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Notes for Class 12th Chemistry Ch-2 “Solutions”

  • The Hetrogeneous Mixtures are called Solutions.
    EXAMPLE : H2O + Sugar , Water + Salt , Alloys .
    1. Binary Solutions :
    The Solutions which has only 2 components i.e. solute + solvent.
    1. Molarity (M)
    2. Molality (m)
    3. Mole fraction
    4. Normality (N)
    5. Formality
    6. W/W = Mass Percentage
    7. V/V = Volume Percentage
    8. W/V = Gram
    9. ppm = Parts per million
    10. ppb = Parts per billion
  • MOl 
    The SI unit of the amount of the substance is mol .
    It has fixed Number (Avagadro Number).
  • No. of moles = Mass in gram /Molar mass
    Mole Fraction of Solute = Moles of Solute / Total Moles
    Mole Fraction of Solvent = Moles of Solvent / Total Moles

    * The Total of all Mole Fraction is always 1.
    It is number of moles of solute
    present in 1 L of solution.
    It is number of moles of solute
    present in 1 Kg of solvent.
    Unit : mol/L Unit : mol/Kg
    Molarity is a Dependable property
    because the Volume of the solution
    changes with temperature.
    It is more prefered quantity because
    its value never changes with change in
    It is the  maximum amount of a solute that can be dissolve in specific amount of solvent at specific temperature and pressure.
    1. Nature of the Solute
    2. Nature of the Solvent
    3. Temperature
    4 . Pressure
    Types of Solubility
    1. Solubility of Solid in a Liquid
    2. Solubility of Gas in a Liquid
    3. Solubility of Liquid in a Liquid
    1. Solubility of Solid in a LiquidQ : Name some readily soluble solids solutes in water but not in organic solvent .
    A : NaCl and Sugar

    Q : Name some solid solutes which are soluble in organic solvent but not in water .
    A :
    Napthalene , Anthracene

    EXCEPTION : Glucose , Sugar , Alcohols are organic compounds having covalent 
    bonding but a readily soluble in water due to hydrogen bonding .

     DISSOLUTION : The process in which solute is added to solvent to increase its concentration in solution.

    CRYSTALLISATION : The process in which some solute particles in the solution collide with the solid solute particles and get separated out .

    At equillibrium ,
    rate of dissolution = rate of crystallisation

    SATURATED SOLUTION : The solution in which no more solute can be dissolve at the same condition of temperature and pressure .

    UNSATURATED SOLUTION : The solution in which more solute can be dissolve at the same condition of temperature and pressure.

    2. Solubility of Gas in a Liquid

    HENRY’S LAW : The solubility of gas in liquid is directly prportional to the partial pressure of  gas , present above the surface of the liquid or solution.
    P ∝ n
    P = KH .x
    where ,
    P = Partial pressure of gas
    KH = Henry’s constant
    x = Mole fraction of the solute { gas }

    K∝ 1/ Solubility
    Hence ,  Solubility ∝ 1/T

  • That is why , the O2 gas has more solubility in water in winter season , where temperature is less.
  • As a result , fish is more comfortable in winter season then summar season .
  • The value to KH  for different gases at the same temperature is different which confirms that Kis a function of nature of the gas.
  • Kfor the same gas at the different Temperaure is different which confirms that KH ∝ T.

