types of solution on basis of number of components:...

Class XII Chemistry Ch. 2: Solutions

9888737919 1

WHAT IS A SOLUTION?

Homogeneous mixtures of two or more chemically non reacting substances. Homogeneous means (single phase)

Solute + solvent = solution

e.g. sugar +water , salt +water, urea + water, liquid medicine. TYPES OF SOLUTION ON BASIS OF NUMBER OF COMPONENTS:

Binary solution: A solution having two components is called a binary solution.

Components of a binary solution: (a) Solute (b) solvent.

SOLUTE: Component of a solution which is generally in small quantity.

SOLVENT: Component of a solution which is in large quantity.

If two component: binary solution If three component: tertiary solution

TYPES OF SOLUTION ON THE BASIS OF SOLVENT:

AQUEOUS SOLUTION: A solution where solvent is water.

NON- AQUEOUS SOLUTION: A solution where solvent is not water like ethanol (C2H5OH), chloroform (CHCl3).

TYPES OF SOLUTION ON THE BASIS OF PHYSICAL STATE:

a. Solid solution: When solvent is in solid state.

b. Liquid solution: When solvent is in: liquid state. c. Gaseous solution: When solvent is in gaseous state.

TYPES OF SOLUTION IN LIQUID LIQUID SOLUTIONS:

a. completely miscible liquids: when both solute and solvent are polar or non polar. E.g. ethyl alcohol + water, e.g. benzene + hexane. b. completely immiscible liquids: when one liquid is polar and other is non polar. E.g.

benzene + water, cyclohexane + water. c. partially miscible liquids:

when two liquids are not exactly similar in nature and also not completely dissimilar in nature. E.g. ether and water, phenol +water, nitrobenzene + hexane.

Q1 Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.

Q2. Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Q3. Give an example of a solid solution in which the solute is a gas.? ADSORPTION


9888737919 2

Solute Solvent Types of solution

Examples

Solid Gas Aerosol or Solid in gas

Iodine vapours in air, Smoke, dust, fumes, camphor in water.

Solid Liquid Solid in liquid NaCl + H2O, Salt or glucose (C6H12O6) in water, sugar (C12H22O11) solution, urea. Paints, starch,

proteins, gold solution, glue, Indian ink, muddy water, milk of magnesia, white of an egg.

Solid Solid Solid in solid Alloys (brass, steel, Stainless steel, 22carot gold, bronze etc) , Coloured glasses, coloured

precious gem stones, rock salt ,.

Liquid Solid Liquid in liquid

or Gel

Hydrates salts Na2CO3.10H2O, Jellies, cheese,

curd, Fe(OH)3, Al(OH)3, butter, boot polish. Tooth polish, Hg in silver.

Liquid Liquid Emulsion Methanol (CH3OH) + water (H2O), Ethanol (C2H5OH) in water, Milk, hair cream, medicinal syrup, cod liver oil.

Liquid Gas Aerosol Humidity, chloroform+ N2, Fog, mist, cloud, mist insecticides, sprays, water vapors.

Gas Solid Solid foam Occluded gases in metals, dried sea foam, Pumice stone, bread, foam rubber, styrene foam,

dissolved gases in minerals, H2 adsorbed on Ni catalyst in hydrogenation of oil.

Gas Liquid Foam Aerated drinks, froth, soap lather, beer, shaving cream, dissolved O2 in water.

Gas Gas Gas in gas Air (O2+ N2)

HOW TO DIFFERENTIATE BETWEEN TWO SOLUTIONS? CONCENTRATION: Amount of solute in given amount of solution / solvent.

METHODS TO EXPRESS CONCENTRATION:

(NOTE: mostly numerical will be asked in this topics) 1. PERCENTAGE:

a. Mass by volume percentage (w/v):

Weight / volume % or Volume / weight % Mass (g) of the solute dissolved in 100mL of solution.

Mathematically:

Mass percentage by volume = Mass of component in solution in (g) x 100 Total volume of solution in (ml)

4% w/v solution means 4grams of solute is dissolved in 100 ml of solution.

3% v/w solution means 3ml of solute is dissolved in 100 grams of solution.

b. Mass by mass percentage (w/w): Weight / Weight %

Mass (g) of the solute dissolved in 100 g of solution.


9888737919 3

Mass percentage of a component = Mass of component in solution(g) x 100

Total mass of solution (g) 2% w / w solution means grams of solute is dissolved in 100 grams of solution.

c. volume by volume percentage (v/v):

Volume (ml) of the solute dissolved in ml of solution.

Volume percentage (v/v) = Volume of a component in mL x 100 Total volume of solution in mL

5% v/v solution means 5ml of solute is dissolved in 100ml of solution.

2. STRENGTH: (g/Lt) Strength = Normality x Equivalent Weight. Mass of solute (g) dissolved in 1L (1dm3) of solution.

1dm3= 1L 1cm3 = 1ml N1V1= N2V2

Strength is amount of solute dissolved in 1lt of solution. (g/Lt) 3. MOLARITY (M) : units: gmol L-1

No. of gmol of solute dissolved in 1L of solution.

