Expressing Concentration of Solutions

 

📘 Expressing Concentration of Solutions

The composition of a solution refers to the relative amount of solute and solvent present in it. This composition can be expressed in two main ways:

• Qualitatively

• Quantitatively

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🔹 Qualitative Description

This is a non-numerical way of expressing the concentration. For example:

• A dilute solution has a relatively small amount of solute.

• A concentrated solution has a relatively large amount of solute.

However, qualitative terms are vague and can lead to confusion, especially in scientific or industrial applications. Hence, quantitative expressions of concentration are preferred.

🔹 Quantitative Description

Several methods are used to express the concentration of a solution quantitatively:

(i) Mass Percentage (% w/w)

Definition:

$$\text{Mass \% of a component} = \left( \frac{\text{Mass of the component in the solution}}{\text{Total mass of the solution}} \right) \times 100$$

Example:
A solution of 10% glucose by mass means:

• 10 g of glucose (solute)

• 90 g of water (solvent)
  Total mass = 100 g

Application:
Mass percentage is commonly used in industrial applications.
E.g., Commercial bleaching solution contains 3.62% NaOCl (sodium hypochlorite) by mass.

(ii) Volume Percentage (% v/v)

Definition:

$$\text{Volume \% of a component} = \left( \frac{\text{Volume of the component}}{\text{Total volume of the solution}} \right) \times 100$$

Example:
A 10% ethanol solution by volume means:

• 10 mL ethanol

• Water added to make the total volume = 100 mL

Application:
Used for liquid–liquid solutions like:

• Antifreeze: A 35% (v/v) ethylene glycol solution lowers the freezing point of water to 255.4 K (-17.6°C).

(iii) Mass by Volume Percentage (% w/v)

Definition:

$$\text{Mass by Volume \%} = \left( \frac{\text{Mass of solute in grams}}{\text{Volume of solution in mL}} \right) \times 100$$

Application:
Commonly used in medicine and pharmacy.
E.g., A 5% w/v solution means 5 g of solute in 100 mL of solution.

(iv) Parts per Million (ppm)

Definition:

$$\text{ppm} = \left( \frac{\text{Number of parts of the component}}{\text{Total number of parts of all components}} \right) \times 10^6$$

Application:
Used for very low concentrations (trace quantities), especially in:

• Environmental chemistry

• Pollution analysis
  E.g., Sea water (1030 g) contains 6 × 10⁻³ g of dissolved O₂

$$\text{ppm} = \frac{6 \times 10^{-3}}{1030} \times 10^6 \approx 5.8\ ppm$$

Variants: ppm can be:

• mass to mass

• volume to volume

• mass to volume

(v) Mole Fraction (χ)

Definition:

$$\chi_i = \frac{n_i}{\sum n_i}$$

Where:

• $n_i$ = number of moles of component i

• $\sum n_i$ = total moles of all components

In a binary mixture of components A and B:

$$\chi_A = \frac{n_A}{n_A + n_B}, \quad \chi_B = \frac{n_B}{n_A + n_B}$$ $$\chi_A + \chi_B = 1$$

Application:
Useful in calculating vapor pressure, colligative properties, and gas mixture compositions.

(vi) Molarity (M)

Definition:

$$\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in litre}}$$

Example:
0.25 M NaOH means 0.25 mol of NaOH is dissolved in 1 litre of solution.

Note:
Temperature-dependent because volume changes with temperature.

(vii) Molality (m)

Definition:

$$\text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$$

Example:
A 1.00 m solution of KCl means 1 mol (74.5 g) of KCl is dissolved in 1 kg of water.

Note:
Independent of temperature because it involves mass, not volume.

📝 Summary Table

Expression	Symbol	Formula	Depends on Temperature?	Typical Use
Mass % (w/w)	—	$\frac{\text{Mass of solute}}{\text{Total mass of solution}} \times 100$	❌ No	Industrial mixtures
Volume % (v/v)	—	$\frac{\text{Volume of solute}}{\text{Total volume of solution}} \times 100$	✅ Yes	Liquid-liquid solutions
Mass/Volume % (w/v)	—	$\frac{\text{Mass of solute}}{\text{Volume of solution in mL}} \times 100$	✅ Yes	Medical & pharmaceutical use
Parts per million (ppm)		$\frac{\text{Part of solute}}{\text{Total parts of solution}} \times 10^6$	❌ No	Environmental concentrations
Mole fraction	$\chi$	$\frac{\text{Moles of component}}{\text{Total moles of solution}}$	❌ No	Gas mixtures, vapor pressure
Molarity	M	$\frac{\text{Moles of solute}}{\text{Volume of solution in litres}}$	✅ Yes	Titrations, lab solutions
Molality	m	$\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$	❌ No	Colligative property calculations

🔍 Conclusion

• Each unit has its specific application depending on the accuracy, temperature dependence, and nature of the solution.

