//M2//QN1//SUB//DL0

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

//X

‘Solution is homogeneous mixture of two or more than two components.’

Types of Solution | Physical State of Solute | Physical State of Solvent
Gaseous Solutions | Gas | Gas
Example: Mixture of oxygen and nitrogen gases.
Liquid | Gas
Example: Chloroform mixed with nitrogen gas
Solid | Gas
Example: Camphor in nitrogen gas
Liquid Solutions | Gas | Liquid
Example: Oxygen dissolved in water, CO 2 dissolved in water
Liquid | Liquid
Example: Ethanol dissolved in water
Solid | Liquid
Example: Glucose dissolved in water, NaCl dissolved in water
Solid Solutions | Gas | Solid
Example: Solution of hydrogen in palladium
Liquid | Solid
Example: Amalgam of mercury with sodium
Solid | Solid
Example: Copper dissolved in gold

//M1//QN2//SUB//DL0

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

//X

of hydrogen (solute) in Palladium (Pd), (solvent).

//M0//QN3//SUB//DL0

What is concentration of solution? Write down different types of units of concentration?

//X

‘The amount of solute present in per unit volume of solution is known as concentration.’
There are 9 types of units of concentration:
(1) % w/w (Mass percentage)
(2) % v/v (Volume percentage)
(3) % w/v (Mass by volume percentage)
(4) ppm (Parts per million)
(5) Mole-Fraction (X)
(6) Molarity (M)
(7) Molality (m)
(8) Normality (N)
(9) Formality (F)

//M2//QN4//SUB//DL0

Explain (% W/W) mass percentage with example.

//X

‘The mass of solute (g) dissolved in 100 gram of solution is known as mass percentage (% w/w).’
Mass % of a component
= × 100
For example, if a solution is described by
10% glucose in water by mass, it means that 10 g of glucose is dissolved in 90 g of water resulting in a 100 g solution.
Concentration described by mass percentage is commonly used in industrial chemical applications.
For example, commercial bleaching solution contains 3.62 mass percentage of sodium hypochlorite
in water.

//M0//QN5//SUB//DL0

Explain (% V/V) volume percentage with

example.

//X

‘The volume of solute (ml) dissolved in 100 ml solution is known as volume percentage (% v/v)’.
Volume % of a component
= × 100
For example, 10% ethanol solution in water means that 10 mL of ethanol is dissolved in water such that the total volume of the solution is 100 mL.
Solutions containing liquids are commonly expressed in this unit.
For example, a 35% (v/v) solution of ethylene glycol, an antifreeze, is used in cars for cooling the engine. At this concentration the antifreeze lowers the freezing point of water to 255.4K (–17.6°C).

//M2//QN6//SUB//DL0

Explain (% W/V) mass by volume percentage with example.

//X

‘The mass of solute (g) dissolved in 100 ml solution is called mass by volume percentage (% W/V)’.
% W/V =
For example 5% W/V aqueous solution of sugar means 5 gram sugar is dissolved in 100 ml solution.

//M0//QN7//SUB//DL0

Explain parts per million (ppm) with example.

//X

‘When a solute is present in trace quantities, it is convenient to express concentration in parts per million (ppm).’
Parts per million
= × 10 6
Parts per million can also be expressed as mass to mass, volume to volume and mass to volume.
A litre of sea water (which weighs 1030 g) contains about 6 x 10–3 g of dissolved oxygen (O 2). Such a small concentration is also expressed as 5.8 g per 10 6 g (5.8 ppm) of sea water.
The concentration of pollutants in water or atmosphere is often expressed in terms of µ g mL–1 or ppm.

//M2//QN8//SUB//DL0//EQ

Explain mole fraction is detail.

//X

‘Mole Fraction of component is ratio of mole of component and total moles of all component in solution’.
Mole fraction is denoted by x.
Mole fraction of a component
=
For example, in a binary mixture, if the number of moles of A and B are n A and n B respectively, the mole fraction of A will be
x A =
For a solution containing i number of components, we have
x i = =
It can be shown that in a given solution sum of all the mole fractions are unity (1)
x 1 + x 2 + ......... + x i = 1
Mole fraction unit is very useful in relating some physical properties of solutions, say vapour pressure with the concentration of the solution and quite useful in describing the calculations involving gas mixtures.