Mole calculations: Mole = Mass in g Molar mass (g/gmol)

Molarity = Number of gmol of solute Volume of solution in litres

For example: If 2 moles of a solute are present in one litre of a solution then the molarity of solution will be (2M).

2 moles means, = (2 x Molar mass) gm of solute dissolved in 1lt of solution.

Molarity equation: M1V1 = M2V2

Q. Calculate the molarity of a solution containing 5 g of NaOH in 450 mL solution. Solution:

Moles of NaOH = 5 g = 0.125 mol 40 g mol-1 Volume of the solution = 450mL.

Molarity: No. of gmol of solute dissolved in 1L of solution or 1000mL. Given: 450ml of solution contains 0.125mol of solute.

1ml of solution contains 0.125 mol 450 1000ml of solution contains 0.125 x 1000 molL-1

450 Molarity = 0.125 mol × (1000 mL) L-1

450mL Molarity = 0.278 M or = 0.278 molL-1 = 0.278 mol dm-3


9888737919 4

4. NORMALITY: geq L-1 No. of gram equivalents of solute dissolved in 1L of solution.

Normality equation: N1V1 = N2V2

On mixing two non reacting solutions , find molarity of of the resultant solutions: N1V1 + N2V2 = N3V3

Or M1V1 + M2V2 = M3(V1+ V2)

Normality = Number of gram equivalent of solute Volume of solution in litres

Normality = Molarity x Mol mass Eq. Mass

For an acid: Normality = Molarity x basicity Basicity= Mol mass Eq. Mass

For H2SO4 basicity =2

Normality of H2SO4 = molarity of H2SO4 x 2

For a base : Normality = Molarity x acidity acidity = Mol mass Eq. Mass

Equivalent mass of an element : Atomic mass of the element Valency of an element

Equivalent mass of an acid : molar mass of the acid

Basicity of acid

Basicity is the number of displaceable H+ ions from one molecule of an acid.

For HCl , H2SO4, HNO3, H3PO4 Basicity 1 2 1 3

Equivalent mass of a base : molar mass of the base

Acidity of base Acidity is the number of displaceable OH- ions from one molecule of the base.

NaOH, KOH, Ca(OH)2 , Ba(OH)2 Acidity 1 1 2 2

Equivalent mass of a salt : Molar mass of the salt

Total positive valency of the metal atoms

Equivalent mass of an ion : Molar mass of the ion

Charge on the ion Equivalent mass of an oxidising / reducing agent : Molar mass or At mass

No. of e- lost / or gained by substance

Q. Calculate the molarity and normality of a solution containing 9.8 g of H2SO4 in 250cm3 of the solution?

Ans Molar mass of H2SO4 = 2x1 + 32 + 4x16 = 98 Moles of H2SO4 = 9.8 = 0.1 moles

98 250cm3 of solution contains moles of H2SO4 = 0.1


9888737919 5

1cm3 of solution contains moles of H2SO4 = 0.1 250

1L= 1000cm3 of solution contains moles of H2SO4 = 0.1 x1000 = 0.4M 250 Basicity =2

For an acid: Normality = Molarity x basicity

Normality = 0.4 x 2 =0.8N Second method: No of g equivalent of H2SO4 = mass of H2SO4 = 9.8

Equivalent mass molar mass / basicity of acid

No of g equivalent of H2SO4 = = 9.8 = 9.8 = 0.2 98/ 2 49

250cm3 of solution contains g equiv of H2SO4 = 0.2 1cm3 of solution contains moles of H2SO4 = 0.2

250 1L= 1000cm3 of solution contains moles of H2SO4 = 0.2 x1000 = 0.8N

250 5. MOLALITY (m):

Molality (m) is the number of gmol of solute present in 1000g or 1kg of solvent.

Mathematically: Molality = Number of moles of solute

Mass of solvent in kilograms(1000g)

NOTE: Molality is inversely proportional to 1 colligative property (Δp, π, ΔTb, ΔTf, s)

Q. Calculate molality of 2.5 g of ethanoic acid (CH3COOH) in 75 g of benzene. Solution:

Solute: ethanoic acid Molar mass of ethanoic acid: C2H4O2: = 12x2 + 1x4 + 16x2 = 60g mol-1 Moles of C2H4O2 = Mass of ethanoic acid = 2.5 g = 0.0417gmol

Molar mass of ethanoic acid 60 g/gmol Solvent: Mass of benzene in kg = 75 g

Molality (m) is the number of gmol of solute present in 1000g or 1kg of solvent.

Molality of C2H4O2 = ?

