Colligative Properties of Solutions: A Comprehensive Study, Schemes and Mind Maps of Chemistry

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• शिक्षिमहर्षी डॉ. बापूजी साळुंखे

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Chapter No. 2

Solutions

Marks without Option : 5

Mark with Option : 7

SUBJECT :CHEMISTRY

1.Types of solutions :-

No. Solute Solvent Example 1 Solid Liquid Sugar in water 2 Solid Solid Metal alloy 3 Solid Gas Iodine in air 4 Liquid Liquid Ethanol in water 5 Liquid Solid Amalgam of mercury with metals Ex-Sodium amalgam 6 Liquid Gas Chloroform in nitrogen 7 Gas Liquid Carbonated water Ex- CO 2 in water 8 Gas Solid Hydrogen in palladium 9 Gas Gas Air Generally we think that a solution is either solid is dissolved in liquid or a mixture of liquid in liquid. There are other types of solutions depends on three states of matter solids, liquids and gases. This

gives nine types of solutions.

2.Capacity of solutions :- 2.1 Saturated solution :- A saturated solution contains maximum amount of solute dissolved in a given amount of solvent at a given temperature. 2.2 Supersaturated solution :- A solution containing greater than the equilibrium amount of solute is said to be supersaturated solution. Example :- The capacity of 100 gm water is to dissolve 36 gms of NaCl, then the solution is saturated. If we dissolve less than 36 gms then it is unsaturated solution. If we dissolve more than 36 gms then it is supersaturated solution.

b) Effect of temperature :-

• Endothermic process :- When the temperature is increased, then solubility of substance increases by Le-chatelier’s principle. Example - KCl dissolves in Water by endothermic process.
• Exothermic process:- When the temperature is increased, then solubility of substance decreases. Example - CaCl 2 dissolves in water by exothermic process.
• Solubilities of NaCl, KCl, NaBr changes slightly with temperature.
• Solubilities of KNO 3 ,NaNO 3 ,KBr increases with temperature due to endothermic process.
• Solubilities of Na 2 SO 4 decreases with increase in temperature due to exothermic process. c) Effect of Pressure :-
• Pressure has no effect on solubilities of solids and liquids.
• However if affects on solubilities of gases in liquids, the solubility of gases increases with increase in pressure. It can be explained by Henry’s law.

Henry’s Law :- It states that the solubility of gas in a liquid is directly proportional to pressure of gas over the solution.

• Thus S  P S=KHP Where KH is proportionality constant or Henry’s law constant
• Units of KH : KH = S/P KH = mol L - 1 bar - 1 When P =1 bar, then KH =S i.e. Proportionality constant is equals to solubility when pressure is 1 bar.
  • Exceptions of Henry’s Law :- Gases like NH 3 and CO 2 doesnot obey Henry’s law because they can reacts with water. NH 3 + H 2 O NH 4 ⁺ + OH¯ CO 2 + H 2 O H 2 CO 3.

4.Vapour Pressure of solution ( liquid in liquid) :- Consider a binary solution of two volatile liquids. Both the liquids vaporize and a equilibrium is established between liquids and its vapour phase. Then the partial vapour pressure are related to their mole fractions and is given by Raoult’s law.

• 1. Raoult Law :- It states that the partial vapour pressure of any volatile component of solution is equal to the vapour pressure of pure component multiplied by mole fraction of solution. Suppose a binary solution of two volatile liquids A 1 & A 2 , P 1 & P 2 are the partial pressures and x 1 & x 2 are the mole fractions, then according to Raoult law P 1 = P 1 O^ x 1 & P 2 = P 2 o x 2 Where P 1 o and P 2 o are the vapour pressures of pure components.
  • According to Dalton’s law of partial pressure P= P 1 + P 2 = P 1 o x 1 + P 2 o x 2 Since x 1 + x 2 = 1, So x 1 = 1- x 2 P = P 1 o (1- X 2 ) + P 2 o x 2 = P 1 o
• P 1 o x 2 + P 2 o x 2 = (P 2 o
• P 1 o ) x 2 + P 1 o As P 2 o & P 1 o are constant then it is equation of straight line y=mx+c. A plot of ‘P’ vs x 2 is a straight line.

