CLASS 12 - Dev Academy

Class

12

Dev Academy Yamunanagar (Haryana)

Solutions

By Mr. Hardev Singh Antaal

CHEMISTRY

Website: www.devacademy22.com

Email id: [email protected] 7404044544

http://www.devacademy22.com/

mailto:[email protected]

Dev Academy – YNR SOLUTIONS

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Solutions

Unit - 2

Introduction A solution is a homogenous mixture of two or more non-reacting compounds whose

composition can be varied within certain limits.

Depending upon the total components present in the solution, A solution may be

binary solution (Two component), ternary solution (Three components),

quaternary solution (Four components)

The components of binary solution are generally referred to as solute and solvent. The

component which is present in large quantity is called Solvent while the component

which is present in lesser quantity is called solute.

Brass (mixture of copper and zinc), German silver (mixture of copper,

zinc and nickel) and Bronze (mixture of copper and tin)

1 ppm of fluoride ions in water prevents tooth decay, while 1.5 ppm

causes the tooth to become mottled and high concentrations of fluoride

ions can be poisonous (for example: Sodium fluoride is used in rat poison.)

Types of Solutions Depending on the physical states of solute and solvent, the solution can be of following

types:

NOTE 1

NOTE 2




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Aqueous solutions and non-aqueous solutions : Aqueous solutions are those in which water is act as solvent

Non-aqueous solutions are those in which water is not act as solvent (usually

benzene, ether, carbon tetrachloride and acetone act as solvent)

Concentrated or Diluted Solution : Concentrated solutions are those which have relatively large quantity of solute.

Diluted solutions are those which have relatively small quantity of solute.

Saturated and Unsaturated solutions : A solution in which no more solute can be dissolved at the same temperature and

pressure is called Saturated solution.

A solution in which more solute can be dissolved at the same temperature and pressure

is called Unsaturated solution.

SECTION -A A 1.: Methods for expressing the concentration of a Solution The concentration of a solution may be defined as the amount of solute present in the

given quantity of the solution. The concentration of solution can be expressed in

various common ways as discussed below :

(a) Mass percentage (b) Volume percentage (c) Mass by volume percentage

(d) parts per million (e) Strength (f) Molarity

(g) Molality (h) Mole fraction (i) Normality

(a) Mass percentage (w/w) : It may be defined as mass of a component in grams present in 100 g of the solution.

Mass % of component = Mass of component in the solution × 100

Total mass of solution

Mass % of A = WA × 100

WA + WB =

WA × 100

Wsol Mass % of B =

WB × 100

WA + WB =

WB × 100

Wsol

[NOTE : Commercial bleaching solution contains 3.62 mass % of sodium

hypochlorite in water]




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(b) Volume percentage (v/v) : It may be defined as volume of a component in ml present in 100 ml of the solution.

Volume % of component = Volume of component in the solution × 100

Total volume of solution

Volume % of A = VA × 100

VA + VB =

VA × 100

Vsol Volume % of B =

VB × 100

VA + VB =

VB × 100

Vsol

[NOTE : 30 % (v/v) aqueous solution of ethylene glycol, an antifreeze, is used in cars

for cooling engine. At this concentration the antifreeze lowers the freezing point of

water to 255.4 K (-17.6 ˚C)]

(c) Mass by volume percentage (w/v) : It can be defined as the mass of component in grams dissolved in 100 ml of solution.

Mass by Volume % of a component (w/v) = Mass of a component in the solution × 100

Total volume of the solution in mL

Mass by Volm % of A = WA × 100

Vsol Mass by Volm % of B =

WB × 100

Vsol

(d) Parts per million (ppm) : It can be defined as the part of a component per million parts of the solution.

➢ ppm of component (w/w) = Mass of component in the solution × 106

Total mass of solution

ppm of A = WA × 106

WA + WB =

WA × 106

Wsol ppm of B =

WB × 106

WA + WB =

WB × 106

Wsol

➢ ppm of component (v/v) = Volume of component in the solution × 106


ppm of A = VA × 106

VA + VB =

VA × 106

Vsol ppm of B =

VB × 106

VA + VB =

VB × 106

Vsol

➢ ppm of component (w/v) = Mass of component in the solution × 106





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Practice problems

Q 1.: If 16 g of oxalic acid is dissolved in 750 ml of solution, what is the mass % of

oxalic acid in solution ? ( density is 1.1 g cm-3)

Q 2.: A litre of public supply water contain 3 × 10-3 g of chlorine. Calculate the mass %

and ppm of chlorine.

Q 3.: A litre of sea water (about 1030 g) contain 6 × 10-3 g of dissolved oxygen.

Calculate the mass % and ppm of dissolved oxygen.

Q 4.: Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when

6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.

Q 5.: Calculate the percentage composition in terms of mass of a solution obtained by

mixing 300 g of a 25 % and 400 g of a 40 % solution by mass.

(e) Strength : It may be defined as the amount of solute in grams present in 1 litre of the solution.

Strength of a solution (g/L) = Mass of the solute in grams

Volume of solution in litres =

WB

Vsol in L =

WB × 1000

Vsol in ml

(f) Molarity : It may be defined as number of moles of solute dissolved in 1 litre of the solution.

It is denoted by ‘M’ and is expressed in mol/L.

Molarity (M) = Number of moles of solute

Volume of the solution in Litres =

nB

Vsol in L =

nB ×1000

Vsol in mL

Moles of solute (nB) = Weight of solute

Molecular mass of solute =

WB

MB

So that Molarity (M) = WB × 1000

MB × Vsol in mL =

Strength

MB

Molarity (M) = WB × 1000

MB × Vsol in mL =

WB × 1000 × Dsol in gm/ml

MB × Wsol in gm =

WB × 1000 × Dsol in gm/ml

MB × (WA+ WB)

= nB × 1000 × Dsol in gm/ml

WA + WB =

nB × 1000 × Dsol in gm/ml

WA + (nB × MB)

= WB × 1000 × Dsol in gm/ml

MB × Wsol in gm =

Mass % of solute × 10 × Dsol

MB

Molarity, Strength and Normality change with temperature because of

expansion or contraction of the liquid. NOTE




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Molarity equation (Dilution Formula) : M1V1 = M2V2

Where M1 and V1 are the molarity and volume of the solution before dilution

M2 and V2 are the molarity and volume of the solution after dilution.

Molarity equation (for Reacting components) : M1 V1

n1 =

M2 V2

n2

Where n1 and n2 are their stoichiometric coefficients in the balanced equation

Molarity of a Mixture : Mmixture = M1 V1 + M2 V2

V1 + V2 or

M1 V1 + M2 V2 …. + Mn Vn

V1 + V2…. + Vn

Where M1 and V1 are the molarity and volume of the solution

M2 and V2 are the molarity and volume of another solution of the same substance.

(g) Molality : It may be defined as the number of moles of solute dissolved in 1000 g (or 1 Kg) of

the solvent. It is denoted by ‘m’ and expressed in mol/Kg.

Molality (m) = Number of moles of solute

Weight of the solvent in Kg =

nB

WA in Kg =

nB × 1000

WA in g

Moles of solute (nB) = Weight of solute

Molecular mass of solute =

WB

MB

So that Molality (m) = WB × 1000

MB × WA in g

Molality = WB × 1000

MB × (Wsol − WB) =

WB × 1000

MB × ((Dsol × Vsol) − WB)

= nB × 1000

(Dsol × Vsol) − WB =

nB × 1000

(Dsol × Vsol) − (nB × MB)

Molality does not change with temperature because mass of the solvent

does not change with change in temperature.

Relationship between Molarity and Molality :

Molality = Molarity × 1000

(density × 1000) − (Molarity × MB) Molarity =

Molality × 1000 × density

1000 + (Molality × MB)

Practice problems

Q 1. : 2.46 g of NaOH (molar mass = 40) is dissolved in water and the solution is made

to 100 mL in the volumetric flask. Calculate the molarity of the solution.

NOTE




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Q 2. : Calculate the molality of a solution containing 20.7 g of K2CO3 dissolved in 500

mL of the solution (assuming density of solution = 1 g/ml)

Q 3. : Calculate the molarity of a solution of CaCl2 if on chemical analysis it is found

that 500 mL of CaCl2 solution contain 1.505 × 1023 Cl- ions.

Q 4. : 100 ml of a solution containing 5 g of NaOH are mixed with 200 ml of 𝑀

5 NaOH

solution. Calculate the molarity of the resulting solution.

Q 5. : The density of a 2.05 M acetic acid in water is 1.02 g/ml. Calculate the molality of

the solution

Q 6. : Battery acid is 4.27 M H2SO4 and has density of 1.25 g/ml. What is the molality of

H2SO4 in the solution?

(h) Mole fraction : Mole fraction of a component is the fraction obtained by dividing number of moles of

that component by the total number of moles of all components present in the solution.

It is denoted by ‘χ’ (chi).

