It is the homogeneous mixtures of two or more than two components .
When the composition and properties are uniform throughout the mixture then it is called homogeneous mixture .
It has following components –
- Solvent .
- Solute .
The component that present in the largest quantity is known as Solvent .
Solvent can be solid , liquid and gas . It determines the physical state of the mixture .
One or more components present in the Solvent is called Solute .
Saturated Solution :-
A mixture in which no more Solute can be dissolved under the given conditions I . e . temperature and pressure is called Saturated Solution .
Unsaturated Solution :-
A mixture in which more Solute can be dissolved under the given conditions I . e . temperature and pressure is called Unsaturated Solution .
Ideal Solution :-
The mixture which obey Raoult’s law over the entire range of concentration is known as Ideal Solution .
- The enthalpy of mixing of the pure components to form the Ideal Solutions is Zero
I . e . ∆H = 0
- Total volume of mixture is equal to sum if volumes of the components I . e . The volume of mixing is zero .
∆V = 0
- If the intermolecular attractive forces between the molecules A – A and B – B are nearly equal to those between A – B . This leads to the formation of Ideal Solution .
A perfectly Ideal Solution is rare but some mixtures are nearly Ideal in behavior .
Examples – Solution of n – hexane and n – heptane , Bromoethane and Chloroethane , Benzene and Toluene etc .
Also read:- Colligative properties
Non – ideal Solution :-
When a mixture does not obey Raoult’s law over the entire range of concentration , then it is called Non – ideal Solution .
- The enthalpy of mixing of both liquid components is changes .
- Volume is increase after dissolution .
- 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 .
The Vapour pressure of such 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 Raoul’s law .
Examples – Solution of Acetone + Ethanol , Water + Ethanol , Water + Methanol , Acetone + Benzene etc .
Vapour Pressure :-
It is defined as the pressure exerted by a Vapour in Thermodynamics equilibrium with solid or liquid at a given temperature in closed system .
- The vapour pressure of any substance increase non-linearly with temperature .
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 . Let the total vapour pressure at this stage be P total , P₁ and P₂ be the partial pressure of the two components 1 and 2 respectively .
These partial pressure are related to the mole fraction X₁ and X₂ .
The French chemist Raoult ( 1886 ) gave the relationship between them . The relationship is known as the Raoult’s law .
Raoult’s law :-
The law states that for a solution of volatile liquids , the partial Vapour pressure of each component in the solution is directly proportional to its mole fraction present in the Solution .
If partial pressure for component 1 = P₁
Mole fraction for component 1 = X₁
P₁ proportional X₁
or P₁ = P₁⁰ X₁
Where P₁⁰ represent the Vapour pressure of component 1
If the partial Pressure for component 2 = P₂
Mole fraction for component 2 = X₂
P₂ proportional X₂
or P₂ = P₂⁰ X₂
Where P₂⁰ represent Vapour pressure of component 2
According to Dalton’s law of partial Pressure , the total pressure over the Solution phase in container will be the sum of the partial Pressures of the components .
P = P₁ + P₂
= X₁ P₁⁰ + X₂ P₂⁰
since X₁ + X₂ = 1
P = ( 1 – X₂ ) P₁⁰ + X₂ P₂⁰
or P = P₁⁰ + ( P₂⁰ – P₁⁰ ) X₂
Expressing Concentration of Solutions :-
Composition of solutions can be described by expressing its Concentration .
It can be expressed either qualitatively or quantitatively .
In qualitatively , Solutions are dilute ( relatively very small quantity of Solute ) while in quantitatively , Solutions are concentrated ( relatively very large quantity of Solute ) .
- In real life we describe the Concentration of Solutions quantitatively in following ways –
1 . Mass percentage
2 . Volume percentage
3 . Mass by Volume percentage
4 . parts per million
5 . Mole fraction
6 . Molarity
7 . Morality
Mass percentage :-
The mass percentage of a component in a Solution is defined as
Mass % of component = mass of the component x 100 / Total mass of Solution
Volume percentage :-
The volume percentage is defined as –
Volume % of a component = Volume of the component X 100 / Total volume of Solution
Mass by Volume percentage ( W / V ) :-
It is defined as the mass of Solute dissolved in 100 ml of the Solution .
Parts of million ( ppm ) :-
When a Solute is present in trace amount , it is convenient to express concentration in parts per million ( ppm ) and is defined as –
Parts per million = Number of parts of the component X 10⁶ / Total no. of parts of all components of the Solution .
Mole fraction :-
It is defined as mole fraction if a component = Number of moles of the component / Total no. of moles of all the components .
Example – If the number of moles if A and B are nA and nB respectively .
