CLASS 12 CHEMISTRY

Chapter 1: Solutions

Comprehensive Lecture Slides � Master Formulas & Numerical Problems

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1. Introduction and Definitions

What is a Solution?

A solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits.

Key Property Homogeneous Mixture: Its composition and physical properties are completely uniform throughout the entire volume of the mixture.

Components of a Binary Solution

Solvent: The component present in the largest quantity. It determines the final physical state of the solution.
Solute: The component(s) present in a lesser quantity, which are dissolved inside the solvent.

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Types of Solutions Matrix

Class	Solute	Solvent	Examples
Gaseous	Gas	Gas	Air
	Liquid	Gas	Chloroform in $N_2$
	Solid	Gas	Camphor in $N_2$
Liquid	Gas	Liquid	Soda Water ($CO_2$)
	Liquid	Liquid	Ethanol in water
	Solid	Liquid	Sugar in water
Solid	Gas	Solid	$H_2$ in Palladium
	Liquid	Solid	Hg in Na (Amalgam)
	Solid	Solid	Copper in Gold

Practice Problem 1 Binary Solution Identification

In an amalgam of mercury with sodium, which substance is the solute and which is the solvent? What type of solution is this?

Since the final state of the amalgam is solid, the solid metal (Sodium) is the solvent, and the liquid metal (Mercury) is the solute.
Type: Liquid in Solid (Solid Solution).

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2. Expressing Concentration (Basics)

Mass & Volume Percentage

Mass Percentage (w/w): Mass of solute per 100 grams of solution.
$$ \text{Mass \%} = \frac{\text{Mass of Solute } (w_2)}{\text{Total Mass of Solution } (w_1 + w_2)} \times 100 $$
Volume Percentage (v/v): Volume of solute per 100 parts by volume of solution.
$$ \text{Volume \%} = \frac{\text{Volume of Solute } (V_2)}{\text{Total Volume of Solution }} \times 100 $$
Mass by Volume Percentage (w/v): Mass of solute dissolved in 100 mL of the solution.
Parts Per Million (ppm): Used when a solute is present in trace quantities.
$$ \text{ppm} = \frac{\text{Parts of Component}}{\text{Total Parts of All Components}} \times 10^6 $$

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Molarity, Molality & Mole Fraction

1. Molarity ($M$)

Number of moles of solute dissolved in exactly 1 Litre of solution.

$$ M = \frac{w_2 \times 1000}{M_2 \times V(\text{in mL})} \quad [\text{mol/L}] $$

2. Molality ($m$)

Number of moles of solute dissolved in exactly 1 kg ($1000\text{ g}$) of the solvent.

$$ m = \frac{w_2 \times 1000}{M_2 \times W_1(\text{in grams})} \quad [\text{mol/kg}] $$

3. Mole Fraction ($x$) For binary mixture of component 1 and 2: $$ x_2 = \frac{n_2}{n_1 + n_2} \quad \text{and} \quad x_1 = \frac{n_1}{n_1 + n_2} $$ Sum is always 1 ($x_1 + x_2 = 1$). Dimensionless.

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Temperature Effects & Conversion

Temperature Dependence

Crucial Concept

Molarity ($M$), Volume %, w/v % depend on the volume of the solution. Since volume expands/contracts with temperature, these values change with temperature.
Molality ($m$), Mass %, Mole Fraction ($x$) depend strictly on the mass of the components. Mass is invariant, making these values **independent of temperature**. Preferred in precise calculations!

Molarity to Molality Interconversion

Using density ($d$) in $\text{g/mL}$ and solute molar mass ($M_2$) in $\text{g/mol}$:

$$ m = \frac{1000 \times M}{(1000 \times d) - (M \times M_2)} $$

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Practice: Concentration Numericals

Practice Problem 2 Mole Fraction Calculation

Calculate the mole fraction of ethylene glycol ($\text{C}_2\text{H}_6\text{O}_2$) in a solution containing $20\%$ of $\text{C}_2\text{H}_6\text{O}_2$ by mass in water.

Assume $100\text{ g}$ of solution: Mass of glycol $= 20\text{ g}$, Mass of water $= 80\text{ g}$.
Molar Mass: Glycol $= 62\text{ g/mol}$, Water $= 18\text{ g/mol}$.
Moles of Glycol ($n_2$) $= 20 / 62 = 0.322\text{ mol}$. Moles of Water ($n_1$) $= 80 / 18 = 4.444\text{ mol}$.
Total Moles $= 0.322 + 4.444 = 4.766\text{ mol}$.
Mole fraction of glycol ($x_2$) $= 0.322 / 4.766 = \mathbf{0.068}$.

