NCERT Class 12 Chemistry Chapter 2 Notes - Download PDF

TYPES OF SOLUTION

Binary Solutions :

A solution which has only 2 components is known as a binary solution. i.e. solute + solvent.

CONCENTRATION OF SOLUTION

The concentration of a solution can be defined in the following ways:
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

MOLARITY (M)

It is a number of moles of solute present in 1 L of solution. Its unit is mol/L.

MOLALITY (m)

It is a number of moles of solute present in 1 Kg of solvent. Its unit is mol/Kg.

MOLE
The SI unit of the amount of substance is mol. It has a fixed Number (Avagadro Number).

No. of moles = Mass in grams/Molar mass

MOLE FRACTION
Mole Fraction of Solute = Moles of Solute / Total Moles
Mole Fraction of Solvent = Moles of Solvent / Total moles

COLLIGATIVE PROPERTY

• In a solution, the properties which depend upon the number of solute particles are known as Colligative properties.
• Four colligative properties are there in our syllabus:
  (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

According to Raoult’s Law, the vapour pressure of the solvent in the solution is less than the pure solvent.
The Lowering of vapour pressure depends only on the concentration of solute particles & is independent of their nature.

P₁ = P₁^ox₁ {Raoult’s law} ………………… (i)
P₂ = P₂^ox₂
The Lowering of vapour pressure,
ΔP = P₁⁰ – P₁
putting value of P₁ from eqⁿ (i)
ΔP = P₁^o – P₁^ox₁
ΔP ⁼ P₁^o (1-x₁)
ΔP = P₁^ox₂
x₂ = ΔP /P₁^o
x_{2 =} P₁⁰ – P₁ /P₁^o {Relative Lowering of Vapour Pressure }
n₂/(n₁ + n₂₎ = P₁⁰ – P₁ /P₁^oFor very dilute solution ,n₂ is very less than n₁ and n₂ can be ignored.
_.’. P₁^o _– P₁ / P₁^o = W₂M₂/W₁M₂
where,
W = Mass in gram
M = Molar mass

(ii) Elevation of Boiling Point
The Boiling point of a solution is higher than the Boiling point of the pure solvent.
Mathematically,
ΔT_b = T_b^o – T_b
where ,
ΔT_{b =} Elevation of Boiling Point
T_b^o = Boiling Point of pure solvent
T_{b =} Boiling Point of solution
also,
ΔT_b ∝ m {molality}
ΔT_{b =} K_b m
On simplifying,
ΔT_{b =} K_b n₂/W₁
ΔT_{b =} K_b W₂/M₂W₁
where,
W_{2 =} mass of solute
M_{2 =} molar mass of solute
W_{1 =} mass of solvent

(iii) Depression in the Freezing Point of Solvent

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 ,
ΔT_f = T_f^o – T_f
T_f^o₌ freezing point of the pure solvent.
T_{f =} freezing point of the pure solvent
For dilute solution ,
ΔT_f ∝ m
ΔT_f = K_f m
ΔT_f = K_f n₂/W₁
ΔT_f = K_f W₂ / M₂ W₁
Here, K_{f =} molal depression constant or cryoscopic constant

(iv) Osmosis & Osmotic Pressure of Solution
The phenomenon in which the solvent molecules flow through the semi-permeable membrane from pure solvent to solution is known as Osmosis.
Semi Permeable Membrane :

• These membranes contain a large network of sub-microscopic holes through which small solvent molecules like water can pass, but passage of bigger molecules like solute is hindered.

HYPERTONIC SOLUTION

• When the cell is placed in a solution having more than 0.9% sodium chloride solution, then exosmosis takes place & the cell shrinks. Such a solution is called a Hypertonic Solution.

HYPOTONIC SOLUTION

• When a cell is placed in a solution less than 0.9% sodium Chloride solution, endosmosis takes place & the cell swells. Such a solution is called a Hypotonic Solution.

OSMOTIC PRESSURE

The pressure that just stops the flow of the solvent is called Osmotic Pressure. It depends on the concentration of the solution.

