First chapter of class XI.

1. P R E P A R E D B Y
J N M A N D A L
P G T C H E M I S T R Y
K V B A L L Y G U N G E
Unit – 1 : Some Basic Concepts
of Chemistry
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2. Marks distribution
 COURSE STRUCTURE CLASS–XI
 (THEORY) (2020-21)
 Total Periods (Theory 119 +Practical 44)
 Time:3Hours TotalMarks70
 Unit No. Title No. of Periods Marks
 Unit I Some Basic Concepts of Chemistry 10 11
 Unit II Structure of Atom 12
 Unit III Classification of Elements and Periodicity in Properties 6 04
 Unit IV Chemical Bonding and Molecular Structure 14 21
 Unit V States of Matter: Gases and Liquids 9
 Unit VI Chemical Thermodynamics 14
 Unit VII Equilibrium 12
 Unit VIII Redox Reactions 4 16
 Unit IX Hydrogen 4
 Unit X s ‐Block Elements 5
 Unit XI Some p ‐Block Elements 9
 Unit XII Organic Chemistry: Some basic Principles and Techniques 10 18
 Unit XIII Hydrocarbons 10
 Total 119 70
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3. (Practical)
 Evaluation Scheme for Examination Marks
 Volumetric Analysis 08
 Salt Analysis 08
 Content Based Experiment 06
 Project Work 04
 Class record and viva 04
 Total 30
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4. Revised syllabus – Unit I
 Unit I: Some Basic ConceptsofChemistry
10Periods
 General Introduction: Importance and scope of
Chemistry.
 Atomic and molecular masses, mole concept and
molar mass, percentage composition, empirical and
molecular formula, chemical reactions,
stoichiometry and calculations based on
stoichiometry.
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5. What do you mean by atomic mass?
 One atomic mass unit is defined as a mass exactly
equal to one twelfth the mass of one carbon - 12
atom.
 And 1 amu = 1.66056×10–24 g
 Mass of an atom of hydrogen= 1.6736×10–24 g
 the average atomic mass of carbon will come out to
be : (0.98892) (12 u) + ( 0.01108) (13.00335 u) + (2
× 10–12) (14.00317 u)
 = 12.011 u[ Relative abundance of C isotopes: 12C -
98.892, 13C - 1.108 , 14C 2 ×10–10
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6. Molecular mass
 Molecular mass is the sum of atomic masses of the
elements present in a molecule.
 Molecular mass of methane,
 (CH4) = (12.011 u) + 4 (1.008 u)
 = 16.043 u
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7. Mole concept
 One mole is the amount of a substance that contains
as many particles or entities as there are atoms in
exactly 12 g (or 0.012 kg) of the 12C isotope.
 We can, therefore, say that 1 mol of hydrogen atoms
= 6.022×1023 atoms
 1 mol of water molecules = 6.022×1023 water
molecules
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8. Molar mass and no. of moles
 The mass of one mole of a substance in grams is
called its molar mass.
 The molar mass in grams is numerically equal to
atomic/molecular/ formula mass in u.
 No of moles(n) = Given mass(g)/Molar mass( g mol-
1)
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9. Empirical and Molecular formula
 An empirical formula represents the simplest
whole number ratio of various atoms present in a
compound whereas the molecular formula shows
the exact number of different types of atoms present
in a molecule of a compound.
 Example: Molecular formula of benzene is C6H6 but
the empirical formula is CH.
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10. Calculation of empirical and molecular formula
 Problem : A compound contains 4.07 % hydrogen, 24.27
% carbon and 71.65 % chlorine. Its molar mass is 98.96 g.
What are its empirical and molecular formulas ?
 Solution: Step 1. Conversion of mass per cent to grams.
 Since we are having mass per cent, it is convenient to use
100 g of the compound as the starting material. Thus, in
the 100 g sample of the above compound, 4.07g
hydrogen is present, 24.27g carbon is present and 71.65 g
chlorine is present.
 Contd. In next slide
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11. Numerical Contd.
 Step 2. Convert into number moles of each element
 Divide the masses obtained above by respective
atomic masses of various
 elements. Moles of hydrogen = 4.07 g/ 1.008 g =
4.04
 Moles of carbon = 24.27 g/ 12.01g = 2.021
 Moles of chlorine = 71.65g/ 35.453 g = 2.021
 Contd….
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12. Numerical contd…..
 Step 3. Divide the mole value obtained above by the
smallest number
 Since 2.021 is smallest value, division by it gives a ratio of
2:1:1 for H:C:Cl . In case the ratios are not whole
numbers, then they may be converted into whole number
by multiplying by the suitable coefficient.
 Step 4. Write empirical formula by mentioning the
numbers after writing the symbols of respective
elements.
 CH2Cl is, thus, the empirical formula of the above
compound.
 Contd….
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13. Contd….
 Step 5. Writing molecular formula
 (a) Determine empirical formula mass
 Add the atomic masses of various atoms present in the
empirical formula.