Practice Problem 3 Molality from Molarity & Density

The density of $3\text{ M}$ solution of $\text{NaCl}$ is $1.25\text{ g mL}^{-1}$. Calculate the molality of the solution.

$3\text{ M}$ means $3\text{ moles}$ of $\text{NaCl}$ in $1\text{ L}$ ($1000\text{ mL}$) of solution.
Mass of $1\text{ L}$ solution $= 1000 \times 1.25 = 1250\text{ g}$.
Mass of $\text{NaCl}$ ($w_2$) $= 3 \text{ mol} \times 58.5 \text{ g/mol} = 175.5\text{ g}$.
Mass of Solvent ($w_1$) $= 1250 - 175.5 = 1074.5\text{ g} = 1.0745\text{ kg}$.
Molality ($m$) $= \text{moles} / \text{solvent kg} = 3 / 1.0745 = \mathbf{2.79\text{ m}}$.

Practice Problem 11 Mass Percentage

A solution contains $10\text{ g}$ glucose in $90\text{ g}$ water. Find mass percentage of glucose.

Total mass $= 10 + 90 = 100\text{ g}$.
$w/w\% = (10 / 100) \times 100 = \mathbf{10\%}$.

Practice Problem 12 Molarity Calculation

Calculate the molarity of a solution containing $4.9\text{ g}$ of $\text{H}_2\text{SO}_4$ in $250\text{ mL}$ of solution.

1. Molar mass of $\text{H}_2\text{SO}_4$ ($M_2$) $= (2 \times 1) + 32 + (4 \times 16) = 98\text{ g/mol}$.
2. Moles of solute ($n_2$) $= 4.9 / 98 = 0.05\text{ mol}$.
3. Volume of solution ($V$) $= 250\text{ mL} = 0.25\text{ L}$.
4. Molarity ($M$) $= \text{moles} / \text{volume in L} = 0.05 / 0.25 = \mathbf{0.20\text{ M}}$.

Practice Problem 13 Molality Calculation

Find molality of a solution prepared by dissolving $9.2\text{ g}$ ethanol ($M=46$) in $200\text{ g}$ water.

Moles of ethanol $= 9.2 / 46 = 0.2\text{ mol}$.
Mass of solvent $= 200\text{ g} = 0.200\text{ kg}$.
Molality ($m$) $= 0.2 / 0.200 = \mathbf{1.0\text{ m}}$.

Practice Problem 14 Temperature effect True/False

State whether each statement is True/False: (i) Molarity decreases on heating. (ii) Molality decreases on heating.

(i) **True** (volume increases with temperature expansion, so $M$ decreases).
(ii) **False** (mass-based quantity, entirely independent of temperature).

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3. Solubility of Solids in Liquids

Core Principles

Solubility: The maximum amount of solute that can be dissolved in a specified amount of solvent at a specific temperature to form a saturated solution.
Like Dissolves Like: Polar solutes (like $\text{NaCl}$) dissolve in polar solvents (like water). Non-polar solutes (like naphthalene) dissolve in non-polar solvents (like benzene).
Effect of Temperature (Le Chatelier's Principle):
- If dissolution is **Endothermic** ($\Delta H > 0$): Solubility **increases** as temperature rises.
- If dissolution is **Exothermic** ($\Delta H < 0$): Solubility **decreases** as temperature rises.
Effect of Pressure: Pressure has **practically no effect** on solid solubility in liquids because solids and liquids are highly incompressible.

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Solubility of Gases: Henry's Law

Henry's Law Statement

"At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of the liquid."

$$ p = K_H \cdot x $$

Where $p$ is partial pressure of gas in vapour phase, $x$ is mole fraction of gas dissolved, and $K_H$ is **Henry's Law Constant**.

Key Conceptual Deductions

At a given pressure, a higher value of $K_H$ implies **lower solubility** ($x$) of the gas in the liquid.
$K_H$ increases with temperature. Thus, **solubility of gases in liquids decreases with rising temperature**.

Conceptual Highlight Aquatic Life Comfort: Aquatic species are significantly more comfortable in cold water than in warm water because cold water holds more dissolved oxygen (due to lower $K_H$ at low temperatures).

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Applications of Henry's Law

1. Carbonated Beverages

To increase the solubility of $CO_2$ gas in carbonated soft drinks, sodas, and champagne, the bottles are sealed tightly under **extremely high pressure**.

2. Scuba Diving & The Bends

At high underwater pressures, atmospheric gases dissolve deeply in a diver's blood. If they ascend rapidly, pressure drops, and dissolved nitrogen forms painful, dangerous bubbles in the bloodstream, blocking capillaries (**"The Bends"**).