Mathematically,
Π = MRT
= (n₂/V) RT
= (W₂ / M₂) RT
where,
M = Molality
R = Universal gas constant = .083 bar L/mol K or 8.134 J/mol K

Most Important Questions of Class 12 Chemistry Chapter 1

Question: $\mathrm{HA}(\mathrm{aq})\rightleftharpoons {\mathrm{H}}^{+}(\mathrm{aq})+{\mathrm{A}}^{-}(\mathrm{aq})$

The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is ${0.20}^{\circ }\mathrm{C}$. The dissociation constant for the acid is
Given :
${\mathrm{K}}_{\mathrm{f}}\left({\mathrm{H}}_2\mathrm{O}\right)=1.8\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{-1}$, molality $\equiv$ molarity

1) $1.38\times {10}^{-3}$

2) $1.1\times {10}^{-2}$

3) $1.90\times {10}^{-3}$

4) $1.89\times {10}^{-1}$

Answer:

$\begin{array}{rl}& \mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}}=\mathrm{i}{\mathrm{K}}_{\mathrm{f}}\mathrm{m}\\ & \mathrm{i}=\frac{\mathrm{\Delta }{\mathrm{T}}_f}{{\mathrm{K}}_{\mathrm{f}}\cdot \mathrm{m}}\\ & \mathrm{i}=\frac{0.20}{1.8\times 0.1}=1.11\\ & \mathrm{i}=1.11\\ & \alpha =\frac{i-1}{\mathrm{n}-1}(\text{ for }\mathrm{HA},\mathrm{n}=2)\\ & \alpha =\frac{1.11-1}{1}=0.11\\ & {\mathrm{K}}_{\mathrm{a}}=\frac{c{\alpha }^2}{1-\alpha }=\frac{0.1\times (0.11{)}^2}{1-0.11}=1.38\times {10}^{-3}\end{array}$

Hence, the correct answer is option (1).

Question: The percentage dissociation of a salt $\left({\mathrm{MX}}_3\right)$ solution at given temperature (van't Hoff factor $\mathrm{i}=$ 2) is__________% (Nearest integer)

Answer:

$\begin{array}{rl}& {\mathrm{MX}}_3\to {\mathrm{M}}^{+3}+3{\mathrm{X}}^{-}\\ & \mathrm{i}=1+(\mathrm{n}-1)\alpha \\ & \mathrm{i}=1+(4-1)\alpha =2\\ & \alpha =\frac{1}{3}=33.33\mathrm{\%}\approx 33\mathrm{\%}\end{array}$

Hence, the answer is 33%.

Approach to Solve Questions of Class 12 Chemistry Chapter 1 Solutions

The approach is crucial to attempt questions effectively. The following are the points that can help you build the strategy

1. Understand the types of solutions
The first step is to identify the solute and solvent and learn to classify the solution based on the states of components (solid in liquid, gas in liquid, etc.). Also, learn to apply the concentration formulas

Molarity (M), Molality (m), Mole fraction (χ), Mass %, Volume %

2. Henry's Law: $P=k_H\cdot x$ for gas solubility.
Raoult's Law- For ideal solutions, use $p_A=x_A\cdot p_A^0$.
Apply to vapor pressure problems.

3. Colligative Properties
Identify which property is involved ( $\mathrm{\Delta }\mathrm{Tf},\mathrm{\Delta }\mathrm{Tb},\pi ,\mathrm{\Delta }\mathrm{P}$ ).
Use-
Boiling Point Elevation: $\mathrm{\Delta }{T}_b=i\cdot K_b\cdot m$
Freezing Point Depression: $\mathrm{\Delta }{T}_f=i\cdot K_f\cdot m$
- Osmotic Pressure: $\pi =i\cdot C\cdot R\cdot T$

4. Van’t Hoff Factor (i)
Learn to determine if solute dissociates or associates.
Adjust formulas using ? for non-ideal cases.

5. Numerical Practice

While solving the numericals, keep units consistent and understand what each symbol represents. You can also draw diagrams for clarity (e.g., colligative property plots). Also focus on ideal vs non-ideal solutions, positive/negative deviations, and azeotropes. Students can also access the NCERT exemplar solutions for better understanding.

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