0                  which gives x = -5

Balancing Redox Reactions Via Oxidation Numbers

The steps involved in this method are as follows:
1.    Identify the elements in the unbalanced equation whose oxidation number are changed.
2.    Balance the number of atoms of each element whose oxidation number is changed.
3.    Find out the total change in oxidation number for each of oxidant and reductant and make them equal  by multiplying by small coefficients.
4.    Balance the remainder of atoms by inspection and add, if necessary H⁺(acidic medium) or OH^-(alkaline medium) or H₂O (to balance oxygen) as the case may be.

Example 7 :  Cu + 8H⁺ + yNO^-₃ ⟶ zCu²⁺ + 4H₂O + 2NO
                     The values of x, y and z are respectively
                    (A) 3, 2, 3,     (B)    2, 3, 3
                    (C) 3, 3, 3    (D)    4, 3, 1
Solution:    (A).

Ion – electron method or Half reaction method

The various steps involved are –
(i)    Write the skeletal equation and indicate the oxidation number (O. N.) of each element.
(ii)    Identify the elements which has undergone change in oxidation state.
(iii)    Divide the skeletal equation into two half reactions i.e. oxidation half reaction and reduction half reaction. In each half reaction balance the atom which undergo change in oxidation number.
(iv)    Add electron to whichever side is necessary to make up the difference in oxidation number in each half reaction.
(v)    Balance oxygen atoms by addition of proper number of H2O molecules to side deficient in ‘O’ atoms.
(vi)    For ionic equations – It involves the balancing of H atoms in each half reaction. (after O is balanced in each half reaction).
    (a) For acidic medium: Add proper number of H+ ions to the side falling short of H atoms.
    (b) For basic medium: Add proper number of H2O molecules to the side falling short of H atoms and equal number of OH- ions to the other side.
(vii)    Equalise the number of electrons lost and gained by multiplying the half reactions with suitable integer and add them to get the final equation.

Note:    It may be remembered that in balanced equation the number of atoms and also the electrical charges must be equal on both sides.

Illustration  8 : wCr₂O₇^-2 + xFe⁺² ⟶ yCr⁺³ + zFe⁺³ + 7H₂O (in acidic medium)
               Balance the reaction by ionic method and find the values of w, x, y 
                   and z respectively?
               (A) 1, 7, 2, 6    (B)    1, 6, 2, 6
               (C) 2, 6, 2, 6    (D)    3, 5, 3, 7
Solution:    (B).
             
        First balance ‘O’ adding equal number of H₂O molecules and then ‘H’ atom by adding H+ to deficient side.

  

CONCEPT OF LIMITING REAGENT

When ever, the ratios of reactant molecules actually used in an experiment are different from those given by co–efficients of the balance equaiton, a surplus of one reagent is left over after the reaction is completed.

Thus the extent to which a reaction takes place, depends on the reactant that is present in limiting amount – the limiting reactant. The other reactant is said to be the excess reactant.

Consider the following chemical reaction
 N2 = 3H2 ⟶ 2NH3
From the reaction, it is clear that 1 mol N₂ reacts with 3 mol H₂ to from 2 mol NH₃.
If we take 1 mol N₂ and 4 mol H₂, even then 2 mol of NH₃ are formed since 1 mol H₂ is left unreacted. So in this case N₂ is limiting reagent.
Now, if we take 2 mol N₂ and 3 mol H₂ even then 2 mol of NH₃ are formed and 1 mol N₂ is left unreacted. So in this case H₂ is limiting reagnet.