Moles of C2H4O2 = 0.0417 gmol Solvent benzene = 75g contains moles of ethanoic acid = 0.0417 gmol

If Solvent benzene 1g it will contains moles of ethanoic acid = 0.0417 gmol 75

If Solvent benzene is 1000g contains moles of ethanoic acid =0.0417 gmol x 1000 g kg-1 75 g Molality = 0.556 mol kg-1


9888737919 6

RELATION BETWEEN MOLARITY AND MOLALITY:

M= m xd where M : molarity, m molality, d density in g/cc, M2 is mol wt of solute. (1+mM2 ) 1000

Q. find the molarity and molality of a 15% solution of H2SO4 (density of H2SO4 = 1.020g

/cm3) atomic mass of H =1, O =16, S =32 Solution: Let mass of solution is 100g

Solute H2SO4 = 15g Moles of H2SO4 = 15 = 0.153 moles

98 Solvent water = 75g

Volume of solution ? Density = mass Volume

1.020 = 100 V

V = 100 = 98.04 cm3 1.020 98.04 cm3 of solution contains moles = 0.153 moles

1cm3 of solution contains moles = 0.153 moles 98.04

1L = 1000cm3 of solution contains moles = 0.153 x1000 = 1.56MolL-1 =1.56M 98.04

Molality = ?

85 g of solvent contains moles = 0.153 moles 1g of solution contains moles = 0.153 moles 85

1000g = 1kg of solution contains moles = 0.153 x1000 = 1.8mol kg-1 =1.8m 85

Using direct formula M= m xd where M : molarity, m molality, d density in g/cc, M2 is mol wt of solute. (1+mM2/1000)

1.56 = m x 1.020 = m x 1.020 ( 1 + (mx98) ) (1 + mx 0.098 )

1000) 1.56 = 1.020m (1 + 0.098m )

1.56 (1 + 0.098m) = 1.020m 1.56 + 0.153m = 1.020m

1.56 = (1.020 – 0.153) m m = 1.56 = 1.8 0.867

6. MOLE FRACTION:

Mole fraction of a component (x) = Number of moles of a component

Total number of moles of all the components If a and b are two components in a binary solution:


9888737919 7

Xa = na Xb = nb

na+ nb na+ nb

always : XA + Xb = 1

we know Moles = Mass in g

Molar mass in g/ gmol Q. Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution containing 20% of

C2H6O2 by mass? Solution : Assume: we have 100g of solution , let %age by w/w basis.

Solute ; ethylene glycol solvent: water Solute 20% therefore ethylene glycol = 20g of ethylene glycol and 80 g of water.

Molar mass of C2H6O2 = 12x2 + 1x6 + 16x2 = 62 g mol-1. Moles of C2H6O2 = 20 g = 0.322 mol 62 g mol-1

Moles of water = 80 g = 4.444 mol 18 g mol-1

Xethylene glycol = nethylene glycol Xwater = nwater nethylene glycol + nwater nethylene glycol + nwater

Xethylene glycol = 0.322 = 0.068 Xwater = 4.444 = 0.932 = 0.322 +4.444 0.068 + 4.444

Mole fraction of water can also be calculated as: Or Xethylene glycol + Xwater = 1 Xwater = 1- 0.068 = 0.932

: 1 – 0.068 = 0.932 Xethylene glycol = 0.068

Xwater = 0.932 7. MASS FRACTION:

Mass fraction of a component (x) = Mass of a component (g) Total mass of all components (g)

If a and b are two components: xa = wa xb = wb wa+ wb wa+ wb

8. PARTS PER MILLION: (ppm)

ppm = Number of parts of component x 106 Total number of parts of all components of solution. Molality, Mole fraction, Mass fraction etc are preferred to molarity, normality

etc. Because molality, mole fraction, mass fraction involves mass of solute and solvent while normality, molarity involves volume of solution. Temperature has no effect on

mass but it has significant effect on volume. Hence molality, mole fraction, mass fraction do not change with temperature whereas molarity, normality has significant effect on volume hence preferred.

SOLUBILITY:

Maximum amount of solute in g that can be dissolved in fixed amount of solvent at a specified temperature.


9888737919 8

FACTORS ON WHICH THE SOLUBILITY OF A SOLID DEPENDS:

(a) Size of solute particles :

solubility surface area of solute. smaller is the size , more is the surface area

exposed to the solvent, and therfore gretaer is the solublility of that solute. E.g. powder sugar easily

dissolves while crystal sugar takes more time comparatively.

(b) Stirring or mixing: solubility mixing.

stirring increases the contact of solute with the solvent thus increases solubility. We use mixer for making the solution very fast.

(c) Temperature:

(i) Solids solubility temperature

Mostly all solids in water solubility increases with temperature but it is not necessary in all cases. In Endothermic process(ΔHsolution= +ve): solubility increases with the increase in

temperature and vice versa. e.g. solubility of potassium nitrate KNO3 increases with the increase in temperature.

In exothermic process (ΔHsolution= -ve): Solubility decrease with the increase in temperature. e.g. solubility of calcium oxide (CaO) decreases with the increase in temperature.

(d) NATURE OF SOLUTE AND SOLVENT Solubility rule: like dissolves like:

A polar solute is dissolved in polar solvent and vice versa. e.g. water (polar) cannot be dissolved in oil which is non polar

NaCl (polar) is easy dissolve in H2O (polar) Naphthalene (non polar) and anthracene (non polar) dissolve readily in benzene (non polar) but sodium chloride (polar) and sugar do not.

DISSOLUTION: process of increase in concentration of solution due to dissolving of

solute in a solvent.