    1. Bends
  • The medical condition in which N2 block the blood capillaries and create very painful and dangerous condition known as bends.
  • In scuba divers under high pressure in the deep sea have high concentration of disslove gases while breathing. {According to Henry’s Law}
  • The Solubility of Atmospheric gases increases in the blood with the increase in the pressure .
  • When scuba divers come to the surface , pressure decreases, which release the dissolve gas and leads to the formation of bubbles in the blood by N2 gas.
  • The toxic effect of high concentration of  Nin the blood is reduced by changing the percentage of N2  from 78 % to 56.2% by replacing it with helium 11.7 % .
    2. ANOXIA 
  • The symptom of medical condition in which a climber become weak and unable to think clearly is known as Anoxia.
    Reason :
  •  At high altitude the partial pressure of O2 is less than the Partial pressure at ground level.
  • This low concentration of Oin blood and tissues of people living at high altitudes create anoxia.
  • In Soft Drinks : To increase the solubility of CO2 gas in soft drinks and soda bottle , the bottles are sealed under igh pressure .
  • In Summer Season , Temperature is more, so the equilibrium will shift in backward direction. Thus less gas is dissolve in water and Fish is not comfortable .
  • In Winter Season, Temperature is less , so the equilibrium will shift in the forward direction.Thus more gas is dissolve in water and Fish is comfortable.RAOULT’S LAW
    Types f Solution for Raoult’s law
    1. Liquid in Liquid
    2. Solid in Liquid
  • The liquid Solvent are generally volatile and the solute may or may not be volatile .
  • 1 is used for Solvent and 2 for Solute .
  • At equilibrium, Rate of evaporation = Rate of condensation .
  • For a Solution of a volatile liquid, the partial vapour pressure  of each component of the solution is directly proportional to its mole fraction present in solution.
    P1 ∝ x1
    P1 = P1o x1 …………………………… (i)
    For the second component,
                                              P2 ∝ x1
    P2 = P2o x2  ……………………………..(ii)
    P1 & P2 = Partial vapour pressure of solution.
    P1o & P2= Vapour pressure of the pure solvent 1 & 2 respectively.
    x1 & x2 = Mole Fraction of component 1 & 2 .
  • On mixing the two liquids & applying Dalton’s Law of partial pressure i.e. The total pressure over the solution phrase in the container will be the sum of the partial pressure of the components of the solution.(Dalton’s law)
    PTotal = P1 + P2PT = P1o x1 + P2o x2
    P= P1(1-x2) + P2o x2
     PT  = P1o – P1o x2 + P2o x2
     PT P1 + x(P2o P1o)

    Q = What is the minimum & maximum value of PTotal ?
    A = P1o is the  minimum value & P2o is the maximum value.

    Q = How can we say that Raoult’s Law is the special case of Henry’s Law ?
    A = The vapour pressure of the volatile components is given by Raoult’s Law i.e. P = Po x
    In solution of gas in liquid is given by Henry’s Law i.e. P = KH x .
    On comparing the both we can say that partial pressure of volatile liquid (Raoult’s Law ) or gas (Henry’s Law) is directly proportional to its mole fraction.
    The only difference is the Proportionality constant KH & Po .
    If one of the component is so volatile that it can exist as a gas then Raoult’s Law become a special case of Henry’s Law where Po = KH .


    When non-volatile, non- electrolyte solid solute  is added to liquid ,the vapour pressure of the solution decreases then the vapour pressure of the pure solvent.

  • Evaporation is the surface phenomenon .
  • In a pure Liquid the entire surface is occupies by the molecules of the liquid.
  • When a non- volatile solute is added then the surface has both solute & solvent .
  • Thereby, the fraction covered by solvent molecules get reduced as a result number of particles escaping from the surface reduced & Vapour pressure is also reduced.

    Q = What will be happen to vapour pressure if
    (i) 1 mol  of urea is added to 1 kg of water ?
    A = The vapour pressure will decrease .

    (ii) if 1 mol of NaCl is added to 1 Kg of water ?
    A = NaCl is electrolyte which breakdown to form 2 ions increasing its quantity . So there is more decrease in vapour pressure .


  • The solutions which obeys the Raoult’s Law over the entire range of concentration are known as Ideal Solution.
    (i) Δmix H = 0
    i.e. The enthalpy of  mixing of the pure component to form solution is 0. It is neither exothermic nor endothermic .
    (ii) Δmix V = 0
    i.e. There is no change in volume when 2 components are mixed to form solution.
    10mL + 10mL = 20mL
    Explanation :
  • The intermolecular forces b/w A-A  & B-B is similar to the intermolecular forces present b/w A-B  .
  • The enthalpy required to break A-A & B-B is equal to the enthalpy released when A-B is formed.
  • The solutions which do not obey the Raoult’s Law over the entire range of concentration are Non-Ideal Solutions.
    Types Of Non-Ideal Solutions :

