The colloidal particles carry an electric charge. The most important property of colloidal solution is that all suspended particles possess either positive or a negative charge. i.e. they carry the same nature of the charge. The mutual forces of repulsion between similarly charged particles prevent them from aggregating and settling under the action of gravity.  This gives stability to the sol.

The dispersion medium carries the opposite charge, hence as a whole, the colloidal solution is electrically neutral.

Origin of the Charge on Colloidal Particles:

A small quantity of electrolyte is always present in the colloidal dispersion. Its presence is necessary for the stability of the sol, as complete removal of the sol causes coagulation of the sol.

Presence of Some Acidic or Basic Groups in Colloidal Solution: 

Colloidal particles may acquire electric charge due to the presence of certain acidic or basic groups in colloidal solution.

For example, protein molecules give rise to the formation of colloidal solutions. Thus the particles of protein sol either have a positive charge or a negative charge depending upon the pH of the medium. A molecule of protein contains a carboxylic acid (COOH) group and also a basic amino (–NH₂) group, it will form a positively charged particle in the acidic medium and a negatively charged particle in the basic or alkaline medium.

In the case of the colloidal solution of proteins, the nature of charge on colloidal particles depends on the pH of the solution. The isoelectric point of a colloid is a pH at which net charge on colloidal particles is zero. Above this pH, the particles are negatively charged and below this pH, particles are positively charged. At isoelectric point, the particles exist in the form of Zwitter ion. Hence they do not migrate under the influence of the electric field.

The isoelectric pH for some proteins is Haemoglobin (pH 4.3-5.3), Casein from human milk (pH 4.1 –  4.7), and Gelatin (pH 4.7).

Due to Self Dissociation:

The dissociation of surface molecules leads to electric charge on colloidal particles of the sol.

For example, Consider an aqueous solution of soap (sodium palmitate) which undergoes dissociation into ions.

The cations (Na⁺) pass into the solvent. Due to the weak attractive forces present in the long hydrocarbon chains, the anions (C₁₅H₃₁ COO^–) have a tendency to form negatively charged aggregates of colloidal dimensions. This type of development of charge is only possible with electrolytes. This is not possible in colloidal solutions of non-electrolytes such as clay, smoke etc.

Electron Capture by Colloidal Particles: 

It is believed that the colloidal solutions prepared by Bredig’s Arc Dispersion Method acquire a charge by electron capture.

Selective or Preferential Adsorption of Ions:

When two or more ions are present in the dispersion medium, then the colloidal particles adsorb preferentially positive or negative ions present in the dispersion medium.

Example: Positively charged Ferric hydroxide sol: 

If FeCl₃ solution is added to the Ferric hydroxide, preferentially Fe³⁺ ions adsorb on Fe(OH)₃ molecules, that is why colloidal particles of ferric hydroxide are positively charged.

FeCl₃ ⇌ Fe³⁺ + 3 Cl^–

Fe(OH)₃ + Fe³⁺ → Fe(OH)₃ / Fe³⁺

Example: Positive charged Siver iodide sol:

When dilute KI is added in excess dilute AgNO₃, the Ag⁺ ions are adsorbed on Agl, and [AgI]Ag⁺ is formed. Thus the colloidal particles have a positive charge.

AgNO₃ + KI  → AgI + KNO₃

AgNO₃ ⇌ Ag⁺ + NO₃^–

AgI + Ag⁺ → AgI / Ag⁺

Example: Negative charged Siver iodide sol: 

When dilute AgNO₃ is added in excess dilute KI, the I^– ions are adsorbed on Agl, and [AgI]I^– is formed. Thus colloidal particles have a negative charge.

AgNO₃ + KI  → AgI + KNO₃

KI ⇌ K⁺ + I^–

AgI + Ag⁺ → AgI / I^–

Example: Negatively charged Arsenious Sulphide Sol: 

It is prepared by passing H₂S gas slowly through the solution of AS₂O₃.

AgNO₃ + KI  → AgI + KNO₃

H2S ⇌ 2H⁺ + S^2-

AS₂S₃ + S^2- → AS₂S₃ / S^2-

Frictional Electrification:

The origin of charge on the colloidal particles may be due to frictional electrification. By mutual rubbing of colloidal particles with molecules of the dispersion medium, the charge is developed on the sol. This view is not satisfactory.

Electrical Double Layer:

Colloidal particles are charged. This charge on the particle is balanced by an opposite charge in the dispersion medium. The charge in the fluid dispersion medium is in the form of free ions. There is a region around each colloidal particle where the charge on particle attracts the free ions from the dispersion medium to form an electrical cloud around it and is called the electrical double layer (Helmholtz electrical double layer).

An electric double layer consists of three regions

• Surface charge – charged ions adsorbed on the particle surface.
• Stern layer – counterions (charged opposite to the surface charge) attracted to the particle surface and closely attached to it by the electrostatic force.
• Diffuse layer – a film of the dispersion medium (solvent) adjacent to the particle. The diffuse layer contains free ions with a higher concentration of the counterions. The ions of the diffuse layer are affected by the electrostatic force of the charged particle. The boundary of this layer is called the slipping plane (shear plane).

The value of the electric potential at the slipping plane is called Zeta potential. The zeta potential is an important parameter for a colloid. Zeta potential depends on the properties of the colloid. For example, adding salt to a colloid shrinks the electrical double layer, and reduces the zeta potential. Zeta potential is given by the relation

Where, ξ = zeta potential

η = coefficient of viscosity

u = velocity of colloidal particles

D = Dielectric constant of the medium = K

Zeta potential and particle size are key indicators of the way colloids behave both in storage and in use. Zeta potential influences the effective size of the particles in the colloid.