Example  9 :  A sample of Ca₃(PO₄)₂ contains 3.1 g phosphorus, the weight of Ca in 
                   the sample is
               (A) 6 g        (B)    4 g
               (C) 2 g        (D)    5 g
Solution:     (A). Mole of phosphorus  
                           Mole of Ca   
                          Weight of Ca  

Example 10 : 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
Solution:     (D). 3BaCl₂ + 2Na₃PO₄  ––⟶  Ba₃(PO₄)₂ + 6NaCl 
        Here Na₃PO₄ is limiting 
        Hence moles of Ba₃(PO₄)₂ = ½ of Na₃PO₄ 
                                 = 1/2 x0.2 = 0.1 mole

Volumetric Stoichiometry

The reactions in solutions are by far the most common. The first step to determine is to express the amount of substance present in a given volume of a solution.
 Solution ––⟶ Solute + Solvent
Qualitative methods to express a solution are dilute, concentrated, saturated and super saturated.
Quantitative method involves the determination of concentration in different way as described below:

(i)    Molarity: It is the number of moles of solute dissolved or present per litre of the solution. It is represented by the symbol M.
        
(ii)    Molality: Number of moles of solute present or dissolve in 1000 g or 1 kg of solvent. It is represented by the symbol m.
        
(iii)    Mole fraction: It is the relative ratio of moles of solute and solution or solvent and solution.
         
    Note: Molarity is temperature dependent while mole fraction and molality is temperature independent.

(iv)    Strength: The strength of a solution is defined as the amount of the solute in grams present per litre of the solution (i.e. g/L).
(v)    Normality: The normality of a solution is defined as the number of gram equivalents of the solute present per litre of the solution, it is represented by N.
        
    E = Equivalent weight
(vi)    Formality: Formality is similar to Molarity. It is especially used for ionic crystals like Na+Cl–.
     

Example 11 :  Which of the following does not depend on temperature?
                         (A) Molality    (B)    Mole fraction
                         (C) Mole    (D)    All of the above

Solution:    (D). 
        Weight is independently temperature.

CALCULATION OF EQUIVALENT WEIGHT

Equivalent weight of an element can be calculated using the composition of the compound of the given element with any other element whose equivalent weight is known by the knowledge of the law of equivalence.
This law states that one equivalent of an element combines with one equivalent of the other.
Hence, the equivalent weight of an element is the weight of its mole combining with one equivalent of another element.
In general, equivalent weight  
We can easily calculate equivalent weight if we know proper idea about n–factor. n–factor varies for element, salt, acid, base and redox as well as disproportionation reaction.
(i)    Equivalent weight of element  
(ii)    Equivalent weight of acid  

Acid	Basicity (maximum)
HCl, HNO3, H3BO3	1
H2SO4, H3PO3, H2C2O4·2H2O	2
H3PO4	3
H4P2O7	4

 

(iii)    Equivalent weight of base  

 

Base	Acidity (maximum)
NaOH, KOH	1
Ca(OH)2, Ba(OH)2	2
AI(OH)3	3

(iv) 

Note: In case of hydrated salt mass of water is included but not the number of H+ ion.

(v)    Equivalent weight of oxidizing and reducing agent
         

(vi)    Equivalent weight of an ion  

Law of Equivalence

For a chemical reaction:
 aA + bB ⟶ cC + dD
       
where n_i and M_i is n factor, molecular mass of i^th species.
If the reaction is carried out in solution, then
        N_AV_A = N_BV_B = N_CV_C = N_DV_D
or    n_AM_AV_A = n_BM_BV_B = n_CM_CV_C = n_DM_DV_D

Normality equation

N₁V₁ = N₂V₂
Gram equivalent of acid = Gram equivalent of base (if volume in L)

Milliequivalents

Milliequivalents = Normality x volume(in ml)
For a given solution, number of equivalents per litre is same as the number of milliequivalents per ml. Moles, millimoles, equivalent, milli equivalent of solute does not change on dilution.

Example 12 : The equivalent weight of potassium permanganate (KMnO₄) in acidic 
             medium is
               (A)158    (B)    79
                (C)31.6    (D)    316
Solution: (C). 
          2KMnO₄ + 3H₂SO₄ ⟶ K₂SO₄ + 2MnSO₄ + 3H₂O + 5O
        The net reaction is
        
        Change in oxidation number = 7 - 2 
                           = 5
        Equivalent weight of KMnO4 = 

 

