Azeotropes (constant boiling mixture)- and types
The liquid mixture, having the Same composition in liquid and vapour phase & boil at a constant temperature, is called azeotrope.
Differentiate between minimum & maximum boiling azeotropes
S. No.
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minimum boiling azeotropes
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Maximum boiling azeotropes
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1
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Azeotropes whose boiling point is less than either of two components present in the liquid mixture.
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Azeotropes whose boiling point is more than either of two components present in the liquid mixture.
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2
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The non-ideal solution which show a large positive deviation from Raoult’s law form minimum boiling azeotropes at specific composition
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The non-ideal solution which show a large negative deviation from Raoult’s law form minimum boiling azeotropes at specific composition
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3
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Example - solution showing positive deviation
Liquid mixture of 95 % ethanol + 5 % water
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Example - solution showing negative deviation
Liquid mixture of 68% Nitric acid +32%water
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Azeotropic matures cannot be separated into their constituents by fractional distillation.
Colligative properties -
Those properties of Ideal solutions which defend only on the no. of particles of the solute (molecules or ions) dissolved in a definite amount of the solvent & do not depend on the nature of solute are called colligative properties.
There are as follow-
1). Relative lowering vapour pressure
2) Elevation of boiling points
3) Depression of freezing point
4) osmotic pressure
1). Relative lowering vapour pressure
We know that when non-volatile solute is dissolved in a solvent, the vapour pressure of the solution becomes lower than that of the pure solvent. The decrease in vapour pressure of solution is called lowering of vapour pressure. Expressed by
∆P = P10 - P1 …........................(1)
By applying Raoult’s law
P1 = P10 x1 .......(2)
∆P = P10 - P10 x1
∆P = P10 - P10 (1-x2) since x1=1-x2
or ∆P = P10 - P10 +P10 x2
or ∆P = P10 x2 ….............(3)
So, ∆P/ P10 = (P10 - P1 )/ P10 = x2 ,…......................... …......(4)
The ratio of ∆P/ P10 = (P10 - P1 )/ P10 is called relative lowering vapour pressure.
Equation (4) may be written as
∆P/ P10 =(P10 - P1 )/ P10
= x2 = n2 /( n1 + n2 )
For dilute solution
(P10 - P1 )/ P10 = x2 = n2 / n1 ( since n2 is lesser than n1)
(P10 - P1 )/ P10 = x2 = n2 / n1 = w2 x M1 /M2 x w1 ….........(5) used to calculate the molar mass of a solute.
Problem- 1