Prevention: Breathing tanks are diluted with Helium ($11.7\% \text{ He}$, $56.2\% \text{ N}_2$, $32.1\% \text{ O}_2$) because Helium has exceptionally low solubility in blood.

3. High Altitudes & Anoxia

At high altitudes, the partial pressure of oxygen is much lower than at sea level. This leads to low concentrations of oxygen in the tissues and blood of climbers, causing physical weakness and cognitive impairment�a condition known as **Anoxia**.

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Practice: Solubility & Henry's Law

Practice Problem 4 Henry's Law Calculations

If $N_2$ gas is bubbled through water at $293\text{ K}$, how many millimoles of $N_2$ gas would dissolve in $1\text{ Litre}$ of water? Assume $N_2$ exerts a partial pressure of $0.987\text{ bar}$. Given $K_H$ for $N_2$ at $293\text{ K}$ is $76.48\text{ kbar}$.

1. Convert $K_H$ to bar: $K_H = 76.48\text{ kbar} = 76,480\text{ bar}$.
2. Find mole fraction ($x$): $x = p / K_H = 0.987 / 76480 = 1.29 \times 10^{-5}$.
3. Water in $1\text{ L} = 1000\text{ g} / 18\text{ g/mol} = 55.5\text{ moles}$.
4. Since $n_{N_2} \ll n_{\text{water}}$, $x \approx n_{N_2} / n_{\text{water}} \implies 1.29 \times 10^{-5} = n_{N_2} / 55.5$.
5. $n_{N_2} = 1.29 \times 10^{-5} \times 55.5 = 7.16 \times 10^{-4}\text{ moles}$.
6. Millimoles $= 7.16 \times 10^{-4} \times 1000 = \mathbf{0.716\text{ mmol}}$.

Practice Problem 15 Temperature & Solubility Relationship

At constant pressure, if the Henry constant ($K_H$) for a gas doubles with temperature, what happens to its solubility (qualitatively)?

From Henry's Law: $x = p / K_H$.
At constant partial pressure ($p$), if $K_H$ doubles, the mole fraction ($x$) is halved.
Therefore, the solubility of the gas **decreases qualitatively by half**.

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4. Vapour Pressure of Liquid Solutions

Vapour Pressure

The pressure exerted by vapours in thermodynamic equilibrium with its liquid phase at a constant temperature in a closed container.

Raoult's Law (Volatile Liquid-Liquid)

"For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction present in the solution."

$$ p_1 = p_1^0 x_1 \quad \text{and} \quad p_2 = p_2^0 x_2 $$

Dalton's Law of Partial Pressures

The total pressure ($P_{total}$) over the solution is the sum of the partial pressures:

$$ P_{total} = p_1 + p_2 = p_1^0 x_1 + p_2^0 x_2 $$ $$ P_{total} = p_1^0 + (p_2^0 - p_1^0)x_2 $$

Mole fraction in vapour phase: $ y_1 = p_1 / P_{total} $ and $ y_2 = p_2 / P_{total} $.

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Vapour Pressure Lowering by Solid Solutes

Adding Non-Volatile Solutes

When a **non-volatile solute** (like sugar, salt, or urea) is added to a pure volatile solvent, the vapour pressure of the resulting solution is **always lower** than that of the pure solvent.

Molecular Explanation In a pure solvent, $100\%$ of the liquid surface is occupied by escape-ready solvent molecules. In a solution, solute particles occupy some of these surface sites. This decreases the surface mole fraction of the solvent, reducing the rate of evaporation and lowering the equilibrium vapour pressure.

Key Takeaway

Vapour Pressure: Pure Solvent > Solution

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Raoult's Law vs Henry's Law

Raoult's Law is a Special Case of Henry's Law

Raoult's Law Formula For a volatile component in solution:

$$ p_i = p_i^0 \cdot x_i $$

Henry's Law Formula For a dissolved gas in liquid:

$$ p = K_H \cdot x $$

Both equations state that the partial pressure of the volatile component is proportional to its mole fraction in the solution. They differ only in the proportionality constant. **Raoult's law is a special case of Henry's law where $K_H = p_i^0$.**

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5. Ideal and Non-Ideal Solutions

Ideal Solutions

Obey Raoult's law precisely across the entire concentration range.

$$ \Delta H_{mix} = 0 \quad \text{and} \quad \Delta V_{mix} = 0 $$ Molecular Cause: A-B forces are identical to A-A and B-B forces.

Examples: Benzene+Toluene; n-Hexane+n-Heptane.

Non-Ideal Solutions

Do not obey Raoult's law. Vapour pressure is either higher or lower.

$$ \Delta H_{mix} \neq 0 \quad \text{and} \quad \Delta V_{mix} \neq 0 $$ Molecular Cause: A-B forces differ from A-A and B-B forces.