Example 13 : The equivalent weight of ferric oxalate in the reaction with KMnO₄/H⁺ is    
                       (A)M/6    (B)    M/3
                       (C) M    (D)    M/2
                       where, M = molecular weight of ferric oxalate.
Solution:     (A).  
        n–factor for Fe2(C2O4)3 = 6x1 = 6
        Equivalent weight  

Example 14 :   1.5 litre of a solution of normality N and 2.5 litres 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:     (B).Eq. of 1.5 litre solution + Eq. of 2.5 litre solution = Eq. of resultant of solution.
        = 1.5 x N + 2.5 x 2 = 4 x 5
          N =   = 10

TITRATIONS

In volumetric analysis, the process of determination of strength of an unknown solution by another solution of known strength under volumetric conditions is known as titration. Titrations are of various types–The fundamental basis of all titration is the law of equivalence which states that at the end point, the milliequivalent of known solution is equal to milliequivalent of given unknown solution which is observed by using an indicator.
                                                                                                                                  

Simple titration

In a simple titration, a solution of substance A of unknown concentration is made to react with a solution of reagent B whose concentration is known, so that the concentration of A may be calculated. 
N₁V₁ = N₂V₂
 (A)    (B)
N2 = 

Example 15 :  1.0 g of mixture of Na₂CO₃ and NaCl is neutralized by 100 ml of a 0.1 
                            N HCl solution. The percentage composition of NaCl in mixture is
                        (A)53%    (B)    47%
                        (C) 50%    (D)    40%
Solution:     (B). 
        meq. of Na₂CO₃ = meq. of HCl (NaCl not react with HCl)
         
        x = 0.53 gm
        Mass of NaCl = 0.47 gm
        % of NaCl = 47%

Back titration

Let us assume that we have a solid C which is impure and we want to calculate the percentage purity. We are given two reagents A and B, where the concentration of A is unknown while that of B is known (N1).
Note:   (i)     A, B and C should be such that A reacts with B, A reacts with C and the product of A and C cannot react with B.
    
(ii) Gram equivalent of A should be either greater than or equal to gram equivalents of C.
     meq. of C = (meq. of A – meq. of B)

Example 16 : 0.5 g of impure CaCO₃ was treated with 50 ml of  HCl. The excess of 
                           HCl was neutralized by 6 ml of  NaOH solution. The percentage 
                          purity of CaCO₃ is
                      (A)84%    (B)    44%
                      (C)21%    (D)    46%

Solution:     (B).  meq.  of excess of HCl = meq.  of NaOH = 0.6
                meq.  of used HCl for CaCO₃ = 5 – 0.6 = 4.4
                meq.  of CaCO₃ = meq.  of used HCl
                  

Double titration

This is done by
(i)      Using two indicators:

Indicator	pH range	Colour
		In acid	In base
HPh (Phenolphthalein)	(8–10)	Colourless	Pink
MeOH (Methyl orange)	(3–4)	Orange	Yellow

When the solution containing NaOH and Na₂CO₃ is titrated using phenolphthalein indicator, following reaction takes place at the end point 
        NaOH + HCl ──⟶ NaCl + H₂O 
        Na₂CO₃ + HCl ──⟶ NaHCO₃ + NaCl 
        meq. of NaOH + meq. of Na₂CO₃ = meq. of HCl                 

                 … (i)
        When MeOH is used Na₂CO₃ convert into NaCl + CO₂ + H₂O
        meq. of NaOH + meq. of Na₂CO₃ = meq. of HCl         

            … (ii)

(ii)    Using two reagents:
    Mixture of (H₂C₂O₄ + H₂SO₄)  (complete reaction)
    Mixture of (H₂C₂O₄ + H₂SO₄)  (complete neutralization) 
    For step (i), H₂SO₄ + H⁺ + KMnO₄ ⟶ Mn₂₊ + CO₂ + H₂O
     Meq. of H₂SO₄ = Mqe. of KMnO₄
               (n=2)        (n=5)
    For step (ii), meq. of H₂SO₄ = meq. of NaOH

