IIt JEE Revision - Emulsions, Surfactants and Micelles

IIt JEE Revision - Emulsions, Surfactants and Micelles
JEE syllabus

Elementary ideas of emulsions, surfactants and micelles (only definitions and examples).


Emulsions:

Emulsion is a liquid dispersed in a liquid.

Any two immiscible liquids form an emulsion.

For example, milk is a naturally occuring emulsion in which particles of liquid fats are dispersed in water.

Since two immiscible liquids do not mix well, the emulsion is generally unstable and separation of liquids may take place on standing for some time.

Emulsifiers are substances which are added to make emulsions more stable.

Emulsifiers reduce the interfacial tension between the two liquids.

Two types of emulsions: Oil in water and water in oil.

Identifications of two types of emulsions:
Dilution test: If on addition of water, the emulsion becomes dilute, it means it is oil in water emulsion.

Dye test: An oil soluble dye is used and if the whole solution becomes coloured it is water in oil emulsion. If only drops become coloured, it is oil in water emulsion.

Any substance which can decrease the surface tension of water to a large extent is known as surfactant. Examples of soap and detergents. Such substances have larger concentrations at the surface of water as compared to the bulk of the solution.


Surfactants in solution are often association colloids, that is, they tend to form aggregates of colloidal dimensions, which exist in equilibrium with the molecules or ions from which they are formed. Such aggregates are termed micelles.

IIT JEE Revision Molecular weight - freezing point.

IIT JEE Revision Molecular weight - freezing point.
Molecular weight determination from depression of freezing point.


The freezing point is the temperature at which the solid and liquid states the substance have the same vapour pressure.

When a non-volatile solute is added to a solvent, the freezing point of the solution is always lower than that of the pure solvent.

The depression in freezing temperature is proportional to the molal concentration of the solution (m).
ΔTf α m Or ΔTf = Kf*m

ΔTf = depression in freezing point.

Kf is the molal depression constant. also called molal cryoscopic constant. It is defined as the depression in freezing point for 1 molal solution i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of solvent.
When m =1; ΔTf = Kf

Depression in freezing point is a colligatvie property as it is directly proportional to the molar concentration of the solute.


To find the molar mass of an unknown substance (nonvolatile compound), a known mass of it is dissolved in a known mass of a solvent and depression in its freezing point (ΔTf)is measured.

weight of solute be Wb g
weight of the solvent be Wa g
Molar mass of the solute be Mb

Molality of the solution, m = Wb*1000/Mb*Wa

Substitute the value of m in ΔTf = Kf*m = Kf*Wb*1000/Mb*Wa

From the above equation Mb can be calculated.

Mb = Kf*Wb*1000/Wa*ΔTf

Example:

Addition of 0.643 g of a compound to 50 ml of benzene (density 0.879 g/ml) lowers the freezing point from 5.51°C to 5.03°C. If Kf for benzene is 5.12 K kg molˉ¹, calculate the molar mass of the compound. (IIT 1992)

The formula of Mb is available above.

weight of solute be Wb g = 0.643 g

weight of the solvent be Wa g = 50*0.879 = 43.95 g
Change in freezing point = 5.51 - 5.03 = 0.48°C

Mb = (5.12 * 0.643 * 1000)/(43.95*0.48)


Mb = [Kf*Wb*1000]/[ΔTf * Wa]

IIT JEE Revision Ch 9. SOLUTIONS - Core Points

IIT JEE Revision Ch 9. SOLUTIONS - Core Points
Jee Syllabus


Solutions:
Raoult's law;
Molecular weight determination from lowering of vapor pressure,
Molecular weight determination from elevation of boiling point
Molecular weight determination from depression of freezing point.
-----------------
Raolt's Law

In the case of a solution of two liquids, A and B, the total vapor pressure Ptot(P total) above the solution is equal to the sum of the vapor pressures of the two components, PA and PB and

PA = PA° * Am
PB = PB° * Bm

Where
PA° = vapour pressure created by 1 mol of liquid A
Am = mole fraction of liquid A in the solution
PB° = vapour pressure created by 1 mol of liquid A
Bm = mole fraction of liquid A in the solution

The pressure exerted by the vapours above the liquid surface in equilibrium with the liquid at a given temperature is called vapour pressure.

If a small amount of non-volatile solute is added to the the solvent, the vapour pressure of the solution becomes less than that of the pure solvent.

Some properties of the solution depend only on the number of solute particles but on the nature of the solute. These are called colligative properties or democratic properties.

The four important ones are:
i) Relative lowering in vapour pressure
ii) elevation in boiling point
iii) depression in freezing point
iv) osmotic pressure

Molecular weight determination from lowering of vapor pressure
Molar mass of a solute can be found from the property of lowering of vapor pressure of a solution.

Mb = (Wb*Ma)/[Wa*(Pa°-Pa)/Pa°]

Wb = weight of solute particles, Wa= weight of solvent
(Pa°-Pa)/Pa° = decrease in vapour pressure of solution
Ma = Molar mass of solvent




Molecular weight determination from elevation of boiling point

Mb = [Kb*Wb*1000]/[ΔTb*Wa]
ΔTb = increase in boiling point of the solution after adding the solute
Kb = molal elevation constant or ebulloscopic constant
= the elevation in boiling point for 1 molal solution, i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of the solvent.


Molecular weight determination from depression of freezing point.

when a non-volatile solute is added to a solvent, the freezing point of the solution is always lower than that of the pure solvent.

The depression in freezing temperature is proportional to the molal concentration of the solution.
ΔTf α m Or ΔTf = Kf*m

Kf is the molal depression constant. also called molal cryoscopic constant. It is defined as the depression in freezing point for 1 molal solution i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of solvent.

Mb = [Kf*Wb*1000]/[ΔTf * Wa]

IIT JEE Revision - Ch 30 Amines - Core Points

IIT JEE Revision - Ch 30 Amines - Core Points
JEE Syllabus

Amines:
Preparation, Properties, Reactions
Characteristic reactions
Basicity of substituted anilines and aliphatic amines,
Preparation from nitro compounds,
Reaction with nitrous acid,
Azo coupling reaction of diazonium salts of aromatic amines,
Sandmeyer and related reactions of diazonium salts;
Carbylamine reaction;


Amines are regarded as derivatives of ammonia in which one, two or all three hydrogen atoms are replaced by alkyl or aryl group.

Preparation from nitro compounds,

Reduction of nitro compound to obtain amine can be done by using either molecular hydrogen and a catalyst (Ni or Pt) or a metal (usually granulated tin) and an acid (HCl)

Reaction with nitrous acid,
Aliphatic primary amine in reactin with nitrous acid forms unstable diazonium salt which on decomposing liberates nitrogen and mixture of alcohols and alkenes.

Basicity of substituted anilines and aliphatic amines,
Nitrogen of amines contains lone pair of electrons, which can be shared with other species and thus these act as Lewis bases.

Azo coupling reaction of diazonium salts of aromatic amines,

Sandmeyer and related reactions of diazonium salts;
The diazonium salt is treated with cuprous chloride or cuprous bromide.

Carbylamine reaction;
The treatment of a primary amine with chloroform and alcoholic potash produces carbylamine (isocyanide) which has most offensive smell. This reaction is not exhibited by secondary and tertiary amines.