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Non-Ideal: Positive Deviation

Core Concept

The total vapour pressure of the mixture is **higher** than expected from Raoult's law calculations.

Thermodynamic Properties $$ \Delta H_{mix} > 0 \quad \text{(Endothermic mixing - Heat absorbed)} $$ $$ \Delta V_{mix} > 0 \quad \text{(Volume expansion - Expansion)} $$

Molecular Cause

The new solute-solvent (A-B) interactions are **weaker** than the original pure solute (B-B) or solvent (A-A) interactions.
Ethanol + Acetone: Acetone molecules break the extensive hydrogen bonding network of pure ethanol, increasing escaping tendency.

Examples: Ethanol + Acetone; $CS_2$ + Acetone.

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Non-Ideal: Negative Deviation

Core Concept

The total vapour pressure of the mixture is **lower** than expected from Raoult's law calculations.

Thermodynamic Properties $$ \Delta H_{mix} < 0 \quad \text{(Exothermic mixing - Heat released)} $$ $$ \Delta V_{mix} < 0 \quad \text{(Volume contraction - Contraction)} $$

Molecular Cause

The new solute-solvent (A-B) interactions are **stronger** than the original pure solute (B-B) or solvent (A-A) interactions.
Chloroform + Acetone: Mixing triggers the formation of a highly stable **hydrogen bond** between the two components, reducing evaporation.

Examples: Chloroform + Acetone; Phenol + Aniline; $HNO_3$ + $H_2O$.

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Azeotropes (Constant Boiling)

What is an Azeotrope?

A liquid mixture which has the **exact same composition** in both the liquid phase and the vapour phase at equilibrium. They boil at a single, constant temperature, behaving like a pure liquid. They cannot be separated by fractional distillation.

Types of Azeotropes

Minimum Boiling Azeotrope: Formed by non-ideal solutions showing large *positive* deviation. Vapour pressure is exceptionally high, so the boiling point is lower than either component.
Example: $95.5\%$ Ethanol + $4.5\%$ Water by volume.
Maximum Boiling Azeotrope: Formed by non-ideal solutions showing large *negative* deviation. Vapour pressure is low, so the boiling point is higher than either component.
Example: $68\%$ Nitric Acid + $32\%$ Water by mass.

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Practice: Raoult's Law & Deviations

Practice Problem 5 Raoult's Law Calculations

Vapour pressures of pure chloroform ($\text{CHCl}_3$) and dichloromethane ($\text{CH}_2\text{Cl}_2$) at $298\text{ K}$ are $200\text{ mm Hg}$ and $415\text{ mm Hg}$ respectively. Calculate the vapour pressure of the solution prepared by mixing $25.5\text{ g}$ of $\text{CHCl}_3$ and $40\text{ g}$ of $\text{CH}_2\text{Cl}_2$ at $298\text{ K}$.

Moles: $\text{CHCl}_3 = 25.5 / 119.5 = 0.213\text{ mol}$. $\text{CH}_2\text{Cl}_2 = 40 / 85 = 0.470\text{ mol}$.
Total Moles $= 0.213 + 0.470 = 0.683\text{ mol}$.
Mole Fractions: $x_2(\text{CH}_2\text{Cl}_2) = 0.470 / 0.683 = 0.688$, $x_1(\text{CHCl}_3) = 1 - 0.688 = 0.312$.
Total Pressure ($P_{total}$) $= p_1^0 x_1 + p_2^0 x_2 = (200 \times 0.312) + (415 \times 0.688) = 62.4 + 285.5 = \mathbf{347.9\text{ mm Hg}}$.

Practice Problem 6 Volume Changes & Deviation

When 50 mL of liquid A and 50 mL of liquid B are mixed, the volume of the resulting solution is found to be 99 mL. What type of deviation from Raoult's law does this solution show?

Expected volume $= 50 + 50 = 100\text{ mL}$. Actual volume $= 99\text{ mL}$.
Change in volume on mixing: $\Delta V_{mix} = 99 - 100 = \mathbf{-1\text{ mL}}$.
Since $\Delta V_{mix} < 0$, the molecules are drawn closer together due to stronger A-B interactions compared to pure A-A and B-B interactions.
Therefore, this solution shows a **Negative Deviation** from Raoult's law.

Practice Problem 16 Binary Ideal Solution Pressure

For a binary ideal solution at $300\text{ K}$, $p_A^0=300\text{ mmHg}$, $p_B^0=150\text{ mmHg}$ and $x_A=0.40$. Find total vapour pressure.