Importance of Charge on Colloidal Particles:

• The mutual forces of repulsion between similarly charged colloidal particles prevent them from aggregating and settling under the action of gravity.
• This gives stability to the sol. In the case of lyophobic sols, charge on colloidal particles is fully responsible for its stability.
• In the case of a lyophilic sol, the stability is due to the charge on colloidal particles and solvation.

Electrophoresis or Cataphoresis:

The unidirectional migration of sol particles or dispersed phase particles or colloidal particles towards the oppositely charged electrode under the influence of the applied electric field is called electrophoresis or cataphoresis.

Cause of Electrophoresis:

All sol particle (colloidal particles) carry the same electric charge either positive or negative.  If an electric potential is applied across two platinum electrodes dipping in a sol, the sol particles move towards oppositely charged electrodes.

Illustration:

Consider a sol of As₂S₃ is taken in a ‘U’ shaped glass tube. The sol particles of the sol are negatively charged. Now the dispersion medium with little quantity of electrolyte is introduced over the colloidal solution. There should be a sharp boundary between the sol and the dispersion medium.

Electric potential is applied across the two platinum electrodes dipped in a sol in two limbs, it is observed that the level of sol drops at the negative electrode and rises at the positive electrode side. This shows that sol particles have migrated to the positive electrode, indicating that the particles are negatively charged.

If colloidal particles are allowed to reach the electrode, their charges are neutralised and coagulation takes place.

Applications of Electrophoresis:

• Electrophoresis is used to detect the nature of charge on colloidal particles.
• It is used in the removal of carbon particles from chimney gases.
• It is used in electro-deposition of rubber on metal, wood or cloth surfaces from latex.
• It is used to bring about coagulation of sol.

Electro-osmosis:

The migration of the dispersion medium of a colloidal solution under the influence of the electric field when the movements of colloidal particles are prevented is called as electro-osmosis.

Cause of Electro-osmosis:

Since the sol as a whole is electrically neutral, dispersion medium has an opposite electric charge as compared with that of the sol particles. If the dispersed phase has a positive charge we say that the dispersion medium has a negative charge.

Illustration:

A sol of As₂S₃ is filled in a glass tube. The sol particles of the sol are negatively charged. Hence the dispersion medium (water) is positively charged. The colloidal solution and pure dispersion medium in a glass tube are separated by a semipermeable membrane.

When an electric potential is applied across the platinum electrodes dipping in each arm, sol particles cannot pass through the semipermeable membrane but dispersion medium (water) move to the negative electrode through the semipermeable membrane.  The level of sol drops at the +ve electrode and rises at -ve electrode. This movement of dispersion medium towards -ve electrode shows that the charge on the dispersion medium is positive.

Applications of Electro-osmosis:

• Electro-osmosis is used in dewatering of moist clay
• It is used in the drying of dye-pastes
• It is used in the removal of water from peat.

Applications of Electrical Properties of Colloids

Sewage Precipitation:

Dirty and muddy water from gutters and drainages is called sewage is in colloidal form (colloidal solution).

Sewage water containing colloidal particles of mud, rubbish etc. is collected in a tank fitted with electrodes.

On applying an electric field, colloidal particles are attracted towards oppositely charged electrodes. As their charge gets neutralised, they settle as a precipitate. The precipitated or coagulated matter called sludge is used as manure while clear water is used for irrigation.

Smoke Precipitation:

Smoke is a colloidal solution of negatively charged carbon particles in the air (aerosol)

These carbon particles may condense water vapour on them and thus cities may have a thick cover of smog (smoke + fog). This smog causes air pollution.

Cottrel’s precipitator is a widely used smoke precipitator. Smoke is passed between metal electrodes at high voltage (about 50,000 V) The charged particles are neutralized at the oppositely charged electrode and get deposited there. The gases free from carbon particles are passed to a chimney or for further purification.

Science > Chemistry > Colloids > Charge on Colloidal Particles

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In this article, we shall study the general, mechanical and optical properties of colloids.

General Properties of Colloids:

Heterogenous Character:

The ultramicroscopic examination indicates that colloidal dispersion is a heterogeneous system consisting of a continuous dispersion medium and discontinuous disperse phase.

Visibility:

Colloidal particles cannot be seen through naked eyes or ordinary microscope due to their very small size. The shortest wavelength in visible spectra is about 4000 Å. Hence we cannot see any object less than 200μm and colloidal particles have sizes less than 200μm.

Recently new techniques like Scanning Electron Microscope (SEM), Transmission Electron Microscope (TEM), and Scanning Transmission Electron Microscope (STEM) are used to determine the size and shape of colloidal particles.

Filterability:

The colloidal particles readily pass through ordinary filter paper. The range of particle size of colloidal substance is in between 5 × 10^-9 m to 2 × 10^-7 m. The pore size of ordinary filter paper is bigger i.e. of order 10^-7 m. So Colloidal particles can pass through it and thus filter paper can be used to separate colloidal particles from coarse suspension.

Sols and true solutions pass through filter paper. The colloids cannot pass (diffuse) through parchment membrane but crystalloids can pass through parchment membrane. The process of separating colloids from other dissolved substance using parchment membrane as the filter is called dialysis. This process is used for purification of colloids.

When the impure sol is placed in specially created ultrafilter, with small pores, the sol particles being bigger than the pores remain behind while dispersion medium and dissolved electrolyte pass through. This process is known as ultra-purification.

Surface Tension and Viscosity:

Lyophilic sols have a higher viscosity and lower surface tension than dispersion medium and lyophobic sols have a nearly same viscosity and surface tension as the dispersion medium.

Molecular Mass:

Colloidal particles are of two types i) Multimolecular and ii) Macromolecular

• Multimolecular colloidal particles are aggregates of a number of small molecules or atoms. e.g. Sulphur sol, gold sol.
• Macromolecular particles are very big molecules or polymers. e.g. Starch, proteins.