    Solutions Showing POSITIVE DEVIATION Solutions Showing NEGATIVE DEVIATION
    1 The Vapour Pressure of such solutions are higher than predicted by Raoult’s Law. The Vapour Pressure of such solutions are lower than predicted by Raoult’s Law.
    2 Δmix H ≠ 0 Δmix H ≠ 0
    3 Δmix V ≠ 0 Δmix V ≠ 0
    4 Δmix H & Δmix V  both are Positive
    10mL +10mL = 22 mL
    Δmix H & Δmix V  both are Negative
    10mL +10mL = 19mL
    5 Δmix H is +ve , therefore, it  endothermic Δmix H is -ve ,therefore, it is exothermic
    6 Explanation : The bond b/w A-B is weaker than bond b/w A-A & B-B  Explanation : The bond b/w A-B is stronger than bond b/w A-A & B-B
    7  Examples :
    (i) Mixture of Ethanol and Acetone
    Explanation :
    The Ethanol molecule have intermolecular Hydrogen bonding & Acetone molecule have dipole-dipole interaction. When both are mixed
    the Hydrogen bond of Ethanol molecule are broken by some acetone molecule & thus the new force is formed b/w Ethanol & Acetone is weaker and thus more molecules will evaporate causing +ve deviation
    The energy required to break the bond b/w ethanol-ethanol & acetone-acetone is more than energy released during bond formation.
    .’. ΔH will be +ve.
    (ii) Carbon disulphide (CS2) and acetone
    Explanation :
    Carbon disulphide have dipole-dipole interaction & acetone also have dipole-dipole interaction. The new bond formed b/w carbon disulphide and acetone is weaker and thus show positive deviation.
    Examples :
    (i) When Phenol and Aniline are mixed
    Explanation :
    Among the phenol & Aniline molecules intermolecular hydrogen bonding is present. On mixing the both, there will be more stronger hydrogen bonding b/w phenolic proton and lone pair of Nitrogen atom on Aniline.
    (ii) Choroform and acetone
    Explanation :
    Both have dipole-diople interaction but on mixing they have intermolecular hydrogen bonding.


  • The binary Mixtures having same composition in Liquid & Vapour phrase.
  • Boils at constant Temperature.
  • They can’t be separated by Fractional distillation.
  • Such Solutions are known as Azeotropes.
    Types of Azeotropes:

    1 The solution showing Positive deviation from Raoult’s Law form Minimum Boiling Azeotropes at specific concentration The  solution showing Negative deviation from Raoult’s Law form Maximum Boling  Azeotropes at specific concentration.
    2 Example :
    Ethanol & water at 95 % (by volume of ethanol form this azeotropes)
    This ethanol water mixture can be obtained by the furmentation of sugar.This on fractional distillation give such solution.
    Example :
    Nitric acid and water at 68 % (by mass of Nitric acid ).They will boil at 393.5 K


  • The properties which depend upon the number of solute particles are called Colligative Property.
  • Four Colligative Property are :
    (i) Relative Lowering of Vapour Pressure of Solvent
    (ii) Elevation of Boiling Point of Solvent
    (iii) Depression in the Freezing Point of Solvent
    (iv) Osmosis & Osmotic Pressure of Solution 

    (i) Relative Lowering of Vapour Pressure 

    Q : Write the mathmatical expression of Relative Lowering of Vapour Pressure by the help of Raoult’s Law .
    A : According to the Raoult’s Law , the vapour pressure of the solvent in the solution is less than the pure solvent.
    The Lowering of vapour pressure depend upon only on the concentration solute particle & is independent of their of nature
    Mathematically ,
    P1 = P1ox1    {Raoult’s law} ………………… (i)
    P2 = P2ox2
    The Lowering of vapour pressure,
    ΔP = P10 – P1
    putting value of P1 from eqn (i)
    ΔP = P1o  – P1ox1
    ΔP P1o (1-x1)
    ΔP  = P1ox2
    x2 = ΔP /P1o
    x2  = P10 – P1 /P1o {Relative Lowering of Vapour Pressure }On simplifying x2,
    n2/(n1 + n2)  = P10 – P1 /P1oFor very dilute solution ,n2 is very less than n1 and ncan be ignored.
    .’. P1o  – P1 / P1o  = W2M2/W1M2
    W = Mass in gram
    M = Molar mass
    (ii) Elevation of Boiling Point
    Boiling point