Since it is binary: $x_B = 1 - x_A = 1 - 0.40 = 0.60$.
Total pressure ($P_{total}$) $= p_A^0 x_A + p_B^0 x_B = 300(0.40) + 150(0.60) = 120 + 90 = \mathbf{210\text{ mmHg}}$.

Practice Problem 17 Deviation Prediction

A solution shows $\Delta H_{mix}<0$ and $\Delta V_{mix}<0$. Predict deviation from Raoult's law.

Since both parameters are negative (exothermic mixing and volume contraction), it means A-B molecular interactions are stronger than pure ones.
The solution shows a **Negative Deviation** from Raoult's law.

Practice Problem 18 Azeotropic Distillation

Why can ethanol-water mixture not be fully separated by fractional distillation?

Because at $95.5\%$ ethanol concentration, it forms a **minimum boiling azeotrope**. At this point, the composition of both liquid and vapor phases becomes identical, meaning they boil together at a constant temperature.

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6. Colligative Properties

What are Colligative Properties?

"Properties of dilute solutions that depend strictly and solely on the **number of solute particles** (ions or molecules) present in the solution, and are completely **independent of their chemical identity or nature**."

1. RLVP: Relative Lowering of Vapour Pressure

2. Elevation of BP: Elevation of Boiling Point ($\Delta T_b$)

3. Depression of FP: Depression of Freezing Point ($\Delta T_f$)

4. Osmotic Pressure: Osmotic Pressure ($\pi$)

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Relative Lowering of Vapour Pressure

RLVP Formula Adding a non-volatile solute lowers vapour pressure. The relative lowering of vapour pressure is equal to the mole fraction of the solute:

$$ \frac{p_1^0 - p_1}{p_1^0} = x_2 = \frac{n_2}{n_1 + n_2} $$

Where $p_1^0$ is pure solvent vapour pressure, $p_1$ is solution pressure, and $x_2$ is solute mole fraction.

Dilute Solutions Simplification

For highly dilute solutions, moles of solute ($n_2$) are negligible compared to moles of solvent ($n_1$). Thus, $n_1 + n_2 \approx n_1$.

Molar Mass Determination We can calculate the molar mass of the solute ($M_2$) directly: $$ \frac{p_1^0 - p_1}{p_1^0} \approx \frac{n_2}{n_1} = \frac{w_2 \times M_1}{M_2 \times w_1} $$

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Elevation of Boiling Point

Boiling Point Elevation ($\Delta T_b$)

A liquid boils when its vapour pressure equals atmospheric pressure. Since adding non-volatile solutes lowers vapour pressure, the solution must be heated to a *higher* temperature to boil.

$$ \Delta T_b = T_b - T_b^0 = K_b \cdot m $$

Where $m$ is the **molality** of the solution, and $K_b$ is the **Ebullioscopic Constant** (Molal Elevation Constant).

Molar Mass ($M_2$) Formula: $ \Delta T_b = \frac{K_b \times w_2 \times 1000}{M_2 \times w_1} $

Unit of $K_b$: $\text{K kg mol}^{-1}$. For water: $K_b = 0.52\text{ K kg mol}^{-1}$.

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Depression of Freezing Point

Freezing Point Depression ($\Delta T_f$)

Freezing point is the temperature at which the solid and liquid phases of a substance have identical vapour pressure. Since a solution has lower vapour pressure, it freezes at a *lower* temperature.

$$ \Delta T_f = T_f^0 - T_f = K_f \cdot m $$

Where $m$ is the **molality**, and $K_f$ is the **Cryoscopic Constant** (Molal Depression Constant).

Molar Mass ($M_2$) Formula: $ \Delta T_f = \frac{K_f \times w_2 \times 1000}{M_2 \times w_1} $

Unit of $K_f$: $\text{K kg mol}^{-1}$. For water: $K_f = 1.86\text{ K kg mol}^{-1}$.

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Osmosis & Osmotic Pressure

Osmosis: Spontaneous flow of solvent molecules through a Semi-Permeable Membrane (SPM) from pure solvent to solution (or from dilute to concentrated solution).

Osmotic Pressure ($\pi$): The excess external pressure that must be applied to the solution side to exactly halt osmosis.

$$ \pi = C R T = \frac{w_2 R T}{M_2 V} $$

Why OP is the Best Method for Polymers

It is measured at **room temperature** (biomolecules like proteins denature/degrade at boiling temperatures).
Its magnitude is **large and easily measurable** even for highly dilute solutions of macromolecules, unlike freezing depression ($\Delta T_f$) or boiling elevation ($\Delta T_b$) which yield unmeasurably tiny differences.
Uses Molarity ($C$) instead of molality.

