Chemistry

SOME BASIC CONCEPT OF CHEMISTRYSESSION -IV

Session Opener

Session Objectives

Session Objectives

1. Equivalent mass

2. Normality

3. Molarity

4. Molality

5. Strength of solution

6. Percentage concentration

Equivalent Mass

Equivalent mass Base

Salt

Acid

Equivalent Mass of AcidEquivalent mass of acid =

Molecular mass of acidNumber of replacable H (Basicity)

Example:Equivalent mass of HCl and H2SO4

HCl H Cl

2 4 4H SO 2H SO

1 35.5Equivalent mass of HCl 36.51

2 42 1 32 4 16Equivalent mass of H SO 492

Equivalent Mass of BaseEquivalent mass of base =

Molecular massNumber of replacable OH (Acidity)

Example:Equivalent mass of NaOH and Ca(OH)2

NaOH Na OH

2Ca(OH) Ca 2OH

23 16 1Equivalent mass of NaOH 401

240 2 16 2 1Equivalent mass of Ca(OH) 372

Equivalent mass of saltEquivalent mass of salt =

Molecular massTotal number of positive ornegative charge

Example:Equivalent mass of NaCl and MgCl2

NaCl Na Cl 23 35.5Equivalent mass of NaCl 58.51

2240 2 35.5Equivalent mass of MgCl 47.52

2MgCl Mg 2Cl

Concentration of solutions(1) Normality

Number of equivalents of solutepresent in one litre of solution.

Equivalent of soluteN Volume of solution in litre

Mass of solute 1000N Equivalent mass of solute volume (in ml)

EquivalentsAlso N V(in litre)

Equivalents = N x V (in litre)Milli equivalents = N x V (in ml)

Question

Illustrative example

Find the normality of H2SO4 having49g of H2SO4 present in 500 ml ofsolution.

Solution:

Mass of solute 1000N Equivalent mass volume (in ml)

49 1000N 98 5002

= 2N

Most important point about equivalentsEquivalent and milliequivalents ofreactants reacts in equal number togive same number of equivalents ormilliequivalents of products separately.

Example:

2 4 2 4 22 Equivalents 2 Equivalents 2 Equivalents

2 moles 1 mole 1 mole 2 mole

2NaOH H SO Na SO 2H O

Question

Illustrative Problem20 ml of 0.1 N BaCl2 is mixed with 30 mlof 0.2 N Al2(SO4)3. How many gram ofBaSO4 are formed?

2 2 4 3 3 4BaCl Al (SO ) AlCl BaSO

Solution:

20 0.11000

By equivalent method, no need of balancing theequation. Because equivalents of reactants andproducts are same.

2 2 4 3 3 4BaCl Al (SO ) AlCl BaSO

Equivalents of BaCl2 = 30 0.21000

= 2 x 10–3

= 6 x 10–3

Equivalents of Al2(SO4)3 =

Solution contd-

Since equivalents of Al2(SO4)3 is in excess, henceequivalents of BaSO4

= equivalents of BaCl2= equivalents of AlCl3= 2 x 10–3

Hence, mass of BaSO4 = Equivalents x equivalent mass3 2332 10 .233g2

If we will discuss thisproblem through mole concept, then we have to balanced the equation.

Molarity

Number of moles of solute present inone litre of solution.

Moles of soluteM Volume (in litre)

Moles of solute 1000M Molecular mass volume (in ml)

Moles = Molarity x volume (in litre)

Milli moles = Molarity x volume (in ml)

Question

Illustrative example

Calculate the molarity of a solution ofNaOH in which 0.40g NaOH dissolvedin 500 ml solution.

0.40M 100040 500

Solution:

= 0.02 M

Relation between normality and molarity

Mass of solute 1000N Molecular mass volume (in ml)n factor

N = M x n factorFor HCl, n = 1 H2SO4, n = 2 H3PO4, n = 3 NaOH, n = 1 Ca(OH)2, n = 2

For monovalent compound (n = 1)Normality and molarity is same.

Question

Illustrative Problem

Calculate molarity of 0.6 N AlCl3 solution.

3AlCl Al 3Cl

Solution:

n = 3

0.6M 0.2M3

Molality

Number of moles of solute present in 1 Kg(or 1000 gram) of solvent. It is representedby m (small letter).

Moles of solutem Mass of solvent

Mass of solute 1000m Molecular mass Mass of solvent (gram)

Question

Illustrative ProblemCalculate the molality of 1 molar solutionof NaOH given density of solution is 1.04gram/ml.

Solution:1 molar solution means 1 mole of solute present perlitre of solution.

1m 1000 1 molal solution.1000

Therefore, mass of 1 litre solution = 1000 x 1.04= 1040 gram

Mass of solute = 1 x 40 = 40gTherefore, mass of solvent 1040 – 40 = 1000g

Strength of solutionAmount of solute present in one litresolution.

