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CBSE Class 12th Chemistry Notes: Solutions (Part – II)

Aug 8, 2016 11:01 IST

    In part I of CBSE Class 12 Chemistry: Chapter Notes on Solutions, we have covered basic concepts like definition of solution, types of solutions, various ways to express concentration of solution, solubility, solubility of solid in a liquid and some conceptual questions.

    Now we will study the following topics and questions based on them:

    • Vapour Pressure of a Liquid Solution
    • Vapour Pressure of a Liquid-Liquid Solution
    • Vapour Pressure of a Solid-Liquid Solution
    • Ideal and Non Ideal Solutions
    • Positive & Negative Deviation from Raoult’s law
    • Azeotropes
      •    Minimum boiling azeotropes
      •    Maximum boiling azeotropes

    The notes given below on above mentioned topics are important for CBSE board exams as well as competitive exams.

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    Vapour Pressure of a Liquid Solution

    Liquid solution are formed when solvent is in liquid phase. Solute may be solid, liquid or gas. On the basis of solute, the liquid solution is classified in 3 types as:

    (i) Solid in liquid

    (ii) Liquid in liquid

    (iii) Gas in liquid

    We will discuss the various properties of liquid in liquid solution and solid in liquid solutions.

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    Vapour Pressure of a Liquid-Liquid Solution

    Let us take a binary solution made up of two volatile liquids 1 and 2. As the liquid start evaporating; a stage will come when the vapour pressure of liquid will be in equilibrium with the corresponding liquid. As we know that vapour pressure of liquid is proportional to its mole fraction.  The quantitative relationship between the vapour pressure and mole fraction in binary solution is given by Roult’s Law.

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    Roult’s Law:

    It states that for the solutions of volatile liquid, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.

    For component 1,

    p1 ∝  x1

    p1 = p1° x1...(1)

    p1 = Vapour pressure of component 1 in solution.

    p1° = Vapour pressure of pure liquid component 1at the same temperature

    x1 = Mole fraction of component 1 in solution.

    Similarly for component 2 :

    P2 = p2° x2 ...(2)

    According to Dalton’s Law of partial pressure, The total pressure (ptotal ) over the solution phase in the container will be the sum of partial pressure of the components in solution. If the solution is made up of two volatile liquids then total pressure above the solution is:

    ptotal = p1 + p2  ...(3)

    ptotal = Total pressure over the solution phase

    p1 = Vapour pressure of component 1 in solution

    p2 = Vapour pressure of component 1 in solution

    Substituting the values of p1 and p2 in eq...(3) we get:

    ptotal = p1° x1 + p2° x2 ...(4)

    As, total mole fraction of components in any solution is 1. Therefore,

    x1 +  x2 = 1

    x1 =(1 - x2)...(5)

    Putting the value of x1 in eq (4) we get:

    ptotal = p1°(1- x2) + p2° x2

    ptotal = p1° - p1° x2 +  p2° x2

    ptotal = p1° +( p2°p1°) x2 ...(6)

    The equation ...(6) is known as the mathematical expression for Roult’s Law for the solution made up of two volatile liquids.

    From this equation ...(6) we can conclude the following points :

    •    ptotal can be related to mole fraction of any of the one component.

    •    ptotal varies linearly with x2

    •    ptotal increases or decreases with increase of x1

    This can be represents on graph as follows assuming p1° < p2°:

    Conclusion from graph:

    • Plot of  p1° vs x1 is linear
    • Plot of  p2° vs xis linear
    • Maximum value of  ptotal = p2°
    • Minimum value of  ptotal = p1°

    Composition of vapour phase at equilibrium is determined by using Dalton’s Law. Let y1 and y2 be the mole fraction of liquid 1 and 2 in solution then partial vapour pressure of each component is written as:

    p1 = y1 ptotal... (7)

    Equation ...(7) gives you the value of partial vapour pressure of each component in vapour phase.

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    Vapour Pressure of a Solid-Liquid Solution

    When a non-volatile solid is added to the solvent to form a solution, then the vapour pressure of solution is found lower than vapour pressure of the pure solvent at same temperature. The decrease in the vapour pressure of the solution is solely depends on the quantity of non-volatile solute present in solution.

    The vapour pressure of such solution is given by using general equation of  Roult’s Law.

    Assume that water is component 1 and non-volatile component is component 2, then vapour pressure of the solution will be equal to the vapour pressure of solvent in solution. Vapour pressure of solvent p1 is proportional to its mole fraction in solution and given as :

    p1   x1

    p1 = p1° x1...(1)

    p1 = Vapour pressure of component 1 in solution.

    p1° = Vapour pressure of pure liquid component 1at the same temperature

    x1 = Mole fraction of component 1 in solution.

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    Ideal and Non Ideal Solutions

    Liquid-liquid solutions can be classified as ideal and non-ideal on the basis of certain properties.

    Ideal Solution

    • Obeys the Roult’s Law over the entire range of concentration.
    • ΔHmixing = 0
    • ΔVmixing = 0
    • Example : solution of n-hexane and n-heptane, solution of bromoethane or chloroethane.

    It can be summarised as: If a Solution formed by mixing the two components A and B, in which intermolecular force of attraction between A and B (A‒B) is nearly equal to intermolecular force of attraction between pure components (A‒A and B‒B) then no heat would be evolved or absorbed in forming the solution. Also volume of the solution will be equal to the total volume of the individual component taken to form the solution.