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Tonicity & Reverse Osmosis

Tonicity in Biological Systems

Isotonic Solutions: Two solutions with identical osmotic pressures. Placed cell stays same ($0.9\% \text{ w/v } \text{NaCl}$ saline is isotonic with blood).
Hypertonic Solutions: Solution with higher concentration. Cell placed in it loses water and shrinks (**Plasmolysis**).
Hypotonic Solutions: Solution with lower concentration. Cell placed in it absorbs water and swells/bursts (**Hemolysis**).

Reverse Osmosis (RO)

If a pressure **greater than the osmotic pressure** ($\pi$) is applied directly to the solution side, the direction of flow reverses: pure solvent molecules are forced *out* of the solution through the semi-permeable membrane.

Global Application Reverse osmosis is used worldwide for the **desalination of seawater** to produce fresh drinking water.

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Practice: Colligative Properties

Practice Problem 7 Boiling Point Elevation

Boiling point of water at $750\text{ mm Hg}$ is $99.63^\circ\text{C}$. How much sucrose ($\text{C}_{12}\text{H}_{22}\text{O}_{11}$) must be added to $500\text{ g}$ of water such that it boils at $100^\circ\text{C}$? ($K_b$ for water $= 0.52\text{ K kg mol}^{-1}$).

1. Required elevation: $\Delta T_b = 100^\circ\text{C} - 99.63^\circ\text{C} = 0.37\text{ K}$.
2. Mass of solvent ($w_1$) $= 500\text{ g}$. Molar mass of sucrose ($M_2$) $= 342\text{ g/mol}$.
3. Apply BP elevation: $\Delta T_b = (K_b \times w_2 \times 1000) / (M_2 \times w_1)$.
4. $0.37 = (0.52 \times w_2 \times 1000) / (342 \times 500) = (1.04 \times w_2) / 342$.
5. $w_2 = (0.37 \times 342) / 1.04 = \mathbf{121.67\text{ g}}$.

Practice Problem 8 Osmotic Pressure & Molar Mass

$200\text{ cm}^3$ of an aqueous solution of a protein contains $1.26\text{ g}$ of the protein. The osmotic pressure of such a solution at $300\text{ K}$ is found to be $2.57 \times 10^{-3}\text{ bar}$. Calculate the molar mass of the protein. ($R = 0.083\text{ L bar K}^{-1}\text{mol}^{-1}$).

1. Volume ($V$) $= 200\text{ cm}^3 = 0.200\text{ L}$. Solute mass ($w_2$) $= 1.26\text{ g}$. Temperature ($T$) $= 300\text{ K}$. Pressure ($\pi$) $= 2.57 \times 10^{-3}\text{ bar}$.
2. Formula: $M_2 = (w_2 R T) / (\pi V)$.
3. $M_2 = (1.26 \times 0.083 \times 300) / (2.57 \times 10^{-3} \times 0.200) = 31.374 / (5.14 \times 10^{-4}) = \mathbf{61,038\text{ g/mol}}$.

Practice Problem 19 RLVP Numerical

Calculate RLVP when the vapour pressure of pure water is $31.8\text{ mmHg}$ and that of the solution is $31.0\text{ mmHg}$.

Formula: $\text{RLVP} = (p_1^0 - p_1) / p_1^0$.
$\text{RLVP} = (31.8 - 31.0) / 31.8 = 0.8 / 31.8 = \mathbf{0.0252}$.

Practice Problem 20 Molality from Freezing Depression

A solution of a non-volatile solute in water shows a freezing point depression of $0.93\text{ K}$. Calculate the molality of the solution. ($K_f \text{ for water} = 1.86\text{ K kg mol}^{-1}$).

Formula: $\Delta T_f = K_f \cdot m$.
Given: $\Delta T_f = 0.93\text{ K}$, $K_f = 1.86\text{ K kg mol}^{-1}$.
Molality ($m$) $= \Delta T_f / K_f = 0.93 / 1.86 = \mathbf{0.50\text{ m}}$ (or $\text{mol/kg}$).

Practice Problem 21 Osmotic Pressure Calculation

A $0.01\text{ M}$ non-electrolyte solution at $300\text{ K}$ has what osmotic pressure? ($R = 0.083\text{ L bar mol}^{-1}\text{K}^{-1}$).

Formula: $\pi = C R T$.
$\pi = 0.01 \text{ mol/L} \times 0.083 \text{ L bar/K mol} \times 300\text{ K} = \mathbf{0.249\text{ bar}}$.

Practice Problem 25 Isotonic Saline Reason

Explain why saline solution ($0.9\%$ w/v NaCl) is given in hospitals as isotonic fluid.