 For CH2Cl, empirical formula mass is 12.01 + 2 × 1.008 +
35.453 = 49.48 g
 (b) Divide Molar mass by empirical formula mass =98.96
g/49.48g = 2 = (n)
 (c) Multiply empirical formula by n obtained above to get
the molecular formula
 Empirical formula = CH2Cl, n = 2. Hence molecular
formula is C2H4Cl2.
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14. STOICHIOMETRY AND
STOICHIOMETRIC CALCULATIONS
 CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g)
 Thus, according to the above balanced chemical reaction,
 • One mole of CH4(g) reacts with two moles of O2(g) to
give one mole of CO2(g) and two moles of H2O(g)
 • One molecule of CH4(g) reacts with 2 molecules of
O2(g) to give one molecule of CO2(g) and 2 molecules of
H2O(g)
 • 22.4 L of CH4(g) reacts with 44.8 L of O2 (g) to give 22.4
L of CO2 (g) and 44.4 L of H2O(g) at STP
• 16 g of CH4 (g) reacts with 2×32 g of O2 (g) to give 44 g of
CO2 (g) and 2×18 g of H2O (g).
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15. Calculate the amount of water (g) produced by the
combustion of 16 g of methane.
The balanced equation for combustion of methane is :
CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g)
 (i)16 g of CH4 corresponds to one mole.
 (ii) From the above equation, 1 mol ofCH4 (g) gives 2
mol of H2O (g).
 2 mol of water (H2O) = 2 × (2+16) = 2 × 18 = 36 g
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16. Limiting Reagent
 The reactant which gets consumed, limits the
amount of product formed and is, therefore, called
the limiting reagent.
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17. 50.0 kg of N2 (g) and 10.0 kg of H2 (g) are mixed to produce NH3 (g).
Calculate the NH3 (g) formed. Identify the limiting reagent in the
production of NH3 in this situation.
 A balanced equation for the above reaction is written as follows :
 N2 (g) + 3H2 (g) ⇌2NH3 (g)
 Calculation of moles :
 moles of N2= 50.0 x 1000 /28= 1785.7
moles of H2
 = 10.00 kg H2 × 1000 /2 = 5×103 mol
 According to the above equation, 1 mol N2 (g) requires 3 mol H2 (g), for the reaction.
 Hence, for 1786 mol of N2, the moles of H2 (g) required would be1786 × 3= 5358mol H2
 But we have only 5000 mol H2. Hence,dihydrogen is the limiting reagent in this
 case.
 So NH3(g) would be formed only fromthat amount of available dihydrogen i.e.,5000 mol
 Since 3 mol H2(g) gives 2 mol NH3(g)
 5000moles H2 gives 2 x5000/3 =3333.3 mol NH3 (g) is obtained.
 If they are to be converted to grams, it is done as follows :
 1 mol NH3 (g) = 17.0 g NH3(g)
 333.3 mol NH3 (g) = 17 x 3333.3 g = 56666 g
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18. Molarity
 Molarity : It is the most widely used unit and is
denoted by M. It is defined as the number of moles of
the solute in 1 litre of the solution. Thus,
 Molarity (M) =No. of moles of solute/ Volume
of solution in lit. Its unit is mol L-1
Formula for numerical:
M = (wB x 1000) / ( MB x VmL.)
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19. Molality
 It is defined as the number of moles of solute
 present in 1 kg of solvent. It is denoted by m.
 Thus, Molality (m) = No. of moles of solute/ Mass of
solvent in kg
 Its Unit is mol kg-1
 m = (wB x 1000) / ( MB x wA [g])
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20. Mole fraction
 It is the ratio of number of moles of a particular
component to the total number of moles of the
solution.
 Mole fraction of component A in a mixture A and B =
χA= nA/(nA+nB) χB = nB/(nA+nB)
 It has no unit
 Remember χA+ χB = 1
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21. % by mass
 Mass per cent= Mass of solute x 100/ Mass of
solution
 For example 10% by mass of an aqueous solution of
glucose means 10 g glucose is dissolved in 100 g
solution.
 Therefore, mass of glucose(solute) = 10 g
 Mass of water ( solvent ) = 100 – 10 = 90 g
 Mass of solution = 100 g
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22. Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in
enough water to form 250 mL of the solution. Atomic masses of Na=23, O=16,
H=1 u.)
 Given: Given mass of solute(B) wB = 4 g
 Volume of solution = V mL = 250 mL
 Molar mass of NaOH = 23+16+1 = 40 g mol-1
 Working formula: M = (wB x 1000) / ( MB x VmL.)
 =4 x 1000 /( 40 x 250) = 0.4 M
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23. The density of 3 M solution of NaCl is 1.25 g mL–1. Calculate molality of
the solution.
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 M = 3 mol L–1
 Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g
 Mass of 1L solution = 1000 × 1.25 = 1250 g
 (since density = 1.25 g mL–1)
 Mass of water in solution = 1250 –175.5 = 1074.5 g
 Molality= No. of moles of solute/Mass of solvent in
kg =3 mol/1.0745kg = 2.79 m

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