• Mass %, ppm, mole fraction, and molality are independent of temperature, making them more reliable in changing environments.

• Molarity, while commonly used, must be handled carefully with temperature-sensitive systems.

Azeotropes – Explained for Class 12 (CBSE Chemistry)

 

📘 Azeotropes – Explained for Class 12 (CBSE Chemistry)

🔹 What is an Azeotrope?

An azeotrope is a mixture of two (or more) liquids that boils at a constant temperature and behaves like a pure substance during boiling.

🧪 Definition:

An azeotrope is a binary mixture that boils at a constant temperature and produces vapour with the same composition as the liquid.

This means you cannot separate the two components by simple distillation — they behave as one compound at that specific composition and boiling point.

🔍 Why does this happen?

Normally, during boiling, the composition of vapour differs from that of liquid (more volatile component escapes faster). But in an azeotropic mixture, this difference disappears at a certain fixed composition — the vapour formed has the same ratio of both components as in the liquid. Hence, no further separation by boiling is possible.

🌡️ Types of Azeotropes

Azeotropes are mainly of two types based on their boiling points in comparison to their pure components:

(i) Minimum Boiling Azeotropes

• These mixtures boil at a lower temperature than either of the pure components.

• They show positive deviation from Raoult’s Law (the components escape more easily due to weaker interactions).

• The total vapour pressure is higher, so they boil at a lower temperature.

🧪 Example:

• Ethanol (95%) + Water (5%)

• Boiling Point ≈ 351 K (78°C)

• This mixture forms an azeotrope that cannot be separated further by simple distillation.

🧠 Reason:

Ethanol and water form weaker hydrogen bonds in the mixture than in the pure liquids, so they escape more easily → higher vapour pressure → lower boiling point.

(ii) Maximum Boiling Azeotropes

• These mixtures boil at a higher temperature than either of the pure components.

• They show negative deviation from Raoult’s Law (the components strongly attract each other).

• The total vapour pressure is lower, so they boil at a higher temperature.

🧪 Example:

• Nitric acid (68%) + Water (32%)

• Boiling Point ≈ 393.5 K (120.5°C)

🧠 Reason:

Strong hydrogen bonding between nitric acid and water lowers the escaping tendency → lower vapour pressure → higher boiling point.

📊 Comparison Table:

Type	Boiling Point	Deviation from Raoult's Law	Vapour Pressure	Example
Minimum Boiling Azeotrope	Lower than components	Positive Deviation	High	Ethanol + Water (95%)
Maximum Boiling Azeotrope	Higher than components	Negative Deviation	Low	Nitric Acid + Water (68%)

🎓 Real-life Importance

• Azeotropes limit the extent of separation by distillation.

• Special methods like azeotropic distillation or adding third components (entrainers) are used in industries to break azeotropes.

• Ethanol production: 95% ethanol + 5% water azeotrope is common in alcohol distillation.

💡 Analogy to Understand

Think of azeotropes as inseparable couples:

• In a normal couple (mixture), one partner might leave early (more volatile component distills first).

• In an azeotrope, both are so balanced in their interactions that they leave together, hand in hand, at the same time (same composition in vapour and liquid).

Raoult’s Law and Henry’s Law – A Comparative Study

 

1. Raoult’s Law

Raoult’s law is applicable to solutions containing volatile components. It states:

"The partial vapour pressure of each volatile component in a solution is directly proportional to its mole fraction."

Mathematical Expression:

For a component 1 in a binary solution:

$$p_1 = x_1 \cdot p_1^0$$

Where:

• $p_1$ = Partial vapour pressure of component 1 in the solution

• $x_1$ = Mole fraction of component 1 in the solution

• $p_1^0$ = Vapour pressure of pure component 1

2. Henry’s Law

Henry’s Law is applicable to gases dissolved in liquids. It states:

"The partial pressure of the gas (p) in the vapour phase is directly proportional to its mole fraction (x) in the solution."

Mathematical Expression:

$$p = K_H \cdot x$$

Where:

• $p$ = Partial pressure of the gas

• $x$ = Mole fraction of the gas in solution

• $K_H$ = Henry’s Law constant

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

• Both laws show a direct proportionality between partial pressure and mole fraction.

• In Raoult’s law, the constant of proportionality is $p_1^0$, while in Henry’s law it is $K_H$.

• Hence, Raoult’s law can be considered a special case of Henry’s law when $K_H = p_1^0$.