Saline is isotonic with blood plasma, meaning their osmotic pressures match. This prevents red blood cells from shrinking (plasmolysis/crenation) or swelling and bursting (hemolysis) when the fluid is injected.

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7. Abnormal Molar Masses

Why Calculation Fails

Colligative properties depend strictly on the number of solute particles. Ionic solutes **dissociate** in polar solvents, and organic solutes **associate** in non-polar solvents, changing the actual count of particles.

Van't Hoff Factor ($i$)

$$ i = \frac{\text{Observed Colligative Property}}{\text{Calculated (Normal) Colligative Property}} $$ $$ i = \frac{\text{Normal Molar Mass}}{\text{Abnormal (Calculated) Molar Mass}} $$

Modified Colligative Formulas

We multiply every colligative formula by $i$:

RLVP: $\frac{p_1^0 - p_1}{p_1^0} = \mathbf{i} \cdot x_2$
Elevation: $\Delta T_b = \mathbf{i} \cdot K_b \cdot m$
Depression: $\Delta T_f = \mathbf{i} \cdot K_f \cdot m$
Osmotic: $\pi = \mathbf{i} \cdot C R T$

Van't Hoff Dissociation & Association Diagram

Vardaan Learning Institute Class 12 � Solutions

Degree of Dissociation & Association

1. Dissociation degree ($\alpha$) If an electrolyte molecule dissociates to give $n$ ions:

$$ \alpha = \frac{i - 1}{n - 1} $$

For complete dissociation: $i = n$. $i > 1$. Calculated molar mass is *lower* than normal.

2. Association degree ($\alpha$) If $n$ simple solute molecules associate to form a single complex:

$$ \alpha = \frac{1 - i}{1 - \frac{1}{n}} $$

For complete dimerization ($n=2$): $i = 0.5$. $i < 1$. Calculated molar mass is *higher* than normal.

Vardaan Learning Institute Class 12 � Solutions

Practice: Van't Hoff Factor & Abnormal Mass

Practice Problem 9 Dissociation calculations

A $0.5\% \text{ w/w}$ aqueous solution of $\text{KCl}$ was found to freeze at $-0.24^\circ\text{C}$. Calculate the Van't Hoff factor and degree of dissociation of $\text{KCl}$. ($K_f \text{ for water} = 1.86\text{ K kg mol}^{-1}$, Molar mass of KCl $= 74.5$).

1. Molality ($m$) $= (0.5 \times 1000) / (74.5 \times 99.5) \approx 0.0674\text{ m}$.
2. Calculated $\Delta T_f$ (no dissociation) $= K_f \times m = 1.86 \times 0.0674 = 0.125^\circ\text{C}$.
3. Observed $\Delta T_f = 0.24^\circ\text{C}$ (given).
4. Van't Hoff factor ($i$) $= Observed / Calculated = 0.24 / 0.125 = \mathbf{1.92}$.
5. KCl gives 2 ions ($n=2$), so degree of dissociation ($\alpha$) $= (i - 1) / (n - 1) = (1.92 - 1) / (2 - 1) = 0.92 = \mathbf{92\%}$.

Practice Problem 10 Association calculations

$2\text{ g}$ of benzoic acid ($\text{C}_6\text{H}_5\text{COOH}$) dissolved in $25\text{ g}$ of benzene shows a freezing point depression of $1.62\text{ K}$. $K_f \text{ for benzene} = 4.9\text{ K kg mol}^{-1}$. Find percentage association if it dimerizes.

1. Normal mass of benzoic acid $= 122\text{ g/mol}$.
2. Calculated Abnormal Mass: $M_{2(abnormal)} = (K_f \times w_2 \times 1000) / (\Delta T_f \times w_1) = (4.9 \times 2 \times 1000) / (1.62 \times 25) = 241.98\text{ g/mol}$.
3. $i = Normal / Abnormal = 122 / 241.98 = 0.504$.
4. Dimerization ($n=2$), so degree of association ($\alpha$) $= (1 - i) / (1 - 0.5) = (1 - 0.504) / 0.5 = 0.992 = \mathbf{99.2\%}$.

Practice Problem 22 Theoretical vs Observed Colligative

If observed colligative property is 2.7 times the theoretical value for $\text{AlCl}_3$, estimate the Van't Hoff factor and infer dissociation behavior.

$i = Observed / Theoretical = \mathbf{2.7}$.
Since $i > 1$, the solute undergoes **dissociation**. Because complete dissociation would yield 4 ions ($Al^{3+}$ and $3Cl^-$) giving $i=4$, an experimental factor of $2.7$ shows partial dissociation ($\alpha \approx 57\%$).

