CS– 1
SOME BASIC CONCEPTS
IN CHEMISTRY
Syllabus :
Matter and its nature, Dalton’s atomic theory;
Concept of atom, molecule, element and compound;
Physical quantities and their measurement in
Chemistry, precision and accuracy, significant
figures, S.I. Units, dimensional analysis; Laws of
chemical combination; Atomic and molecular
masses, mole concept, molar mass, percentage
composition, empirical and molecular formulae;
chemical equations and stoichiometry.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS – 2
CONCEPTS
C1
C2
In this chapter we will discuss the calculations based on chemical equations. It has been classified into two
parts :
1.
Mole Concept
2.
Equivalent Concept
MOLE CONCEPT :
In mole concept we deal with different types of relations like weight-weight, weight-volume, or
volume-volume relationship between reactants or products of the reaction.
Mole concept is based on balanced chemical chemical reaction. Some basic definitions used in mole
concept are as follows :
Limiting Reagent : A reagent which is consumed completely during the chemical reaction.
Number of moles of a substance(n) 
weight of substance
atomic or molecular weight
Also, Number of moles of a substance(n) 
Given number of molecules
Avogadro number
In gas phase reaction number of moles of a gas (n) =
1.
2.
C3
PV
,
RT
At STP/NTP one mole of any gas contains 22.4 L i.e. at 273 K and 1 atm pressure.
In aq. solution
n = MV [M - molarity, V - volume of solution]
Practice Problems :
Chlorine can be produced by reacting H2SO4 acid with a mixture of MnO2 and NaCl. The reactions
follows the equation : 2NaCl + MnO2 + 3H2SO4  2NaHSO4 + MnSO4 + Cl2 + H2O what volume of
chlorine at STP can be produced from 100 g of NaCl ? (At. wt. Na = 23, Cl = 35.5)
(a)
19.15 lt
(b)
30 lt
(c)
29 lt
(d)
5 lt
A solution contains 5 g of KOH was poured into a solution containing 6.8 g of AlCl3, find the mass of
precipitate formed [At. wt. : H-1, Al-27, Cl-35.5, K-39]
(a)
2.3 g
(b)
23 g
(c)
32 g
(d)
0.32 g
[Answers : (1) a (2) a]
EQUIVALENT CONCEPT :
Volumetric Analysis is based on acid base titration and redox titration mainly.
Important Definitions :
No. of moles of solute
Vol. of solution in L
No. of gramequiva lents of solute
Normality ( N ) 
Vol. of solution in L
Molarity(M) 
No. of gram equivalents of solute(neq ) 
Equivalent weight 
Weight of solute
Equivalent weight
Molecular weight (or ) Atomic weight (or ) Ionic weight
n factor
neq = nmol × n-factor
neq = Normality × Volume (L)
Number of moles(nmol) = Molarity × Volume (L)
Normality = Molarity × n-factor
Calculation of ‘n’ Factor for Different Class of Compounds
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS– 3
1.
n = basicity
n=3
n=2
n=1
2.
n = acidity
e.g. Ammonia and all amines are monoacidic bases
3.
Salt : (Which does not undergo redox reactions)
n factor = Total cationic or anionic charge
4.
Oxidizing Agents or Reducing Agents : ‘n’ factor = change in oxidation number Or
number of electron lost or gained from one mole of the compound.
Note :
In a balance equation n factor ratio of two compounds is reverse of their molar ratio.
Practice Problems :
1.
[Na+] in a solution prepared by mixing 30.00 mL of 0.12 M NaCl with 70 mL of 0.15 M Na2SO4 is
(a)
2.
Mn2O3
0.141 M
(b)
6.20 g
MnO2
(b)
–
3
(c)
0.210 M
(d)
0.246 M
(c)
MnO4–
(d)
MnO42–
7.75 g
(c)
10.5 g
(d)
21.0 g
–
When BrO ion reacts with Br ion in acid solution Br2 is liberated. The equivalent weight of KBrO3
in this reaction is
(a)
5.
