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1
Deptt. Of Applied Sciences Govt. Polytechnic College For Girls Patiala
Presented By- Dr. Raman Rani Mittal
M.Sc., M.Phil, Ph.D. (Chemistry)
2
SOLUTIONS &
MOLE CONCEPT
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Contents
Mole Concept• Atomic & Gram atomic mass•Molecular mass & Gram molecular
mass•Mole concept & its importance
Solutions•Methods of expressing
concentrations of solutions
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Mole Concept
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Atomic & Molecular Mass• Matter is made up of atoms & molecules• Mass of matter is due to atoms & molecules• Mass of an atom is called as atomic mass &
mass of molecule is termed as molecular mass• The actual mass of an individual atom or
molecule is extremely small. This mass can not be expressed in grams.
• To express the masses of atoms & molecules, the unit called Atomic Mass Unit (a.m.u.) is introduced.
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• The atomic mass unit (amu) may be defined as one-twelfth of the actual mass of an atom of carbon (Carbon-12 isotope).
• The atomic mass of an element tell us as to how many times an atom of the element is heavier than 1/12th of an atom of Carbon (C-12)
Mass of an atom of an element
1/12 x mass of an atom of carbon(C-12)
Atomic mass =
7
• For example, atomic mass of oxygen is 16 a.m.u. It means that an atom of oxygen is 16 times heavier than 1/12th of mass of carbon atom (C-12)
The atomic mass of an element is the average relative mass of its atoms as compared with an atom of carbon taken as 12 a.m.u.
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Gram Atomic MassGram atomic mass
It is the quantity of an element whose mass in grams is numerically equal to its atomic mass.
In simple words, atomic mass of an element expressed in grams is the gram atomic mass or it is also called gram atom
For example, the atomic mass of oxygen is 16 a.m.u. and thus gram atomic mass of oxygen is 16 g.
Mass in grams
Gram atomic mass
Number of gram-atom
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Molecular mass
Molecular mass of a substance ( element or compound) may be defined as the average relative mass of a molecule of the substance as compared with mass of an atom of carbon (C12)taken as 12 a.m.u.
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In other words, Molecular mass expresses as to how
many times a molecule of the substance is heavier than 1/12th of the mass of an atom of carbon.
For example, A molecule of CO2 is 44 times heavier
than 1/12th of the mass of carbon atom. Hence, molecular mass of CO2 is 44
a.m.u.
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The molecular mass is obtained by adding together the atomic masses of various
atoms present in a molecule.For example,
The molecular formula of carbon dioxide is CO2.
Hence , its molecular mass is= Atomic mass of carbon + 2 x (Atomic
mass of oxygen)
= 12 + 2 x 16= 44 a.m.u.
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Gram Molecular Mass
Gram molecular mass is the quantity of a substance whose mass in grams is numerically equal to its molecular mass.
In other words, molecular mass of a substance expressed in grams is called gram molecular mass. It is also known as gram molecule.
For example, the molecular mass of oxygen is 32 and therefore, its gram molecular mass is 32 g.
Mass in grams Gram molecular mass
No. of gram molecule
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Example : Calculate the molecular mass of calcium carbonate.
Solution : Molecular mass of calcium carbonate
(CaCO3)
= 1 x atomic mass of Ca + atomic mass of C+ 3 x atomic mass of
O= 1 x 40 + 12 + 3 x 16 = 40 + 12 + 48= 100 amu
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Mole Concept• In chemistry we use a unit mole to count
particles (atoms, ions or molecules). • A mole is a collection of 6.023 x 1023 particles
irrespective of their nature.• The no. 6.023 x 1023 is called Avogadro’s
number and is denoted by No.
1 Mole = 6.023 x 1023 particles
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For example, 1 mole of atoms = 6.023 x 1023 atoms1 mole of molecules = 6.023 x 1023 molecules
1 mole of ions = 6.023 x 1023 ions 1 mole of electrons = 6.023 x 1023 electrons 1 mole of protons = 6.023 x 1023 protons
While using the term mole, it is important to indicate the kind of particles involved.
1 mole of H atoms = 6.023 x 1023 atoms of H1 mole of H molecules = 6.023 x 1023 molecules of
HydrogenThe mole is also related to the mass of the substance
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1. Mole & Gram Atomic Mass• Mass of one mole of atoms of any element in
grams is equal to its gram atomic mass or one gram atom.One mole of atoms = 6.023 x 1023atoms
= Gram atomic mass of the element
For example, The mass of 6.023 x 1023 atoms of
oxygen is 16 grams.
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2. Mole & Gram Molecular MassMass of one mole of molecules of any substance in grams is equal to its gram molecular mass.
One mole of molecules = 6.023 x 1023 molecules = Gram molecular mass
For example,• The mass of 6.023 x 1023 molecules of sulphur
dioxide (1 mole) is equal to 64 grams.• Similarly, the mass of 6.023 x 1023 molecules of
carbon dioxide (CO2) is equal to 44 grams.