//M2//QN9//SUB//DL0//EQ

Explain molarity in detail.

//X

Molarity (M) is defined as number of moles of solute dissolved in one litre of solution.
Molarity =
For example, 0.25 mol L–1 (or 0.25 M) solution of NaOH means that 0.25 mol of NaOH has been dissolved in one litre (or one cubic decimetre) of the solution.
Unit of molarity is , Molar

//M2//QN10//SUB//DL0//EQ

Explain molality in details.

//X

Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent.
Molality (m) =
For example, 1.00 mol kg–1 (or 1.00 m) solution of KCl means that 1 mol (74.5 g) of KCl is dissolved in 1 kg of water.
Unit of molality is , or Molal (m).

//M0//QN11//SUB//DL0//EQ

What is solubility? Explain solubility of a solid in a liquid.

//X

Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature.
Solubility depends upon
(1) Nature of solute and solvent
(2) Temperature
(3) Pressure
Nature of solute and solvent: Every solid does not dissolve in a given liquid. While sodium chloride and sugar dissolve readily in water, naphthalene and anthracene do not. On the other hand, naphthalene and anthracene dissolve readily in benzene but sodium chloride and sugar do not.
It is observed that polar solutes dissolve in polar solvents and non-polar solutes in non-polar solvents.
In general, a solute dissolve in a solvent if the intermolecular interactions are similar in the two or we may say like dissolves like.
Effect of temperature: The solubility of a solid in a liquid is significantly affected by temperature changes.
Solute + Solvent Solution
This, being dynamic equilibrium, must follow Le Chateliers Principle.
If in a nearly saturated solution, the dissolution process is endothermic (D sol H > 0), the solubility should increase with rise in temperature.
If it is exothermic (D sol H < 0) the solubility should decrease.
Effect of pressure: Pressure does not have any significant effect on solubility of solids in liquids.
It is so because solids and liquids are highly incompressible and practically remain unaffected by changes in pressure.

//M0//QN12//SUB//DL0

Why do gases always tend to be less soluble in liquids as the temperature is raised?

//X

Solubility of gases in liquids decreases with an increase in temperature. This is because dissolution of gases in liquids is an exothermic process.
Gas + Liquid → Solution + Heat
Therefore, when the temperature is increased, heat is supplied and the equilibrium shifts backwards, thereby decreasing the solubility of gases.

//M0//QN13//SUB//DL0//EQ

Explain solubility of gas in a liquid solvent.

//X

Many gases dissolve in water. Oxygen dissolves only to a small extent in water. It is this dissolved oxygen which sustains all aquatic life. On the other hand, hydrogen chloride gas (HCl) is highly soluble in water.
Solubility of gases in liquids is greatly affected by pressure and temperature.
Effect of pressure: The solubility of gases increase with increases of pressure.
(Pressure ∝ solubility of gas)
Consider a system as shown in Fig. (a). The lower part is solution and the upper part is gaseous system at pressure p and temperature T.
Assume this system to be in a state of dynamic equilibrium, i.e., under these conditions rate of gaseous particles entering and leaving the solution phase is the same.
Now increase the pressure over the solution phase by compressing the gas to a smaller volume Fig. (b).
This will increase the number of gaseous particles per unit volume over the solution and also the rate at which the gaseous particles are striking the surface of solution to enter it.
The solubility of the gas will increase until a new equilibrium is reached resulting in an increase in the pressure of a gas above the solution and thus its solubility increases.
Effect of temperature: Solubility of gases in liquids decreases with rise in temperature.

//M0//QN14//SUB//DL0//EQ

Explain Henry's law and write it's application.OR State Henry's law and mention some important applications.