Example 17 :  40 ml of 0.05 M solution of sodium sesquicarbonate 
                   (Na₂CO₃.NaHCO₃.2H₂O) is titrated against 0.05 M HCl. X ml of HCl is 
                   used when phenolphthalein is the indicator and y ml of HCl is used 
                   when methyl orange is the indicator in two separate titrations. Hence 
                   (y–x) is
                  (A) 80 ml            (B) 30 ml
                  (C) 120 ml    (D) 180 ml

Solution: (A).  Millimole of mixture = Millimole of Na₂CO₃      
        =40 x 0.05 = 2 millimole = millimole of NaHCO₃
        Millimole of Na₂CO₃ = Millimole of HCl
        2 = 0.05 x
        x = 40 ml
        meq.  of Na₂CO₃ + meq.  of NaHCO₃ = meq.  of HCl
         2 x 2 + 1 x2 = 0.5g
         
        x = 120 – 40 = 80 ml

Redox titration

It is a common reaction used in redox titration.
(i)  2KMnO₄ + 3H₂SO₄ ──⟶K₂SO₄ + 2MnSO₄ + 3H₂O +5O
     (n = 5)         
(ii) K₂Ce₂O₇ + 4H₂SO₄ ──⟶ K₂SO₄ + Cr(SO₄)₃ + 4H₂O + 3O
     (n = 6)
(iii) 2[Fe(SO₄).(NH₄)₂SO₄.6H₂O]+ H₂SO₄ + O ⟶Fe₂(SO₄)₃ + 2(MnO)₂SO₄ + 13H₂O
    Valency of KMnO4 in different medium:
     
Example 18 :  0.185 g of an iron wire containing 99.8% iron is dissolved in an acid 
                            to form ferrous ions. The solution requires 29.3 ml of K₂Cr₂O₇ 
                            solution for complete reaction. The normality of the K₂Cr₂O₇ solution 
                            is
                         (A) 0.02    (B)    0.025
                         (C) 0.01    (D)    0.05
Solution:     (C).
         Fe₂₊ + K₂  O₇ + H⁺ ──⟶Cr³⁺ + Fe³⁺
                   n=1      n=6
        meq. of Fe2+ = meq.  of K₂Cr₂O₇
         
        N = 0.01 (approx)

Iodine titration

Titrations involving compounds of iodine are known as Iodometric titrations. Iodine molecules, I2 gives blue colour with starch. Thus, the completion of reaction can be detected when blue colours disappears at the end of the reaction.

(i)    Iodometric: When an iodine containing sample is treated with oxidising reagent (KMnO4, K2Cr2O7, CuSO4, H2O2 and O3). Iodine so liberated is then treated with hypo solution, meq. of which is known to us.
    i.e. 2KI + H₂O₂ ──⟶2KOH + I₂   
    I₂ + 2Na₂S₂O₃ ──⟶2NaI + Na₂S₄O₆
    meq. of hypo = meq. of I₂ = meq. of H₂O₂

(ii)    Iodimetric: In this titration, I₂ is directly treated with a reducing agent.
     I₂ + 2Na₂SO₃ ──⟶Na₂SO₄ + NaI
     I₂ + Na₃AsO₃ ──⟶Na₃AsO₄ + NaI
     I₂ + 2Na₂S₂O₃ ──⟶ Na₂S₄O₆ + 2NaI
    meq. of I2 = meq. of hypo or reducing agent

VOLUME STRENGTH OF H2O2
 
X volume strength of H₂O₂ means X litre of O₂ is liberated by 1 L of H₂O₂ on decomposition at STP.

Example  19:  A ‘10–volume’ H₂O₂ solution is equal to
                        (A)   3% (W/W) H₂O₂    (B)    30 gm/L H₂O₂
                        (C)   1.76 N                    (D)    All are true

Solution:     (D). 
        Volume strength = N x 5.6

         

        W = E.N.V (in L)

     

   

Some useful reactions for stoichiometry
(i)    2NaHCO₃  Na₂CO₃ + CO₂ ↑ +H₂O ↑         
(ii)    Ag₂CO₃  2Ag + CO₃ +   
(iii)    2(NH₄)₂SO₄  N₂ ↑ + 6H₂O↑ + 3SO₂ ↑ + 4NH₃ ↑