Mole fraction of solvent (χA) = nA

nA + nB Mole fraction of solute (χB) =

nB

nA + nB

The sum of mole fraction of all components in solution is always equal to 1,

as shown below:

χA + χB = nA

nA + nB +

nB

nA + nB = 1

So that : χA = 1 − χB or χB = 1 – χA

Relationship between Molality and Mole fraction of the solute (χB) :

As we know: Molality = nB × 1000

WA in g so that nB =

Molality × WA

1000 ------ (i)

Similarly: χB = nB

nA + nB ∴ nA >>> nB , hence, nA + nB ≈ nA

So that : χB = nB

nA We can also write this equation as: nB = nA × χB -----(ii)

Comparing equations (i) and (ii), we get: nA × χB = Molality × WA

1000

Now: χB = Molality × WA

1000 × nA (nA =

WA

MA) χB =

Molality × MA

1000




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Practice problems

Q 1.: Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution containing

20 % of C2H6O2 by mass.

Q 2.: 2.82 g of glucose (molar mass = 180) is dissolved in 30 g of water.

Calculate (a) molality (b) mole fraction of glucose and water.

Q 3.: A solution is 25 % water, 25 % ethanol and 50 % acetic acid by mass.

Calculate the mole fraction of each component.

Q 4.: A solution of glucose in water is labelled as 10 % (w/w). The density of the

solution is 1.20 g/ml. Calculate the (a) molality (b) molarity and

(c) mole fraction of each component in solution.

Q 5.: A sugar syrup of weight 214.2 g contains 34.2 g of sugar (C12H22O11).

Calculate : (a) Molal concentration (b) mole fraction of sugar in solution

(i) Normality : It may be defined as the number of gram equivalents of the solute dissolved in 1

litre of the solution. It is denoted by ‘N’ and expressed in gm.eq. L–1.

Normality (N) = Number of gram equivalent of solute

Volume of solution in Litre =

Number of gram equivalent of solute × 1000

Volume of solution in mL

Number of gram equivalent of solute = Weight of solute in gm

Equivalent mass of solute =

WB

Eq.wt.B

∴ Normality (N) = WB × 1000

Eq.wt.B × VSol in mL Eq.wt. of B =

MB

n_factor

Normality (N) = WB × 1000 × n_factor

MB × VSol in mL = Molarity × n_factor =

Strength × n_factor

MB

How to determine n_factor of solute ?

n_factor of acid = Basicity of the acid (or number of H+)

n_factor of base = Acidity of the base (or number of OH-)

n_factor of ion = Charge on the ion

n_factor of salt = Total cationic charge or Total anionic charge per molecule

n_factor of an element = Valency of the element

n_factor of oxidising or reducing agent = No. of es− gain or loss by 1 molecule

NOTE 1




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Relationship between Normality and Mass percentage :

Normality (N) = WB × 1000

Eq.wt.B × VSol in mL =

WB × 1000 × 𝐷𝑠𝑜𝑙

Eq.wt.B × WSol =

Mass % of solute × 10 × 𝐷𝑠𝑜𝑙

Eq.wt.B

Normality equation (Dilution Formula) : N1V1 = N2V2

Where N1 and V1 are the normality and volume of the solution before dilution

N2 and V2 are the normality and volume of the solution after dilution.

Normality of a Mixture : Nmixture = N1 V1 + N2 V2

V1 + V2 or

N1 V1 + N2 V2 …. + Nn Vn

V1 + V2…. + Vn

Where N1 and V1 are the normality and volume of the solution

N2 and V2 are the normality and volume of another solution of the same substance.

Normality equation for acid and base represented in term of molarity:

Normality of Acid = Molarity of Acid × Number of H+ i.e. NA = MA × nA

Normality of Base = Molarity of Acid × Number of OH- i.e. NB = MB × nB

Normality equation NAVA = NBVB becomes MA nA VA = MB nB VB

Practice problems

Q 1.: A commercially available sample of sulphuric acid is 15 % H2SO4 by weight

(density = 1.10 g/mL). Calculate (a) molarity (b) normality (c) molality

Q 2.: Concentrated sulphuric acid has a density 1.9 g/mL and is 99 % H2SO4 by mass.

Calculate the molarity of sulphuric acid.

Q 3.: A solution is prepared by adding 60 g of methyl alcohol to 120 g of water.

Calculate the mole fraction of methanol and water.

Q 4.: One litre solution of N/2 HCl is heated in a beaker. It was observed that when the

volume of the solution is reduced to 600 mL, 3.25 g of HCl is lost. Calculate the

normality of the new solution.

Q 5.: Calculate the molality and mole fraction of the solute in aqueous solution

containing 3 g of urea per 250 g of water.




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SECTION -B

B 1.1: Solubility of Solid in Liquid

When a solid solute is added continuously to a liquid solvent, the solute keeps on

dissolving and the concentration of the solution keeps on increasing. This process is

called dissolution. Ultimately, a stage is reached when no more solute dissolves at the

given temperature. The solution at this stage is said to be saturated solution.

At this stage rate of dissolution becomes

equal to rate of crystallization.

Crystallization is the process when soluble

solute particles in solution collides on the

surface of solid solute particles and get

precipitated out.

The maximum amount of solute that can be dissolved by the solvent at a particular

temperature is called its solubility.

[The solubility is defined as the maximum amount of the solute in grams which can

dissolves in 100 g of the solvent at a given temperature to form a saturated solution.]

B 1.2: Factors Affecting the Solubility of a Solid in a Liquid

(a) Nature of the solute and the solvent :

Every solid does not dissolve in a given liquid. In general, a solid dissolve in a liquid

(solvent) if the intermolecular interactions are similar in solute and solvent. This is

expressed by saying “Like dissolves like”. This means that, non-polar substances are

more likely to be soluble in non-polar solvents while ionic (or polar) substances are

more likely to be soluble in polar solvents.

Compounds like NaCl and sugar dissolve more readily in polar solvents (like water)

but are very little soluble or almost insoluble in non-polar solvents (like benzene, CS2 ,

CCl4 , ether etc.). Similarly, non-polar compounds like naphthalene and anthracene

are soluble in non-polar solvents (like benzene, ether, CCl4 , CS2 etc. ) but very little

soluble in water.

[NOTE : The solubilities of NaCl and sugar in water are 5.3 and 3.8 moles per litre

respectively.]

In case of ionic compounds, the ions are being attracted to the oppositely charged poles

of the solvent molecules by electrostatic attraction. The ion thus moving freely in the

solution are said to hydrated. During the hydration of an ion a significant amount of

energy is released (called hydration enthalpy) which splits ionic compounds into ions.

The energy required for splitting is called lattice enthalpy.

A substance dissolves if hydration enthalpy is greater than the lattice enthalpy.




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(b) Effect of Temperature :

The solubility may increase or decrease or show irregular behaviour with increase in

temperature. We observe following three behaviours:

i) The solubility of solids increases with increase in

temperature: The solubility of most of substances such as

NaNO3 , NH4Cl , KCl , AgNO3 , NaCl, KI and sugar etc. increases

with rise in temperature. This is because the dissolution

process for these substances is endothermic (∆𝒔𝒐𝒍𝑯 > 0).

Therefore, according to Le-Chatelier's principle, an increase in

temperature will result in an increase in the solubility of a gas

if the process is endothermic. [or equilibrium will shift to a

direction in which the heat is absorbed]

Solvent + Solute + Heat ⇌ Solution

ii) The solubility of solids decreases with increase in

temperature: The solubility of some substances like Li2SO4 ,

Ce2(SO4)3 , Na2CO3 and CaO etc. decrease with rise in

temperature. This is because dissolution process is

exothermic (∆𝒔𝒐𝒍𝑯 < 0).

iii) The solubility shows irregular behavior with increase in

temperature: The solubility of some substances (like Na2SO4)

increases up to a certain temperature and then decreases as

the temperature is further raised. The temperature at which

solubility starts decrease is called transition temperature.

For example, the solubility curve of Na2SO4 shows a sharp break at 32.8 ˚C. This is due

to change in one solid form into another solid form. Below transition temperature, only

hydrated Na2SO4.10H2O exists while above this temperature, anhydrous Na2SO4 exists.

(c) Effect of Pressure :

The effect of pressure on the solubility of solids in liquid is very small or insignificant.

This is because solids and liquids are incompressible and practically remain unaffected

by changes in pressure.

Example: Change of 500 atm in pressure increase the solubility of NaCl in water only

by 2.3 %.

Solubility Curves




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B 2.1: Solubility of Gases in Liquid : Almost all gases are soluble in water in varying amounts. The existence of aquatic life in

lakes, rivers etc. is due to the dissolution of oxygen gas of air in water.

Solubility may be defined as the volume of the gas measured at S.T.P. dissolved per

unit volume of the solvent. This method of expressing the concentration is called

absorption coefficient of the gas and is usually represented by ‘∝’.

Solubility of a gas in a liquid at a particular temperature is also expressed in term of

molarity and mole fraction.

B 2.2: Factors Affecting the Solubility of a Gas in a Liquid

(a) Nature of the gas and the solvent:

Typically, those gases that are easily liquefied (or have high critical temperatures) are

more soluble in the solvent. Furthermore, gases that can react chemically or are capable

of forming ions in aqueous solution are more soluble in water than in any other solvent.