//X

Henry give a quantitative relation between pressure and solubility of a gas in a solvent which is known as Henry's law.
At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution.
If we use the mole fraction of a gas in the solution as a measure of its solubility, then it can be said that the mole fraction of gas in the solution is proportional to the partial pressure of the gas over the solution.
p = K H x
Where, p = Partial pressure of gas
K H = Henry's constant
x = Solubility of gas
If we draw a graph between partial pressure of the gas versus mole fraction of the gas in solution, then we should get a plot of the type as shown in Fig.
Different gases have different K H values at the same temperature.
Higher the value of K H at a given pressure, the lower is the solubility of the gas in the liquid.
If temperature increases the value of K H increases so solubility decreases.
Applications of Henry’s Law:
(1) To increase the solubility of CO 2 in soft drinks and soda water, the bottle is sealed under high pressure.
(2) Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure underwater. Increased pressure increases the solubility of atmospheric gases in blood. When the divers come towards surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. This blocks capillaries and creates a medical condition known as bends, which are painful and dangerous to life.
To avoid bends, as well as, the toxic effects of high concentrations of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen).
(3) At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers. Low blood oxygen causes climbers to become weak and unable to think clearly, symptoms of a condition known as anoxia.

Values of Henry's Law Constant
Gas (in water) | Temperature/K | K H / kbar
He | 293 | 144.97
H 2 | 293 | 69.16
N 2 | 293 | 76.48
N 2 | 303 | 88.84
O 2 | 293 | 34.86
O 2 | 303 | 46.82
Argon | 298 | 40.3
CO 2 | 298 | 1.67
Formaldehyde | 298 | 1.83 × 10–5
Methane | 298 | 0.413
Vinyl chloride | 298 | 0.611

//M0//QN15//SUB//DL0//EQ

Explain Raoult's Law for volatile solute and Volatile solvent and derive formula for total vapour pressure with graph. OR Derive equation of Raoult's law for vapour pressure of liquid-liquid solution and give its conclusion.

//X

Let us consider a binary solution of two volatile liquids and denote the two components as 1 and 2.
When taken in a closed vessel, both the components would evaporate and eventually an equilibrium would be established between vapour phase and the liquid phase.
Suppose p 1 and p 2 partial vapour pressure of component 1 and 2 and x 1 and x 2 are Mole-Fraction of component 1 and 2 respectively.
The French chemist, Francois Marte Raoult gave the quantitative relationship between them. The relationship is known as the Raoult's law.
For a solution of volatile liquids the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
For component: 1.
\ p 1 = p 0 1 . x 1
where p 0 1 is the vapour pressure of pure
component 1.
Similarly for component 2
\ p 2 = p 0 2 . x 2
where p 0 2 is the vapour pressure of pure component 2
According to Dalton's law of partial pressures, Total pressure over the solution phase in the container will be the sum of the partial pressures of the components of the solution
p Total = p 1 + p 2
= p 0 1 . x 1 + p 0 2 . x 2
= p 0 1 (1 – x 2) + p 0 2 . x 2
= p 0 1 – p 0 1 . x 2 + p 0 2 . x 2
p Total = p 0 1 + x 2 (p 0 2 – p 0 1)
Following conclusions can be drawn from above equation:
(i) Total vapour pressure over the solution can be related to the mole fraction of any one component.
(ii) Total vapour pressure over the solution varies linearly with the mole fraction of component 2.
(iii) Depending on the vapour pressures of the pure components 1 and 2, total vapour pressure over the solution decreases or increases with the increase of the mole fraction of component 1.
A plot of p 1 or p 2 versus the mole fractions x 1 and x 2 for a solution gives a linear plot as shown in Fig.

These lines (I and II) pass through the points for which x 1 and x 2 are equal to unity.
Similarly, the plot (line III) of p total versus x 2 is also linear.
The minimum value of p Total is p 0 1 and the maximum value is p 0 2, assuming that component – 1 is less volatile than component – 2 ie., (p 0 1 < p 0 2).
The composition of vapour phase in equilibrium with the solution is determined by the partial pressures of the components.
If y 1 and y 2 are the mole-fractions of the component 1 and 2 respectively in vapour phase then,
Using Dalton's law of partial pressure,
p 1 = y 1 . p Total
p 2 = y 2 . p Total
In general,
p i = y i p Total

//M0//QN16//SUB//DL0//EQ

Explain Raoult's Law as a special case of Henry's Law.

//X

According to Raoult's law, the vapour pressure of a volatile component in a given solution is given by
p 1 = . x 1
According to Henry's Law solubility of gaseous solute in liquid solvent is given by
p = K H . x
If we compare the equations for Raoult's law and Henry'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 constant K H differs from .
Thus, Raoult's law becomes a special case of Henry's law in which K H becomes equal to .