As the colloidal particles are aggregates of a number of molecules or a large molecule themselves their molecular mass is very high.

Colligative Properties:

Colloidal particles are bigger aggregates. The colligative properties depend on the number of particles. Due to less number of particles compared to true solution colligative properties are lower. Hence the values of colligative properties like osmotic pressure., depression in freezing point and elevation in boiling points are of small order compared to the values shown by true solutions at the same concentration.

Colour:

Many sols are coloured. Sol particles are able to scatter light rays. Colour of the sol depends upon the wavelength of scattered light by the sol particles and which again depends on the size of the sol particles. The colour of colloidal solution also changes with the way the observer receives the light.

Let us consider silver sol (colloidal solution of the same substance) having different types of particles. It is found that the sols show different colours. 

Notes:

• The blue sky is due to the blue light scattered by small dust particles in the atmosphere. The atmosphere is a colloidal system consisting of dust particles suspended in air.
• The red sky is due to red light scattered by larger dust particles in the atmosphere.
• Depending on the size of the dust and water particles, different colours are seen in the cloud.
• Fine gold sol is red but as particle size increases it becomes blue or purple.

Mechanical Properties of Colloids:

Brownian Movement:

The English Botanist Robert Brown, in 1927 observed that colloidal particles exhibit continuous random motion in all directions in a straight line.  He found such movement when pollen grains were suspended in water. The phenomenon of continuous zig-zag movement of colloidal particles in straight line paths in a random direction is known as a Brownian movement.

Explanation:

Colloidal particles are surrounded by a large number of dispersion medium molecules which constantly bombard the colloidal particles. On unequal bombardment, the colloidal particles get pushed in certain directions. Since colloidal particles possess like charge, they repel each other.

Factors Affecting Brownian Movement:

• Brownian movement depends on the viscosity of the dispersion medium. Brownian movement is more in less viscous solution.
• Brownian movement depends on the size of the particle. If the particles are of smaller size. The Brownian movement is more rapid.

Applications Brownian Movement:

• Due to the Brownian movement colloidal particles hardly settle down and prevent aggregation of colloidal particles. Thus colloidal solution becomes stable.
• Avogadro’s number can be calculated by Brownian movement.  

Optical Properties of Colloids:

Tyndall effect:

When an intense beam of light is passed through the colloidal solution (taken in a glass vessel) placed in a dark the path of light through the colloidal solution is clearly visible due to the scattering of light by sol particles.  This effect is known as Tyndall effect.

This fact was first noted in 1857 by Faraday and then studied in  details by Tyndall in 1868. True solutions do not exhibit Tyndall effect.

The emitted light emerges in the form of a bright cone called Tyndall cone. Through ultramicroscope, each colloidal particle appears a bright point against the dark background, due to the scattering. Thus the colloidal particles become self-luminous. As a result, the path of the beam of light through colloidal solution becomes clearly visible. The nature of scattering depends on the size of the sol particle and the refractive indices of sol particle.

Explanation:

Colloidal particles are not large enough like suspension particles to reflect the light nor they are small enough, like true solution particles to allow the light to pass through them. Due to the intermediate size of colloidal particles, they scatter part of the absorbed light, from their surfaces in all directions. Thus the cause of Tyndall effect is a scattering of light by colloidal particles.

Conditions to be Satisfied for Viewing Tyndall Effect:

• The diameter of the dispersed particle is not much smaller than the wavelength of light used.
• There should be a large difference between the magnitudes of refractive indices of the dispersed phase and the dispersion medium.

Applications of Tyndall Effect:

• Tyndall effect is useful to distinguish colloidal solution from the true solution
• To test the purity of gases in the manufacture of H₂SO₄ by the contact process.
• Count the number of colloidal particles in colloidal sols using ultra-microscope.

Science > Chemistry > Colloids > Properties of Colloids

The post Properties of Colloids appeared first on The Fact Factor.

In this article, we shall study the phenomenon of osmosis and osmotic pressure.

Semipermeable Membrane:

A semipermeable membrane is a membrane which allows the solvent molecules, but not the solute molecules through it.
Examples: Cellulosde, cellulose nitrate, animal bladder.

Osmosis:

The spontaneous and unidirectional flow of solvent molecules through a semipermeable membrane, into a solution OR flow of solvent from a solution of low concentration to a solution of higher concentration through a semipermeable membrane, is called osmosis.

Everyday Examples of Osmosis:

• Raw mangoes when placed in a concentrated solution of common salt lose water through osmosis and ultimately shrivel into a pickle.
• Flowers revive and regain their freshness when placed in freshwater because of osmosis.
• Carrots get limed due to the loss of water to the atmosphere. But, when limped carrots are placed in water, they become firm due to the inflow of water because of osmosis.
• People consuming more salt and excessive salty food suffer from edema which is swelling and puffiness produced in the body due to retention of water in tissue cells and intracellular spaces.
• The preservation of meat and fishes against bacteria is done by salting it. The bacteria on meat or fishes lose water through osmosis and ultimately die.
• The preservation of fruits against bacteria is done by adding sugar to it. The bacteria on meat or fishes lose water through osmosis and ultimately die.
• Plants absorb water from the soil through roots due to osmosis because the root hair cells have higher osmotic pressure than that of soil water.
• Red blood cells burst when kept in water due to endosmosis.
• Blooming, opening, and closing of flowers are governed by osmosis.
• Dead bodies swell underwater due to endosmosis.