Example:

➢ Gases like H2, O2, N2 etc. dissolve in water in very small amounts whereas the

gases like SO2, H2S, HCl, NH3 etc. are highly soluble.

➢ O2, N2 and CO2 are more soluble in ethyl alcohol than in water while gases such as

NH3 and H2S are more soluble in water than in ethyl alcohol.

(b) Effect of temperature :

Gases dissolve in a liquid by exothermic process.

Therefore, according to Le-Chatelier's principle, an increase in temperature will result

in a decrease in the solubility of a gas if the process is exothermic. [or equilibrium will

shift to a direction in which the heat is absorbed]

Solvent + Solute ⇌ Solution + Heat

This is why aquatic species are more comfortable in cold water rather than warm

water. However, it may be noted that there are certain gases such as hydrogen and

inert gases whose solubility increase slightly with increase of temperature especially in

the non-aqueous solvents like alcohols, acetone etc.

(c) Effect of pressure :

The solubility of gases increases with increase of pressure. This behaviour is also in

accordance with Le-Chatelier’s principle.




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To understand this, consider a gas in dynamic equilibrium

with a solution. The lower part represents the solution and the

upper part is gaseous system. Now increase the pressure over

the solution. This will increase the number of gaseous particles

per unit volume over the solution. Hence, more molecules will

dissolve and solubility of a gas increase until a new equilibrium

is reached.

William Henry gave quantitative relationship between the solubility of a gas in a

solvent and the pressure of a gas on a solution, known as Henry's law.

According to this law, the mass of a gas dissolved per unit volume of the solvent at a

given temperature is directly proportional to the pressure of the gas in equilibrium with

the solution.

Where : m = mass of the gas dissolved in a unit volume of the solvent

P = pressure of gas in equilibrium and K = proportionality constant.

Henry’s Law may also be stated as follow:

The solubility of a gas in a liquid at a particular temperature is directly proportional to

the pressure of the gas on that solution at equilibrium.

Dalton, a contemporary of Henry, also concluded

independently that if a mixture of gases is in

equilibrium with a liquid at a particular

temperature, the solubility of any gas in the

mixture is directly proportional to the partial

pressure of that gas in the mixture.

If we express the solubility of a gas in terms of mole fractions, then Henry's law can be

written as:

Thus, most commonly used form of

Henry’s law may be defined as :

The pressure of a gas over a solution in

which the gas is dissolved is directly

proportional to the mole fraction of

the gas dissolved in the solution.




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Mathematical Modifications to Henry's Law

According to Henry’s Law : Pgas = KH . χgas ------- (i)

χgas = ngas

nsolution =

ngas

nsolvent + ngas

[ As we know, nsolvent ≫ ngas , so that, nsolvent + ngas ≈ nsolvent ]

Therefore : χgas = ngas

nsolvent (A) ---------- (ii) [ nA =

WA

MA ]

χgas = ngas

WAMA

= ngas × MA

W𝐴 ---------- (iii) [ngas =

Wgas

Mgas ]

χgas =

Wgas

Mgas × MA

W𝐴 =

Wgas × MA

Mgas × WA -----------(iv)

Substituting the value of equation (ii) in equation (i), we get: Pgas = KH × ngas

nsolvent (A)

Substituting the value of equation (iii) in equation (i), we get: Pgas = KH × ngas × MA

W𝐴

Substituting the value of equation (iv) in equation (i), we get: Pgas = KH × Wgas × MA

Mgas × WA

Properties of Henry’s constant (KH) :

i) KH is a function of nature of gas. Different gases have different KH values at

same temperature.

ii) Greater the value of KH, lower the solubility of the gas.

iii) The value of KH increases with increase in temperature.

Limitation of Henry’s law :

i) Pressure should be low. At high pressure, the law becomes less accurate

ii) The temperature is not too low.

iii) The gas should not highly soluble in solvent.

iv) The gas neither reacts chemically with the solvent nor dissociates or

associates in solvent.




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Applications of Henry’s law :

(i) In the production of carbonated beverages:

To increase the solubility of CO2 in soft drinks, soda water, beer or

champagne, the bottles are sealed under high pressure. When the

bottle is opened under normal atmospheric conditions, the pressure

inside the bottle falls and the excess CO2 bubbles out of the solution.

(ii) In deep sea diving (Scuba diving):

Scuba divers depends upon compressed air for

breathing under water. According to Henry’s law,

increased pressure increases the solubility of

atmospheric gases in blood. Oxygen is used up for

metabolism, but nitrogen will remain dissolved in blood.

When the divers come towards the surface, the pressure

gradually decreases. N2 comes out of the body quickly

forming bubbles in the blood stream. These bubbles

restrict blood flow and can even burst or blocks

capillaries, which cause painful and dangerous condition

called the bends or decompression sickness.

To avoid bends as well as the toxic effects of high concentration of nitrogen in blood,

most divers these days use air diluted with helium gas (about 11.7 % He, 56.2 % N2 and

32.1 % O2).

(iii) At high altitudes:

At high altitudes the partial pressure of O2 is

less than that of the ground level. This results

in low concentration of O2 in the blood and

tissue of the climbers. They feel weak and are

unable to think properly, a disease called

anoxia.

(iv) Combination of haemoglobin and oxygen in lungs:

Haemoglobin combines with oxygen to form oxyhaemoglobin, which is under high

partial pressure in the lungs. In various tissues, where the partial pressure of oxygen is

low, this oxyhaemoglobin releases oxygen for use in cellular activities.




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Question 1. If N2 gas is bubbled through water at 293 K, how many milli moles of N2

gas would dissolve in 1 litre of water. Assume that N2 exerts a partial pressure of 0.987

bar. The KH for N2 at 293 K is 76.48 k bar.

Answer : Given Values: 𝑃𝑁2 = 0.987 bar , KH = 76.48 k bar =76480 bar

Vsolvent(H2O) = 1 Litre or 1000 ml (As we know DH2O = 1 gm/ml)

WH2O = DH2O × VH2O = 1 gm/ml × 1000 ml = 1000 gm , MH2O = 2+16 =18

Calculate moles of N2 gas:

Pgas = KH × ngas × MA

W𝐴 OR ngas =

Pgas × WA

𝐾𝐻 × M𝐴

ngas = 0.987 𝑏𝑎𝑟 × 1000 gm

76480 𝑏𝑎𝑟 × 18 = 7.16 × 10–4 mol

Now Milli moles of N2 = 7.16 ×10–4 ×103 = 7.16 × 10-1 m mol or 0.716 m mol

Question 2. At what partial pressure, oxygen will have solubility of 0.05 g L-1 in water at 293 K? Henry’s constant (KH) for O2 in water at 293 K is 34.36 k bar. Assume the density of the solution to be same as that of the solvent.

Answer : Given Values: KH = 34.36 k bar = 34.36 × 103 bar

MH2O = 2+16 = 18 , MO2 = 16+16 = 32

Solubility of gas = 0.05 g/L [i.e. 0.05 gm of O2 in 1 L of solution]

WO2= 0.05 gm Vsol = 1 L or 1000 ml Dsol = DH2O = 1 gm/ml

Wsol = Dsol × Vsol = 1000 gm WH2O = Wsol − WO2 = 1000 − 0.05 ≈ 1000

Calculate partial pressure of O2 gas:

Pgas = KH × Wgas × MA

Mgas × WA =

34.36 × 103 × 0.05 × 18

32 × 1000 = 0.966 bar

Question 3. Air contain O2 and N2 in the ratio of 1 : 4. Calculate the ratio of solubilities in term of mole fraction of O2 and N2 dissolved in water at atmospheric pressure and room temperature at which Henry’s constant for O2 and N2 are 3.30 × 107 torr and 6.60 × 107 torr respectively. Answer : At 1 bar pressure,

Partial pressure of O2 (𝑃𝑂2) =

1

5 × 1 bar = 0.2 bar

Partial pressure of N2 (𝑃𝑁2) =

4

5 × 1 bar = 0.8 bar

Applying Henry’s law 𝑃𝑂2 = KH (O2) × 𝜒𝑂2

or 𝜒𝑂2 =

𝑃𝑂2

KH (O2)

𝑃𝑁2 = KH (N2) × 𝜒𝑁2

or 𝜒𝑁2 =

𝑃𝑁2

KH (N2)

∴ 𝜒𝑂2

𝜒𝑁2 =

𝑃𝑂2

KH (O2) ×

KH (N2)

𝑃𝑁2

= 𝑃𝑂2

𝑃𝑁2

× KH (N2)

KH (O2) =

0.2 𝑏𝑎𝑟

0.8 𝑏𝑎𝑟 ×

6.60 ×107 𝑡𝑜𝑟𝑟

3.30 ×107 𝑡𝑜𝑟𝑟 =

1

2

𝜒𝑶𝟐 : 𝜒𝑵𝟐

= 1 : 2




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Question 4. H2S, a toxic gas with rotten egg like smell, is used for qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant.

