75. Mole Concept and Stoichiometry-2

7/29/2019 75. Mole Concept and Stoichiometry-2

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Chemistry


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Session Opener


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Session Objectives

Problems related to

1. Mole concept

2. Stoichiometry


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Simple Titrations

Find out the concentration of a solution

with the help of a solution of knownconcentration.

1 1 2 2N V N V

For mixture of two or more substances

N1V1 + N2V2+ = NVWhere V=(V1 + V2+ ..)


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Normality of mixing two acids

1 1 2 2

1 2

N V +N VN=

V +V

Normality of mixing acid and bases

1 1 2 2

1 2

N V -N VN=

V +V

2 2 1 1

1 2

N V -N V

or N= V +V


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Questions


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Illustrative example 1

Find the molality of H2SO4 solution

whose specific gravity(density) is1.98 g/ml and 95% mass by volume H2SO4.

100 ml solution contains 95 g H2SO4.

95

98Moles of H2SO4 =

Mass of solution = 100 1.98 = 198 g

Mass of water = 198 95 = 103 g

Molality =

95 1000

98 103= 9.412 m

Solution:

Molality =moles of solute

mass of solvent in kg


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A sample of H2SO4 (density 1.787 g/ml)is 86% by mass. What is molarity of acid?

What volume of this acid has to beused to make 1 L of 0.2 M H2SO4?

d10%

M= Molecular mass

1.7871086

= =15.68 molar98

Let V1 ml of this H2SO4 are used to prepare 1 L of 0.2 M

H2SO4.M1V1 = M2V2

15.68 V1 = 0.2 1000

10.21000

V = =12.75 ml15.68

Solution:


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Illustrative example 3A mixture is obtained by mixing 500ml0.1M H2SO4 and 200ml 0.2M HCl at 25

0C.

Find the normality of the mixture.

2 2 1 1

1 2

N V +N VWe know, N =

V + V

For the mixture, 500 0.1 2 200 0.2 1N 0.2700

Solution:


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500 ml 0.2 N HCl is neutralized with250 ml 0.2 N NaOH. What is thestrength of the resulting solution?

Equivalents of HCl

3

500 0.2 10 eqv

Equivalents of NaOH -3=2500.210 eqv

Equivalence of excess HCl

3 3(500 0.2 10 250 0.2 10 eqv)

Normality of HCl (excess)-3 3

50010 10= =0.067 N750

Strength of HCl = .067 36.5 g/litre

= 2.44 g/litre

Solution:

2HCl NaOH NaCl H O


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Solution

1 1 2 2

1 2

N V -N VN=

V +V

0.2 1 500 - 0.2 1 250N =

500 + 250

N = 2.44 NStrength of HCl = .067 36.5 grams/litre

= 2.44 grams/litre

Normality of HCl (excess),


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Calculate the empirical formula of a mineral

having the following compositionCaO=48.0% ;P2O5=41.3% ; CaCl210.7%

9

3

1

48.0/56=0.857

41.3/142=0.291

10.7/111=0.096

56

142

111

48.0

41.3

10.7

CaO

P2O5

CaCl2

Simple

ratio

Mol.

Mass

%tageConstituents Relative no. of const.

%=Mol. mass


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1.5 g of an impure sample of sodiumsulpate dissolved in water was treatedwith excess of barium chloride solutionwhen 1.74 g of BaSO4 were obtained asdry precipitate. Calculate the percentagepurity of the sample.

2 4 2 42 23 32 16 4 137 32 4 16

142gm 233gm

Na SO BaCl BaSO 2NaCl

4 2 4

4

233 gm of BaSO is produced from 142 gmNa SO

1421.74 gm of BaSO is produced from 1.74 1.06gm233

2 4

2 4

The mass of pure Na SO in 1.5gm of impure sample is 1.06gm

1.06%tage purity of Na SO 100 70.67%

1.5


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Solution

3 2 22(23 14 32) 138gm2[23 14 48]170gm

2NaNO 2NaNO O

3

3

170gm of NaNO gives residue 138gm

138(5 x) gm of NaNO gives residue (5 x)gm

1700.812(5 x)gm

28Actual residue obtained (5 ) 100 3.6 gm

100

3 2

3

0.674x 0.812(5 x) 3.6

or 0.318x 0.46

or x 3.33 gm

Pb(NO ) in mixture 3.33gm

and NaNO in mixture 5 3.33 1.67 gm


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What volume of oxygen isrequired to effect complete combustion of

200 cm3 of acetylene and what would bethe volume of CO2 formed?

2 2 2 2 22 volume 5 volume 4 volume

2C H 5O 4CO 2H O

2 2 2

According to Gay lussac's law of gaseous volumes

2 volume of C H require 5 volume of O for complete combustion

3 3

2 2

5

200 cm of C H require 200 500 cm at STP2

2

3 32 2 2

2 volume of produce CO 4volume

4200cm of C H produce CO 200 400 cm at STP

2


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The formula weight of an acid is 82.0in a titration.100 cm3 of a solution of

this acid containing 39.0 g of the acidper litre were completely neutralisedby 95.0 cm3 of aqueous NaOH containing40.0 g of NaOH per litre. What is the basicity of the acid?