  • The Temperature at which the vapour pressure of liquid becomes equal to the atmospheric pressure is known as Boiling point.
  • The Boiling point of solution is higher than the Boiling point of the pure solvent.
    Mathematically ,
    ΔTb = Tbo – Tb
    where ,
    ΔTb = 
    Elevation of Boiling Point
    Tbo  = Boiling Point of pure solvent
    b = Boiling Point of solution
    ΔT∝ m {molality}
    ΔTb = Km
    On simplifying,
    ΔTb = Kb n2/W1
    ΔTb = Kb W2/M2W1
    W2 = mass of solute
    M2 = molar mass of solute
    W1 = mass of solvent
    (iii) Depression in the Freezing Point of Solvent
  • The temperature at which the vapour pressure of the substance in its liquid phrase is equal to its vapour pressure in solid state is called its Freezing Point .Q : What would happen if non-volatile solid is added to the solvent ?
    A : When a non-volatile solid is added to the solvent, its vapour pressure decreases . As a result ,it would become equal to the vapour pressure of the solid solvent at low temperature and this is known as freezing point of the solvent.
    Mathematically ,
    ΔTf = Tfo – Tf
    Tfo freezing point of the pure solvent.
    Tf = freezing point of the pure solvent
    For dilute solution ,
    ΔTf  ∝ m
    ΔTf  = Kf m
    ΔT= Kf n2/W1
    ΔT= Kf W2 / M2 W1
    here, Kf = molal depression constant or cryoscopic constant(iv) Osmosis & Osmotic Pressure of Solution
  • The phenomenon in which the solvent molecules will flow through the semi permeable membrane from pure solvent to solution.
    Semi Permeable Membrane :
  • These membranes contains a network of sub microscpic holes or pores.
  • Small solvent molecules like water can pass through these holes but  passage of bigger molecules like solute hindered.
    EXAMPLE : 
  • Natural : Animal & vegatable orgin
  • Synthetic : Cellophane 

    Q : Why raw mangoes are kept in brine (salt water ) ?
    A : Raw mangoes shrivel when kept in brine .

    Q: What would happen if your blood cell containing 0.9 % concentration of is placed in
    (i) in solution less than 0.9 % concentration
    (ii) in solution more than 0.9 % concentration
    (iii) containing 0.9 % concentration
    A :
    (i)The water will flow from outside into the blood cell i.e. Endoosmosis.
    (ii) The water will flow from inside the blood cell to outer solution i.e. Exoosmosis .
    (iii) No osmosis take place .


  • When we place the cell in the solution more than 0.9% sodium Chloride solution, Exoosmosis take place & the cell will shrink .Such a solis called Hypertonic Solution.


  • When we place the cell in the solution less than 0.9% sodium Chloride solution,Endoosmosis take place & the cell will swell .Such a soln  is called Hypotonic Solution.Third case , where solution is having same concentration are separated by semi permeable membrane then no osmosis take place & solution is called Isotonic.OSMOTIC PRESSURE :

  • The pressure that just stop the flow of the solvent is called Osmotic Pressure .
  • It depends upon the concentration of the solution.
  • Mathmatically,
    Π = MRT
    = (n2/V) RT
    = (W2 /  M2) RT
    M = Molaity
    R = Universal gas constant = .083 bar L / mol K or 8.134 J/mol K
    Q : What is edema ?
    A : The swelling of the body organs is known as edema .
    REASON : People taking lots of salt or salty food experiences water retention in tissue cell and intercellular spaces because of osmosis.

    Q: How the prevention of meat & candid fruits done ?
    A : By salting meat is preserved & by adding sugar the fruits are preserved .
    REASON :Through the process of osmosis , a bacterium on  the salted meat or on the candid fruit lose water ,shrivelled and die .