Answer : Solubility of H2S gas = 0.195 m = 0.195 mole in 1 kg of the solvent (water)

1 kg of the solvent (water) = 1000 g = 1000 𝑔

18 𝑔/𝑚𝑜𝑙 = 55.55 moles

Mole fraction of H2S gas in the solution (𝜒) = 0.195

0.195 + 55.55 =

0.195

55.745 = 0.0035

Pressure at STP = 0.987 bar

Applying Henry’s law, 𝑃𝐻2𝑆 = KH × 𝜒𝐻2𝑆

KH = 𝑃𝐻2𝑆

𝜒𝐻2𝑆 =

0.987 𝑏𝑎𝑟

0.0035 = 282 bar

Question 5. Henry’s law constant for CO2 in water is 1.67 × 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K. Answer : KH = 1.67 × 108 Pa, 𝑃𝐶𝑂2

= 2.5 atm = 2.5 × 101325 Pa

Applying Henry’s law 𝑃𝐶𝑂2 = KH × 𝑥𝐶𝑂2

𝑥𝐶𝑂2 =

𝑃𝐶𝑂2

𝐾𝐻 =

2.5 × 101325 Pa

1.67 × 108 Pa = 1.517 × 10-3, i.e.

𝑛𝐶𝑂2

𝑛𝐻2𝑂 + 𝑛𝐶𝑂2

≈ 𝑛𝐶𝑂2

𝑛𝐻2𝑂 = 1.517 × 10-3

For 500 mL of soda water, water present ≈ 500 mL = 500 g = 500

18 = 27.78 moles

𝑛𝐶𝑂2

27.78 = 1.517 × 10-3 or 𝑛𝐶𝑂2

= 42.14 × 10-3 mole = 42.14 m mol

Weight = moles × molecular wt. = 42.14 × 10-3 × 44 g = 1.854 g

B 3.: Solid Solutions : Solid solutions are those solutions in which both solute and solvent are solids.

Solid solutions are of two types :

Substitutional solid solutions are those in which

constituent particles of one substance replaced by the

particles of another substance in a crystal lattice.

Example: Brass, Bronze etc.

Interstitial solids solutions are those which are formed

by placing particles of one kind into lattice voids of other

substance. Example: Tungsten carbide (WC) in which

tungsten atoms are arranged in FCC and carbon atom

occupy octahedral voids.

Similarly, Steel is interstitial carbide of iron.




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SECTION -C

C 1.: Vapour Pressure of Liquid Solutions Liquid solutions are the solutions in which solvent is liquid and the solute can be gas,

liquid or solid.

Vapour Pressure of Liquid :

When a liquid is allowed to evaporate in a closed vessel, a part

of the liquid evaporates. It fills the available space with vapours

and level of liquid decreases. This process is called

vaporisation.

Some molecules of vapour strike the surface of liquid and get

condensed. The process of condensation acts in opposite

direction to the process of evaporation. Ultimately, a stage is

reached when the rate of evaporation becomes equal to rate of

condensation and equilibrium gets established between liquid

and vapour phase.

The pressure exerted by the vapours above the liquid surface at the time of equilibrium at

a given temperature is called vapour pressure.

Factors Affecting Vapour Pressure :

(A) Nature of liquid : The liquids, which have weaker intermolecular forces, have

greater vapour pressure because more molecules tend to escape into vapour phase.

Example : Dimethyl ether and ethyl alcohol have greater vapour pressure than water

because of weaker intermolecular forces as compared to water

(B) Temperature : The vapour pressure of liquid increases with increase in

temperature. This is because more molecules will have larger kinetic energies and

escape readily into vapour phase resulting in higher vapour pressure.

C 2.: Vapour Pressure of Liquid-Liquid Solutions

When a binary solution of two volatile liquids is placed in a

closed vessel, both components would evaporate and

eventually an equilibrium would be established between liquid

and vapour phase.

The total vapour pressure (PTotal) in the container will be the

sum of the partial vapour pressures (PA and PB) of both

components of the solution.

PTotal = PA + PB




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A French chemist, F.M. Raoult in 1886 gave a quantitative relationship between the

partial vapour pressures (PA or PB) of components and their mole fractions in

solution. This relationship is known as Raoult’s law which states that :

For a solution of volatile liquids, the partial vapour pressure of each component in the

solution is directly proportional to their mole fraction in solution.

Then, according to Raoult’s law :

PA ∝ 𝜒𝐴 and PA = PAₒ . 𝜒𝐴

PB ∝ 𝜒𝐵 and PB = PBₒ . 𝜒𝐵

Where :

PA and PB are the partial vapour pressures of components

in solution.

𝑃𝐴ₒ and 𝑃𝐵

ₒ are the vapour pressures of components in

Pure state.

𝜒𝐴 and 𝜒𝐵 are the mole fractions of components in

solution

According to Dalton’s law of partial pressures, the total vapour pressure (PTotal) over

the solution in a closed vessel will be sum of the partial vapour pressures (PA and PB)

of both components of solution. This is given as :

PTotal = PA + PB

Substituting the values of PA and PB, we get

PTotal = [PAₒ . 𝜒𝐴] + [PB

ₒ . 𝜒𝐵]

As we know, 𝜒𝐴 + 𝜒𝐵 = 1 or 𝜒𝐴 = 1−𝜒𝐵

PTotal = [PAₒ (1 − 𝜒𝐵)] + [PB

ₒ . 𝜒𝐵]

PTotal = PAₒ − PA

ₒ . 𝜒𝐵 + PBₒ . 𝜒𝐵

PTotal = PAₒ + (PB

ₒ − PAₒ ) 𝜒𝐵

Following conclusions can be drawn from above equation :

❖ Total vapour pressure over the solution can be related to the mole fraction of any

one component [𝜒𝐵 and 𝜒𝐴 = (1 - 𝜒𝐵)].

❖ Total vapour pressure of the solution varies linearly with mole fraction of

components because PAₒ and PB

ₒ are constant.

❖ Depending on the vapour pressure of the pure components, total vapour pressure

over the solution decrease or increase with increase of the mole fraction of

components.




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The composition of vapour phase in equilibrium can be determined by the

partial pressure of components. If YA and YB are the mole fractions of

components A and B respectively in vapour phase then, using Dalton’s law

of partial pressures :

YA = 𝑃𝐴

𝑃𝐴 + 𝑃𝐵 =

𝑃𝐴

𝑃𝑇𝑜𝑡𝑎𝑙 and YB =

𝑃𝐵

𝑃𝐴 + 𝑃𝐵 =

𝑃𝐵

𝑃𝑇𝑜𝑡𝑎𝑙

Konowaloff’s rule :

At any fixed temperature, mole fraction of the more volatile component is

always greater in the vapour phase than in the solution phase.

Raoult’s law as a special case of Henry’s Law :

If we compare the equations for Raoult’s law and Henry’s law.

P = KH . 𝜒𝐵 (According to Henry’s law)

PA = PAₒ . 𝜒𝐴 (According to Raoult’s law)

It can be seen that the partial pressure of the volatile component or gas is directly

proportional to its mole fraction in solution. Only the proportionality constants are

different KH (for henry’s law constant) and 𝑃𝐴ₒ (for Raoult’s law). Thus, Raoult’s law

becomes a special case of Henry’s law. Further, it is observed that if the solute obeys

Henry’s law, the solvent obeys Raoult’s law.

Similarities and Dissimilarities between Raoult’s law and Henry’s law :

Similarities : (i) Both apply to the volatile component of the solution.

(ii) Both state that the vapour pressure of any component in the solution

is proportional to the mole fraction of that component in the solution.

Dissimilarities : (i) The two laws differ in the proportionality constant.

(ii) In Raoult’s law, constant is equal to vapour pressure of pure

component P˚ while in Henry’s law, constant is equal to an

experimentally determined value.

C 3.: Vapour Pressure of Solid-Liquid Solutions When the solute is non-volatile, only the solvent molecules are present in the vapour

phase. Therefore, the vapour pressure of solution is due to solvent only.

However, the vapour pressure of the solution is found to be less than that of the pure

solvent.

NOTE 1

NOTE 2

NOTE 3

NOTE 4




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This can be explained as follows :

We know that evaporation is a surface phenomenon. The

vapour pressure depends on the escape of solvent

molecules from the surface of the liquid.

In the case of solution, some of the non-volatile solute

particles occupy a certain surface area. As a result, lesser

number of solvent molecules will escape into vapours.

Hence, there will be a lowering in vapour pressure of

solution.

Therefore, it indicates that the extent to which the vapour pressure is lowered is

dependent on the amount of solute present, i.e., more concentration of solution, lower

will be the vapour pressure.

The total vapour pressure of solution will be equal to the partial vapour pressure of

solvent.

PTotal = PA

According to Raoult’s law, the partial vapour pressure of volatile component in the

solution is directly proportional to the mole fraction in it.