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3. Mole concept for Ionic compoundsMass of one mole of formula units of any ionic compound in grams is equal to its gram formula mass.One mole of formula units = 6.023 x 1023 formula units
= Gram formula massFor example,
The mass of 6.023 x 1023 formula units of NaCl or 6.023 x 1023 Na+ ions and 6.023 x 1023 Cl- ions (one mole of NaCl) is equal to 58.5 g or
1 formula mass of NaCl.
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4.Mole in terms of VolumeOne mole of any gas at S.T.P. (0oC and 760 mm pressure) occupies 22.4 litres. This volume is known as molar volume.
For example,1 mole (2 grams) of H2 gas = 22.4 litres at S.T.P.
1 mole (28 grams) of N2 gas = 22.4 litres at S.T.P.
1 mole (32 grams) of O2 gas = 22.4 litres at S.T.P.
1 mole (44 grams) of CO2 gas = 22.4 litres at S.T.P.
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One mole of a gas = 22.4 litres at S.T.P. = 6.023 x 1023 molecules
= Gram molecular mass For example,
1 mole H2 gas = 22.4 litres at S.T.P.
= 6.023 x 1023 molecules of H2
= 2 g
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To calculate number of moles
(a)For Elements
No. of atoms of the element 6.023 x 1023
Mass of the element in grams
Gram atomic mass
Number of moles
22
Example: Calculate the number of moles in 22 g of CO2.
Solution: Molecular mass of CO2 = 12+2x16
= 44 amu Gram molecular mass of CO2 = 44 g
Since, 44 g of CO2 make = 1 mole
18 6.023 x
1023g = 0.5 mole
. . ∙ 22 g of CO2 will make
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(b) For Compounds
No. of molecules of the compound 6.023 x 1023
Mass of the compound in grams
Gram molecular mass
No. of moles
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(c) For an assembly of things
Number of things 6.023 x 1023
(d) 1 mole = gram molecular mass = 6.023 x 10 23
= 22.4 L at S.T.P.
Number of moles
25
Importance of Mole concept
1. To calculate mass of one atom of element
Gram atomic mass
6.023 x 10 23
Mass of an atom of element =
26
Example : What is the mass of single atom of Hydrogen ?
Solution : Atomic mass of Hydrogen = 1.008 amu
Gram atomic mass of hydrogen = 1.008 gThus 6.023 x 1023 atoms of
hydrogen have mass = 1.008 g 1.008 6.023 x
1023
= 1.66 x 10-
24g
Hence, a single atom of H will have mass =
27
Example : What is the mass of single atom of Carbon?
Solution : Atomic mass of Carbon =12 amu
Gram atomic mass of carbon = 12 gThus 6.023 x 1023 atoms of Carbon
have mass = 12 g
12 6.023 x 1023
= 1.99 x 10- 23g
Hence, a single carbon atom will have mass =
28
2. To calculate mass of one molecule of the substance.
Gram molecular mass
6.023 x 10 23
Mass of one molecule of substance
29
Example: What is the mass of a single molecule of hydrogen?
Solution: Molecular mass of H2 = 2 amu
Gram molecular mass of H2 = 2 g
Thus, 6.023 x10 23 molecules of H2
have mass = 2g
2 6.023 x10 23 = 3.3 x 10 - 24 g
. . A ∙ single molecule of H2 will have mass =
30
Example: Calculate the mass of a single molecule of water?
Solution: Molecular mass of H2O = 2x1 + 16
=18 amu Gram molecular mass of H2O = 18 g
Thus, 6.023 x10 23 molecules of H2O
have mass = 18g
18 6.023x1023 = 2.99 x 10-
23g
. . A ∙ single molecule of H2O will have mass =
31
3. To calculate the number of atoms in a given mass of the element.
No. of atoms in a given mass of element
Mass of element in grams Gram atomic mass x 6.023x1023
32
Example: Calculate the number of atoms in 20g of calcium (At. Mass of Ca = 40).
Solution: Atomic mass of Ca = 40 amu Gram atomic mass of Ca = 40 g
Thus 40 g of Ca contains = 6.023 x1023 atoms 6.023 x1023
40 = 3.01 x
1023atoms
. . ∙ 20g of Ca will contain x20
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4. To calculate the number of molecules in a given mass of substance.
No. of molecules in a given mass of substance
Mass of substance in grams Gram molecular massX 6.023x1023
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Example: Calculate the number of molecules of methane in 0.80 g of methane.
Solution: Molecular mass of methane (CH4)
= 1x12 + 4x1 = 16 amu
Gram molecular mass of CH4= 16 g
Thus 16g of CH4 contain = 6.023 x1023 molecules
6.023 x1023 16
= 3.01x1022 molecules
. . ∙ 0.80g of CH4 will contain x 0.80
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5. To calculate the volume occupied by a given mass of the gas at S.T.P.
Volume occupied by a given mass of a gas at S.T.P.
Mass of gas in grams Gram molecular mass of gasx22.4
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Example: Calculate the volume occupied by 16 g of oxygen at S.T.P.
Solution: Molecular mass of O2 = 32 amu
Gram molecular mass of O2 = 32 g
Now 32 g of O2 at S.T.P. occupy volume
= 22.4 L. . ∙ 16g of O2 at S.T.P. would occupy volume
22.4 32 = 11.2
litres
x16 =
37
SOLUTIONS
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Solutions
• Solution is a homogeneous mixture of two or more pure substances
• Composition of solution can be varied within certain limits.
