Class-12th, Chemistry

CH-1 Solution

Solution: A solution is homogeneous mixture composed of two or more substance/component.. E.g-Seawater, Lemonade, fe+Mg

component

solute

Solvent

Solution = Solute+solvent. Type of Solution

Solute

Solvent

Gaseous Solutions

Gas

Cias

Common Examples

Mixture of oxygen and nitrogen gases

Liquid

Gas

Solid

Gas

Chloroform mixed with nitrogen gas

Camphor in nitrogen gas

Liquid Solutions

Gas

Liquid

Oxygen dissolved in water

Liquid

Liquid

Ethanol dissolved in water

Solid

Liquid

Glucose dissolved in water

Solid Solutions

Cias

Solid

Solution of hydrogen in palladium

Liquid

Solid

Amalgam of mercury with sodium

Solid

Solid

Copper dissolved in gold1.2 Classification

Solutions which contain two components in it are called Binary Solutions.

Substances which are used to prepare a solution are called as Components.

The component that is present in the largest quantity is known as Solvent. Solvent determines the physical state in which solution exists.

The other component present in lesser quantity in the solution is termed as Solute.

Each component may be solid, liquid or in gaseous stateImethods of expressing strength of solution/ Strengthor concentration

It is the amount of solute dissolved in per unit Solution or solvent is called strength of solution. Strength unit = grams or Litre. 1. Mass Percentage (mass/mass%or m/m/% or w/w%): It may be defined as the number of parts of mass of solute per hundred parts by mass of solution

mass %of solute = mass of solute /Total mass of son X100

% by mass w /wt. of solute/ wt. of solution olute x 100

X % by mass means 100gm solution contains xgm solute and (100-x) gm solvent ] 2. volume Percentage (%v/v)= It represents volume of a component in 100mL of solution.

volume %of solute = volume of solute / Total volume of soln×100

  1. MASS by volume percentage (% w/v): It represents mass of solute in grams present in 100mL of solution.

mass by volume% = Mass of solute in gram/volume of solution in mlx 100

  1. parts per million (PPM):-

The amt. or mass of solute present in millionth part in the mass of soln.

PPM = Mass of solute/Total mass of solution x 10⁶ .

  1. Mole per kg concentration (molality,m)÷ It is the number of moles of solute per kilogram of solvent.

molality (m) = moles of solute/ mass of solvent in kg

unit of molality = mol/kg or molal. 6. Molarity (M)÷ It represents motes of solute present in IL of solution

Molarity (M) = Moles of solute /volume of soln in L .. 7.Mole Fraction (x)

“It represents the moles of a solute present in one mole of solution”

Mole fraction = No. of moles of the component/ Total no. of moles all the components

For example, in a binary mixture, if the number of moles of A and B are n_{A} and n_{B} respectively, the mole fraction of A will be

x_{A}=\frac{n_{A}}{n_{A}+n_{B}} . The sum of all mote fraction in a solution always equal one. 8. Normality(N)

It represents no. of equivalents of solute present in 1L of solution.

Normality(N) = No. of equivalents of solute/ volume of solution in L

Equivalents moles = Given mass /equivalent mass

Equivalent mass = molar mass /n-factor

Some important!

Normality = z x molarity

mass%, ppm, mole fraction and molality are independent of temperature, whereas molarity and normality are a function of temperature. This is because volume depends on temperature and the mass does not. . VAPOUR PRESSURE

2.1 Definition

Vapour pressure of a liquid/solution is the pressure exerted by the vapours in equilibrium with the liquid/solution at a particular temperature.

Vapour pressure directly proportional escaping tendency

2.2 Vapour pressure of liquid solutions and Raoult’s Law:

(Raoult’s law for volatile solutes)

Raoult’s 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.

Consider a solution containing two volatile components land 2 with mole fractions x, and x, respectively. Suppose. at a particular temperature, their partial vapour pressures are P_{1} and p_{2} and the vapour pressure in pure state are p_{1}^{\circ} p_{2}^{0}.

Thus, according to Raoult’s Law, for component 1 and p_{1}=p_{1}^{0}x_{1} Similarly, for component 2 p_{2}=p_{2}^{0}x_{2} P_{1}\times X_{1}

According to Dalton’s law of partial pressure, the total pressure (p_{yotal}) over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as:

Substituting the values of p, and P_{2} we get p_{ustal}=x_{1}p_{1}^{0}+x_{2}p_{2}^{0} =p_{1}^{a}+(p_{2}^{0}-p_{1}^{0})x_{2} p_{total}=p_{1}+p_{2} =(1-x_{2})p_{1}^{0}+x_{2}p_{2}^{0} . 2.4 Ideal and Non-ideal solutions

Ideal solutions:

An ideal solution is the solution in which Each component obeys Raoult’s law under all condition of temperatures and concentration.

properties of Ideal solution:

(i) These solution obey Raoult’s law

(ii) AHmixing = 0 (enthalpy change in mixing)

(iii) Vmixing = 0 (no change in volume upon mixing solute &solvent)

(iv) no heat is absorbed or realeased during dissolution

(v) They do not form azetrops

(vi) No. Positive or negative deviation are seen.