PA = PAₒ . 𝜒A

Now: PTotal = PA = PAₒ . 𝜒A

The above relationship may also be put forward in different ways as :

PA = PAₒ . 𝜒A (∴ 𝜒A = 1 − 𝜒B)

PA = PAₒ (1 − 𝜒B)

PA = PAₒ − PA

ₒ 𝜒B

PAₒ𝜒B = PA

ₒ − PA

𝜒𝐁 = 𝐏𝐀

ₒ − 𝐏𝐀

𝐏𝐀ₒ

In this expression, PAₒ − 𝑃𝑇𝑜𝑡𝑎𝑙 = ∆𝑃𝐴expresses the lowering of vapour pressure while

PAₒ − 𝑃𝑇𝑜𝑡𝑎𝑙

PAₒ is called relative lowering in vapour pressure and 𝜒𝐵 represents mole

fraction of the solute in solution.

Hence, the above expression may be expressed in words as follows :

The relative lowering of vapour pressure of a solution containing a non-volatile solute

is equal to the mole fraction of the solute in the solution at a given temperature.




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QUESTION 1. : The vapour pressure of ethyl alcohol at 298 K is 40 mmHg. Its mole

fraction in a solution with methyl alcohol is 0.80. What is its vapour pressure in

solution if the mixture obeys Raoult’s law?

ANSWER : According to Raoult’s law PA = PAₒ . 𝜒𝐴

Vapour pressure of pure ethyl alcohol (PAₒ ) = 40 mmHg

Mole fraction of ethyl alcohol (𝜒𝐴) = 0.80

Partial vapour pressure of ethyl alcohol (PA) = 40 × 0.80 = 32 mmHg

QUESTION 2. : Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2)

at 298 K are 200 mmHg and 415 mmHg 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

(ii) Mole fraction of each component in vapour phase.

ANSWER : i) vapour pressure of the solution

MCH2Cl2= 85 g/mol WCH2Cl2

= 40 g nCH2Cl2=

WCH2Cl2

MCH2Cl2

= 40 g

85 g/mol = 0.47 mol

MCHCl3= 119.5 g/mol WCHCl3

= 25.5 g nCHCl3=

WCHCl3

MCHCl3

= 25.5 g

119.5 g/mol = 0.213 mol

Total number of moles in solution = 0.47 + 0.213 = 0.683 mol

𝜒CH2Cl2 =

nCH2Cl2

nsolution =

0.47 𝑚𝑜𝑙

0.683 𝑚𝑜𝑙 = 0.688

𝜒CHCl3 = 1 – 𝜒CH2Cl2

= 1 – 0.688 = 0.312

By using equation : PTotal = PA + PB

PA (CH2Cl2) = PAₒ . 𝜒𝐴 = 415 × 0.688 = 285.5 mmHg

PB (CHCl3) = PBₒ . 𝜒𝐵 = 200 × 0.312 = 62.4 mmHg

PTotal = PA + PB = 285.5 + 62.5 = 347.9 mmHg

ii) Mole fraction in vapour phase

YA (CH2Cl2) = 𝑃𝐴

𝑃𝑇𝑜𝑡𝑎𝑙 =

285.5

347.9 = 0.82 YB (CHCl3) =

𝑃𝐵

𝑃𝑇𝑜𝑡𝑎𝑙 =

62.4

347.9 = 0.18

PRACTICE PROBLEMS

Q.1. : The vapour pressure of a pure liquid A is 40 mmHg at 310 K. The vapour

pressure of this liquid in solution with liquid B is .32 mmHg Calculate the mole fraction

of A in the solution if the mixture obeys Raoult’s law.




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Q.2. : The vapour pressure of benzene and toluene at 298 K are 75 mmHg and 22

mmHg respectively. 23.4 g of benzene and 64.4 g of toluene are mixed. If the two forms

an ideal solution, calculate the mole fraction of benzene in the vapour phase assuming

that vapours are in equilibrium with the liquid mixture at this temperature.

Q.3. : The vapour pressure of ethyl acetate and ethyl propionate are 72.8 and 27.7

mmHg respectively. A solution is prepared by mixing 25 g of ethyl acetate and 50 g of

ethyl propionate. Assuming the solution to be ideal, Calculate the vapour pressure of

the solution.

Q. 4 : Benzene and toluene form nearly ideal solution. At certain temperature, the

vapour pressure of the pure benzene is 150 mmHg and of pure toluene is 50 mmHg. For

this temperature, calculate the vapour pressure of solution containing equal weights of

two substances. Also calculate their composition in the vapour phase.

Q. 5 : At 298 K, the vapour pressure of pure benzene is 0.256 bar and vapour pressure

of pure toluene is 0.925 bar. If the mole fraction of benzene in solution is 0.40, find the

total vapour pressure of solution. Also find the mole fraction of toluene in vapour

phase.

Q.6 : The vapour pressure of pure liquids A and B are 450 and 700 mmHg respectively,

at 350 K. Find out the composition of the mixture if total vapour pressure is 600 mmHg.

Also find the composition of the vapour phase.

C 4.: Ideal and Non-Ideal Solutions :

A. Ideal Solutions An ideal solution may be defined as the solution which obeys Raoult’s law

under all conditions of temperatures and concentrations.

Such solutions are formed by mixing the two components which are

identical in molecular size and have almost identical intermolecular forces.

So that intermolecular interactions between components A – B are of same

magnitude as in pure components A – A and B – B.

The ideal solutions have the following characteristics :

i. Heat change on mixing is zero (∆Hmixing = 0)

ii. Volume change on mixing is zero (∆Vmixing = 0)

iii. Obeys Raoult’s law (PA = PAₒ . 𝜒𝐴 and PB = PB

ₒ . 𝜒𝐵)

FACT




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Examples of ideal solution :

i. Benzene + Toluene iii. n-Hexane + n-Heptane

ii. Ethyl chloride + Ethyl bromide iv. Chlorobenzene + Bromobenzene

B. Non – Ideal Solutions The solutions which do not obey Raoult’s law under all conditions of

temperatures and concentrations are called non-ideal solutions.

Such solutions are formed by mixing the two components which are not

identical in molecular size and have different intermolecular forces. So that

intermolecular interactions between components A – B are of different

magnitude as in pure components A – A and B – B.

The non-ideal solutions have the following characteristics :

i. Heat change on mixing is not equal to zero (∆Hmixing ≠ 0)

ii. Volume change on mixing is not equal to zero (∆Vmixing ≠ 0)

iii. Do not obey Raoult’s law (PA ≠ PAₒ . 𝜒𝐴 and PB ≠ PB

ₒ . 𝜒𝐵)

Types of non-ideal solution :

I. Non-ideal solution showing positive deviations from Raoult’s law.

II. Non-ideal solution showing negative deviations from Raoult’s law.

I. Non-ideal solution showing positive deviation from Raoult’s law

When a solution is formed by mixing two components A and B,

sometime the partial pressure of components is found to be

more than expected on the basis of Raoult’s law. As a result,

total vapour pressure of the solution is greater than

expected. Such behavior of solutions is described as a positive

deviation from Raoult’s law.

The intermolecular interactions between components A – B are

of less magnitude as in pure components A – A and B – B.

The boiling points of such solutions are relatively lower as compared to those of pure

component. The composition of such solution remains unchanged on distillation, hence

behave like minimum boiling azeotropes.

Characteristics :

i. Heat change on mixing is always positive (∆Hmixing = positive)

ii. Volume change on mixing is also always positive (∆Vmixing = positive)

iii. Partial pressure is found to be more than expected

(PA > PAₒ . 𝜒𝐴 and PB > PB

ₒ . 𝜒𝐵)

FACT




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Examples :

i. Acetone + Carbon disulphide iv. Acetone + Ethyl alcohol

ii. Acetone + Benzene v. Methyl alcohol + Water

iii. Ethyl alcohol + Water

II. Non-ideal solution showing negative deviation from Raoult’s law

When a solution is formed by mixing two components A and B,

sometime the partial pressure of components is found to be

less than expected on the basis of Raoult’s law. As a result,

total vapour pressure of the solution is lower than expected.

Such behaviour of solution is described as a negative

deviation from Raoult’s law.

The intermolecular interactions between components A – B are

of greater magnitude as in pure components A – A and B – B.

The boiling points of such solutions are relatively higher as compared to those of pure

component. Such solutions behave like maximum boiling azeotropes.

Characteristics :

i. Heat change on mixing is always negative (∆Hmixing = negative)

ii. Volume change on mixing is also always negative (∆Vmixing = negative)

iii. Partial pressure is found to be less than expected

(PA < PAₒ . 𝜒𝐴 and PB < PB

ₒ . 𝜒𝐵)

Examples :

i. Chloroform + Acetone iv. Acetone + Aniline

ii. Chloroform + Benzene v. HCl + Water

iii. Chloroform + Diethyl ether

C 5.: Azeotropes : Azeotropes are the binary solution having the same composition in both liquid and

vapour phase, and boil at a constant temperature. In such cases, it is not possible to

separate the components by fractional distillation method.

There are two types of Azeotropes :

Minimum boiling azeotropes :

When the boiling point of the azeotrope is less than the boiling points of the pure

components, then it is called minimum boiling azeotropes or positive azeotropes.