(b)
The anion nitrate can be converted into ammonium ion. The equivalent mass of NO3– ion in this
reaction would be
(a)
4.
0.135 M
The equivalent mass of MnSO4 is half of its molar mass when it is converted to
(a)
3.
Acids :
H3PO4
H3PO3
H3PO2
Bases :
M/8
(b)
M/3
(c)
M/5
(d)
M/6
The number of moles of KMnO4 that will be needed to react completely with one mole of ferrous
oxalate in acidic solution is
(a)
3/5
(b)
2/5
(c)
4/5
(d)
1
[Answers : (1) d (2) b (3) b (4) c (5) a]
C4
C5
C6
LAW OF CHEMICAL EQUIVALENTS : In a chemical reaction the equivalents of all the species
(reactants or products) are equal to each other provided none of these compounds is in excess.
N1V1 = N2V2 (when normalities and volumes are given)
Relation between percentage weight by weight (x), density (d) and strength in gm / litre (s) S = 10 xd
BACK TITRATION
This is a method in which a substance is taken in excess and some part of its has to react with another
substance and the remaining part has to be titrated against standard reagent.
BASIC PRINCIPLE OF TITRATIONS :
In voltmetric analysis, a given amount (weight or volume) of an unknown substance is allowed to react with
a known volume of a standard solution slowly. A chemical reaction takes place between the solute of an
unknown substance and the solute of the standard solution. The completion of the reaction is indicated by
the end point of the reaction, which is observed by the colour change either due to the indicator or due to the
solute itself. Whether the reactions during the analysis are either between an acid and or base or between
O.A. and R.A., the law of equivalence is used at end point.
Following are the different important points regarding this process :
(i)
In case of acid base titration at the equivalence point
(neq)acid = (neq)base
(ii)
In case of redox titration
(neq)oxidant = (neq)reductant
(iii)
If a given volume of solution is diluted then number of moles or number of equivalence of
solute remains same but molarity or normality of the solution decreases.
(iv)
If a mixture contains more than one acids and is allowed to react completely with the base then
at the equivalence point, (neq) acid1 + (neq) acid2 + ... = (neq) base
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS – 4
(v)
Similarly if a mixture contains more than one oxidising agents then at equivalence point,
(neq) O.A1 + (neq) O.A2 +... = (neq) reducing agent.
(vi)
If it is a difficute to solve the problem through equivalence concept then use the mole concept.
Practice Problems :
1.
5 ml of N-HCl, 20 ml of N/2-H2SO4 and 30 ml of N/3 – HNO3 are mixed together and the volume
made to 1 litre.
(i)
The normality of the resulting solution is
(a)
N/5
(b)
N/10
(c)
N/20
(d)
N/40
(d)
2.5 g
(d)
2.
(ii)
The wt. of pure NaOH required to neutralize the above solution is
(a)
10 g
7
(b)
3
(c)
2
(d)
5
100
mL
3
(b)
500
mL
3
(c)
300
mL
3
(d)
100 mL
100 mL of a mixture of NaOH and Na2SO4 is neutralised by 100 mL of 0.5 M H2SO4. Hence amount
of NaOH in 100 mL mixture is
0.2 g
(b)
0.4 g
(c)
0.6 g
(d)
1.0 g
3 mol of a mixture of FeSO4 and Fe2(SO4)3 required 100 mL of 2 M KMnO4 solution is acidic
medium. Hence mol fraction of FeSO4 in the mixture is
(a)
6.
1g
100 mL of 1 M KMnO4 oxidised 100 mL of H2O2 in acidic medium (when MnO is reduced to Mn2+);
volume of same KMnO4 required to oxidise 100 mL of H2O2 in basic medium (when MnO4–. is
reduced to MnO2) will be
(a)
5.
(c)
–
4
(a)
4.
2g
0.7 g of a sample of Na2CO3.xH2O were dissolved in water and the volume was made to 100 ml, 20 ml
of this solution required 19.8 ml of N/10 HCl for complete neutralization. The value of x is
(a)
3.