Crystallisation: process of separating out of solute in form of crystal from solution. State of dynamic equilibrium is reached when


9888737919 9

rate of dissolution = rate of crystallisation . Solute + Solvent Solution

At this stage the concentration of solute in solution will remain constant under the given conditions, i.e., temperature and pressure.

(e) PRESSURE: no effect on solubility of solid.

FACTORS ON WHICH THE SOLUBILITY OF A GAS DEPENDS:

For gases: solubility of solute decreases with temperature.

solubility of gas 1 Temperature

Gas+ solvent solution + heat Dissolution is an exothemic process, thus solubility decreases with increase of

temperature (Lechatlier principle). e.g. hot water taste flat because dissolved O2 decreses with temperature as solubility of O2 decreases with temperature.

Henry’s law:

For an ideal gas , at constant temperature, mass of gas dissolved in a fixed volume of liquid is directly proprtional to the presure of gas on the surface of the liquid.

Mathematically: m P also m = K P where K is henry’s constant which depends on the gas to be dissolves, temperature, type of the solvent used for dissolvoing gas.

Higher the KH value lower is the solubility. With the increase in temperature, KH value increases therefore solubility decreases.

Gas Temperature (K) KH (Kbar-1) O2 293 34.86 O2 303 46.82

Henry’s law can also be stated as :

The partial pressure of gas in vapour phase (p) is proportional to the mole fraction of the gas in the solution.

pA = KH xA where xA is mol fraction of gas in solution, KH is Henry’s constant. e.g. for dissolving CO2 in coca cola soft drinks, high presure is used as CO2 is having

poor solubility in aqueous solvent.

Effect of pressure on the solubility of a gas. The concentration of dissolved gas is proportional to the pressure on the gas above the

solution. Clausius Clapeyron equation: d lnC =ΔH

dT RT2


9888737919 10

lnC2 = ΔH (1 - 1) C1 RT2 (T1 T2)

LIMITATIONS OF HENRY’S LAW: (a) Gas must be ideal in nature i.e. low pressure and high temperature.

(b) Gas should not undergo compound formation with the solvent or association or dissociation in the solvent.

APPLICATIONS OF HENRY’S LAW: (a) Production of carbonated drinks like coke etc to increase the solubility of CO2 in cold

drinks at high pressure. (b) In deep sea diving e.g. He is used in artificial respiration to see divers (scuba divers).

11.7% He + 56.2% N2 + 32.1% O2 (c) For climbers or high altitude living people as. As pressure decreases concentration of gas decreases.

At constant temperature: density of gas pressure of gas (Boyles law PV =constant). i.e. Lesser amount of O2 is available in mountains for breathing due to low pressure,

therefore mountain trackers take O2 cylinder with them. Q. If N2 gas is bubbled through water at 293 K, how many millimoles of N2 gas would

dissolve in 1 litre of water? Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N2 at 293 K is 76.48 kbar.

Ans. The solubility of gas is related to the mole fraction in aqueous solution. By Henry’s law. pN2 = K xn2

x (Nitrogen) = p (nitrogen) = 0.987 = 1.29x10-5 K 76.48 x 1000

Voume of water = 1 litre Moles of water = 1000 = 55.5 mol of it,

18 if n represents number of moles of N2 in solution,

x (Nitrogen) = n = 1.29x 10–5

n+ 55.5 mol (n in denominator is neglected as it is < < 55.5) Thus n = 1.29 10-5 x 55.5 mol = 7.16 x 10-4 mol = 7.16×10-4 mol × 1000 m mol =

0.716 mmol

TYPES OF SOLUTION ON THE BASIS OF SOLUBILITY:

UNSATURATED SOLUTION: Solution which contains less amount of solute than is required to saturate it at fixed

temperature. i.e. we can add more solute to the unsaturated solution at the same T.


9888737919 11

Check for Unsaturated solution: when we add solute to the unsaturated solution and

stirr, it will dissolve more solute which means solution is unsaturated solution. SATURATED SOLUTION:

When no more of solute can be dissolved in a given solvent at the same temperature and pressure.

Or When we add crystal of solute and stirr it will not dissolve and remains as crystals. Solution which is in dynamic equilibrium with undissolved solute is the saturated solution and contains the maximum amount of solute dissolved in a given amount of solvent.

Thus, the concentration of solute in such a solution is its solubility.

SUPER SATURATED SOLUTION: When we heat the saturated solution excess of solute can be dissolved in a given

amount of solvent. A super saturated solution at a particular temperature is one that is more concentrated (contains more solute) than its saturated solution at that temperature.

CHECK: If a crystal of solute is added to this solution, the excess of solute crystallizes. VAPOR PRESSURE (P0):

Pressure exerted by vapour present in equilibrium with the solution at a particular temperature.

Factors on which vapour pressure depends: (a) Nature of liquid: lesser is the intermolecular forces, more is the vapour pressure.

(b) Temperature: higher the temperature, higher is the vapour pressure. log (P2/P1) = ΔHvap (T2-T1) 2.303R (T1T2)

Where P1, P2 are the vapour pressure at temperature T1 and T2 respectively.

