diploma_i_applied science(chemistry)_u-ii(a) preparation of solution

Course: DiplomaSubject: Applied science(Chemistry)

Unit: II(A)

Preparation of Solution

Some DefinitionsSome DefinitionsA solution is a A solution is a

______________________________ mixture of 2 or more mixture of 2 or more substances in a single phase. substances in a single phase.

One constituent is usually One constituent is usually regarded as the regarded as the SOLVENTSOLVENT and the others as and the others as SOLUTESSOLUTES..

DEFINITIONS:SOLUTES -- substances that are

dissolved(sugar)

SOLVENTS -- substance in which solutes are dissolved (usually water)

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Examples:

DefinitionsDefinitionsSolutions can be classified as Solutions can be classified as saturatedsaturated or or

ununsaturatedsaturated..A A saturatedsaturated solution contains the solution contains the

maximum quantity of solute that maximum quantity of solute that dissolves at that temperature.dissolves at that temperature.

An An unsaturatedunsaturated solution contains less solution contains less than the maximum amount of solute than the maximum amount of solute that can dissolve at a particular that can dissolve at a particular temperaturetemperature

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Mole:A mole is defined as the quantity of a substance that has the same

number of particles as are found in 12.000 grams of carbon-12. mol = weight of sample (g) / molar weight (g/mol) Avogadro's number, is 6.022x1023. The mass in grams of one mole of a compound is equal to the

molecular weight of the compound in atomic mass units. One mole of a compound contains 6.022x1023 molecules of the

compound. The mass of 1 mole of a compound is called its molar

weight or molar mass. The units for molar weight or molar mass are grams per mole.

Mole:For example, 2 H2 + O2  2 H→ 2O

2 mol of dihydrogen (H2) and 1 mol of dioxygen (O2) react to form 2 mol of water (H2O). The mole may also be used to express the number of atoms, ions, or other elementary entities in a given sample of any substance.

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/ml

Since, One litre water = 1,000 g of water

Number of water molecules 1000

Avogadro 's number18

= ×

= 55.55 NA

Hence, answer is (d)

Solution:

Molecular weight:The molecular weight of a molecule is calculated by adding

the atomic weights(in atomic mass units or amu) of the atoms in the molecule.

For example, the molecular weight of methane, molecular formula CH4, is calculated as follows.

atomic mass total mass

C 12.011 12.011

H 1.00794 4.03176

CH4 16.043

Problem 1: Find out the Molecular weight of H2O.

Atomic weight of H = 1.0079 amu Atomic weight of O = 15.9994 amu

Problem 2: Find out the Molecular weight of H2O2.

Problem 3: Al2(SO4)3

atomic weight of Al is 26.98 amu atomic weight of S is 32.06 amu. atomic weight of O is 16.00 amu.

Problem 4:  (NH4)2SN = 14.0067 amu,H = 1.0079 amuS = 32.06 amu

Problem 5: Fe2O3

Fe = 55.847O = 15.9994 amu

Problem 6: KClO4

K = 39.098 amu, Cl = 35.453 amu, O = 15.9994 amu

Equivalent weight:the mass of a substance especially in grams that combines

with or is chemically equivalent to eight grams of oxygen or one gram of hydrogen :  the atomic or molecular weight divided by the valence.

Eq. wt. of acid = molecular weight of acid Basicity of acid Eq. wt. of Base = molecular weight of Base Acidity of base

Equivalent Mass of Acid

Equivalent 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 36.5

1+= =

2 42 1 32 4 16

Equivalent mass of H SO 492

× + + ×= =

Equivalent Mass of Base

Equivalent 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 40

1+ += =

240 2 16 2 1

Equivalent mass of Ca(OH) 372

+ × + ×= =

Equivalent mass of salt

Equivalent mass of salt = Molecular mass

Total number of positive ornegative charge

Example:Equivalent mass of NaCl and MgCl2

NaCl Na Cl+ −→ +23 35.5

Equivalent mass of NaCl 58.51

+= =

2240 2 35.5

Equivalent mass of MgCl 47.52

+ ×= =

2MgCl Mg 2Cl++ −→ +

Problem 1: Find out Equivalent wt. of HCl.Problem 2: H2SO4

Problem 3: HNO3

Problem 4: NaOHProblem 5: KOHProblem 6: Na2CO3

Problem 7: K2CO3

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)=

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

98500

2

×=×

= 2N

Problem 1:  Determine the normality of given 100ml of NaOH solution.

Problem 2: Determine the Normality of given 250ml of HCl solution.

Problem 3: Determine the normality of given 750 ml of Ca(OH)2.

Problem 4: Determine the normality of given 900 ml of H3PO4.

Problem 5: Determine the amount required to prepare 1N KOH solution.

Problem 6: Determine the amount required to prepare 5N HCl solution.

Problem 7: Determine the amount required to prepare 3N NaCl solution.

Problem 8: Determine the amount required to prepare 0.1 N NaOH solution.

Problem 9: Determine the amount required to prepare 0.01 N KNO3 solution.

Problem 10: Determine the amount required to prepare 0.001 H2SO4 N solution.

Molarity

Amount of molecular weight of the solute present inone liter of solution.

M = weight of solute per liter Molecular wt. of solute

M = Wt. of Solute x 1000 Mole. Wt. of Solute X Vol of solution

example

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

0.40M 1000

40 500= ×

×

Solution:

= 0.02 M

Problem 1:  Determine the Molarity of given 100ml of NaOH solution.

Problem 2: Determine the Molarity of given 250ml of HCl solution.

Problem 3: Determine the Molarity of given 750 ml of Ca(OH)2.

Problem 4: Determine the Molarity of given 900 ml of H3PO4.