Mass of soluteStrength Volume of solution (in litre)

Strength Molarity Molecular mass

Strength Normality Equivalent mass

Question

Illustrative ProblemCalculate strength of 0.01 N of NaOHsolution.

Solution:

Strength = Normality x equivalent mass= 0.01 x 40 = 0.4 gram/litre

Concentration in terms of percentage

Mass of solute% by mass 100Volume of solution

W %w

Volume of solute% by volume 100Volume of solution

V %v

Question

Illustrative ProblemCalculate the concentration of 1 molalsolution of NaOH in terms of percentageby mass.

Solution:1 molal solution means 1 mole (or 40g) NaOH presentin 1000g of solvent.Total mass of solution = 1000 + 40 = 1040gTherefore, 1040g solution contains 40g NaOH

Therefore, 100g solution contains 40 1001040

= 3.84% by mass.

Class exercise

Class exercise 10.115 g of pure sodium metal was dissolved in 500 ml distilled water. The molarity of the solution would be (Na = 23)(a) 0.010 M (b) 0.00115 M

(c) 0.023 M (d) 0.046 M

Mass of solute 1000M Molecular mass of solute Volume in ml

0.115 1000 0.01M23 500

Hence, answer is (a)

Solution:

Class exercise 2The number of moles of oxygen in one litre of air containing 21% oxygen by volume, in standard conditions, is(a) 0.186 mole (b) 0.21 mole

(c) 2.10 mole (d) 0.0093 mole

21% oxygen by volume means 21 ml oxygen is present in 100 ml of solution.1,000 ml of solution will contain 210 ml.Since at STP 22,400 ml of gas = 1 mole,

210210 ml of oxygen 0.0093 mole22400

Hence, answer is (d)

Solution:

Class exercise 3The vapour density of a gas is 11.2. The volume occupied by 11.2 g of the gas at STP will be(a) 11.2 L (b) 22.4 L

(c) 1 L (d) 44.8 L

Molecular mass = 2 × Vapour density = 2 × 11.2 = 22.4

Since 22.4 g contains 22.4 L of gas at STP,22.4×11.2=11.2L of gas22.411.2g contains

Hence, answer is (a)

Solution:

Class exercise 4The number of water molecules in one litre of water is(a) 18 (b) 18 × 1000

(c) NA (d) 55.55 NA

For water d = 1 g/mlSince, One litre water = 1,000 g of waterNumber of water molecules 1000 Avogadro's number18

= 55.55 NA

Hence, answer is (d)

Solution:

Class exercise 5Which is not affected by temperature?(a) Normality

(b) Molarity and molality

(c) Molarity

(d) Molality

Molality involves mass of solute and solvent which are not affected by temperature.

Solution:

Class exercise 6Calculate the molality and mole fraction of the solute in aqueous solution containing 3.0 g of urea (molecular mass = 60) in 250 g of water.

Mass of solute 1000Molality Molecular mass of solute mass of solvent in gram

3 1000 0.260 250

Mole fraction of urea =Moles of urea 3 / 60 0.003593 250Total moles60 18

Mole fraction of water = 1 – 0.00359 = 0.996

Solution:

Class exercise 7Calculate the molarity and normality of a solution containing 0.5 g ofNaOH dissolved in 500 ml.

Mass of solute 1000Molarity Molecular mass Volume in ml

0.5 1000 0.025 M40 500

Mass of solute 1000Normality N Equivalent mass Volume in ml

0.5 1000 0.025 N40 5001

Or for monovalent compound like NaOH normality and molarity are same.

Solution:

Class exercise 8Calculate the mol fraction of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass.

95% of ethanol by mass means 95 g ethanol present in 100 g of solution.Hence, mass of water = 100 – 95 = 5 g

Moles of C2H5OH =9546

= 2.07 moles

Moles of water(H2O)= 5 0.28mol18

Solution:

Solution

Mole fraction of C2H5OH = 0.28 0.880.28 2.07

Mole fraction of water = 1 – 0.88 = 0.12

Class exercise 9A solution contains 25% of water, 25% of ethanol and 50% of acetic acid by mass. Calculate the mole fraction of each component.

25x + 25x + 50x = 100x = 1Mass of water = 25 gMass of ethanol = 25 gMass of acetic acid = 50 g

Moles of water = 25 1.388 moles18

Solution:

Solution

Moles of ethanol = 25 0.543 moles46

Moles of acetic acid = 50 =0.833 moles60Mole fraction of ethanol = 1.388 0.5022.764

Mole fraction of acetic acid = 1 – 0.503 –0.196 = 0.301

Class exercise 1020 ml of 10 N HCl are diluted with distilled water to form one litre of the solution. What is the normality of thediluted solution?

N1V1 = N2V2

2

20 100010 N1000 1000

N2 = 0.2 N

Solution:

Thank you