Experiment Exhibiting Phenomenon of Osmosis OR Abbe Nollet Experiment:

A wide-mouthed thistle funnel with a narrow long stem was taken. Then pig’s bladder (semipermeable membrane) is tied tightly around the wide mouth of the funnel with the help of a thread or rubber band. Now dilute sugar solution is carefully poured into the stem of the funnel to a certain level. The wide mouth of the funnel containing sugar solution is now kept in a beaker with the help of an Iron stand. Now three-fourths of the beaker is filled with pure water. The apparatus is left undisturbed for some time.

After a few hours we find the level of sugar solution increases from its initial level. This indicates that there is a net flow of solvent molecules into the solution through the semipermeable membrane. We have to apply excess sufficient pressure from the stem side on the solution to stop this migration of solvent molecule and this excess pressure is called osmotic pressure.

Osmotic Pressure:

The excess of pressure on the side of a solution that stops the net flow of solvent into the solution through a semipermeable membrane is called osmotic pressure. The equilibrium is reached when hydrostatic pressure of the column is equal to that of osmotic pressure. Osmotic pressure is not created by the solution but it comes into existence when the solution is separated from the solvent by a semipermeable membrane. If the pressure applied to the solution is greater than the osmotic pressure of the solution then solvent starts passing from solution to solvent. This phenomenon is called reverse osmosis. This process is used for purification of seawater and hard water.

Types of Solutions on the Basis of Osmosis:

Isotonic Solutions:

Two or more solutions having the 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, called normal saline solution and it is safe to inject intravenously.

Hypertonic Solution:

A solution having osmatic pressure higher than that of another solution is called a hypertonic solution.

For example, the osmotic pressure associated with the fluid inside the blood cell is less than sodium chloride solution having a concentration of more than 0.9% (mass/volume). Thus the solution of sodium chloride is hypertonic. In this case, water will flow out of the cells and cells would shrink.

Hypotonic Solution:

A solution having osmatic pressure lower than that of another solution is called a hypotonic solution.

For example, the osmotic pressure associated with the fluid inside the blood cell is more than sodium chloride solution having a concentration of less than 0.9% (mass/volume). Thus the solution of sodium chloride is hypotonic. In this case, water will flow into the cells and cells would swell.

Laws of Osmotic Pressure:

van’t Hoff’s Theory of Osmotic Pressure:

He found that the solute particles in dilute solutions possess kinetic energy and move in random directions in the solutions. Thus they have similar behaviour as that of gas molecules.

On collision against the semipermeable membrane, the solute molecules exert osmotic pressure equal to the pressure which the solute molecules would exert if it were gas molecule at the same temperature and occupying the same volume as that of solution. Thus the gas laws are equally applicable to dilute solutions.

van’t Hoff’s Boyle’s Law of Solution:

At constant temperature, the osmotic pressure (π) of a dilute solution is directly proportional to its molar concentration (C) or inversely proportional to volume (V) of the solution.

Explanation:

van’t Hoff’s Charle’s Law of Solution:

The concentration remaining constant, the osmotic pressure (π) of a dilute solution is directly proportional to absolute temperature (T) of the solution.

Explanation:

van’t Hoff’s General Solution Equation:

By van’t Hoff Boyle’s law at a constant temperature, the osmotic pressure (π) of a dilute solution is inversely proportional to volume (V) of the solution.

 By van’t Hoff Charles law, The concentration remaining constant, the osmotic pressure (π) of a dilute solution is directly proportional to absolute temperature (T) of the solution.

Where k is proportionality constant called general solution constant. van’t Hoff further proved that this constant k is equal to universal gas constant R

Equations (a) and (b) represents general solution equation.

van’t Hoff’s Avogadro’s Law of Solution:

Two solutions of equal concentrations of different solutes exert same osmotic pressure at the same temperature OR equal volumes of isotonic solutions contain an equal number of solute particles at the given temperature.

Explanation:

Related Topics

Solutions and Their Colligative Properties

• Solutions and Their Types
• Solutions of Solids and Liquids
• Concentration of Solution
• Numerical Problems on Percentage by Mass and Volume
• Numerical Problems on Mole Fraction
• Numerical Problems on Molarity
• Numerical Problems on Molality
• Short Cuts For Above Numerical Problems
• Solutions of Gases in Liquid
• Ideal and Non-ideal Solutions
• Lowering of Vapour Pressure
• Numerical Problems on Lowering of Vapour Pressure
• Elevation in Boiling Point and Depression in Freezing Point

For More Topics of Chemistry Click Here

The post Osmosis and Osmotic Pressure appeared first on The Fact Factor.

In this article, we shall study two colligative properties of solutions, namely elevation of boiling point and depression in freezing point due to addition of solute.

Elevation of Boiling Point:

Boiling Point of a Liquid:

Boiling point is defined as the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure. The boiling point of a liquid is a characteristic property and can be treated as criteria for the purity of liquid.  It increases with the increase in external pressure. Liquids having greater intermolecular forces have high boiling points.

Elevation of Boiling Point of a Liquid:

The vapour pressure of the solution of non-volatile solute is always less than the vapour pressure of the pure solvent.

At the boiling point of pure solvent, the solution will not boil because its vapour pressure of the solution is less than the vapour pressure of the pure solvent. Thus vapour pressure of the solution is less than the external pressure. To boil the solution we have to the increases vapour pressure of the solution to make it equal with external pressure. It is achieved by increasing the temperature of the solution. Thus there is an elevation of the boiling point of the liquid.

Let T_b0 be the boiling point of pure solvent and Tb be the boiling point of the solution. The increase in the boiling point Δ T_b  = T_b   – T_b0     is known as the elevation of boiling point. The elevation of boiling point (ΔTb) is directly proportional to the lowering of vapour pressure (Δp).

Thus   Δ T_b  α  Δp.