How vapour pressure varies If solute is added to pure solvent? When a non-volatile solute is dissolved in a volatile solvent, the vapour pressure of solution is less than that of pure solvent.

P10-Ps = n2 = x2 n2 is no of moles of solute, n1 is no. of moles of solvent , x2 is

P10 (n2+n1)

mole fraction of solute.

P0 is vapour pressure of solvent Ps is vapour pressure of solution

P10-Ps

= i n2 n2 no. of moles of solute, n1 moles of solvent P1

0 n1 n2<<<<<n1 (for dilute solutions)

P10-Ps

= i w2/ M2 w2 is mass of solute, M2 is molar mass of solute P1

0 w1/M1 w1 mass of solvent, M1 is molar mass of solvent w2<<<<<w1 (for dilute solutions)

P0 is vapour pressure of solvent Ps is vapour pressure of solution

RAOULT’S LAW: According to Raoult’s law:

For a solution of volatile liquids the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.

PA = p0A xA PB = po

BxB

Using Dalton’s law of partial pressure the total pressure of solution is calculated. Ptotal = PA+ PB = PA

0 xA+ PB0 xB we know XA+ XB = 1


9888737919 12

= PA0(1-xB)+ PB

0 xB

saturated vapour pressure : closed system :

Equilibrium when number of particles breaking away from the surface is exactly the

same as the number sticking on to the surface again.

At equilibrium, more particles in the vapour Fewer particles in the vapour

Saturated vapour pressure is more Saturated vapour pressure is lower. Q. Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 298 K are

200 mm Hg and 415 mm Hg respectively. (i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at 298 K and, (ii) mole

fractions of each component in vapour phase. (i) Molar mass of CH2Cl2 = 12x1 + 1x2 + 35.5x2 = 85 g mol-1 Molar mass of CHCl3 = 12x1 + 1x1 + 35.5x3 = 119.5 g mol-1

Moles of CH2Cl2 = 40 g = 0.47 mol 85 g mol-1

Moles of CHCl3 = 25.5 g = 0.213 mol 119.5 g mol-1 Total number of moles = 0.47 + 0.213 = 0.683 mol

x CH2Cl2 = 0.47 mol = 0.688

0.683 mol x CHCl3 = 1.00 – 0.688 = 0.312

Using equation Ptotal = PCH2Cl2+ PCHCl3 = P CH2Cl20 x CH2Cl2+ P CHCl3

0 x CHCl3 = (415) 0.688 + 200 x 0.312

= 285.5 + 62.4 = 347.9 mm Hg (ii) Using the relation , yi = pi

ptotal, mol fraction of the components in gas phase (yi).

p CH2Cl2 = 0.688 x 415 mm Hg = 285.5 mm Hg p CHCl3 = 0.312 x 200 mm Hg = 62.4 mm Hg

y CH2Cl2 = 285.5 mm Hg/347.9 mm Hg = 0.82 y CHCl3 = 62.4 mm Hg/347.9 mm Hg = 0.18

Note: Since, CH2Cl2 is a more volatile component than CHCl3, [p0

CH2Cl2 = 415 mm Hg

and p0CHCl3 = 200 mm Hg] and the vapour phase is also richer in CH2Cl2 [y CH2Cl2 = 0.82

and y CHCl3 = 0.18], it may thus be concluded that at equilibrium, vapour phase will be always rich in the component which is more volatile.

COMPARISION OF RAOULT’ LAW AND HENRY’S LAW, It is observed that


9888737919 13

(a) The partial pressure of volatile component or gas is directly proportional to its mole fraction in solution.

(b) both apply to the volatile component of the solutions. DISSIMILARITIES:

In proportionality constant: In case of Henry’s Law the proportionality constant is KH and it is different from p1

0 which is partial pressure of pure component.

Raoult’s Law becomes a special case of Henry’s Law when KH becomes equal to p10

in Henry’s law.

IDEAL AND NON IDEAL SOLUTIONS: Liquid –liquid solutions can be classified into ideal and non-ideal solutions on basis of

Raoult’s Law.

Ideal solutions Non- ideal solutions

The solutions that obey Raoult’s Law over the entire range of concentrations

are known as ideal solutions.

When a solution does not obey Raoult’s Law over the entire range of concentration,

then it is called non-ideal solution.

ΔHmix =0 and ΔVmix=0 ΔHmix ≠0 and ΔVmix ≠0

The intermolecular attractive forces between solute molecules and solvent

molecules are nearly equal to those present between solute and solvent molecules i.e. A-A and B-B interactions

are nearly equal to those between A-B

The intermolecular attractive forces between solute molecules and solvent

molecules are not equal to those present between solute and solvent molecules i.e. A-A and B-B interactions are not equal to

those between A-B

Benzene +toluene,

n- hexane + n heptanes, chlorobenzene + bromobenzene.