Non-ideal solution:

When a solution does not obey Raault’s law over the Entire range of cento concentration, then it is called non-ideal solution

properties of Non-ideal slution

(i) These solution do not follow Raoult’s law

(ii) Hmixing #0

(iii) AVmixing ≠0

(iv) Heat is absorbed or released during dissolution ΔΗ#Ο

(v) They form azeotropes

(vi) positive or negative deviation are seen.
Solubility
(Solubility of a solid in liquid)

Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent.

Factors affecting the solubility of a solid in liquid:

  1. Nature of solute and solvent:

Like dissolves like for example, while sodium Chloride and sugar dissolve readily in water, naphthalene and anthracene do not on the other hand naphthalene and anthracene dissolve readily in benzene but sodium chloride and sugar do not ..

  1. Temperature:

In a nearly saturated solution If (Δsol. H>0), the solubility increase with rise in temperature and If (Δsol.H<0) the solubility decrease with rise in temperature.

Effect of pressure

Does not have any significant effect as solids and liquids are highly incompressible. 3.2 Henry’s law

Henry’s law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.

The most commonly used form of Henry’s law states that “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution”. This is expressed as:

P ∝ C (or) P = kH.x

Where,

‘P’ denotes the partial pressure of the gas in the atmosphere above the liquid.
X” denotes the concentration of the dissolved gas.
‘kH’ is the Henry’s law constant of the ga

Characteristics of KH

K_{H} is a function of the nature of the gas.

Higher the value of K_{H} at a given pressure, the lower is the solubility of the gas the liquid.

K_{H} values increase with increase of temperature indicating that the solubility of gases increases with decrease of temperature. Applications of Henry’s law

  1. In the production of carbonated beverages.
  2. In the deep sea diving.
  3. For climbers or people at high altitudes.

Raoult’s Law as a special case of Henry’s Law

According to Raoult’s law, p_{i}=x_{i}p_{i}^{0}

In the solution of a gas in a liquid, one of the components is so volatile that it exists as a gas. Its solubility according to Henry’s law,

p=K_{H}x.

Thus, Raoult’s law becomes a special case of Henry’s law in which K_{H} becomes equal to p_{i}^{0} .

colligative properties

Colligative properties are those properties Which depends upon the no. of particles or no. of moles of solute irrespectively.

There are four colligative properties:

  1. Relative Lowering of vapour pressure
  2. Elevation in Boiling Point
  3. Depression in Freezing point
  4. Osmotic pressure
  5. Relative Lowering of vapour pressure

After adding the solute, the vapour pressure of the solution is found to be lower than that of the pure liquid at a given temperature. It is called Relative lowering of vapour pressure.

According to Raoult’s Law :

p_{t}=x_{1}p_{1}^{0}

The reduction in the vapour pressure of solvent (\Delta p_{1}) is given as:

\Delta p_{1}=p_{1}^{0}-p_{1}=p_{1}^{0}-p_{1}^{0}x_{1}

=p_{1}^{0}(1-x_{1})

Knowing that x_{2}=1-x_{y} equation reduces to \Delta p_{t}=x_{2}p_{t}^{0} Equation can be wirtten as \frac{\Delta p_{1}}{p_{1}^{0}}=\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=x_{2}

The expression on the left hand side of the equation as mentioned earlier is called relative lowring of vapour pressure and is equal to the mole fraction of the solute. The above equation can be written as: \frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}&\frac{n_{2}}{n_{1}+n_{2}}(since~x_{2}\frac{n_{2}}{n_{1}+n_{2}}) Here n_{1} and n_{2} are the number of moles of solvent and solute respectively present in the solution. For dilute solutions n_{2}<<n_{1} hence neglecting n_{2} in the denominator we have \frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{n_{2}}{n_{1}} \frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{w_{2}\times M_{1}}{M_{2}\times w_{1}} or

Here w_{1} and W_{2} are the masses and M_{1} and M_{2} are the molar masses of the solvent and solute respectively. (2)Elevation in Boiling Point

Boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure.

On addition of non-volatile solute the vapour pressure of the solvent decreases and therefore, to boil the solution the required temperature will be higher. So, there will be a rise in the boiling point of the solution.

The increase in boiling point \Delta T_{b}=T_{b}-T_{b}^{0} where T_{b}^{0} is the boiling point of pure solvent and \Gamma_{b} is the boiling point of solution is known as elevation of boiling point.

Expression:

\Delta T_{b}=K_{b}m

K_{b} is called Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic Constant). Calculation of molar mass of solute:

m= W/M w/1000 1000xw, Mxw

Substituting the value of molality in equation we get

\Delta T_{b}=\frac{K_{b}\times1000\times w_{2}}{M_{2}\times w_{i}}

M_{z}=\frac{1000\times W_{z}\times K_{b}}{\Delta T_{b}\times W_{1}}

K_{b}

It is defined as the elevation in boiling point when the molality of the solution is unity.