4. Vapour Pressure of Solutions

(a) Pure Solvent

• Molecules escape from the surface and exert vapour pressure at equilibrium.

(b) Solution with a Non-Volatile Solute

• When a non-volatile solute is added, the vapour pressure decreases.

• Fewer solvent molecules are available at the surface to escape into vapour phase.

Key Point:

• The reduction in vapour pressure depends only on the quantity of solute and not on its nature.

5. Ideal and Non-Ideal Solutions

(A) Ideal Solutions

Definition: Solutions that obey Raoult’s law over the entire concentration range.

Characteristics:

• $\Delta_{\text{mix}} H = 0$ (No heat is evolved or absorbed)

• $\Delta_{\text{mix}} V = 0$ (No change in volume on mixing)

• Intermolecular forces: A–A ≈ B–B ≈ A–B

Examples:

• Benzene and Toluene

• n-Hexane and n-Heptane

(B) Non-Ideal Solutions

Definition: Solutions that deviate from Raoult’s law.

They show either:

(i) Positive Deviation

• $p_{\text{solution}} > p_{\text{expected}}$

• A–B interactions < A–A and B–B interactions

• More molecules escape → Higher vapour pressure

Examples:

• Ethanol + Acetone

• Acetone + Carbon Disulphide

(ii) Negative Deviation

• $p_{\text{solution}} < p_{\text{expected}}$

• A–B interactions > A–A and B–B interactions

• Fewer molecules escape → Lower vapour pressure

Examples:

• Chloroform + Acetone (Hydrogen bonding between components)

• Phenol + Aniline

6. Azeotropes

Definition: Binary mixtures that boil at a constant temperature and whose liquid and vapour phases have the same composition.

Types of Azeotropes:

(i) Minimum Boiling Azeotropes

• Show positive deviation

• Boil at a lower temperature than both components

Example:

• Ethanol (95%) + Water → Boiling point: ~351 K

(ii) Maximum Boiling Azeotropes

• Show negative deviation

• Boil at a higher temperature than both components

Example:

• Nitric acid (68%) + Water → Boiling point: 393.5 K

7. Graphical Representations

(i) Raoult’s Law for Ideal Solutions

• Linear plot of vapour pressure vs. mole fraction

(ii) Non-Ideal Solutions

• Positive Deviation: Upward curve (convex)

• Negative Deviation: Downward curve (concave)

Summary Table

Type	Deviation	A-B Interaction	Vapour Pressure	Example
Ideal Solution	None	A–B ≈ A–A ≈ B–B	Matches Raoult’s law	Benzene + Toluene
Non-Ideal (+ve)	Positive	A–B < A–A or B–B	Greater than expected	Ethanol + Acetone
Non-Ideal (–ve)	Negative	A–B > A–A or B–B	Less than expected	Chloroform + Acetone
Min. Boiling Azeotrope	Positive	Weak A–B	Boils below both liquids	Ethanol (95%) + Water
Max. Boiling Azeotrope	Negative	Strong A–B	Boils above both liquids	Nitric Acid (68%) + Water

Conclusion

Raoult’s law and Henry’s law both describe the relationship between vapour pressure and mole fraction, although applicable in different contexts (liquids vs. gases). Understanding deviations from Raoult’s law helps us classify solutions as ideal or non-ideal and explains the formation of azeotropes, which are critical in processes like distillation.

 

🌡️ Vapour Pressure of Liquid Solutions

Introduction:
Liquid solutions are formed when the solvent is a liquid. The solute in such solutions can be a gas, a liquid, or a solid. In this section, we will discuss binary solutions, i.e., solutions containing two components. These include:

• (i) Solutions of liquids in liquids

• (ii) Solutions of solids in liquids

Such solutions may contain one or more volatile components, with the solvent generally being volatile.

🔹 Vapour Pressure of Liquid–Liquid Solutions

Let us consider a binary solution made of two volatile liquids, denoted by components 1 and 2.

When this mixture is placed in a closed container, both liquids evaporate. Eventually, an equilibrium is established between the vapour phase and the liquid phase.

Let:

• $p_1$ = Partial vapour pressure of component 1

• $p_2$ = Partial vapour pressure of component 2

• $p_{\text{total}}$ = Total vapour pressure of the solution

• $x_1$ = Mole fraction of component 1 in the liquid phase

• $x_2$ = Mole fraction of component 2 in the liquid phase

📘 Raoult’s Law:

Given by: François-Marie Raoult (1886)

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

So,

$$p_1 \propto x_1 \quad \Rightarrow \quad p_1 = p_1^0 \cdot x_1 \tag{1.12}$$
$$p_2 \propto x_2 \quad \Rightarrow \quad p_2 = p_2^0 \cdot x_2 \tag{1.13}$$

Where:

• $p_1^0$ and $p_2^0$ = Vapour pressures of pure components 1 and 2 respectively.