Ethyl bromide + ethyl chloride Chloro benzene+ bromo benzene

Ethanol +water, CS2+ acetone ,

propanone+ CHCl3

Pure solvent Ideal solution

Non Ideal solution


9888737919 14

TYPES OF NON IDEAL SOLUTIONS

Non ideal solution showing positive

deviation (+ve deviation )

Non ideal solution showing negative

deviation (-ve deviation )

The vapour pressure of a solution is

higher than that predicted by Raoult’s Law

The vapour pressure of a solution is

lower than that predicted by Raoult’s Law

The intermolecular attractive forces between solute-solvent molecules are weaker than those between solute-

solute and solvent-solvent molecules i.e. A-B < A-A and B-B interactions

The intermolecular attractive forces between solute-solvent molecules are stronger than those between solute-

solute and solvent-solvent molecules i.e. A-B > A-A and B-B interactions

e.g ethyl alcohol + cyclo hexane or CS2+ acetone.

e.g. chloroform + acetone.

ΔV mixing = +ve ΔV mixing = -ve

ΔH mixing = +ve ΔH mixing = -ve

Examples : Acetone +carbon disulphide

Acetone + benzene Ethyl alcohol + water

Carbon tetrachloride + benzene Acetone + ethyl alcohol Methyl alcohol + water

Carbon tetrachloride + chloroform Carbon tetrachloride + toluene

Examples : Chloroform + benzene

Chloroform + di ethyl ether Acetone + aniline

HCl + water HNO3 + water Acetic acid + pyridine


9888737919 15

AZEOTROPIC MIXTURES/ AZEOTROPES : Constant boiling mixtures: Binary mixtures having same composition in liquid and vapour phase and boil

at constant temperature like a pure liquid. Liquids forming azeotrope cannot be separated by fractional distillation.

TYPES OF AZEOTROPES: (a) Minimum boiling azeotrope (+ve deviation from ideal behaviour):

The solutions which show a large positive deviation from Raoult’s law form minimum boiling azeotrope at a specific composition.

(b) Maximum boiling azeotrope (-ve deviation from ideal behaviour). The solutions that show large negative deviation from Raoult’s law form maximum

boiling azeotrope at a specific composition.

EXAMPLES OF AZEOTROPES:

Maximum bp azeotropes (-ve deviation) Minimum bp azeotropes (+ve dev-) HNO3+ H2O bp 393.5K Ethanol + Water bp = 351.1K

HCl + water C2H5OH + C6H6 HBr +H2O C2H5OH+ CCl4 CHCl3 + (CH3)2CO CH3COOC2H5 + H2O

CHCl3 + CH3COOCH3 (CH3)2CO+ CH3OH HI+ H2O

COLLIGATIVE PROPERTIES:

Properties of solution which depends on only the number of solute particles in fixed amount of solvent but not on the nature of solute are called colligative properties.

There are four colligative properties: a. Relative lowering of vapour pressure


9888737919 16

(p1o - p1 ) = x2 p1

o vapour pressure of pure solvent

p10 p1 vapour pressure of solution.

x2 mol fraction of solute b. Elevation of boiling point.

bTb- Tb0

bKb x m m is molality Kb is molal elevation constant or ebullioscpic constant.

c. Depression of freezing point: bb

ΔTf = Kf x m Kf is molal depression constant or cryoscopic constant M is molality.

d. Osmotic pressure: CRT

Osmotic pressure () Molarity (C) of the solution R = 0.082atm L /K .gmol T is temperature in Kelvin

USES OF COLLIGATIVE PROPERTY: Colligative properties help in calculation of molar mass of solutes.

Colligative property no. of particles

no. of molecules in case of non electrolyte.

no. of ions (in case of electrolytes)

no. of moles of solute

mol fraction of solute Relative lowering of vapour pressure:

Difference in the vapour pressure of pure solvent (P10) and solution (ps) represents

lowering in vapour pressure (p1o - ps ).

Dividing lowering in vapour pressure by vapour pressure of pure solvent is called relative lowering of vapour pressure (p1

o - ps )

p10

Relative lowering of vapour pressure is directly proportional to mole fraction of solute (x2). Hence it is a colligative property.

Q. The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar

mass 78 g mol-1). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?

Ans. The various quantities known to us are as follows: w2 = 0.5 g; w1 = 39 g p1

0 = 0.850 bar; ps = 0.845 bar; M1 = 78 g mol-1 ,

Let M is the moles of solute= M g/ gmol moles of solute = 0.5/ M

moles of solvent = 39/ 78 p1

o - ps = 0.850 bar – 0.845 bar = n solute = 0.5 g × 78 g mol

Po 0.850 bar n solvent M2 × 39 g

Therefore, M2 = 170 g mol–1 Elevation of boiling point:

The difference in boiling points of solution ( ΔTb ) and pure solvent (Tb0 ) is called

elevation in boiling point

bTb- Tb0


9888737919 17

For a dilute solution elevation of boiling point is directly proportional to molal concentration of the solute in solution. Hence it is a colligative property.

b bb

ΔTb =Kb x 1000x w2 w2 is wt of solute, w1 is wt of solvent , M2 mol wt of solute

M2w1 Kb is molal elevation constant or ebullioscpic constant , Kb T2

Q. 18 g of glucose, C6H12O6, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? Kb for water is 0.52 K kg mol-1.

Sol. Kb for water is 0.52 K kg mol-1 Moles of glucose = 18 g/ 180 g mol-1 = 0.1 mol

Number of kilograms of solvent = 1 kg =1000g Thus molality of glucose solution = 0.1 mol kg-1

For water, change in boiling point ΔTb = Kb m = 0.52 K kg mol-1x 0.1 mol kg-1 = 0.052 K Since water boils at 373.15 K at 1.013 bar pressure, therefore, the boiling point of

solution will be 373.15 + 0.052 = 373.202 K. The boiling point of benzene is 353.23 K.