     
(iv)     

(v)     

(vi)    2CuSO₄ + 4KI ──⟶Cu₂I₂ + I₂ + 2K₂SO₄     
(vii)    O₃ + KI + H₂O ──⟶2KOH + I₂ + O₂     
(viii)    2MnO^-₄ + 5C2O^-₄ + 16H+ ──⟶ 2Mn²⁺ + 10CO₂ + 8H₂O     
(ix)    BaCO₃ + HCl ──⟶ BaCl₂ + CO₂ + H₂O     
(x)    BaCl₂ + H₂CrO₄ ──⟶ BaCrO₄ + 2HCl     
(xi)    MnO₂ + 4HCl ──⟶ MnCl₂ + 2H₂O + Cl₂ 
(xii)    Cr₂O^--₇ + 3C₂O₄ + 14H⁺ ──⟶ 6CO₂ + 2Cr³⁺ + 7H₂O     
(xiii)    3Br₂ + 6OH^- ──⟶ BrO^-₃ + 5Br^- + 3H₂O     
(xiv)    Ca(HCO3)2 + CaO ──⟶ 2CaCO3 + H2O     
(xv)    3BaCl₂ + 2Na₃PO₄ ──⟶ Ba₃(PO₄)₂ + 6NaCl     
(xvi)    BaCrO₄ + 6Kl + 8H₂SO₄ ──⟶ 3I₂ + 2BaSO₄ + 3K₂SO₄ +Cr2(SO₄)₃ + 8H₂O      
(xvii)     2KMnO₄ + 10FeSO₄ + 8H₂SO₄ ──⟶2MnSO₄ + 5Fe(SO₄)₃ + 8H₂O + K₂SO₄
(xviii)     Fe₂S₃ + O₂ ──⟶ 2FeS + SO₂

(xix)    Kl + I₂ ──⟶ Kl₃ (brown colour)     
(xx)    Kl₃ + 2Na₂S₂O₃ ──⟶ 2Nal- + Na₂S₄O₆ + Kl     
(xxi)    2N₃H + I₂ ──⟶ 3N₂ + 2HI     
(xxii)     
(xxiii)    PCI₅ + H₂O ──⟶ POCI₃ + HOCI
        POCI₃ + 3H₂O──⟶H₃PO₄ + 3HCI

ANSWER TO EXCERCISE

Exercise 1.                D 
Exercise 2.         i)     A 
                           ii)     A

Exercise 3:                D

Exercise 4:                B

Exercise 5.                C

Exercise 6.                B

Exercise 7.                D

Exercise 8.               C

Exercise 9.               B

Exercise 10.    (1)    A 
                         (2)    A

Exercise 11.            B

Exercise 12.            A
  
Exercise: 13.          D 

Exercise 14.           C
 

POINTS TO PONDER

1.    Stoichiometry is that branch of chemistry which deals with quantitative interpretation of chemical reaction.

2.    Law of conservation of mass states that, matter can neither be created nor be destroyed.

3.    For ideal gases molecular weight of the gas is twice its molecular weight.

4.    Weight of 1 mol of a substance is equal to its molecular weight.

5.    1 mole of an ideal gas occupy a volume of 22.4 litre at standard temperature and pressure (STP).

6.    Standard temperature and pressure refers to 0C and 1 atm pressure respectively.

7.    Limiting reagent is completely consumed during chemical reaction.

8.    The law of equivalence states that are equivalent of an element or compound combines with one equivalent of the other.