Experiments have shown that for dilute solutions the elevation of boiling point ( Δ T_b ) is directly proportional to the molal concentration of the solute in a solution. Thus, the elevation of boiling point also depends on the number of solute molecules rather than their nature.

Δ T_b   α  m

Δ T_b  =   K_b  m       ………….. (1)

Here m (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality, K_b  is called Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic Constant). The unit of K_b  is K kg mol-1.

The molal elevation of boiling point constant is defined as the elevation of boiling point produced when one mole of solute is dissolved in 1 kg of solvent.

Now

The experimental method to determine the molecular mass of non-volatile solute by determining boiling points of pure solvent and solution of known concentration is called ebullioscopy.

Depression of Freezing Point:

Freezing Point of a Liquid:

The freezing point of a liquid is a temperature at which the vapour pressure of solid is equal to the vapour pressure of the liquid.

Depression of Freezing Point of a Liquid:

The lowering of the vapour pressure of a solution causes a lowering of the freezing point compared to that of the pure solvent.

We know that at the freezing point of a substance, the solid phase is in dynamic equilibrium with the liquid phase. A solution will freeze when its vapour pressure equals the vapour pressure of the pure solid solvent as is clear from the graph.

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 a lower temperature. Thus, the freezing point of the solvent decreases.

Let T_f0  be the freezing point of pure solvent and Tf be the freezing point of the solution. The increase in the freezing point Δ T_f  = T_f  – T_f0 is known as depression of freezing point.

The depression of freezing point (Δ T_f) is directly proportional to the lowering of vapour pressure (Δp).

Thus, Δ T_f  α  Δp.

Experiments have shown that for dilute solutions the depression of freezing point (Δ T_f) is directly proportional to the molal concentration of the solute in a solution. Thus, the depression of freezing point also depends on the number of solute molecules rather than their nature.

Δ T_f   α  m

Δ T_f  =  K_f  m       ………….. (1)

Here m (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality,   K_f is called Freezing Point Elevation Constant or Molal Elevation Constant (cryoscopic Constant). The unit of   K_f is K kg mol-1.

The molal elevation of freezing point constant is defined as the depression of freezing point produced when one mole of solute is dissolved in 1 kg of solvent.

Now,

The experimental method to determine the molecular mass of non-volatile solute by determining freezing points of pure solvent and solution of known concentration is called cryoscopy.

Problems:

1. Which of the following aqueous solutions will have maximum depression in freezing point. a) 0.5 M Li2SO4 b) 1 M NaCl  c) 0.5 M Al2(SO4)3 d) 0.5 M BaCl2
2. A solution containing 0.73 g of camphor (molar mass 152 g mol-1) in 36.8 g of acetone (boiling point 56.3° C) boils at 56.55° C. A solution of 0.564 g of unknown compound in the same weight of acetone boils at 56.46° C. calculate molar mass of the unknown compound.
3. 1.0 x 10-3 kg of urea when dissolved in 0.0985 kg of a solvent, decreases the freezing point of the solvent by 0.211 K. 1.6 x 10-3 kg of another non-electrolyte solute when dissolved in 0.086 kg of the same solvent depresses the freezing point by 0.34 K. Calculate the molar mass of another solute.
4. Which of the following aqueous solutions will have minimum elevation in boiling point. a) 0.1 M KCl b) 0.05 M NaCl  c) 1 M AlPO4 d) 0.1 M MgSO4
5. The boiling point of a solvent is 80.2° C. When 0.419 g of the solute of molar mass 252.4 g mol-1, is dissolved in 75 g of above solvent, the boiling point of the solution is found to be 80.26° C. Find molal elevation constant.

Related Topics

Solutions and Their Colligative Properties

• Solutions and Their Types
• Solutions of Solids and Liquids
• Concentration of Solution
• Numerical Problems on Percentage by Mass and Volume
• Numerical Problems on Mole Fraction
• Numerical Problems on Molarity
• Numerical Problems on Molality
• Short Cuts For Above Numerical Problems
• Solutions of Gases in Liquid
• Ideal and Non-ideal Solutions
• Lowering of Vapour Pressure
• Numerical Problems on Lowering of Vapour Pressure
• Osmosis and Osmotic Pressure

For More Topics of Chemistry Click Here

The post Boiling Point Elevation and Freezing Point Depression appeared first on The Fact Factor.

In this article, we shall study to solve problems based on relative lowering of vapour pressure and to calculate the molecular mass of a solute.

Example – 01:

The vapour pressure of a pure liquid at 298K is 4 x 10⁴ N/m². When a non-volatile solute is dissolved the vapour pressure becomes 3.65 x 10⁴ N/m². Calculate relative vapour pressure, lowering of vapour pressure and relative lowering of vapour pressure

Given: Vapour pressure of pure liquid = p^o = 4 x 10⁴ N/m², vapour pressure of solution = p = 3.65 x 10⁴ N/m², temperature = T = 298 K,

To Find: Relative vapour pressure = p/p^o =? Lowering of vapour pressure = p^o – p =? and relative lowering of pressure = (p^o – p)/p^o = ?

Solution:

Relative vapour pressure = p/p^o = (3.65 x 10⁴ N/m²)/(4 x 10⁴ N/m²) = 0.9125

Lowering of vapour pressure = p^o – p = 4 x 10⁴ N/m² – 3.65 x 10⁴ N/m²

Lowering of vapour pressure = 0.35 x 10⁴ N/m² = 3.5 x 10³ N/m²

Relative lowering of pressure = (p^o – p)/p^o = (3.5 x 10³ N/m²)/(4 x 10⁴ N/m²) = 0.0875

Ans:  Relative vapour pressure = 0.9125, Lowering of vapour pressure = 3.5 x 10³ N/m², Relative lowering of pressure = 0.0875

Example – 02:

The vapour pressure of a solution containing 13 × 10^-3 kg of solute in 0.1 kg of water at 298 K is 27.371 mm Hg. calculate the molar mass of the solute. Given that the vapour pressure of water at 298 K is 28.065 mm Hg.