Let requirement wt. of acid =E

Normality of acid 39/EEquivalent wt. of NaOH 40 /1

Normality of NaOH 40 /40 1

Solution:


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Solution

1 1 2 2

(acid) (NaOH)

N V N V 39 100 1 95

EE 41.0

Formula wt. 82.0Basicity of acid 2

Eq. wt 41.0


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Calculate approximate molecular mass

of dry air containing 78% N2 and 22% O2.Solution:

78 22Molecular mass 28 32 28.88

100 100


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Calculate atomic mass of elementX and Y given thatComposition of X=9.76%Relative no. of atoms of X=0.406Composition of Y=26.01Relative no. of atoms of Y=1.625

Solution:

%tage of element 9.76Atomic mass of X 24

Relative no. of Atoms of X 0.406

%tage of element 26.01Atomic mass of Y 16Relative no. of Atoms in Y 1.625


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The molality of a solution of ethyl alcohol(C2H5OH) in water is 1.55m.How many

grams of ethyl alcohol are dissolved in 2kgof water?

Solution:

2

2

1000gm of H O contain 1.55mol of ethyl alcohol

1.55 20002000gmof H O contain 3.10 mol

1000

Mass of ethyl alcohol 3.10 46 142.6gm


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You are given 1 litre of 0.15 M HCl and1 litre of 0.40 M HCl. What is the maximumvolume of 0.25 M HCl which you can makefrom these solutions without adding any water?

Solution:

Volume of 0.25 M HCl cant be more than 2 litresbecause no water is added

X litre of 0.40M HCl be added to 1L 0.15 M HCl

1 1 2 2 3 3M V M V M V 0.15 1 0.40 X 0.25 (1 X)

0.15X 0.100.10

X 0.667litre0.15

Total volume of 0.25M solution 1 X

1 0.667 1.667


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How many ml of H2SO4 of density

1.8 g/ml containing 92.5% by volumeof H2SO4 should be added to 1 litre of40% solution of H2SO4 ( density 1.30 g/ml)in order to prepare 50% solution of H2SO4(density 1.4 g/ml).

Molarity(M ) of solution containing 92.5% of H SO1 2 4

vol. of H SO Density 1000 92.5 1.8 10002 4 16.99M98 100 98 100

2 2 4Molarity(M ) of solution containing 40% of H SO

40 1.3 10005.31M

98 100

2 4Molarity(M) of solution containing 50% of H SO

50 1.4 10007.14M

98 100


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Solution

1 1 2 2M V M V MV

16.99 V 5.31 1 7.14(1 V)

9.85V 1.83

1.83V 0.186L

9.85

V 0.186L or 186 ml

1

2

Let VL of solution with molarity(M ) is added to 1L of solution

with molarity(M ) to prepare(1 V)L of solution with molarityM,

then


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Gastric juice contains about 3.0 gof HCl per litre. If a person produces

about 2.5 litre of gastric juice per day,how many antacid tablets each containing400 mg of Al(OH)3 are needed to neutralizeall the HCl produced in one day?

3 3 23(35.1 1) 27 3(16 1)109.5 78g

3HCl Al(OH) AlCl 3H O

Amount of HCl produced ina day 2.5 3 7.5g

3Now 109.5g of HCl require Al(OH) 78g

378

7.5 g of HCl will require Al(OH) 7.5 5.34g109.5

5.34Number of tablets required 13.35 14 tablets

0.4

Solution:


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Calculate the number of Cl ions in

100 ml of 0.001 M HCl solution.

Solution:

HCl is a strong acid, it ionises completely

1000ml of 0.001M HCl solution contains Cl 0.001 mole

4

0.001 100100 ml of 0.001M HCl solution contains Cl

1000

1 10 mole

Conc. of HCl is equal to that of Cl

ions

23 4

19

No. of Cl ions 6.023 10 1.0 10

6.023 10


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Helena kabr se farar

Group 1 elements (Alkalimetals)

He

Li

Na

K

Rb

Cs

Fr


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Be

Mg

Ca

Sr

Ba

Ra

Bear mugs can serve bar rats

Group 2 elements(Alkaline earth metals)


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B

Al

Ga

In

Tl

Bob Allen Gave Indians Tennis Lessons

Group 13 elements (Boronfamily)


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C

Si

Ge

Sn

Pb

Can sily or Genius snatch lead

Group 14 elements (Carbon family)


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N

P

As

Sb

Bi

Never put arsenic in silver bullet bear

Group 15 elements (Nitrogen family)


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O

S

Se

Te

Po

Oh, she sells tie Poles

Group-16 elements (Oxygen family)


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F

Cl

Br

I

At

Fat Clyde bribed Innocent Atul

Group-17 elements (Halogen family)


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Group-18 elements (The NobleGas)

He

Ne

Ar

Kr

Xe

Rn

He needs our crazy Xerox repairman


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Thank you

75. Mole Concept and Stoichiometry-2

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