In case of non-ideal solutions showing positive deviation from Raoult’s law, at one of

the intermediate compositions, the vapour pressure is the highest and boiling point is




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the lowest. At this intermediate composition solution behaves as minimum boiling

azeotrope.

Example : 95.4 % solution of ethanol with water boil at 351.15 K.

Maximum boiling azeotropes :

When the boiling point of the azeotrope is more than the boiling point of the pure

components, then it is called maximum boiling azeotropes or negative azeotropes.

In case of non-ideal solutions showing negative deviation from Raoult’s law, at one of

the intermediate compositions, the vapour pressure is the lowest and boiling point is

the highest. At this intermediate composition solution behaves as maximum boiling

azeotrope.

Example : 68 % solution of nitric acid with water boil at 393.5 K.




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SECTION -D Colligative Properties : Those properties of ideal solutions which depend only on the number of solute

particles but do not depend on the nature of solute are called colligative properties.

The important colligative properties are : 1. Relative lowering of vapour pressure 2. Elevation in boiling point 3. Depression in freezing point. 4. Osmotic pressure.

1. Relative lowering of vapour pressure : The vapour pressure of solvent in solution is less than that of pure solvent. This is

because of non-volatile solute occupy some surface area; hence the vapour pressure

decrease. Lowering of the vapour pressure depends only on the concentration of solute

particles but not on nature of solute.

Expression for relative lowering in vapour pressure :

When the solute is non-volatile, only the solvent molecules are present in the vapour

phase. Therefore, the vapour pressure of solution is due to solvent only.

PTotal = PA According to Raoult’s law, the partial vapour pressure of volatile component in the

solution is directly proportional to the mole fraction of that component in solution.

PA = PAₒ . χA (∴ χA = 1 − χB)

PA = PAₒ (1 − χB)

PA = PAₒ − PA

ₒ χB

PA ₒ χB = PA

ₒ − PA

𝛘𝐁 = 𝐏𝐀

ₒ − 𝐏𝐀

𝐏𝐀ₒ

Determination of molecular masses of solutes from lowering of vapour pressure :

PA

ₒ − PA

PAₒ = χB -----------(i)

nB

nA + nB = χB -----------(ii)

Compare equation (i) and (ii)

➢ PA

ₒ − PA

PAₒ =

nB

nA + nB

For a dilute solution, nA ≫ nB , hence, nA + nB ≅ nA

PAₒ − 𝑃𝐴 is the lowering of vapour pressure while

PAₒ − 𝑃𝐴

PAₒ is called relative lowering in vapour pressure




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➢ PA

ₒ − PA

PAₒ =

nB

nA [nA =

WA

MA , nB =

WB

MB ]

➢ PA

ₒ − PA

PAₒ =

WBMB

WAMA

=

WB × MA

MB × WA [ ∴ 𝐏𝐀

ₒ − 𝐏𝐀 = ∆𝐏𝐀 ]

➢ ∆𝐏𝐀

𝐏𝐀ₒ =

𝐖𝐁 × 𝐌𝐀

𝐌𝐁 × 𝐖𝐀 Or 𝐌𝐁 =

𝐖𝐁 × 𝐌𝐀 × 𝐏𝐀ₒ

∆𝐏𝐀 × 𝐖𝐀

PRACTICE PROBLEMS Q.1. : Vapour pressure of water at 293 K is 17.51 mm. Lowering of vapour pressure of a sugar solution is 0.0614 mm. Calculate (i) Relative lowering of vapour pressure (ii) Mole fraction of water (iii) Vapour pressure of the solution. Q.2.: The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile and non-electrolyte solid weighing 0.5 g is added to 39.0 g of benzene (molar mass 78 g/mol). The vapour pressure of the solution then is 0.845 bar. What is the molar mass of the solid substance? Q.3.: The vapour pressure of 2.1 % of an aqueous solution of non-electrolyte at 373 K is 755 mm. Calculate the molar mass of solute. (Consider that vapour pressure of water at 373 K is 760 mm). Q.4.: The vapour pressure of water is 92 mm at 323 K. 18.1 g of urea is dissolved in 100 g of water. The vapour pressure is reduced by 5 mm. Calculate the molar mass of urea. Q.5.: 20 g of solute was added to 100 g of water at 25 ˚C. The vapour pressure of water and that of the solution were 23.76 mmHg and 22.41 mmHg respectively at that temperature. Calculate the relative molecular mass of the solute.

2. Elevation in Boiling Point:

The boiling point of liquid is the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure. Example : Vapour pressure of water at 100 ˚C (373 K) is 1.013 bar or 760 mm, which is equal to 1 atm. This is the reason that water boil at 373 K, because its vapour pressure becomes equal to one atmospheric pressure.

It is found that the boiling point of the solution is always higher

than that of pure solvent (is called elevation in boiling point).

This is because vapour pressure of the solution is lower than

that of pure solvent. Hence, solution has to be heated more to

make its vapour pressure equal to atmospheric pressure.




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Expression for elevation in boiling point :

The elevation in boiling point (∆Tb) depends on molal concentration (m) of solute in

solution.

∆𝐓𝐛 ∝ m or ∆𝐓𝐛 = 𝐊𝐛 × m

Where Kb is called the molal elevation constant or ebullioscopic constant and ‘m’ is the

molality.

If molality of the solution is 1, then ∆𝐓𝐛 = 𝐊𝐛.

Hence, molal elevation constant can be defined as the elevation in boiling

point when the molality of the solution is unity (i.e. 1 mole of solute in 1 kg of

solvent).

Calculation of molecular mass of the solute from elevation in boiling point :

∆Tb = Kb × m --------(i) Molality (m) = WB × 1000

MB × Wsolvent in g -----------(ii)

Put the value of equation (ii) in to (i)

∆Tb = Kb × WB × 1000

MB × Wsolvent in g or Tb − Tb

° = Kb × WB × 1000

MB × Wsolvent in g

MB = Kb × WB × 1000

∆Tb × Wsolvent in g or MB = Kb ×

WB × 1000

(Tb− Tb° ) × Wsolvent in g

PRACTICE PROBLEMS

Q.1. : The boiling point of benzene is 353.23 K. 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.

Q.2.: 18 g of glucose, C6H12O6, is dissolved in 1 kg of water. At what temperature will

the water boil at 1.013 bar pressure ? Kb for water is 0.52 K kg/mol.

Q.3.: A solution of glycerol (C3H8O3), in water was prepared by dissolving some

glycerol in 500 g of water. This solution has a boiling point of 100.42 ˚C. What mass of

glycerol was dissolved to make the solution? Kb for water = 0.52 K kg/mol.

3. Depression in Freezing Point : Freezing point of a substance is the temperature at which the

solid and liquid forms of the substance have same vapour

pressure. It is found that the freezing point of the solution is

always lower than that of pure of solvent. This is because

vapour pressure of the solution is lower than that of pure

solvent. Hence, solution has to be cooled more to make its

vapour pressure equal to solid phase.

∆𝐓𝐛 = 𝐓𝐛 − 𝐓𝐛°

NOTE

Solvent Kb (K Kg/mol)

H2O 0.52

C6H6 2.53

C2H5OH 1.20




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Expression for depression in freezing point :

The depression in freezing point (∆Tf) depends upon molal concentration (m) of solute

in solution.

∆𝐓𝐟 ∝ m or ∆𝐓𝐟 = 𝐊𝐟 × m

Where Kf is called the molal depression constant or cryoscopic constant and m is the

molality.

If molality of the solution is 1, then ∆𝐓𝐟 = 𝐊𝐟.

Hence, molal depression constant can be defined as the depression in

freezing point when the molality of the solution is unity (i.e. 1 mole of

solute in 1 kg of solvent).

Calculation of molecular mass of the solute from depression in freezing point :

∆Tf = Kf × m --------(i) Molality (m) = WB × 1000

MB × Wsolvent in g -----------(ii)

Put the value of equation (ii) in to (i)

∆Tf = Kf × WB × 1000

MB × Wsolvent in g or Tf

° − Tf = Kf × WB × 1000

MB × Wsolvent in g

MB = Kf × WB × 1000

∆Tf × Wsolvent in g or MB = Kf ×

WB × 1000

(Tf°− Tf) × Wsolvent in g

Applications of Depression of freezing point :

(i) In making antifreeze solutions:

The running of a vehicle (car or bus) in sub-zero weather even when the

radiator is full of water, which has freezing point 0 ˚C or 273 K, has been

possible due to the fact that depression in freezing point of water takes place

by addition of some quantity of ethylene glycol (called antifreeze) in water.

(ii) In melting of ice on roads:

In winter, where snowfall occurs heavily, the ice deposited on roads is molten

by scattering common salt (NaCl) or CaCl2 on the roads. This is because salt-

ice mixture has very low freezing point and hence ice keeps on melting.

Calculation of molal depression constant from enthalpy of fusion :

Kf = 𝑹 × (𝑻𝒇

𝒐)𝟐

𝟏𝟎𝟎𝟎 × 𝒍𝒇 =

𝑹 × (𝑻𝒇𝒐)

𝟐 × 𝑴𝑨

𝟏𝟎𝟎𝟎 × ∆𝒇𝒖𝒔𝑯 ∴ 𝒍f =

∆𝒇𝒖𝒔𝑯

𝑴𝑨

R = Gas constant , MA = Molar mass of solvent.