Given: mass of solute W₂ = 13 × 10^-3 kg, mass of solvent (water) = W₁ = 0.1 kg, vapour pressure of pure solvent (water) = p^o = 28.065 mm of Hg, vapour pressure of solution = p = 27.371 mm of Hg, temperature = T = 298 K, Molecular mass of solvent (water) = M₁ = 18 g mol^-1

To Find: Molecular mass of solute = M₂ =?

Solution:

For dilute solution relative lowering of vapour pressure is given by

Ans: Molecular mass of solute is 94.63 g mol^-1

Example – 03:

The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile solute of a mass 2.175 × 10^-3 kg is added to 39.0 × 10^-3 kg of benzene. The vapour pressure of a solution is 600 mm Hg. What is the molar mass of the solute? Given C= 12, H = 1.

Given: mass of solute W₂ = 2.175 × 10^-3 kg, mass of solvent = W₁ = 39.0 × 10^-3 kg, vapour pressure of pure solvent (benzene) = p^o = 640 mm of Hg, vapour pressure of solution = p = 600 mm of Hg,

To Find: Molecular mass of solute = M₂ =?

Solution:

Molecular mass of solvent (benzene C₆H₆) = M₁ = 12 x 6 + 1 x 6 = 78 g mol^-1

For dilute solution relative lowering of vapour pressure is given by

Ans: Molecular mass of solute is 69.6 g mol^-1

Example – 04:

In an experiment, 18.04 g of mannitol were dissolved in 100 g of water. The vapour pressure of water was lowered by 0.309 mm Hg from 17.535 mm Hg. Calculate the molar mass of mannitol.

Given: mass of solute (mannitol) W₂ = 18.04 g, mass of solvent (water) = W₁ = 100g, vapour pressure of pure solvent (water) = p^o = 17.535 mm of Hg, decrease in vapour pressure of solution = p^o – p = 0.309 mm of Hg, Molecular mass of solvent (water) = M₁ = 18 g mol^-1

To Find: Molecular mass of solute (mannitol) = M₂ =?

Solution:

For dilute solution relative lowering of vapour pressure is given by

Ans: Molecular mass of solute is 184.3 g mol^-1.

Example – 05:

A solution is prepared from 26.2 × 10^-3 kg of an unknown substance and 112.0 × 10^-3 kg acetone at 313 K. The vapour pressure of pure acetone at this temperature is 0.526 atm. Calculate the vapour pressure of solution if the molar mass of a substance is 273.52 × 10^-3 kg mol^-1. Given C = 12, H = 1, O = 16.

Given: mass of solute W₂ = 26.2 × 10^-3 kg, mass of solvent = W₁ = 112.0 × 10^-3 kg, vapour pressure of pure solvent (acetone = p^o = 0.526 atm, Molecular mass of solute = M₂ = 273.52 × 10^-3 kg

To Find: vapour pressure of solution = p =?

Solution:

Molecular mass of solvent (acetone (CH₃)₂CO) = M₁ = 12 x 3 + 1 x 6 + 16 x 1

Molecular mass of solvent (acetone (CH₃)₂CO) = M₁ = 58 g mol^-1 = 58 × 10^-3 kg

For dilute solution relative lowering of vapour pressure is given by

Ans: Vapour pressure of solution is 0.500 atm.

Example – 06:

The vapour pressure of water at 20 °C is 17 mm Hg. Calculate the vapour pressure of a solution containing 2.8 g of urea (NH₂CONH₂) in 50 g of water.

Given: Temperature of water = 20 °C = 20 + 273 = 293 K, vapour pressure of pure solvent (water) = p^o =  17 mm of Hg, mass of urea (solute)(NH₂CONH₂) = W₂ = 2.8 g, mass of water (solvent) = W₂ = 50 g

To Find: Vapour pressure of solution = p =?

Solution:

The molecular mass of solute urea (NH₂CONH₂) = M₂

M₂ = 14 g x 2 + 1 g x 4 + 12 g x 1 + 16 g x 1 = 60 g mol^-1.

The molecular mass of water M₁ = 18 g mol^-1.

Ans: Vapour pressure of solution = 16.714 mm of Hg.

Example – 07:

At 300 K the vapour pressure of water is 1.2 x 10⁴ Pa. 0.8 x 10^-2 kg of oxalic acid (molecular mass = 126) is dissolved in 700 cm³ of water at the same temperature, find the vapour pressure of the solution.

Given: Temperature of water = 300 K, vapour pressure of pure solvent (water) = p^o =  1.2 x 10⁴ Pa, mass of solute (oxalic acid) = W₂ = 0.8 x 10^-2 kg, volume of water (solvent) = 700 cm³, molecular mass of water = M₁ = 18 g mol^-1, molecular mass of solute (oxalic acid) = M₂ = 126 g mol^-1.

To Find: Vapour pressure of solution = p =?

Solution:

Volume of water = 700 cm³

Mass of water = W₁ = 700 cm³ x 1 g cm^-3 = 700 g = 0.7 kg

Ans: Vapour pressure of solution = 1.198 x 10⁴ Pa.

Example – 08:

Calculate the decrease in the vapour pressure when 1.81 g x 10^-2 kg of a solute (molecular mass = 57) is dissolved in 0.1 kg of water. Vapour pressure of water is 1.223 x 10⁴ Pa.