Practice Problem 23 Degree of Dissociation Calculation

A salt $\text{AB}_2$ dissociates into $\text{A}^{2+}$ and $2\text{B}^-$. If the experimental Van't Hoff factor is $2.2$, calculate the degree of dissociation ($\alpha$) of the salt.

Formula: $\alpha = \frac{i - 1}{n - 1}$.
Here, salt $\text{AB}_2$ dissociates into $1$ $\text{A}^{2+}$ ion and $2$ $\text{B}^-$ ions, so number of ions ($n$) $= 1 + 2 = 3$.
Given: $i = 2.2$.
$\alpha = \frac{2.2 - 1}{3 - 1} = \frac{1.2}{2} = \mathbf{0.60}$ (or $\mathbf{60\%}$).

Practice Problem 24 Association Degree Calculation

For dimerization, if $i=0.80$, find the degree of association.

For dimerization, $n=2$.
$\alpha = (1 - i) / (1 - 1/2) = (1 - 0.80) / 0.5 = 0.20 / 0.5 = \mathbf{0.40}$ ($40\%$ association).

Vardaan Learning Institute Class 12 � Solutions

8. NCERT Deep-Dive & Mind Map

Deep-Dive Core Additions

Saturated: Dynamic equilibrium exists between dissolved solute and undissolved crystals.
Supersaturated: Metastable state holding more solute than maximum solubility at that temperature. Seeding/shaking triggers rapid crystallization.
Ideal Dilute Solution: Solvent obeys Raoult's law; solute obeys Henry's law.

NCERT Key Note Ideal dilute solutions are frequently asked as assertion-reason questions. Remember that the solvent obeys Raoult's law while the solute obeys Henry's law.

Solutions Chapter Visual Mind Map Diagram

Vardaan Learning Institute Class 12 � Solutions

Congratulations!

Chapter 1: Solutions is Complete

Quick Recall Checklist

Henry's Law: $p = K_H \cdot x$ (solubility drops as temperature rises).
Raoult's Law: $P_{total} = p_1^0 x_1 + p_2^0 x_2$.
Colligative: Depend *only* on particle count.
Van't Hoff Factor ($i$): Electrolytes require $i$ multiplier in all colligative equations.

Student Success Action Plan

1. Re-solve all 25 practice numerical problems step by step.

2. Pay high attention to temperature-dependent units vs invariant units.

3. Master the signs ($\Delta H$, $\Delta V$, $P_{obs}$) for deviations.

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CLASS 12 CHEMISTRY

Chapter 1: Solutions

1. Introduction and Definitions

What is a Solution?

Components of a Binary Solution

Types of Solutions Matrix

2. Expressing Concentration (Basics)

Mass & Volume Percentage

Molarity, Molality & Mole Fraction

Temperature Effects & Conversion

Temperature Dependence

Molarity to Molality Interconversion

Practice: Concentration Numericals

3. Solubility of Solids in Liquids

Core Principles

Solubility of Gases: Henry's Law

Key Conceptual Deductions

Applications of Henry's Law

Practice: Solubility & Henry's Law

4. Vapour Pressure of Liquid Solutions

Raoult's Law (Volatile Liquid-Liquid)

Dalton's Law of Partial Pressures

Vapour Pressure Lowering by Solid Solutes

Adding Non-Volatile Solutes

Vapour Pressure: Pure Solvent > Solution

Raoult's Law vs Henry's Law

Raoult's Law is a Special Case of Henry's Law

5. Ideal and Non-Ideal Solutions

Ideal Solutions

Non-Ideal Solutions

Non-Ideal: Positive Deviation

Core Concept

Molecular Cause

Non-Ideal: Negative Deviation

Core Concept

Molecular Cause

Azeotropes (Constant Boiling)

What is an Azeotrope?

Types of Azeotropes

Practice: Raoult's Law & Deviations

6. Colligative Properties

What are Colligative Properties?

Relative Lowering of Vapour Pressure

Dilute Solutions Simplification

Elevation of Boiling Point

Boiling Point Elevation ($\Delta T_b$)

Depression of Freezing Point

Freezing Point Depression ($\Delta T_f$)

Osmosis & Osmotic Pressure

Why OP is the Best Method for Polymers

Tonicity & Reverse Osmosis

Tonicity in Biological Systems

Reverse Osmosis (RO)

Practice: Colligative Properties

7. Abnormal Molar Masses

Van't Hoff Factor ($i$)

Modified Colligative Formulas

Degree of Dissociation & Association

Practice: Van't Hoff Factor & Abnormal Mass

8. NCERT Deep-Dive & Mind Map

Deep-Dive Core Additions

Congratulations!

Chapter 1: Solutions is Complete

Slide Index