(b)
1
3
(b)
2
3
(c)
2
5
(d)
3
5
5.3 g of M2CO3 is dissolved in 150 mL of 1 N HCl. Unused acid required 100 mL of 0.5 N NaOH.
Hence equivalent weight of M is
(a)
23
(b)
12
(c)
24
(d)
13
[Answers : (1) c (2) c (3) b (4) b (5) a (6) a]
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS– 5
INITIAL STEP EXERCISE
1.
2.
1 g of the carbonate of a metal was dissolved in
25 ml of N-HCl. The resulting liquid required 5 ml
of N-NaOH for neutralization. The eq. wt. of the
metal carbonate is
8.
6.
0.08 M KCl and 0.01 M HCl
Molality of 18 M H2SO4 (d = 1.8 gmL–1) is
(b)
30
(a)
36 mol kg–1
(b)
200 mol kg–1
(c)
20
(d)
None
(c)
500 mol kg–1
(d)
18 mol kg–1
If 0.5 mol of BaCl2 is mixed with 0.20 mol of Na3PO4,
the maximum amount of Ba 3(PO4)2 that can be
formed is
(a)
0.70 mol
(b)
0.50 mol
(c)
0.20 mol
(d)
0.10 mol
9.
When one gram mol of KMnO4 reacts with HCl,
the volume of chlorine liberated at NTP will be
(a)
11.2 litres
(b)
22.4 litres
(c)
44.8 litres
(d)
56.0 litres
34 g of hydrogen peroxide is present in 1120 ml of
solution. This solution is called
(a)
10 vol solution (b)
20 vol solution
(c)
30 vol solution (d)
32 vol solution
12.
5.
(d)
50
11.
4.
0.08 M KCl and 0.01 M KOH
(a)
10.
3.
(c)
To prepare a solution that is 0.50 M KCl starting
with 100 mL of 0.40 M KCl
(a)
add 0.75 g KCl
(b)
add 20 mL of water
(c)
add 0.10 mol KCl
(d)
evaporate 10 mL water
1 g equiv. of a substance is the weight of that amount
of a substance which is equivalent to
(a)
0.25 mol of O2
(b)
0.50 mol of O2
(c)
1 mol of O2
(d)
8 mol of O2
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
The density of 1 M solution of NaCl is 1.0585 g/mL.
The molality of the solution is
(a)
1.0585
(b)
1.00
(c)
0.10
(d)
0.0585
Which is false about H3PO2
(a)
it is tribasic acid
(b)
one mole is neutralised by 0.5 mol
Ca(OH)2
(c)
NaH2PO2 is normal salt
(d)
it disproportionates to H3PO3 and PH3 on
heating.
In hot alkaline solution, Br2 disproportionates to
Br– and BrO3–
3Br2 + 6OH–  5Br– + BrO3– + 3H2O
hence equivalent weight of Br 2 is (molecular
weight = M)
7.
(a)
M
6
(b)
M
5
(c)
3M
5
(d)
5M
5
When 80 mL of 0.20 M HCl is mixed with 120 mL
of 0.15 M KOH, the resultant solution is the same
as a solution of
(a)
0.16 M KCl and 0.02 M HCl
(b)
0.08 M KCl
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS – 6
FINAL STEP EXERCISE
1.
2.
3.
4.
5.
6.
8 g of sulphur are 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 mol
(b)
0.5 mol
(c)
0.24 mol
(d)
0.25 mol
One mole of a mixture of CO and CO2 requires
exactly 20 gram of NaOH in solution for complete
conversion of all the CO2 into Na2CO3. How many
grams of NaOH would it require for conversion into
Na2CO3 if the mixture (one mole) is completely
oxidised to CO2
(a)
60 grams
(b)
80 grams
(c)
40 grams
(d)
20 grams
One gram of a mixture of Na2CO3 and NaHCO3
consumes y gram equivalents of HCl for complete
neutralisation. One gram of the mixture is strongly
heated, the cooled and the residue treated with HCl.