//M0//QN17//SUB//DL0//EQ

Explain vapour pressure of solutions of solids in liquids.

//X

Liquids at a given temperature vapourise and under equilibrium conditions the pressure exerted by the vapours of the liquid over the liquid phase is called vapour pressure
In a pure liquid [Fig. a] the entire surface is occupied by the molecules of the liquid. If a non-volatile solute is added to a solvent to give a solution [Fig. b] the vapour pressure of the solution is solely from the solvent alone.
This vapour pressure of the solution at a given temperature is found to be lower than the vapour pressure of the pure solvent at the same temperature.
In the solution, the surface has both solute and solvent molecules thereby the fraction of the surface covered by the solvent molecules gets reduced.
Consequently, the number of solvent molecules escaping from the surface is correspondingly reduced, thus, the vapour pressure is also reduced.
The decrease in the vapour pressure of solvent depends on the quantity of non-volatile solute present in the solution, irrespective of its nature.
For example, decrease in the vapour pressure of water by adding 1.0 mol of sucrose to one kg of water is nearly similar to that produced by adding 1.0 mol of urea to the same quantity of water at the same temperature.
Raoult's law:

- ‘In binary solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.’
- In Binary solution p 1 is vapour pressure of the solvent and x 1 is mole-fraction.

\ p 1 = . x 1
The proportionality constant is equal to the vapour pressure of pure solvent, p 0 1
A plot between the vapour pressure and the mole fraction of the solvent is linear.

//M3//QN18//SUB//DL0

What is ideal solution? Explain with example.

//X

‘The solutions which follows Raoult's law over the entire range of concentration are known as ideal solutions’.
The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero,
∆mix H = 0 , ∆mix V = 0
It means that no heat is absorbed or evolved when the components are mixed.
The volume of solution would be equal to the sum of volumes of the two components.
At molecular level, ideal behaviour of the solutions can be explained by considering two components A and B. In pure components, the intermolecular attractive interactions will be of types A-A and B-B, where as in the binary solutions in addition to these two interactions, A-B type of interactions will also be present.
If the intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B, this leads to the formation of ideal solution.
Examples of ideal solution: n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene, etc.

//M4//QN19//SUB//DL0//EQ

What are non–ideal solutions? Explain

non-ideal solutions with positive and negative deviation.OR What is meant by positive and negative deviations from Raoult's law and how is the sign of D mixH related to positive and negative deviations from Raoult's law?

//X

‘The solutions which does not follows Raoult's law over the entire range of concentration are known as non ideal solutions’.
The vapour pressure of such a solution is either higher or lower than that predicted by Raoult's law
If it is higher, the solution exhibits positive deviation and if it is lower, it exhibits negative deviation from Raoult's law.
For non-ideal solution D mix H ≠ 0 and D mix V ≠ 0
Positive deviation: ‘In positive deviation vapour pressure of solution is higher than that of Raoult's law’.
In positive deviation D mix H > 0, D mix V > 0
In case of positive deviation from Raoult's law, A-B interactions are weaker than those between A-A or B-B.
In this case the intermolecular attractive forces between the solute-solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules.
This means that in such solutions, molecules of A (or B) will find it easier to escape than in pure state. This will increase the vapour pressure and result in positive deviation.
Mixtures of ethanol and acetone behave in this manner. In pure ethanol, molecules are hydrogen bonded. On adding acetone, its molecules get in between the host molecules and break some of the hydrogen bonds between them. Due to weakening of interactions, the solution shows positive deviation from Raoult's law.
In a solution formed by adding carbon disulphide to acetone, the dipolar interactions between solute-solvent molecules are weaker than the respective interactions among the solute-solute and solvent-solvent molecules. This solution also shows positive deviation.
Negative deviation: ‘In negative deviation the vapour pressure of solution is lower than that of Raoult's Law’.
In negative deviation D mix H < 0, D mix V < 0
In case of negative deviations from Raoult's law, the intermolecular attractive forces between A-A and B-B are weaker than those between A-B and leads to decrease in vapour pressure resulting in negative deviations.
Mixture of phenol and aniline is show negative deviation. the intermolecular hydrogen bonding between phenolic proton and lone pair on nitrogen atom of aniline is stronger than the respective intermolecular hydrogen bonding between similar molecules.
Mixture of chloroform and acetone forms a solution with negative deviation from Raoult's law. This is because chloroform molecule is able to form hydrogen bond with acetone molecule as shown.
This decreases the escaping tendency of molecules for each component and consequently the vapour pressure decreases resulting in negative deviation from Raoult's law.