    Non-Ideal Solution

    • Does not obey the Roult’s Law over the entire range of concentration.
    • ΔHmixing ≠ 0
    • ΔVmixing ≠ 0
    • Example: Solution of chloroform and acetone

    It can be summarised as: If a Solution formed by mixing the two components A and B , in which intermolecular force of attraction between A and B (A‒B) is not equal to intermolecular force of attraction between pure components (A‒A and B‒B). This new interaction (A‒B) is either less than or more than the interaction of the pure components (A‒A and B‒B). This leads to the positive or negative deviations from Roult’s Law.

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    Positive Deviation from Roult’s Law

    • The vapour pressure of solution formed by mixing two components is higher than predicted from Roult’s Law
    • The new intermolecular interactions formed by mixing the component A and B (A‒B) are weaker than the intermolecular interactions of pure component (A‒A and A‒B)
    • Example: mixture of ethanol and acetone, solution of carbon disulphide and acetone.

    Graph representing the positive deviation:

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    Negative Deviation from Roult’s Law

    • The vapour pressure of solution formed by mixing two components is lower than predicted from Roult’s Law
    • The new intermolecular interactions formed by mixing the component A and B (A-B) are stronger than the intermolecular interactions of pure component (A-A and A-B)
    • Example: solution of phenol and aniline, chloroform and acetone.

    Graph representing the negative deviation:

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    Azeotropes

    These are mixture of two liquids having same composition in liquid as well as vapour phase and boil at the constant temperature. This liquid mixture cannot be separated into pure component even on fractional distillation.

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    Types of Azeotropes

    Minimum boiling azeotrope:

    The solution which show large positive deviation from Raoult’s Law.  Example: solution of 95% ethanol in water.

    Maximum boiling azeotropes:

    The solution which show large negative deviation from Raoult’s Law.  Example: solution of 68% nitric acid and 32% water by mass.

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    Intext Questions:

    Question 1: What type of deviation is shown by a mixture of ethanol and acetone? Give reason. [CBSE 2014]

    Solution 1: A mixture of ethanol and acetone shows positive deviation from Raoult's Law. The new intermolecular interaction formed by mixing the component ethanol and acetone is weaker than the intermolecular interactions of pure component pure ethanol. Pure ethanol possesses hydrogen bonding. When acetone is mixed, the molecule of acetone takes the space in between the molecules of ethanol which results in breaking of some of the hydrogen bonds. Due to weakening of interactions, the vapour pressure of solution formed by mixing ethanol and acetone is higher than predicted from Roult’s Law. Hence the solution shows positive deviation from Raoult’s law.

    Question 2: State Raoult's law for the solution containing volatile components. What is the similarity between Raoult's law and Henry's law? [CBSE 2014]

    Solution 2: Raoult's law :  It states that for the solutions of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.

    For component 1,

    p1   x1

    p1 = p1° x1...(1)

    p1 = Vapour pressure of component 1 in solution.

    p1° = Vapour pressure of pure liquid component 1at the same temperature

    x1 = Mole fraction of component 1 in solution.

    Similarly for component 2,

    P2 = p2° x2 ...(2)

    Similarity between Raoult's law and Henry's law:
    Both laws state that the partial pressure of the volatile component is directly proportional to its mole fraction in the solution. In case of Roult’s law it is liquid and in case of Henry’s law it is gas.

    Equation for Roult’s law:

    p = p° x

    Equation for Henry’s law :

    p= KH. x

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    Question 3: Some liquids on mixing form 'azeotropes'. What are 'azeotropes'? [CBSE 2014]

    Solution 3: Azeotropes: These are mixture of two liquids having same composition in liquid as well as vapour phase and boil at the constant temperature. This liquid mixture cannot be separated into pure component even on fractional distillation. Azeotropic solution boils at constant temperature, regardless of difference in boiling points of respective components.

    Question 4: Define an ideal solution and write one of its characteristics. [CBSE 2014, 2012]

    Solution 4: The solutions that obey Raoult’s law over the entire range of concentration are called ideal solutions. Examples: n-hexane and n-heptane
    Characteristics of ideal solutions:
    Enthalpy of mixing (ΔmixH) of the pure components to form the solution is zero. Volume of mixing (ΔmixV) is also equal to zero.

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    Question 5: State Raoult’s law for a solution containing volatile components. How does Raoult’s law become a special case of Henry’s law? [CBSE 2013]

    Solution 5: Similar to answer 2

    Raoult’s Law as a Special Case of Henry’s Law

    According to Raoult’s law, the vapour pressure of a volatile component in a given solution is p = p° x

    According to Henry’s law, the partial vapour pressure of a gas(volatile component) in a liquid is p = KH. x

    It can be observed that in both the equations, the partial vapour pressure of the volatile component varies directly with its mole fraction. The only difference is the proportionality constants which is KH in Henry’s law and p° in Roult’s Law. Thus Raoult’s law becomes a special case of Henry’s law in which KH is equal to p°.

    Question 6: Define the following terms: (i) Ideal solution (ii) Azeotrope [CBSE 2013]

    Solution 6: (i) Ideal solution ; same as answer 4
    (ii) Azeotrope; same as answer 3

    Question 7: Non-ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused? Explain with one example for each type. [CBSE 2010]

    Solution 7: Follow the above theory.

    Click here, to get the CBSE Class 12th Chemistry Notes: Solutions (Part – I)

    x1 +  x2 = 1

    x1 =(1- x2)...(5)

    Putting the value of x1 in eq (4) we get:

    ptotal = p1°(1- x2) + p2° x2

    ptotal = p1° - p1° x2 +  p2° x2

    ptotal = p1° +( p2°-  p1°) x2...(6)

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