How many gram equivalents of HCl would be
required for complete neutralization ?
ANSWERS (INITIAL STEP
EXERCISE)
(a)
2 y gram equivalent
1.
a
7.
c
(b)
y gram equivalents
2.
d
8.
c
(c)
3y/4 gram equivalents
(d)
3y/2 gram equivalents
3.
d
9.
a
If equal volumes of 1 M KMnO4 and 1 M K2Cr2O7
solutions are allowed to oxidise Fe (II) to Fe(III),
then Fe (II) oxidised will be
4.
a
10.
b
5.
a
11.
b
(a)
more by KMnO4
6.
c
12.
a
(b)
more by K2Cr2O7
(c)
equal in both cases
(d)
none of these
0.5 g of fuming H2S2O7(Oleum) is diluted with
water. This solution is completely neutralized by
26.7 ml of 0.4 N NaOH. The percentage of free SO3
in sample is
(a)
30.6%
(b)
40.6%
(c)
20.6%
(d)
50%
One mole of N2H4 loses 10 mol of electrons to form
a new compound Y. Assuming that all the nitrogen
appears in the new compound. What is the
oxidation state of nitrogen in Y.
(a)
–1
(b)
–3
(c)
+3
(d)
+5
Einstein Classes,
ANSWERS (FINAL STEP EXERCISE)
1.
2.
3.
4.
5.
6.
d
a
b
b
c
c
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS– 7
AIEEE ANALYSIS [2002]
1.
Number of atoms in 558.5 gram Fe
(at. wt. of Fe = 55.85 g mol–1) is
With increase of temperature, which of these
changes ?
(a)
half than in 8 g He
(a)
mole fraction
(b)
twice that in 60 g carbon
(b)
Fraction of solute present in water
(c)
molality
(d)
weight fraction of solute
(c)
(d)
2.
558.5 × 6.023 × 10
6.023 × 10
3.
23
22
When KMnO 4 acts as an oxidising agent and
ultimately forms [MnO4], MnO2, Mn2O3, Mn2+ then
the number of electrons transferred in each case
respectively is
(a)
3, 5, 7, 1
(b)
1, 5, 3, 7
(c)
4, 3, 1, 5
(d)
1, 3, 4, 5
4.
In a compound C, H and N atoms are present in
9:1:3. 5 by weight of compound is 108. Molecular
formula of compound is
(a)
C2H6N2
(b)
C3 H 4 N
(c)
C9H12N3
(d)
C6H8N2
AIEEE ANALYSIS [2003]
5.
25 ml of a solution of barium hydroxide on
titration with a 0.1 molar solution of hydrochloric
acid gave a titre value of 35 ml. The molarity of
barium hydroxide solution was
(a)
0.35
(b)
0.07
(c)
0.14
(d)
0.28
AIEEE ANALYSIS [2004/2005/2006]
6.
7.
8.
6.02 × 1020 molecules of urea are present in 100 ml
of its solution. The concentration of urea solution
is
(a)
0.001 M
(b)
0.01 M
Two solutions of a substance (non electrolyte) are
mixed in the following manner 480 ml of 1.5 M first
solution and 520 mL of 1.2 M second solution. What
is the molarity of the final mixture ?
(c)
0.02 M
(d)
0.1 M
(a)
1.20 M
(b)
1.50 M
(Avogadro constant, NA = 6.02 × 1023 mol–1)
(c)
1.344 M
(d)
2.70 M
[2004]
[2005]
To neutralise completely 20 mL of 0.1 M aqueous
solution of phosphorus acid (H3PO3), the volume
of 0.1 M aqueous KOH solution required is
9.
10.
If we consider that 1/6, in place of 1/12, mass of
carbon atom is taken to be the relative atomic mass
unit, the mass of one mole of a substance will
(a)
10 mL
(b)
20 mL
(a)
decrease twice
(c)
40 mL
(d)
60 mL
(b)
increase two fold
[2004]
(c)
remain unchanged
Excess of KI reacts with CuSO4 solution and then
Na 2S 2O 3 solution is added to it. Which of the
statements is incorrect for this reaction ?