𝑻𝒇𝒐 = Freezing point of pure solvent in Kelvin,

𝒍f = Latent heat of fusion per gram of solvent,

∆𝐟𝐮𝐬𝐇 = Enthalpy of fusion per mole of solvent.

NOTE

∆𝐓𝐟 = 𝐓𝐟° − 𝐓𝐟

NOTE 1

Solvent Kf (K Kg/mol)

H2O 1.86

C6H6 5.12

C2H5OH 1.99

R = 8.134 J K-1 mol-1 if 𝑙f or

∆𝑓𝑢𝑠𝐻 are in joules

R = 2 cal deg-1 mol-1 if 𝑙f or ∆𝑓𝑢𝑠𝐻 are in calories




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Calculation of molal elevation constant from enthalpy of vaporization :

Kb = 𝑹 × (𝑻𝒃

𝒐)𝟐

𝟏𝟎𝟎𝟎 × 𝒍𝒃 =

𝑹 × (𝑻𝒃𝒐)

𝟐 × 𝑴𝑨

𝟏𝟎𝟎𝟎 × ∆𝒗𝒂𝒑𝑯 ∴ 𝒍b =

∆𝒗𝒂𝒑𝑯

𝑴𝑨

R = Gas constant , MA = Molar mass of solvent.

𝑻𝒃𝒐 = Boiling point of pure solvent in Kelvin,

𝒍b = Latent heat of vaporization per gram of solvent,

∆𝐯𝐚𝐩𝐇 = Enthalpy of vaporization per mole of solvent.

PRACTICE PROBLEMS

Q.1.: A solution containing 34.2 g of cane-sugar (C12H22O11) dissolved in 500 cm3 of

water froze at -0.374 ˚C. Calculate the freezing point depression constant of water.

Q.2.: 1.00 g of non-electrolyte solute dissolved in 50 g of benzene lowered the freezing

point of benzene by 0.40 K. Kf of benzene is 5.12 K kg/mol. Find Molar mass of solute.

Q.3.: 45 g of ethylene glycol (C2H6O2) is mixed with 600 g water. Calculate : (a)

freezing point of depression (b) freezing point of solution. (Kf for H2O = 1.86 K kg/mol)

Q.4. : Calculate the temperature at which a solution containing 54 g of glucose in 250 g

of water will freeze. (Kf for water = 1.86 K kg/mol)

Q.5.: What mass of ethylene glycol (Molecular mass = 62.0 g/mol) must be added to

5.50 kg of water to lower the freezing point of water from 0 ˚C to -10.0 ˚C. (Kf H2O is

1.86 K kg/mol).

4. Osmotic Pressure:

Osmosis and Diffusion :

Diffusion is the spontaneous process of movement of both solute and solvent from

their higher concentration to lower concentration.

Osmosis is the spontaneous process of movement of solvent only, from its higher

concentration to lower concentration through semi-permeable membrane.

Demonstration of Osmosis :

When grapes placed in concentrated solution of NaCl, the grapes start shrink due to

exosmosis. This is because the concentration of H2O (solvent) is high in grapes sap

than in NaCl solution. Hence, movement of water takes place from grapes sap to NaCl

solution and grapes start shrink.

Similarly, when kismis placed in pure water, the kismish start swell up due to

endosmosis. This is because the concentration of H2O (solvent) is low in kismish sap.

Hence, movement of water takes place from pure water to kismish sap and kismish

start swell up.

NOTE 2

R = 8.134 J K-1 mol-1 if 𝑙b

or ∆𝑣𝑎𝑝𝐻 are in joules

R = 2 cal deg-1 mol-1 if 𝑙b

or ∆𝑣𝑎𝑝𝐻 are in calories




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Semi-permeable Membrane :

It can be defined as the membrane which is selectively permeable to substances and

allow solvent molecules to pass through it but do not allow solute molecule.

It is of two types :

Natural semi-permeable membranes: They are vegetable membranes or animal

membranes which are found in outer skin of plants and animals.

Artificial semi-permeable membranes: They are synthesized artificially.

Common examples are: parchment membrane, cellophane membrane, copper

ferrocyanide etc.

Osmotic Pressure :

It can be defined as the additional pressure applied on

the solution to prevent the entry of the solvent into the

solution through semi-permeable membrane. It is

represented by ‘𝝅’.

Reverse Osmosis (R.O.) :

If a pressure higher than the osmotic pressure is applied

on the solution, the solvent will flow from the solution

into the pure solvent through the semi-permeable

membrane. The process is called reverse osmosis (RO).

Expression for Osmotic Pressure :

Osmotic pressure (𝝅) of a solution is found to be directly proportional to molar

concentration (C) of the solution and its temperature (T).

𝛑 ∝ 𝐂 ----------- (i) 𝛑 ∝ 𝐓 ----------- (ii)

Compare the equation (i) and (ii)

𝛑 ∝ 𝐂𝐓 or 𝛑 = 𝐑𝐂𝐓

Where ‘R’ is a constant and its value is found to be same as that of the ‘Gas constant’.

The above equation is usually written as :

𝛑 = 𝐂𝐑𝐓 C is molar concentration = 𝐧

𝐕

Where ‘n’ is the number of moles of solute and V is the volume of solution in Litre.

Now : 𝛑 =𝐧

𝐕𝐑𝐓 or 𝛑𝐕 = 𝐧𝐑𝐓

This equation is called vant’s Hoff equation for dilute solutions.




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Determination of Molecular mass from Osmotic Pressure :

𝛑𝐕 = 𝐧𝐑𝐓 Where :- ‘n’ is moles of solute = 𝑾𝑩

𝑴𝑩

𝛑𝐕 = 𝐖𝐁

𝐌𝐁𝐑𝐓 If volume of solution in mL then equation become :

𝛑 = 𝐖𝐁 × 𝐑𝐓

𝐌𝐁 × 𝐕 𝛑 =

𝐖𝐁 × 𝐑𝐓 × 𝟏𝟎𝟎𝟎

𝐌𝐁 × 𝐕

𝐌𝐁 = 𝐖𝐁 × 𝐑𝐓

𝛑 × 𝐕 𝐌𝐁 =

𝐖𝐁 × 𝐑𝐓 × 𝟏𝟎𝟎𝟎

𝛑 × 𝐕

Hypertonic, Hypotonic and Isotonic solutions :

➢ The solution which has high osmotic pressure with respect to another solution is

called hypertonic solution. It contains relatively large quantity of solute.

➢ The solution which has low osmotic pressure with respect to another solution is

called hypotonic solution. It contains relatively small quantity of solute.

➢ The solution which has same osmotic pressure with respect to another solution is

called isotonic solution. It has equimolar concentration of solute.

Biological Importance of Osmosis :

Plants absorb water from the soil through their roots due to osmosis.

In animals, water moves into different parts of the body due to osmosis.

Bursting of RBCs when placed in water is also due to osmosis.

Opening and closing of stomata of plant leaves is also due to osmosis.

Raw mangoes shrink to pickle when placed in common salt is due to osmosis.

Wilted flowers revive when placed in fresh water is due to osmosis.

Explanation of Some Phenomenon on the Basis of Osmosis :

o People taking a lot of salt in food develop swelling of their

tissues, a disease is called edema. This is due to retention

of water in the tissue cells because of osmosis.

o Raw mangoes shrink into pickle when placed in common

salt solution. This is due to outflow of water from the cells

of mangoes because of osmosis.

o Wilted flower revives again when placed in fresh water.

This is due to flow of water into the flower cells because

of osmosis.




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By study of osmosis, it helps us to conclude the following :

• When RBCs are placed in hypotonic

solution, having salt concentration less than

0.9 %, they will start swell up due to

endosmosis.

• When RBCs are placed in hypertonic

solution, having salt concentration more

than 0.9 %, they will start shrink due to

exosmosis.

• When RBCs are placed in isotonic solution,

having salt concentration 0.9 %, the cell

never swell and nor shrink because no

osmosis takes place.

When water enters into the cell, it starts swell up due to endosmosis, the

cell wall of the cell then exert pressure to oppose the flow of water into the

cell, this exerted pressure is called turgor pressure and the cell is celled

turgid cell.

When water going out the cell, it starts shrink due to exosmosis, the

protoplasm of the cell attains small spherical shape, this phenomenon is

called plasmolysis or crenation and the cell is called plasmolysed cell.

Remember that osmosis takes place in the direction of :

High solvent concentration to Low solvent concentration

Low solute concentration to high solute concentration

OR

Less concentrated solution to More concentrated solution

Low osmotic pressure to high osmotic pressure

PRACTICE PROBLEMS

Q.1.: 200 cm3 of an aqueous soln of protein contains 1.26 g of protein. Osmotic

pressure of this solution at 300 K is found to be 2.57 × 10-3 bar. Calculate molecular

mass of protein.