Given: Vapour pressure of pure solvent (water) = p^o = 1.223 x 10⁴ Pa, mass of solute = W₂ = 1.81 x 10^-2 kg, mass of water (solvent) = 0.1 kg, molecular mass of water = M₁ = 18 g mol^-1, molecular mass of solute = M₂ = 57 g mol^-1.

To Find: Decrease in vapour pressure of solution = p^o – p =?

Solution:

Ans: Decrease in vapour pressure is 699 Pa

Example – 09:

A solution containing 30 g of a non-volatile solute in exactly 90 g of water has a vapour pressure of 21.85 mm of Hg at 25 ^oC. Further 18 g of water is then added to the solution. The new vapour pressure becomes 22.15 mm of Hg at 25 ^oC. Calculate the molecular mass of the solute and the vapour pressure of water at 25 ^oC.

Given: Temperature of water = 25 ^oC = 25 + 273 = 298 K

For solution 1: Mass of solute = W₂ = 30 g, mass of solvent = W₁= 90 g, solution vapour pressure = p = 21.85 mm of Hg

For solution 2: Mass of solute = W₂ = 30 g, mass of solvent = W₁= 90 g + 18 g = 108 g, solution vapour pressure = p = 22.15 mm of Hg

To Find: Molecular mass of solute = M₂ =? vapour pressure of pure water solution = p^o =?

Solution:

5p^o – 109.25 = 6p^o – 132.9

p^o = 132.9 -109.25 = 23.65 mm of Hg

Substituting in equation (1)

Ans: Molecular mass of solute is 72.83 u and vapour pressure of pure water solution 23.65 mm

Related Topics

Solutions and Their Colligative Properties

• Solutions and Their Types
• Solutions of Solids and Liquids
• Concentration of Solution
• Numerical Problems on Percentage by Mass and Volume
• Numerical Problems on Mole Fraction
• Numerical Problems on Molarity
• Numerical Problems on Molality
• Short Cuts For Above Numerical Problems
• Solutions of Gases in Liquid
• Ideal and Non-ideal Solutions
• Lowering of Vapour Pressure
• Elevation in Boiling Point and Depression in Freezing Point
• Osmosis and Osmotic Pressure

For More Topics of Chemistry Click Here

The post Numerical Problems on Lowering of Vapour Pressure appeared first on The Fact Factor.

In this article, we shall learn the meaning of colligative properties, colligative properties of solutions, and the lowering of vapour pressure due to the addition of solute in a solvent.

Colligative Properties:

Colligative properties are those properties of dilute solutions that depend only on the number of solute particles (atoms or molecules, ions or aggregates of molecules) in the solution and not on the nature of solute particles. These properties are important because they are used to determine molar masses of non-electrolyte solutes. Similarly, they are related to one another. If one is measured then other properties can be calculated. These properties can be observed clearly if the solution is very dilute, the solute is non-volatile and the solute does not dissociate or associate in the solution. There are four colligative properties of solutions

• Lowering of vapour pressure.
• Elevation of the boiling point of the
  solvent in the solution.
• Depression in the freezing point of
  the solvent in the solution.
• Osmotic pressure

The Concept of Vapour Pressure:

If a container is partially filled with a liquid, a portion of the liquid evaporates to fill the remaining volume of the container with vapour. The molecules of liquid evaporated are in continuous random motion, they collide with the walls of the container and with each other. Thus they create pressure on the walls of the container and on the liquid. At the same time some molecules which have left liquid return back to the liquid, the process is called condensation. After some interval of time, an equilibrium is established between the two phases of substance. At this stage, the rate of evaporation is equal to the rate of condensation.

The pressure exerted by the vapours of the liquid on the surface of the liquid when equilibrium is established between liquid and its vapour is called vapour pressure of the liquid. The temperature at which vapour pressure of the liquid is equal to the external pressure is called boiling temperature at that pressure.

Relation Between Vapour Pressure and Temperature:

Vapour pressure of liquid increases with the increase in the temperature.

Clausius Clapeyron Equation:

Using this relation we can find vapour pressure at some another temperature when its value at some temperature is known.

Lowering of Vapour Pressure:

Liquids at a given temperature vapourize and under equilibrium conditions, the pressure exerted by the vapours of the liquid over the liquid phase is called vapour pressure [Fig (a)].

In a pure liquid, 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 that 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.

Reasons for Lowering of Vapour Pressure:

The evaporation of liquid and evaporation of the liquid is a surface phenomenon. It is directly proportional to the surface area available for evaporation. Due to the addition of non-volatile solute the surface area of liquid decreases. It results in a decrease in the rate of evaporation and vapour pressure decreases.

By Graham’s law, the rate of evaporation is inversely proportional to the square root of the density of the liquid. When a solute is added to a solvent, the density of the resulting solution is more than the pure solvent. Hence the rate of evaporation decreases which results in the decrease of the vapour pressure of the solution.

Relative Lowering of Vapour Pressure:

The relative lowering of vapour pressure for the given solution is the ratio of vapour pressure lowering of solvent from the solution to the vapour pressure of the pure solvent.
Mathematically, the relative lowering of vapour pressure is given by

Where,
p₁⁰= Vapour pressure of the pure solvent
Δp= Lowering of vapour pressure
p= Vapour pressure of the solution

Raoult’s Law:

The partial vapour pressure of any volatile component of a solution is the product of vapour pressure of that pure component and the mole fraction of the component in the solution.

