Method of preparation. An emulsion is prepared by shaking strongly the mixture of the two liquids or by passing the mixture through a colloid mill, called the homogenizer. The emulsions thus prepared from the pure liquids are usually not stable and the two liquids separate out on standing. To get a stable emulsion, small quantities of certain other substances are added during preparation. The substances thus added (tJ stabilize the emulsions are called emulsifiers or emulsifying agents. The substances commonly used as emulsifying agents are soaps of various kinds, long chain sulphonic acids or lyophilic colloids (proteins, gum, agar etc.).
Role of emulsifier. The role of emulsifier can be explained taking the example of soap as an emulsifier. Soaps are sodium or potassium salts of higher fatty acids, e.g., sodium palmitate (C15H31COONa), sodium stearate (C₁₇H₃₅COONa), etc. A molecule of soap consists of two parts, the hydrocarbon part (C₁₅H₃₁ -, C₁₇H₃₅ - , etc.) which is soluble in oil and the polar group (COO-Na+) which is soluble in water
         R                           COO-Na+
Hydrophobic part         Hydrophilic part
(Soluble in oil)            (Soluble in water)
Thus, if a drop of oil is surrounded by soap solution, the R - part of the soap remains in the oil and the COO^-Na⁺ part remains in water as shown in Fig. 5.31. Thus, soap molecules are concentrated over the surface of the drop of oil. As a result, the interracial tension between oil and water decreases and hence they are intermixed into each other to form the emulsion.Thus, micelles are formed Some other commonly used stabilizing agents are proteins, gums and agar
Types of Emulsions. If the term "oil" is used for any liquid which is immicible with water and is capable of forming an emulsion with water, then the various types of emulsions may be classified into two types (fig. 5.32). These are:
(i) Emulsion of oil-in-water (o/w) in which oil is the dispersed phase and water is the dispersion medium.
For example, milk is an emulsion of liquid fat dispersed
in water. Another well known example is that of vanishing cream
(ii)Emulsion of water-in-oil (w/o) in which water is the dispersed phase and oil is the dispersion medium.For example, cod liver oil is an emusion of water in oil, i.e., water is the dispersed phase and oil is the dispersion medium. Two other common examples of this type are butter and cold cream.
Evidently, the type of emulsion depends upon the relative amounts of the two liquids. If water is in excess, it is oil-in-water emulsion.JIf oil is in excess, it is water-in-oil emulsion. The type of emulsion formed also depends upon the nature of the emulsifying agent. For example, the presence of soluble soaps as the emulsifying agents generally favours the formation of emulsions of oil-in-water whereas the presence of insoluble soaps (containing non-alkali metal atoms) favours the formation of emulsions of water-in-oil.Some very commonly used emulsifying agents for o/w emulsions are : proteins, gums, natural and synthetic soaps whereas for w/o emulsions are: heavy metal salts of fatty acids, long chain alcohols, lamp black, etc
To test the type of emulsion. The type to which the given emulsion belongs can be tested experimentally by anyone of the following methods:
(i) Microscopic method. To a small amount of the given emulsion, a few drops of water are added. The water will mix completely if the emulsion is oil-in-water type. Likewise, for the water-in-oil emulsion, the miscibility with a few drops of oil can be tested. The miscibility can be seen under a microscope. 
(ii) Conductance method. This method is based upon adding a small amount of an electrolyte to the emulsion. If the conductance increases, the emulsion is oil-in-water type and if there is no significant change, it is water-in-oil type.
(iii) Dye method. A small amount of an oil-soluble dye is added to the emulsion. If it is water-in-oil type, it becomes deeply coloured, otherwise it remains colourless.

Surface area per gram of gel = 561.8 m².

At P = 0.5 atm, \(\frac { x }{ m } kP1/n\ =\ 2\times { (0.5) }^{ 1 }=1.0\)

\(\frac { No.of\quad sites\quad occupied }{ No.of\quad { N }_{ 2 }\quad molecules } =\frac { 1.2046\times { 10 }^{ 17 } }{ 6.023\times { 10 }^{ 16 } } =2\)

V₂ = 59.6 cm³