Q.2.: 10 g of a substance were dissolved in water and solution was made up to 250 cm3.

Osmotic pressure of solution was found to be 8 × 105 Nm-2 at 288 K. Find molar mass of

solute.

NOTE 2

NOTE 1




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Q.3.: A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water

has an osmotic pressure of 0.335 torr at 25 ˚C. Assuming that the gene fragment is a

non-electrolyte, calculate its molar mass.

Q.4.: The osmotic pressure of blood is 8.21 atm at 37 ˚C. How much glucose would be

used for an injection that is at the same osmotic pressure as blood ?

(R= 0.0821 atm mol-1 K-1]

Methods used for the measurement of different colligative properties are:

• Relative lowering in vapour pressure – Ostwald and Walker method

• Elevation in Boiling point – Landsberger method

• Depression in Freezing point – Beckmann method

• Osmotic pressure – Berkeley and Hartley method OR Morse and Frazer method

SECTION - E Abnormal Molecular Masses When the molecular mass of a substance determined by studying any of the colligative

properties comes out to be different than the theoretically expected value, then the

substance is said to show abnormal molecular mass.

The abnormal molecular masses are observed in any one of the following cases :

o When the solute undergoes dissociation in the solution.

o When the solute undergoes association in the solution.

o When the solution is non-ideal.

To calculate the extent of association or dissociation, Van’t Hoff introduced a factor ‘i'

called Van’t Hoff factor. It is defined as the ratio of the experimental value of the

colligative property to the calculated value of the colligative property.

𝑖 = 𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑜𝑟 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑐𝑜𝑙𝑙𝑖𝑔𝑎𝑡𝑖𝑣𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦

𝑁𝑜𝑟𝑚𝑎𝑙 𝑜𝑟 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑐𝑜𝑙𝑙𝑖𝑔𝑎𝑡𝑖𝑣𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦 =

𝐶𝑜

𝐶𝑐

As we know that, colligative property is inversely proportional to the molecular mass of

the solute, we can also write:

𝑖 = 𝑁𝑜𝑟𝑚𝑎𝑙 𝑜𝑟 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠

𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑜𝑟 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 =

𝑀𝐶

𝑀𝑂




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Modified Expressions of Colligative Properties :

Relative Lowering in Vapour pressure:

∆𝐏𝐀

𝐏𝐀ₒ = 𝐢

𝐖𝐁 × 𝐌𝐀

𝐌𝐁 × 𝐖𝐀 OR

𝐏𝐀ₒ − 𝐏𝐀

𝐏𝐀ₒ = 𝐢 𝐱𝐁

Elevation in Boiling point:

∆𝐓𝐛 = 𝐢 × 𝐊𝐛 × m OR ∆𝐓𝐛 = 𝐢 × 𝐊𝐛 × 𝐖𝐁 × 𝟏𝟎𝟎𝟎

𝐌𝐁 × 𝐖𝐬𝐨𝐥𝐯𝐞𝐧𝐭 𝐢𝐧 𝐠

Depression in Freezing point:

∆𝐓𝐟 = 𝐢 × 𝐊𝐟 × m OR ∆𝐓𝐟 = 𝐢 × 𝐊𝐟 × 𝐖𝐁 × 𝟏𝟎𝟎𝟎

𝐌𝐁 × 𝐖𝐬𝐨𝐥𝐯𝐞𝐧𝐭 𝐢𝐧 𝐠

Osmotic pressure:

𝛑 = 𝐢 𝐖𝐁 × 𝐑𝐓 × 𝟏𝟎𝟎𝟎

𝐌𝐁 × 𝐕

E 1.: Calculation of the degree of Dissociation : When an electrolyte is dissolved in a solvent, it dissociates into ions. The fraction of the

total number of electrolyte molecules that undergoes dissociation is called degree of

dissociation.

Degree of dissociation (α) = No of moles dissociated

Total number of moles taken

Consider that, A molecules dissociates in solvent to give ‘n’ ions

A n1B + n2C + …… ∴ 𝐧𝟏 + 𝐧𝟐 + …. = 𝐧

Initial conc. : 1 0 moles

Final conc. : 1 – α n ∝ moles

Hence,

Total number of moles of solute and ions after dissociation = (1−∝) + n ∝ = 1+ (n −1)∝

Thus, observed colligative property = 1 + (n − 1)∝

And , Normal or calculated colligative property = 1 (as 1 mole of solute ‘A’ was taken)

∴ van’t Hoff factor ‘i’ = Observed colligative property

Calculated colligative property =

1 + (n−1)∝

1 OR ∝ =

i−1

n−1

As we know that, i’ = Calculated Molecular mass

Observed Molecular mass =

MC

MO ∴ ∝ =

MCMO

− 1

n − 1 =

MC − MO

MO (n−1)




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E 2.: Calculation of the degree of Association :

In some cases, ‘n’ molecules of the solute ‘A’ associate to form large associated molecule

‘An’. The fraction of the total number of solute molecules which exist in the form of

associated molecule is called degree of association.

Degree of association (α) = No of moles associated

Total number of moles taken

Example : Molecules of ethanoic acid (acetic acid) dimerise in benzene due to hydrogen bonding. This normally happens in solvents of low dielectric constant. In this case the number of particles is reduced due to dimerization.

Consider that, ‘n’ number of simple molecules of solute ‘A’ associate to form associated

molecule An.

𝐧 A 𝐀𝐧

Initial conc. : 1 0 moles

Final conc. : 1 – α ∝/n moles

Hence, total number of moles in solution = (1− ∝) + ∝

n (observed colligative property)

And, total number of moles of solute was taken = 1 (calculated colligative property)

∴ van’t Hoff factor ‘i’ = Observed colligative property

Calculated colligative property =

(1− ∝) + ∝

𝐧

1

∝ = (1 − i)n

n−1 As we know that, ′i’ =

Calculated Molecular mass


MC

MO

∴ ∝ = (1 −MC

MO )

n

n − 1 = (

Mo − Mc

MO )

n

n − 1

Question .: 2 gm 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?

Answer .: Given Values : WB = 2 gm , WA = 25 gm , ∆𝐓𝐟 = 1.62 K

Kf = 4.9 K kg mol-1




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Let’s find out the experimental or observed molecular mass of the solute

𝐌𝐁 = 𝐊𝐟 × 𝐖𝐁 × 𝟏𝟎𝟎𝟎 𝐠 𝐤𝐠−𝟏

∆𝐓𝐟 × 𝐖𝐀 𝐢𝐧 𝐠 =

𝟒.𝟗 𝐊 𝐤𝐠 𝐦𝐨𝐥−𝟏 × 𝟐 𝐠 × 𝟏𝟎𝟎𝟎 𝐠 𝐤𝐠−𝟏

𝟏.𝟔𝟐 𝐊 × 𝟐𝟓 𝐠 = 241.98 g mol–1

Now, the calculated or normal molecular mass of the solute is

C6H5COOH = 12 x 7 + 6 + 16 x 2 = 122 g mol–1

Calculate Van’t Hoff factor ′i’ = Calculated Molecular mass


MC

MO =

122 𝐠 𝐦𝐨𝐥−𝟏

241.98 𝐠 𝐦𝐨𝐥−𝟏 = 0.504

Calculate Degree of association (∝): ∝ = (1 − i)n

n−1

As we know, benzoic acid forms dimer in solution, so that, the value on ‘n’ is 2.

2 C6H5COOH ⇄ (C6H5COOH)2

Hence, Degree of association will be; ∝ = (1 − i)2

2−1 = (1 − i) 2

∝ = (1−0.504) × 2 = 0.496 × 2 = 0.992

Therefore, percentage association of benzoic acid in benzene is 99.2 %.

PRACTICE PROBLEMS

Q.1.: 0.5 gm KCl was dissolved in 100 gm water and the solution originally at 20 °C,

froze at −0.24 °C. Calculate the percentage ionization of salt.

Kf for water = 1.86 K Kg mol-1.

Q.2.: 0.6 mL of acetic acid having density of 1.06 gm/ml is dissolved in 1 Litre of water.

The depression in freezing point observed for this strength of the acid was 0.0205 °C.

Calculate the Van’t Hoff factor and the dissociation constant of the acid. Kf for water =

1.86 K Kg mol-1.

Q.3.: Depression in freezing point of 0.10 molal solution of HF is −0.201 °C. Calculate

the percentage degree of dissociation of HF. Kf for water = 1.86 K Kg mol-1.

Q.4.: Calculate the freezing point depression expected for 0.0711 molal aqueous

solution of Na2SO4. If this solution actually freezes at −0.32 °C, What would be the value

of Van’t Hoff factor ? Kf for water = 1.86 K Kg mol-1.

Q.5.: Calculate the boiling point of a solution prepared by adding 15 gm of NaCl to 250

gm of water (Kb for water = 0.52 K Kg mol-1).

Q.6.: What mass of NaCl must be dissolved in 65 gm of water to lower the freezing

point by 7.5 °C ? Kf for water = 1.86 K Kg mol-1. Assuming that Van’t Hoff factor for NaCl

is 1.87.


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CLASS 12 - Dev Academy