Explanation:

Let us consider a solution containing two volatile components say A₁ and A₂, with mole fractions x₁ and x₂ respectively. Let  p₁^o and  p₂^o be the vapour pressures of the pure components A₁ and A₂ respectively, then by Raoult’s law

p₁ = p₁^ox₁   and   p₂ = p₂^o x₂

The total vapour pressure of the solutions of two volatile components is the sum of partial vapour pressures of the two components

p_T = p₁  +  p₂

p_T = p₁^ox₁  +  p₂^o x₂

But x₁ + x₂ = 1

∴   x₂ = 1 – x₁

∴   p_T = p₁^ox₁  +  p₂^o (1 – x₁)

∴   p_T = p₁^ox₁  +  p₂^o – p₂^ox₁

∴   p_T =  p₂^o + ( p₁^o – p₂^o)x₁

The solution which obeys Raoult’s law over the entire range of concentration is called an ideal solution. If a solution does not obey Raoul’s law are called non-ideal solutions.

Raoult’s Law for a Solution of Non-Volatile Solute:

Let us consider a solution containing two volatile component A₁ and non-volatile component A₂, with mole fractionsx₁ and x₁ respectively.

Let p₁⁰ and p₂⁰ be the vapour pressures of the pure components A₁ and A₂ respectively. Now component A₂ is non-volatile, hence it will not contribute to vapour pressure. Thus p₂⁰ = 0. We have

p_T =  p₂^o + ( p₁^o – p₂^o)x₁

∴   p_T =  0 + ( p₁^o -0)x₁

∴   p_T = p₁^ox₁

Thus vapour pressure of a solution of non-volatile solute is the product of vapour pressure p₁⁰ of pure solvent and mole fraction x₁ of the solvent, which is Raoult’s law.

The equation shows that vapour pressure of the solution p < p₁⁰ ,  i.e. there is a lowering of the vapour pressure of the solution mathematically, it is given by

Δp = p₁^o –  p

∴   Δp = p₁^o –  p₁^ox₁

∴   Δp = p₁^o( 1 –  x₁)

∴   Δp = p₁^ox₂

In a solution containing several non-volatile solutes, the lowering of the vapour pressure depends on the sum of the mole fraction of different solutes.
Now, the relative lowering of vapour pressure is given by

This relation proves that the lowering of vapour pressure is colligative property because it depends on the concentration of non-volatile solute.

Raoult’s Law as a Special Case of Henry’s Law:

By Raoult’s law, we have

p = p₁^ox    …………… (1)

By Henry’s law, we have

p = K_Hx    …………… (2)

If we compare the two 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 of solute in the solution. Only the proportionality constant K_H differs from p₁⁰. Thus, Raoult’s law becomes a special case of Henry’s law in which K_H is equal to p₁⁰.

Relation Between Molar Mass of Solute and Lowering of Vapour Pressure:

Let W₂ g of the solute of molar mass M₂ be dissolved in W₁ g of the solvent of molar mass M₁. The numbers of moles of solvent and solute are given by n₁ = W₁/M₁ and n2 = W₂/M₂ respectively

The mole fraction of solute is given by

Now for dilute solutions n₂ << n₁. Hence n₂ can be neglected.

This is the relation between the molar mass of solute and the lowering of vapour pressure.

Measurement of Lowering of Vapour Pressure:

Barometric Method:

Raoult introduced the liquid or the solution into the Torricellian vacuum of a barometer tube and measured the depression of the mercury level.

This method is neither practicable nor accurate as the lowering of vapour pressure is too small.

Manometric Method:

This method is generally used for aqueous solutions.

The bulb of the apparatus is charged with the liquid or solution. The air in the connecting tube is then removed with a vacuum pump. When the stopcock is closed, the pressure inside is due only to the vapour evaporating from the solution or liquid.  The manometric liquid can be mercury or n-butyl phthalate which has a low density and low volatility.

Ostwald and Walker’s Dynamic Method (Gas Saturation Method):

In this method, the relative lowering of vapour pressure can be determined directly. The measurement of the individual vapour pressures of a solution and solvent is thus eliminated.

Apparatus: The apparatus used is shown in the figure. It consists of two sets of bulbs. The first set of three bulbs is filled with the solution to half of their capacity and second set of another three bulbs is filled with the pure solvent. Each set is separately weighed accurately. Both sets are connected to each other and then with the accurately weighed set of guard tubes filled with anhydrous calcium chloride or some other dehydrating agents like P₂O₅, concentrated H₂SO₄etc. The bulbs of the solution and pure solvent are kept in a thermostat maintained at a constant temperature.

Procedure: A current of pure dry air is bubbled through. The air gets saturated with the vapours in each set of bulbs. The air takes up the amount of vapours proportional to the vapour pressure of the solution first and then it takes up more amount of vapours from the solvent which is proportional to the difference in the vapour pressure of the solvent and the vapour pressure of the solution, i.e. p₀ – p. The two sets of bulbs are weighed again. The guard tubes are also weighed.

Calculation:

The loss in the mass in the solution bulbs  ∝  p

The loss in the mass in the solvent bulbs ∝ (p₀ – p)    

The total loss of the mass in both sets of bulbs ∝ [p_s + (p₀ – p)] ∝ p₀

The total loss in the mass of both sets of bulbs is equal to gain in mass of guard tubes.

Thus measuring quantities on R.H.S. relative lowering of vapour pressure can be found.

Related Topics

Solutions and Their Colligative Properties

• Solutions and Their Types
• Solutions of Solids and Liquids
• Concentration of Solution
• Numerical Problems on Percentage by Mass and Volume
• Numerical Problems on Mole Fraction
• Numerical Problems on Molarity
• Numerical Problems on Molality
• Short Cuts For Above Numerical Problems
• Solutions of Gases in Liquid
• Ideal and Non-ideal Solutions
• Numerical Problems on Lowering of Vapour Pressure
• Elevation in Boiling Point and Depression in Freezing Point
• Osmosis and Osmotic Pressure

For More Topics of Chemistry Click Here

The post Lowering of Vapour Pressure of Solution appeared first on The Fact Factor.