• Solution results by dissolving a solute in a solvent. Solute + Solvent Solution
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• The substance which is dissolved & is present in lesser quantity is called solute.
• The substance in which solute is dissolved & is present in greater quantity is called solvent.
For example, 2g of sugar is dissolved into 50 ml of
water to form a solution.In this case sugar is solute & water is
solvent.
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Methods of expressing Concentration of a solution
The concentration of solution is defined as the amount of solute present in the given quantity of the solution.It can be expressed in the following ways:
1. Strength2. Molarity3. Molality4. Mole fraction5. Normality6. Mass percentage7. Volume percentage
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1. Strength: It is the amount of solute in grams dissolved per litre of the solution.Thus,
Mass of solute in grams Volume of solution in litres
If ‘a’ grams of solute is dissolved in V ml of a given solution, then
a x 1000 VIt is expressed in g/L.
Strength of solution =
Strength =
42
2. Molarity (M):The number of moles of solute dissolved per litre of the solution is called molarity.
Number of moles of soluteVolume of solution in litres
It is convenient to express volume in ml.So that
Number of moles of soluteVolume of solution in ml
Molarity =
M = x1000
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And number of moles of solute can be calculated from the given mass of solute which is dissolved
Given massGram molecular mass
Thus,Given mass x 1000Gram mol. mass x V
Molarity changes with temperature.
Moles of solute =
M =
44
Example: Calculate the molarity of the solution containing 0.5g of NaOH dissolved in 500 ml
Solution: Given mass of NaOH = 0.5g Mol.mass of NaOH = 23 + 16 + 1= 40
0.5 40
Volume of solution, V = 500ml No. of moles of NaOH x
1000
V0.0125 x 1000
500
No. of moles of NaOH = = 0.0125
Now, Molarity =
= = 0.025 M
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3. Molality: The number of moles of solute dissolved per kg of the solvent is called molality.
No. of moles of soluteMass of solvent in kgNo. of moles of soluteMass of solvent in grams
If ‘a’ grams of the soute is dissolved in W gram of the solvent, then
a 1000mol. Mass of solute W
Molality does not changes with temperature.
Molality (m) =
= x1000
m = x
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Example: Calculate the molality of an aqueous solution containing 4g of urea (Mol.mass = 60) in 500 g of water.
Solution: Given mass of urea = 4gMol. mass of urea = 60Mass of water , W = 500g Given mass 1000 Mol. mass W 4 1000 60 500
= 0.133 m
Since, Molality (m) =
=
x
x
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4. Mole fraction (x): It is the ratio of the number of moles of a component to the total number of moles of all the components (solute & solvent) in the solution.Suppose a solution contains
nA moles of solute (A) and
nB moles of solvent (B). Then,
nA
nA + nB
nB nA + nB
Mole fraction of solute (xA) =
Mole fraction of solvent (xB) =
48
The sum of mole fractions of all the components is always equal to one.
nA nB
nA + nB nA + nB
Mole fraction being a ratio, is dimensionless property.
xA + xB =
+ = 1
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Example: A solution containing 23g of ethanol & 90g of water. What is the mole
fraction of ethanol & water in solution? Solution:
Mass of ethanol (C2H5OH) = 23g
Molecular mass of ethanol = 2x12 + 1x6 + 16 = 46 23 46
Mass of water = 90g Molecular mass of water = 18
. . Number of moles of ethanol = ∙ = 0.5
50
90 18
= 5. .∙ Total number of moles = 0.5 +
5 = 5.5 0.5 5.5
= 0.09
Mole fraction of water = 1 – 0.09 = 0.91
. . Number of moles of water =∙
. . ∙ Mole fraction of ethanol =
51
5. Normality (N): It is the number of moles of gram
equivalents of the solute dissolved per litre of the given solution.
No. of gram equivalents of solute
Volume of solution in litres
No. of g.equivalents of solute Volume of solution in ml
Normality (N) =
x1000Or =
52
Gram equivalents of solute can be calculated as
Mass of solute aEquivalent mass Eq. mass
where ‘a’ is the mass of the solute in grams present in V ml of a given solution
a 1000 Eq. mass V
Like molarity, normality of a solution also changes with temperature.
Gram equivalents = =
Thus, N = x
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6. Mass percent(w/w): It is equal to the weight of solute present
per 100 g of the solution. Weight of solute Weight of solution Weight of solute in grams Wt. of solute + Wt. of solvent
Mass (weight) percent can be expressed as w/w.
For example, a 5%(w/w) solution of NaCl meansa solution containing 5g of NaCl in 100g of the solution or 95 g of the solvent.
x100
x100
%Mass =
=
54
7. Volume %age: It is equal to the volume of the component
present per 100 parts of the volume of the solution.For example,
VA and VB are volumes of the components A & B respectively in a solution, then
Volume of A Vol. of A + Vol. of B
It is expressed as v/v.
Volume %age of A = X 100
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THANK YOU