9.    Equivalent weight calculated by dividing molecular mass or atomic mass with n-factor.

10.    Titration is the process of determining strength of an unknown solution by another solution of known strength.
 

SOLVED EXAMPLES

1.    The oxidation number of Cr is +6 in
    (A)    FeCr₂O₄        (B)    Fe₂(CrO₄)₂
    (C)    Cr₂(SO₄)₃   (D)    [Cr(OH)₄]^–

Sol.    (B). 2(+2) + 2[x + 4(-2)] = 0
    Here x is oxidation number of Cr
    +4 + 2x – 16 = 0
    2x = +12
    x = +6

2.    Oxidation number of chromium pentoxide (CrO₅)
    (A)    +5    (B)    +10
    (C)    +8    (D)    +6

Sol.       (D). 
                        Butterfly Structure
    x + 1 (–2) + 4 (–1) = 0
    x = +6

3.    Which of the following process is used to remove temporary as well as permanent hardness in water?
    (A)    Boiling the water with a calculated quantity of Na₂CO₃ solution and then with HCl
    (B) Treating the water with a Ca(OH)₂ solution and then with HCl
    (C)    Boiling the water with carbon
    (D)    Boiling the water with anhydrous CaCl₂

Sol.    (A).

4.    1.578 g Ba(OH)2.xH2O neutralizes 40 ml of an   solution. The value of x is 
    (Ba = 137.4)
    (A)    3    (B)    6    
    (C)    4    (D)    8
Sol.    (D). meq.  of Ba(OH)₂.xH₂O = meq.  of HNO₃
      

5.    Find the ratio of number of molecules contained in 1 gm of NH3 and 1 gm N₂
    (A) 20: 17    (B) 28: 17
    (C) 17: 28    (D) 14: 17

Sol.     (B).    Ratio of moles  
6.    What is the relation between volume strength and normality of H₂O₂ solution? 
    (A) Volume strength = 5.6 ´ normality     (B) normality = 5.2 ´ volume strength 
    (C) Volume strength = 1.78 ´ normality     (D) normality = 2.8 ´ volumes strength 

Sol.     (A).     

7.    The weight of Na₂CO₃ required to completely neutralize 45.6 ml of 0.235 N H₂SO₄ acid will be 
    (A) 0.47 g     (B) 0.57 g
    (C) 0.67 g    (D) 0.77 g

Sol.    (B).     meq.  of Na₂CO₃ = meq.  of H₂SO₄
                  0.57 gm (approx)

8.    20 ml solution containing NaOH and Na₂CO₃ required 32.5 ml 0.2 N H₂SO₄ for the end point using phenolphthalein as indicator. In another titration 20 ml solution of the mixture required 52.5 ml 0.2 N H₂SO₄ for the endpoint using MeOH as indicator. Hence, molarity of solution in respect of Na₂CO₃ was 
    (A) 01.     (B) 0.2
    (C) 0.05     (D) None of these 

Sol.    (B). In first case using Phenolphthalein,
        meq. of NaOH + meq. of Na₂CO₃ = meq. of H₂SO₄
             n=1       …(1)
    In second case using Methyl orange,
        meq. of NaOH + meq. of Na₂CO₃ = meq. of H₂SO₄
             n=2       …(2)
    From (1) and (2) we get, 

9.  A 10.0 g samples of a mixture of calcium chloride and sodium chloride is treated with Na₂CO₃ to precipitate the 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)    32.7%
    (C)    21.8%    (D)    11.07%

Sol.  (B). CaCl₂ + Na₂CO₃ ⟶CaCO₃ + 2NaCl
                       113                           100
            CaCO₃  CaO + CO₂
               100                 56
       56 gm CaO Formed by = 100 g CaCO3
      1.62 gm will formed by  CaCO₃
      100 gm of CaCO₃ is formed by = 113 g CaCl₂
          will be formed by   = 3.27 gm
      10.0 gm contains = 3.27 g CaCl₂
      1 gm contains  
      100 gm contains  of CaCl₂

10.  1g of Ca was burnt in excess of O₂ and the oxide was dissolved in water to make up one litre solution. The normality of solution is 
    (A) 0.04     (B) 0.4
    (C) 0.05    (D) 0.5

Sol.  (C). 2Ca + O₂ ⟶2CaO
                    80          112
     CaO + H₂O ⟶ Ca(OH)₂
              56          74
    80 gm gives = 112 gm CaO
    1.0 gm gives  
    56 gm CaO gives = 74 g Ca(OH)₂

  