Q. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. Kb for benzene is

2.53 K kg mol-1? Solution

W2 = 1.80 W1 = 90 M2: molar mass of solute

Kb (benzene) = 2.53 K kg mol-1 The elevation (ΔTb) in the boiling point = 354.11 K – 353. 23 K = 0.88 K ΔTb =Kb x 1000x w2 w2 is wt of solute, w1 is wt of solvent , M2 mol wt of solute

M2w1 Kb is molal elevation constant ΔTb = Kb m

0.88 = 2.53 x (1.8/M2) x1000

90 we get M2 = 2.53 K kg mol-1 × 1.8 g × 1000 g kg–1 = 58 g mol-1

0.88 K × 90 g Therefore, molar mass of the solute, M2 = 58 g mol-1

Depression of freezing point: The lowering of vapour pressure of solution causes a lowering of freezing point compared

to that of pure solvent.The difference in freezing point of the pure solvent (Tf0) and

solution ( Tf ) is called the depression in freezing point.

ff


9888737919 18

For a dilute solution depression in freezing point is a colligative property because it is directly proportional to molal concentration of solute.

ΔTf = Kf x1000x w2 M2w1 w2 is mass of solute,

M2 mol wt of solute, w1 mass of solvent.

Kf is molal depression constant or

cryoscopic constant Kf T2 ΔTf = Kf x1000x w2

M2w1

APPLICATIONS OF DEPRESSION IN FREEZING POINT: (a) In making antifreeze solutions (ethylene glycol in water) for car radiators in hilly areas.

(b) melting of ice/ snow on roads by adding NaCl or CaCl2

Q. What are anti-freeze solutions? Give one example.

Ans. Substance which depress the freezing point of the solution/ water. i.e. after adding depressant water would not freeze at 273K and can handle sub zero temperature without solidification/ freezing. e.g. ethylene glycol is added in car radiators or NaCl,

CaCl2 are scattered on hilly roads to melt the ice.

Q. 45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of the solution. Depression in freezing point is related to the molality, therefore, the molality of the

solution with respect to ethylene glycol. Moles of ethylene glycol = 45 g = 0.73 mol

62 g mol-1 Mass of water in g = 600g

Hence molality of ethylene glycol = 0.73 x 1000 = 0.73 mol = 1.2 mol kg-1 600 0.60 kg Therefore freezing point depression, ΔTf = Kf x m = 1.86 K kg mol-1x1.2 mol kg-1 =2.2 K

Freezing point of the aqueous solution = 273.15 K – 2.2 K = 270.95 K

Q. 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K

kg mol–1. Find the molar mass of the solute. Ans. ΔTf = Kf x1000x w2


9888737919 19

M2w1

M2 = Kf x 1000x w1

w2 M2 = 5.12 K kg mol-1× 1000 x1.00 g = 256 g mol-1 0.40 × 50 g

Thus, molar mass of the solute = 256 g mol-1

OSMOSIS: The phenomenon of flow of solvent molecules through a semi permeable membrane from pure solvent to solution is called osmosis.

REVERSE OSMOSIS:

The process of movement of solvent through a semi permeable membrane from the solution to the pure solvent by applying excess pressure on the solution side is called

reverse osmosis.

REVERESE OSMOSIS

EXCESS PRESSURE APPLIED MUST BE APPLIED ON THE SOLUTION TO PREVENT

OSMOSIS. OSMOTIC PRESSURE:

The excess pressure that must be applied to solution to prevent the passage of solvent into solution through a semi

permeable membrane is called osmotic pressure. Osmotic pressure is a colligative property as it depends on the Number of solute particles and not on their identity.

For a dilute solution,

osmotic pressure () is directly proportional to the molarity (C)

of the solution i.e. CRT R = 0.082atm L /K .mol

Osmotic pressure can also be used to determine the molar mass of solute using the equation M2 =iw2 RT w2 is mass pof solute, M2 mol wt of solute, V volume of solution.

V

i n2RT V ISOTONIC SOLUTION.

Two solutions having same osmotic pressure at a given temperature are called isotonic solution.

n1 = n2 or w1 = w2

V1 V2 m1V1 m2V2

Human blood is isotonic 0.9% solution of NaCl


9888737919 20

HYPERTONIC SOLUTION: If a solution has more osmotic pressure than other solution it is called hypertonic

solution. A solution of NaCl with concentration > 0.9% is hypertonic solution.

HYPOTONIC SOLUTION:

If a solution has less osmotic pressure than other solution it is called hypotonic solution. A solution of NaCl with concentration < 0.9% is hypotonic solution.

RBCs will shrink in this condition called plasmolysis.