//M0//QN20//SUB//DL0

What are azeotropes? Explain its types.

//X

‘Some liquids on mixing, from azeotropes which are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature’.
In such cases, it is not possible to separate the components by fractional distillation.
There are two types of azeotropes.
Minimum boiling azeotrope: The solutions which show is large positive deviation from Raoult's law form minimum boiling azeotrope at a specific composition.
For example, ethanol-water mixture (obtained by fermentation of sugars) on fractional distillation gives a solution containing approximately 95% by volume of ethanol. Once this composition, known as azeotrope composition, has been achieved, the liquid and vapour have the same composition, and no further separation occurs.
Maximum boiling azeotrope: The solutions that show large negative deviation from Raoult's law form maximum boiling azeotrope at a specific composition.
Nitric acid and water is an example of this class of azeotrope. This azeotrope has the approximate composition, 68% nitric acid and 32% water by mass, with a boiling point of 393.5K.

//M0//QN21//SUB//DL0

What are colligative properties? Write colligative properties of solution.

//X

‘Colligative properties of solution depend on the number of solute particles and does not depend on the nature of solute’.
There are 4 types of colligative properties.

(1) Relative lowering of vapour pressure of the solvent.

(2) Depression of freezing point of the solvent.

(3) Elevation of boiling point of the solvent.

(4) Osmotic pressure of the solution.

//M0//QN22//SUB//DL0//EQ

Write Raoult's Law for non-volatile solute and volatile solvent and derive its formula.

//X

The vapour pressure of a solvent in solution is less than that of the pure solvent.
Raoult established that the lowering of vapour pressure depends only on the concentration of the solute particles and it is independent of their identity.
A relation between vapour pressure of the solution, mole fraction and vapour pressure of the solvent,
p 1 = . x 1
The reduction in the vapour pressure of solvent (D p 1) is given as :
D p 1 = – p 1
\ D p 1 = – . x 1
\ D p 1 = (1 – x 1)
\ D p 1 = x 2
The lowering of the vapour pressure is directly proportional to mole-fraction of solute.
\ = x 2
\ = x 2
\ =
Where, n 1 = Moles of solvent
n 2 = Moles of solute
For dilute solutions n 2 << n 1
=
\ =
Where, w 1 = Weight of solvent
w 2 = Weight of solute
M 1 = Molar mass of solvent
M 2 = Molar mass of solute
= Vapour pressure of pure solvent
p 1 = Vapour pressure of solution

//M0//QN23//SUB//DL0//EQ

What is boiling point? What is elevation of boiling point? Explain molal elevation constant and derive its formula.

//X

The temperature at which, the vapour pressure of solution is equal to the atmospheric pressure, such temperature is known as boiling point.
For example, water boils at 373.15K (100°C) because at this temperature the vapour pressure of water is 1.013 bar (1 atmosphere).
Vapour pressure of the solvent decreases in the presence of non-volatile solute.
The vapour pressure of an aqueous solution of sucrose is less than 1.013 bar at 373.15K. In order to make this solution boil, its vapour pressure must be increased to 1.013 bar by raising the temperature above the boiling temperature of the pure solvent (water).
Thus, the boiling point of a solution is always higher than that of the boiling point of the pure solvent.
Let be the boiling point of pure solvent and T b be the boiling point of solution. The increase in the boiling point D T b = T b – is known as elevation of boiling point.
Experiments have shown that for dilute solutions the elevation of boiling point (D T b) is directly proportional to the molal concentration of the solute in a solution.
D T b ∝ m
\ D T b = K b . m ....(1)
Kb is called Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic constant).
Molal elevation constant: “Increase in boiling point of a solution prepared by dissolving one gram molar mass of non volatile solute in one kg of solvent is called as molal elevation constant.”