ASSIGNMENT PROBLEMS

1.    Simplest formula of the compound containing 60 % of element A (At. wt.10) and 40% of element B (At.wt.20) is 
    (A) A₃B            (B) A₂B₃
    (C) A₃B₂    (D) AB₂

2.    The molality of a H₂SO₄ solution is 9. The weight of the solute in 1 kg H₂SO₄ solution is
    (A)    900.0 g    (B)    468.65 g
    (C)    882.0 g    (D)    9.0 g

3.    An aqueous solution of 6.3 g oxalic acid dihydrate is made upto 250 ml. The volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution is
    (A)    40 ml    (B)    20 ml
    (C)    10 ml    (D)    4 ml
4.    100 cc of 0.3 (N)^– HCl solution were mixed with 200 cc of 0.6 (N) H₂SO₄ solution. What is the normality of H₂SO₄ in the final solution?
    (A) 0.4    (B) 0.5
    (C) 0.9    (D) 0.6 

5.    The eq. wt. of iodine in I₂ + 2S₂O^2-₃ ⟶ 2I– + S₄O^2-₆  is equal to its 
    (A)     mol.wt.    (B) mol.wt./2
    (C) mol.wt./4    (D) none of these 

6.    The normality of 100 ml of '20 vol' H₂O₂ solution is 
    (A) 20/5.6 N    (B) 20 x 5.6 N
    (C) 5.6/20 N    (D) 5.6 N

7.    1 litre of 18 molar H₂SO₄ has been diluted to 100 litres. The normality of the resulting solution is 
    (A)    0.09 N    (B) 0.18 N
    (C) 1800 N    (D) 0.36 N

8.    0.311 gram sample of crude NaOH required 46.1 ml of 0.122 N H₂SO₄ for complete neutralisation, the percentage of NaOH in the sample is
    (A)    72.3    (B)    7.23
    (C)    36.15    (D)    23.7

9.    If 8.3 ml of a sample of H₂SO₄ (18 M) is diluted by 991.7 ml of water. Then approximate normality of the resulting solution is
    (A)    0.4    (B)    0.2
    (C)    0.1    (D)    0.3

10.    In the reaction 
    Cl₂ + OH^– ⟶ Cl^– + ClO^3– + H₂O
    (A)    Chlorine is oxidised                                    (B)    Chlorine is reduced 
    (C)    Chlorine is reduced as well as oxidized     (D)    Chlorine is neither reduced nor oxidised 

11.    A self contained breathing apparatus uses KO₂, to convert the carbondioxide and water in exhaled air into oxygen, as shown by the equation.
    4KO₂(s) + 2H₂O(g) + 4CO₂(g) ⟶ 4KHCO₃(s) + 3O₂(g)
    How many molecules of oxygen gas will be produced from 0.0468 g of carbondioxide that is exhaled in a typical breath?
    (A)    4.8 x 10²⁰    (B)    6.4 x 10²⁰
    (C)    8.5 x 10²⁰    (D)    1.9 x 10²¹

12.    The weight of MnO₂ (87%) pure that must be taken so that on treatment with conc. HCl, the liberated Cl₂ will liberate 12.7 g of I₂ from KI.
    (A)    10 g    (B)    12 g
    (C)    13 g    (D)    20 g
13.    An aqueous solution of 6.3 g oxalic acid dihdyrate is made upto 250 ml. The volume of 0.1 NaOH required to completely neutralise 10 ml of this solution is
    (A)    40 ml    (B)    20 ml
    (C)    10 ml    (D)    4 ml

14.    A person needs 1.71 g of cane sugar (C₁₂H₂₂O₁₁) to sweeten his tea. What would be the number of carbon atoms consumed through sugar in his tea?
    (A)    3.6x10₂₂           (B)7.2x10²¹
    (C)    5x10²¹            (D)6.6x10²²

15.    In a reaction, 4 moles of electrons are transferred to 1 mole of HNO₃ .The possible product obtained due to reduction is
    (A)    0.5 mole of     (B)    0.5 mole of  
    (C)    1 mole of     (D)    1 mole of  
 

ANSWERS TO ASSIGNMENT PROBLEMS

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