Q. 200 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57x10-3 bar. Calculate the

molar mass of the protein.

Ans. The various quantities known to us are as follows: = 2.57x10-3 bar, V = 200 cm3 = 0.200 litre (1lt 1000cm3)

T = 300 K R = 0.083 L bar mol-1 K-1

CRT

Let M2 is the molar mass of solute

wRT M2

M2 = wRT = 1.26 g × 0.083 L bar K-1 mol -1× 300 K = 61,022 g mol-1

2.57×10-3 bar × 0.200 L

ABNORMAL MOLAR MASS: Molar mass that is either lower or higher than expected or normal molar mass is called as abnormal molar mass.

VANT HOFF FACTOR:

van’t Hoff factor (i)accounts for the extent of dissociation or association .

icalculated molar mass(MC) = Observed colligative property observed (abnormal) molar mass (Mc) Calculated colligative property

= Total number of moles of particles after association/dissociation Total number of moles of particles before association/dissociation

For Association: i < 1 For dissociation : i> 1

Inclusion of van’t Hoff factor modifies the equations for colligative properties as: P1

0-P1 =i n2 P1

0 n1

ΔTb =i Kb m


9888737919 21

ΔTb =i Kb x 1000x w2 /M2w1 ΔTf = iKf m

ΔTf = i Kf x1000x w2 M2w1

i n RT V

For association: i = 1- (1-1/n)

= 1-i (1-1 )

n

for dissociation : i= (1- + n ) 1

= i-1 Where is degree of dissociation, n is no. of ions produced by complete (n-1) dissociation of 1 mole of substance.

Q. 2 g of benzoic acid (C6H5COOH) dissolved in 25 g of benzene shows a depression in

freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol–1. What is the percentage association of acid if it forms dimer in solution? The given quantities are: w2 = 2 g; Kf = 4.9 K kg mol-1; w1 = 25 g,

ΔTf = 1.62 K ΔTf = Kf x1000x w2

M2w1 Substituting these values in above equation M2 = 4.9 K kg mol-1 × 2 g × 1000

25 g × 1.62 K Observed molar mass of benzoic acid M2 = 241.98 g mol-1

Thus, experimental molar mass of benzoic acid in benzene is = 241.98 g mol-1 Now consider the following equilibrium for the acid:

If represents the degree of association of the solute then we would have (1 – ) mol of benzoic acid left in unassociated form and correspondingly as associated moles of

benzoic acid at equilibrium. Therefore, total number of moles of particles at equilibrium as below: 2C6H5COOH (C6H5COOH)2

T=0 1 0

T= equilibrium 1- /2

Therefore total no. of moles after association = 1- + /2 = 1- /2 Thus, total number of moles of particles at equilibrium equals van’t Hoff factor i.


9888737919 22

i = 1- /2 = 0.504 1

= (1-0.504) x 2 = 0.992 or %age dissociation = 99.2% Therefore, degree of association of benzoic acid in benzene is 99.2 %.

Q. 0.6 mL of acetic acid (CH3COOH), having density 1.06 g mL-1, is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205°C.

Calculate the van’t Hoff factor and the dissociation constant of acid. Sol. Number of moles of acetic acid = 0.6 mL = 0.0106 mol = n

1.06 g mL x60 g mol

Molality = 0.0106 mol x 1000 mL = 0.0106 mol kg–1

1 g mL-1

Using equation ΔTf = Kf x m = 1.86 K kg mol-1x 0.0106 mol kg-1 = 0.0197 K

van’t Hoff Factor (i) = Observed freezing point = 0.0205 K = 1.041 Calculated freezing point 0.0197 K

Acetic acid is a weak electrolyte and will dissociate into two ions: acetate and hydrogen

ions per molecule of acetic acid. CH3COOH H+ + CH COO-

If is the degree of dissociation of acetic acid, then we would have C(1 – )

moles of undissociated acetic acid, C moles of CH3COO- and C moles of H+ ions, Inital moles 1 0 0

At equilibrium 1-

Thus total moles of particles are: C(1 – + + ) = C(1 + ) i = 1.041

Thus degree of dissociation of acetic acid = = 1.041– 1.000 = 0.041

Then [CH3COOH] = C(1 – ) = 0.0106 (1 – 0.041),

[CH3COO-] = C = 0.0106 x 0.041,

[H+] = C = 0.0106 x 0.041

Dissociation constant Ka = [C ][ C ] = [0.0106 x 0.041] x [0.0106 x 0.041]

[CH3COOH ] 0.0106 (1.00 - 0.041) Ka = 1.86 x 10-5

DIFFERENCE BETWEEN OSMOSIS AND DIFFUSION:

OSMOSIS DIFFUSION

Semi permeable membrane is used. No semi permeable membrane is used.

Flow of solvent from semi permeable membrane

Solvent and solute molecules move directly to each other.

Molecules move from lower concentration to higher concentration.

Molecules move from higher concentration to lower concentration.

Applies to liquid solutions only. It takes place in liquid and gases solution .

Can be reversed by applying osmotic

pressure.

It cannot be reversed.

---xxx---

types of solution on basis of number of components:...

Documents