- Unit of Kb = K⋅kg.Mol–1
- If w 2 gram of solute of molar mass M 2 is dissolved in w 1 gram of solvent, then molality, m of the solution is given by the expression:

m =
Substituting value of molality in eq. (1)
D T b =
M 2 =
Where, w 2 = Weight of Solute
w 1 = Weight of Solvent
M 2 = Molar mass of Solute

//M0//QN24//SUB//DL0//EQ

How to find out Kb and Kf ?

//X

Kf =
K b =
Where, R = Gas constant
M 1 = Molar mass of solvent
T f = Freezing point of pure solvent
T b = Boiling point of pure solvent.
D fus H = Fusion enthalpy
D vap H = Vapourisation enthalpy

//M4//QN25//SUB//DL0//EQ

What is depression of freezing point? What is molal depression constant (Kf)? Derive its formula of finding molar mass of solute.

//X

Freezing point of a substance may be defined as, ‘the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase’.
A solution will freeze when its vapour pressure equals the vapour pressure of the pure solid solvent.
According to Raoult's law, when a non-volatile solid is added to the solvent its vapour pressure decreases and now it would become equal to that of solid solvent at lower temperature.
Thus, the freezing point of the solvent decreases.
Let be the freezing point of pure solvent and be its freezing point when non-volatile solute is dissolved in it. The decrease in freezing point.
D T f = – T f is known as depression in
freezing point.
Depression of freezing point (D T f) for dilute solution is directly proportional to molality (m) of the solution.
D T f ∝ m
\ D T f = K f . m .....(1)
Kf is known as molal depression constant or freezing point depression constant or cryoscopic constant.
The unit of K f is k.kg.mol–1
m =
Substituting this value of molality in eq. (1)
D T f = K f .
\ M 2 = K f .
Where, w 2 = weight of solute
w 1 = weight of solvent
M 2 = Molar mass of solute

//M0//QN26//SUB//DL0

What is semi–permeable membrane? Give examples.

//X

‘The membrane which allows only the small molecule of solvent to pass but cannot pass solute molecule is known as semi-permeable membrane’.
Examples: Pig's bladder, parchment, cellophane etc.
These membranes appear to be continuous sheets or films, yet they contain a network of submicroscopic holes or pores.

//M0//QN27//SUB//DL0//EQ

What is osmosis? What is osmotic pressure? Derive it's formula.OR Explain osmosis and osmatic pressure. Derive formula to find the molecular mass of solute on the basis of osmatic pressure of solution.

[June 2024]

//X

‘If semi–permeable membrane is placed between the solvent and solution as shown fig. the solvent molecules will flow through the membrane from pure solvent to the solution. This process of flow of the solvent is called osmosis’.
The flow will continue till the equilibrium is attained.
The flow of the solvent from its side to solution side across a semi-permeable membrane can be stopped if some extra pressure is applied on the solution.
This pressure that just stops the flow of solvent is called osmotic pressure of the solution.
The osmotic pressure of a solution is the excess pressure that must be applied to a solution to prevent osmosis.
Osmotic pressure is a colligative property as it depends on the number of solute molecules and not on their identity.
Osmotic pressure is proportional to the molarity,
C of the solution at a given temperature T.
p = CRT
Where, p = Osmotic pressure
R = gas constant
\ p =
p =
Where, w 2 = Weight of solute
M 2 = Molar mass of solute
V = Volume of solution (L)
T = Temperature

//M0//QN28//SUB//DL0//EQ

Which method is most suitable to determine molecular mass of polymer?

//X

Measurement of osmotic pressure provides another method of determining molar masses of solutes.
This method is widely used to determine molar masses of proteins, polymers and other macro molecule.
The osmotic pressure method has the advantage over other methods as pressure measurement is around the room temperature and the molarity of the solution is used instead of molality.
As compared to other colligative properties, its magnitude is large even for very dilute solutions.
The technique of osmotic pressure for determination of molar mass of solutes is particularly useful for biomolecules as they are generally not stable at higher temperatures and polymers have poor solubility.

//M0//QN29//SUB//DL0//EQ

What is isotonic, hypertonic and hypotonic solution?