The unit of K_{b} is K kg mol

Determination of K_{b}

K_{b}=\frac{R\times M_{1}\times T_{b}^{3}}{1000\times\Delta_{vup}}

where: R= gas constant (8.314 JK/mol),

T_{f}= freezing temperature in K,

M_{1}= Molar mass of solvent in Kg/mol,

\Lambda_{vap}H= enthalpy of vapourisation of solvent in J/mol.

The vapour pressure curve for solution lies below the curve for pure water. The diagram shows that \Delta T_{b} denotes the elevation of boiling point of a solvent in solution. (3) Depression in freezing point

The freezing point of a substance may be defined as the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase.

When a non-volatile solid is added to the solvent its vapour pressure decreases and now it would become equal to that of solid solvent at lower temperature. Thus, the freezing point of the solvent decreases.

\Delta T_{t}=T_{t}^{0}-T_{t} where T_{T}^{0} is the freezing point of pure solvent and T_{t} is its freezing point when non-volatile solute is dissolved is known as depression in freezing point.

Expression:

\Delta T_{f}=K_{f}m K_{r} is known as Freezing Point Depression Constant or Molal Depression Constant or Cryoscopic Constant.

Calculation of molar mass of solute:

m=\frac{w_{2}/M_{2}}{w_{1}/1000} Substituting this value of molality in equation we get: \Delta T_{r}=\frac{K_{r}\times w_{2}/M_{2}}{w_{1}/1000} \Delta T_{z}=\frac{K_{1}\times w_{2}\times1000}{M_{2}\times w_{1}} M_{2}=\frac{K_{f}\times w_{3}\times1000}{\Delta T_{f}\times w_{1}}

K_{r}

It is defined as the depression in freezing point when the molality of the solution is unity. The unit of K_{r} is K kg mol¹.

Determination of K_{r}

K_{t}=\frac{R\times M_{t}\times T_{t}^{2}}{1000\times\Delta_{fot}} where: R= gas constant (8.314 JK/mol), T_{t}= freezing temperature in K,M_{1}= Molar mass of solvent in Kg/mol,

\Delta_{fins}H= enthalpy of fusion of solvent in J/kg. (4)Osmosis

When a pure solvent and solution are kept with a semi- permeable membrane between them then the solvent particles pass through the membrane from the solvent side to the solution side. This phenomenon is called “Osmosis”.

The semi-permeable membrane is a membrane that allows only small molecules to pass through and blocks the larger solute molecules.

Osmosis

O Water

O Sugar

Selectively Permeable Membrane

ㅇㅇ

Low Sugar Concentration

High Sugar Concentration

High Water Concentration

Osmotic pressure:

Low Water Concentration

The osmotic pressure of a solution is the excess pressure that must be applied to a solution to prevent osmosis, i.e., to stop the passage of solvent molecules through a semipermeable membrane into the solution. osmotic pressure

  • osmotic pressure is denoted by I
  • mathematically

\Pi=CRT

\pi= osmotic pressure

c= concentration of solution

r= Gas constant

T= Temperature

osmotic pressurelr) – Depend upon two factor

(ⅰ) ra concentration(molarity)

(ii) ra Temperature

T↑ solvent molecule K.E↑ motion ↑ osmosist osmotic ↑ Pressure.

π=RCT Here R=gas constant. Concentration=no.of moles of solute / volume

C=\frac{n_{B}}{V} \Pi=CRT

\Pi=\frac{n_{B}}{V}RT

\Pi V=n_{B}RT

\pi V=\cap_{B}RT

We known that n_{B}=\frac{WB}{MB} \pi V=\frac{WB}{N1B}RT MB=\frac{WB\times R\times T. Isotonic solutions:

Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.

The solution with lower concentration or lower osmotic pressure is known as “Hypotonic” with respect to more concentrated solution.

The solution with higher concentration or higher osmotic pressure is known as “Hypertonic” with respect to dilute solution. Reverse osmosis:

If a pressure larger than the osmotic pressure is applied to the solution side, the solvent will flow from the solution into the pure solvent through the semi permeable membrane. This phenomenon is called reverse osmosis.

Application:

Desalination of sea water:

When pressure more than osmotic pressure is applied, pure water is squeezed out of the sea water through the membrane.

4.5 Abnormal Molar Masses

When the molecular mass of a substance determined by studying any of the colligative properties comes out to be different than the theoretically expected value, the substance is said to show abnormal molar mass.

Abnormal Molar Masses are observed:

  1. When the solute undergoes association in the solution.
  2. When the solute undergoes dissociation in the solution.

van’t Hoff Factor:

To calculate extent of association or dissociation, van’t Hoff introduced a factor i, known as the van’t Hoff Factor.

i= Normal molar mass Abnormal molar mass

i= Observed colligative property Calculated colligative property

Total no. of moles of particles after association (dissociation) = No. of moles of particles before association (dissociation).

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