(d)
be a function of the molecular mass of
the substance
(a)
Cu2I2 formed
(b)
CuI2 is formed
(c)
Na2S2O3 is oxidised
(d)
evolved I2 is reduced
[2004]
Einstein Classes,
[2005]
11.
The oxidation state of chromium in the final
product formed by the reaction between Kl and
acidified potassium dichromate solution is
(a)
+4
(b)
+6
(c)
+2
(d)
+3
[2005]
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS – 8
12.
The number of hydrogen atom(s) attached to
phosphorus atom in hypophosphorous acid is
13.
(a)
zero
(b)
two
How many moles of magnesium phosphate,
Mg3(PO4)2 will contain 0.25 mole of oxygen atoms
?
(c)
one
(d)
three
(a)
2.5 × 10–2
(b)
3.125 × 10–2
(c)
0.02
(d)
1.25 × 10–2
[2005]
[2006]
AIEEE ANALYSIS [2007]
14.
15.
In the reaction, 2Al(s) + 6HCl(aq)  2Al3+(aq) + 6Cl–(aq)
+ 3H2(g),
(a)
67.2 L H2(g) at STP is produced for every
mole Al that reacts
(b)
11.2 L H2(g) at STP is produced for every
mole HCl(aq) consumed
(c)
6 L HCl(aq) is consumed for every 3L H2(g)
produced
(d)
33.6 L H2(g) is produced regardless of
temperature and pressure for every mole
Al that reacts
The density (in g mL–1) of a 3.60 M sulphuric acid
solution that is 29% H2SO4
(Molar mass = 98 g mol–1) by mass will be
(a)
1.22
(b)
1.45
(c)
1.64
(d)
1.88
ANSWERS AIEEE ANALYSIS
1.
b
2.
d
3.
b
4.
d
5.
b
6.
b
7.
c
8.
b
9.
c
10.
a
11.
d
12.
b
13.
b
14.
b
15.
a
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS– 9
TEST YOURSELF
1.
Number of moles of a substance can be calculated
by
molecular weight of substance
weight of substance
(a)
2.
3.
4.
(b)
given number of molecules
Avogadro number
(c)
PT
RV
(d)
all
9.
Change in number of moles when temperature
increase from 270C to 520C, pressure increase from
5 atm to 5.44 atm and volume remain the same is
(a)
1.09
(b)
4.18
(c)
2.14
(d)
no change
n-factor in the 2NH3  N2 is
(a)
3
(b)
2
(c)
1
(d)
0
The molarity of 58.5 g NaCl in 5 litre solution is
(a)
11.7
(b)
643.5
(c)
0.2
(d)
63.5
What weight of silver is present in 4.5 g Ag2S ?
[At. weight of Ag = 107.9, S = 32]
5.
6.
7.
8.
(a)
3.91
(b)
3.50
(c)
2.41
(d)
1.50
The number of molecule present in 4.48 litre
solution at STP is
(a)
1.2 × 10–23
(b)
4.7 × 10–23
(c)
27.98 × 10–23
(d)
30.11 × 10–23
N
NaOH is titrated with 30 ml of HCl,
50
then normality of HCl is
If 20 ml of
(a)
N
60
(b)
N
75
(c)
N
80
(d)
N
65
The normality of 127.4 g of H 2SO 4 in 2 litre
solution is
(a)
2.8
(b)
2.6
(c)
1.4
(d)
1.3
Which of the following relation is wrong ?
ANSWERS
(a)
neq = nmol × n-factor
(b)
neq = Normality × Volume
1.
b
6.
b
(c)
Number of moles = Molarity × volume
2.
a
7.
d
(d)
Molarity = Normality × n-factor
3.
c
8.
d
4.
a
9.
d
5.
a
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111