//X

Isotonic solution: ‘Two solutions having same osmotic pressure at a given temperature are called isotonic solutions’.
When such solutions are separated by semipermeable membrane, no osmosis occurs between them.
For example, the osmotic pressure associated with the fluid inside the blood cell is equivalent to that of 0.9% (mass/volume) sodium chloride solution,
Hypertonic solution: ‘The solution which possess more osmotic pressure with respect to other solution is known as hypertonic solution’.
For example, if we place the cells in a solution containing more than 0.9% (mass/volume) sodium chloride, water will flow out of the cells and they would shrink. Such a solutio n is called hypertonic.
Hypotonic solution: ‘The solution which possess less osmotic pressure with respect to other solution is known as hypotonic solution’.
For example, If the salt concentration is less than 0.9% (mass/volume), the solution is said to be hypotonic. In this case, water will flow into the cells placed in the solution and they would swell.

//M0//QN30//SUB//DL0//EQ

Explain reverse osmosis for purification of water.

//X

‘The direction of osmosis can be reversed if a pressure larger than the osmotic pressure is applied to the solution side. That is, now the pure solvent flows out of the solution through the semi permeable membrane. This phenomenon is called reverse osmosis.’
Reverse osmosis is used in desalination of sea water.
When pressure is more than osmotic pressure is applied, pure water is squeezed out of the sea water through the membrane. A variety of polymer membranes are available for this purpose.
The pressure required for the reverse osmosis is quite high. A workable porous membrane is a film of cellulose acetate placed over a suitable support. Cellulose acetate is permeable to water but impermeable to impurities and ions present in sea water.
These days many countries use desalination plants to meet their potable water requirements.

//M0//QN31//SUB//DL0//EQ

Explain abnormal molar masses. Also explain association and dissociation of solute.

//X

Ionic compounds when dissolved in water dissociate into cations and anions.
For example, if we dissolve one mole of KCl
(74.5 g) in water, we expect one mole each of K+ and Cl– ions to be released in the solution. If this happens, there would be two moles of particles in the solution.
one mole of KCl in one kg of water would be expected to increase the boiling point by
2 × 0.52 K = 1.04 K
If we did not know about the degree of dissociation, we could be led to conclude that the mass of
2 mol particles is 74.5 g and the mass of one mole of KCl would be 37.25 g.
When there is dissociation of solute into ions, the experimentally determined molar mass is always lower than the true value.
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 dimerisation.
Association of molecules is depicted as follows:
It can be undoubtedly stated here that if all the molecules of ethanoic acid associate in benzene, then D T b or D T f for ethanoic acid will be half of the normal value.
The molar mass calculated on the basis of this D T b or D T f will, therefore, be twice the expected value. Such a molar mass that is either lower or higher than the expected or normal value is called as abnormal molar mass.

//M0//QN32//SUB//DL0//EQ

Explain Van't Hoff factor.

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Van't Hoff introduced a factor i, known as the Van't Hoff factor, to account for the extent of dissociation or association. This factor i is defined as:
i =
=
i =
Here abnormal molar mass is the experimentally determined molar mass and calculated colligative properties are obtained by assuming that the non-volatile solute is neither associated nor dissociated.
In case of association, value of i is less than unity while for dissociation it is greater than unity.
Inclusion of Van't Hoff factor modifies the equations for colligative properties as follows:
Relative lowering of vapour pressure of solvent,
= i.
Elevation of Boiling point, D T b = i . K b . m
Depression of Freezing point, D T f = i . K f . m
Osmotic pressure of solution, p = i n 2 RT/V
i for several strong electrolytes. For KCl, NaCl and MgSO 4, i values approach 2 as the solution becomes very dilute. As expected, the value of
i gets close to 3 for K 2 SO 4.

Salt | *Values of i | Van't Hoff Factor i for complete dissociation of solute
0.1 m | 0.01 m | 0.001 m |
NaCl | 1.87 | 1.94 | 1.97 | 2.00
KCl | 1.85 | 1.94 | 1.98 | 2.00
MgSO 4 | 1.21 | 1.53 | 1.82 | 2.00
K 2 SO 4 | 2.32 | 2.70 | 2.84 | 3.00

Class 12 Chemistry (Part 1) 003