Composition of vapour phase : The composition of vapour in equilibrium with the solution can be determined by Dalton’s law of partial pressures. If we take y 1 and y 2 as the mole fractions of two components in the vapour, then P 1 = y 1 P and P 2 = y 2 P where P 1 and P 2 are the partial pressures of two components in the vapour and P is the total vapour pressure.

3.1 : Deviations from Raoults Law :-

Positive Deviations 1.The solution in which solute – solvent intermolecular attaractions are weaker than solute-solute and solvent-solvent molecules, exhibits positive deviations

2. The vapour pressure of such solutions are higher than those of pure components.
3. Example - Ethanol and acetone, CS 2 and acetone, Chloroform and ethanol, Chloroform and carbon tetrachloride.

3.2 : Deviations from Raoults Law :- Negative Deviations

1. The solution in which solute – solvent intermolecular attaractions are stronger than solute-solute and solvent-solvent molecules, exhibits negative deviations
2. The vapour pressure of such solutions are lower than those of pure components
3. Example - Phenol and aniline, Chloroform and acetone, Chloroform and benzene, Nitric acid and water.

1. Vapour pressure lowering :- 1.1 : Vapour pressure : When a liquid in a closed container is in equilibrium with its vapour. Then pressure exerted by vapours on liquid surface is called vapour pressure. 1.2 : Vapour pressure lowering :- ➢ When nonvolatile, non ionizable solid is dissolved in liquid solvent, the vapour pressure of solution is lower than that of pure solvent. ➢ In other words the vapour pressure of a solvent is lowered by dissolving a non-volatile solute into it. ➢ When the solute is non-volatile it does not contribute to the vapour pressure above the solution. ➢ If P 1 o is the vapour pressure of pure solvent and P 1 is the vapour pressure of solution then P 1 o  P 1 The vapour pressure lowering is P = P 1 o

• P 1 ---------(1)

1.3 : Relative lowering of V.P. :- The ratio of vapour pressure lowering of solvent divided by V.P. of pure solvent is called relative lowering of vapour pressure. Thus 𝛥𝑃 𝑃 1

0 =^

𝑃 1 0 −𝑃 1 𝑃 1 0 1.4 : Raoults law for non volatile solute :- It states that the vapour pressure of solvent over the solution is equal to the vapour pressure of pure solvent multiplied by mole fraction of solution. Thus P 1 = P 1 o x 1 For a binary solution, ( x 1 + x 2 = 1), x 1 = 1 - x 2  P 1 = P 1 o (1 - x 2 ) = P 1 o

• P 1 o x 2  P 1 o
• P 1 = P 1 o x 2  P = P 1 o x 2 ----------(2) From this equation it shows that P depends upon X2. i.e. number of solute particles. Thus P is a colligative property.

Problem :

1. The vapour pressure of pure benzene ( molar mass 78gm/mol) at a certain temperature is 640 mm of Hg. A non-volatile solute of mass 2.315 gm is added to 49 gm of benzene. The vapour pressure of the solution is 600 mm of Hg. What is the molar mass of solute?  Given : P 1 o^ = 640 mm of Hg, P 1 = 600 mm of Hg, W 1 = 49 gm, W 2 = 2.315 gm, M 1 = 78 gm/mol M 2 =?  𝑝 1 0 −𝑃 1 𝑃 1 0

𝑤 2 𝑀 1 𝑀 2 𝑊 1 M 2 = 𝑤 2 𝑀 1 𝑊 1

𝑝 1 0 𝑃 1 −𝑃 1 0 = 2.315  78  600 49 (640-600) = 2708. 49 M 2 = 55.27 gm/mol

2. Boiling point elevation :- The temperature at which vapour pressure of liquid equals to atmospheric pressure is called boiling point. The boiling point of solvent is elevated by dissolving a nonvolatile solute into it. Thus the solution containing nonvolatile solute boils at higher temperature than its pure solvent. If Tb o is boiling point of pure solvent and Tb is boiling point of solution then Tb Tb o Tb = Tb - Tb o **Comparision with –

1. Vapour pressure :-** The vapour pressure of solution and vapour pressure of pure solvent are plotted as a function of temperature.