The mole fraction of A will be
XA = nA / nA + nB
The mole fraction of B will be
XB = nB / nA + nB
XA + XB = ( nA / nA + nB ) + ( nB / nA + nB ) = 1
Q 1. Calculate the mole fraction benzene containing 30 % by mass in Carbon tetrachloride .
moles if Carbon tetrachloride ( CCl₄ ) = 70 gm / 154 gm mol⁻¹ = 0 ·45 mol .
Moles of benzene = 30 gm / 78 gm mol⁻¹ = 0 · 38 mol .
or XB = nB / nA + nB
= 0 · 38 / 0 · 45 + 0 · 38
= 0 · 38 / 0 · 83 = 0 · 457
XA = 1 – 0 · 457 = 0 · 543
Molarity ( M ) :-
It is defined as number of moles of Solute dissolved in one litre of liquid .
Molarity = Moles of Solute / Volume of Solutions in litre
I . e . M = n / V
Molarity is a function of temperature . This is because Volume depends on temperature and the mass does not .
Molality ( m ) :-
It is defined as the number of moles of the Solute per kilogram ( Kg ) of the Solvent and us expressed as
Molality = Moles of Solute / mass of Solvent in Kg
I . e . m = n / W
Note :- Mass % ,ppm , mole fraction and Molality are independent of temperature .
Q 2 . Calculate molality of 2 · 5 gm of ethanoic acid ( CH₃COOH ) in 75 gm of benzene .
moles of ethanoic acid ( n ) = 2 · 5 gm / 60 gm mol⁻¹ = 0 · 0417 mol⁻¹
mass of benzen in kg ( W ) = 75 x 10⁻³ kg
Molality of ethanoic acid ( m ) =?
m = n / W
m = 0 · 0417 x 1000 / 75 = 0 · 556 mol kg⁻¹
It is defined as the amount of substance ( solute ) that dissolves in a unit volume of a liquid substance ( solvent ) to form a saturated solution under specified conditions of temperature and pressure .
Solubility = mass of Solute in gm X 100 / mass of Solvent .
- Solubility of one substance into another substance depends on the following factors –
1 . Nature of the substances .
2 . Temperature .
3 . Pressure .
Nature of the Substances :-
Strong solute – solvent attractions equate to greater solubility while weak solute – solvent attraction equate to lesser solubility .
Polar solute tend to dissolve best in polar solvent . non – polar solute tend to dissolve best in non – polar solvent .
Example – NaCl and sugar dissolve in water .
Effect of temperature :-
The solubility of a solid in a liquid is affected by temperature changes .
In general , if in a nearly saturated solution ,the dissolution process is endothermic , the solubility should increase with rise in temperature and if it is exothermic the solubility should decrease .
- Solubility of gases in liquids decreases with rise in temperature . As dissolution is an exothermic process , the solubility should decrease with increase of temperature .
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 .
Henry’s law :-
At constant temperature , Solubility of gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid .
which is known as Henry’s law .
If we use the mole fraction of a gas in the liquid as a measure of its solubility , then it can be said that the mole fraction of a gas in the liquid is proportional to the partial pressure of the gas over the liquid .
I . e . P proportional X
P =K X
Where P – partial pressure of the gas , X – mole fraction of the gas and K is constant called Henry’s law constant .
The formula is called Henry’s law constant .
Different gases have different K values at the same temperature . So K is a function of nature of the gas .
Q 3 . If N₂ gas is bubbled through water at 273 K . How many milli moles if N₂ gas would dissolve in 1 litre of water . Assume that N₂ exerts a partial pressure of 0 · 987 bar . Give that Henry’s law constant for N₂ at K is 76 · 48 k bar .
X = P / K = 0 · 987 bar / 76480 bar = 1 · 29 x 10⁻⁵
As 1 litre of water contains 55 · 5 mol of it therefore if n represent number of moles of N₂
X = n / n + 55 ·5 = n / 55 · 5 = 1· 29 x 10⁻⁵
( n in denominator is neglected as it is << 55 · 5 )
Thus n = 1 · 29 x 10 ⁻⁵ x 55 · 5 mol = 7 · 16 x 10⁻⁴ mol = 0 · 716 m mol.
Applications of Henry’s law :-
1 . Increase the solubility of CO₂ in cold drinks and soda water , the bottle is sealed under high pressure .
2 . At high altitudes the partial pressure of oxygen at high altitudes less than the ground level . This results in low concentration of oxygen in the blood and tissues of the Peoples .
3 . In scuba diving – during scuba diving when the diver breaths in compressed air from the supply tank , more Nitrogen dissolves in the blood because the pressure at that Depth is far greater the surface atmospheric pressure .