Problem 5: Determine the weight required to prepare 5M KOH solution.

Problem 6: Determine the weight required to prepare 2M HCl solution.

Problem 7: Determine the weight required to prepare 1M H2SO4 solution.

Problem 8: Determine the amount required to prepare 0.1 M H3PO4 solution.

Problem 9: How many grams of KCl would you need to add to 2 liters of water to make a .50 M solution?

Relation between normality and molarity

Mass of solute 1000N

Molecular massvolume (in ml)

n factor

×=×

N = M x n factor

For 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.

Illustrative Problem

Calculate molarity of 0.6 N AlCl3 solution.

3AlCl Al 3Cl+++ −→ +

Solution:

n = 3

0.6M 0.2M

3∴ = =

Molality

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

M = Number of mole X 1000 weight of solution

Mass of solute 1000m

Molecular mass Mass of solvent (gram)×=

×

Problem 1:  Determine the Molality of given 100gm of NaOH.

Problem 2: Determine the Molality of given 250gm of HCl solution.

Problem 3: Determine the Molality of given 750 gm of Ca(OH)2.

Problem 4: Determine the Molality of given 900 gm of H3PO4.

Problem 5: Determine the gms of solute required to prepare 5m KOH solution.

Problem 6: Determine the gms of solute required to prepare 2m HCl solution.

Problem 7: Determine the gms of solute required to prepare 1m H2SO4 solution.

Problem 8: Determine the gms of solute required to prepare 0.1 m H3PO4 solution.

gm/liter:unit of measurement of mass concentration shows how many grams of a certain substance are present in

one liter of a usually liquid or gaseous mixture.

Concentration in terms of percentage

Mass of solute% by mass 100

Volume of solution= ×

Volume of solute% by volume 100

Volume of solution= ×

V%

v=

= w/v (%)

1. What is the weight/volume percentage concentration of 250mL of aqueous sodium chloride solution containing 5g NaCl?

Calculate the weight/volume (%) = mass solute ÷ volume of solution x 100     mass solute (NaCl) = 5g     volume of solution = 250mLw/v (%) = 5g ÷ 250mL x 100 = 2g/100mL (%)

2. 10.00g BaCl2 is dissolved in 90.00g of water. The density of the solution is 1.09g/mL. Calculate the weight/volume percentage concentration of the solution.

Calculate the volume of the solution: density = mass solution ÷ volume solution     mass solution = 10g BaCl2 + 90g water = 100.00g     volume solution = mass ÷ density = 100g ÷ 1.09g/mL = 91.74mL

Calculate w/v (%) = mass solute ÷ volume solution x 100     mass solute (BaCl2) = 10.00g     volume solution = 91.74mLw/v (%) = 10.00 ÷ 91.74mL x 100 = 10.90g/100mL (%)

Numerical based on v/v%For example wine contains about 12% v/v ethanol, which

means there are 12 mL of ethanol in every 100 mL of wine.

% w/w:-

1. As an example consider 5 g sugar dissolved in 20 g of water. What is the w/w% concentration of sugar in this solution?

2. Now suppose we have 450 g of an NaCl solution that is 35 NaCl w/w %. What is the mass of NaCl ?

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= ×

0.1151000 0.01M

23 500= × =

×

Hence, answer is (a)

Solution:

ppmParts per million is a commonly used unit of concentration

for small values.One part per million is one part of solute per one million

parts solvent. ppm = 1g/m3 = 1mg/L = 1μg/mL 1ppm = 1mg/kg = 1μg/g

1. 150mL of an aqueous sodium chloride solution contains 0.0045g NaCl. Calculate the concentration of NaCl in parts per million (ppm).

ppm = mass solute (mg) ÷ volume solution (L)mass NaCl = 0.0045g = 0.0045 x 1000mg = 4.5mg 

volume solution = 150mL = 150 ÷ 1000 = 0.150Lconcentration of NaCl = 4.5mg ÷ 0.150L = 30mg/L =

30ppm

2. What mass in milligrams of potassium nitrate is present in 0.25kg of a 500ppm KNO3(aq)?

ppm = mass solute (mg) ÷ mass solution (kg)Re-arrange this equation to find the mass of solute: 

mass solute (mg) = ppm x mass solution (kg)Substitute in the values: 

mass KNO3 = 500ppm x 0.25kg = 125mg

3. A student is provided with 500mL of 600ppm solution of sucrose. What volume of this solution in millilitres contains 0.15g of sucrose?

ppm = mass solute (mg) ÷ volume solution (L)Re-arrange this equation to find volume of solution: 

volume solution (L) = mass solute (mg) ÷ ppmSubstitute in the values: 

volume solution (L) = (0.15g x 1000mg/g) ÷ 600 = 0.25LConvert litres to millilitres: volume solution = 0.25L x

1000mL/L = 250mL

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/ml

Since, One litre water = 1,000 g of water

Number of water molecules 1000

Avogadro 's number18

= ×

= 55.55 NA

Hence, answer is (d)

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 1000

MolalityMolecular mass of solute mass of solvent in gram

= ×

31000 0.2

60 250= × =

×

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 1000

MolarityMolecular mass Volume in ml

= ×

0.5 10000.025 M

40 500= × =

( )Mass of solute 1000

Normality NEquivalent mass Volume in ml

= ×

0.51000 0.025 N

40500

1

= × =×

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 =95

46= 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 10

20 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 N

1000 1000× = ×

N2 = 0.2 N

Solution:

References1. Essentials of Physical chemistry by Bahl & Tuli2. http://www.ssc.education.ed.ac.uk/bsl/chemistry/soluted.html3. http://www.chem4kids.com/files/matter_solution.html