🧪 Total Vapour Pressure:

According to Dalton’s Law of Partial Pressures, the total pressure is the sum of partial vapour pressures:

$$p_{\text{total}} = p_1 + p_2 \tag{1.14}$$

Substituting from equations (1) and (2):

$$p_{\text{total}} = x_1 p_1^0 + x_2 p_2^0 \tag{1.15}$$

Since $x_1 + x_2 = 1$, we can write:

$$p_{\text{total}} = p_1^0 + (p_2^0 - p_1^0)x_2 \tag{1.16}$$

📊 Interpretation of Equation :

• (i) Total vapour pressure depends on mole fraction of any one component.

• (ii) The relation is linear with respect to mole fraction $x_{2.}$

• (iii) If $p_1^0 < p_2^0$, then total vapour pressure increases with increasing $x_2$.

🧩 Graphical Representation:

A plot of:

• $p_1$ vs. $x_1$

• $p_2$ vs. $x_2$

• $p_{\text{total}}$ vs. $x_2$

gives straight lines, intersecting at the points where mole fractions = 1 (i.e., pure components).

The line for $p_{\text{total}}$ lies between those of $p_1$ and $p_2$.

🌫️ Composition of Vapour Phase at Equilibrium:

Let:

• $y_1$ = Mole fraction of component 1 in vapour phase

• $y_2$ = Mole fraction of component 2 in vapour phase

Using Dalton’s law:

$$p_1 = y_1 \cdot p_{\text{total}} \tag{1.17}$$
$$p_2 = y_2 \cdot p_{\text{total}} \tag{1.18}$$

In general, for any component $i$:

$$p_i = y_i \cdot p_{\text{total}}$$

This shows that mole fraction in vapour phase ( $y_i$ ) is not necessarily equal to mole fraction in liquid phase ( $x_i$ ).

🔍 Important Points to Remember:

• Raoult’s law applies to ideal solutions.

• For non-ideal solutions, deviations occur due to intermolecular interactions.

• Mole fractions in liquid and vapour phases differ unless the solution is ideal and forms an azeotrope.

• Vapour pressure is a colligative property, depending on the number of particles, not their nature.

📘 Application in Daily Life:

• Perfumes and essential oils use volatile liquid mixtures.

• Distillation techniques (like in petrochemicals) rely on vapour pressure principles.

• Meteorology and humidity studies use vapour pressure data of water.

✍️ Practice Questions:

1. State Raoult’s law and write its mathematical form.

2. Derive the expression for total vapour pressure in a binary liquid solution.

3. Explain why total vapour pressure changes with mole fraction.

4. A solution has $x_1 = 0.6$, $p_1^0 = 80 \, \text{kPa}$, $p_2^0 = 60 \, \text{kPa}$. Calculate $p_{\text{total}}$.

5. Define ideal and non-ideal solutions with examples.

Solubility of a Gas in a Liquid

 📘 Class 12 CBSE Chemistry Study Material

Chapter: Solutions
Topic: Solubility of a Gas in a Liquid

Many gases dissolve in liquids to form homogeneous mixtures. This is an important phenomenon not just in chemistry but also in environmental science, biology, and industry.

Some common examples include:

• Oxygen dissolving in water, which is vital for aquatic life.

• Hydrogen chloride (HCl), which is highly soluble in water and forms hydrochloric acid.

However, the solubility of gases is highly affected by pressure and temperature, unlike solids, which are mostly unaffected by pressure.

Dynamic Equilibrium in Gas-Liquid Solutions

Consider a container partly filled with a liquid and some gas above it, both at constant pressure (p) and temperature (T). Initially, gas molecules start dissolving in the liquid. Eventually, an equilibrium is reached when:

Rate of gas molecules entering the solution = Rate of gas molecules escaping from the solution

This is called dynamic equilibrium, represented as:

                         Gas(g) ⇌ Gas (in solution)

When this equilibrium is disturbed (e.g., by changing pressure or temperature), the system shifts to restore balance, as predicted by Le Chatelier’s Principle.

Effect of Pressure on Gas Solubility

As pressure increases, more gas molecules are forced into the solution.

➤ Henry’s Law

Henry’s Law gives a quantitative relationship between pressure and gas solubility.

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

Mathematically:

p = K_H. x     …… (1.11)

Where,

• p = partial pressure of the gas above the liquid

• x = mole fraction of the gas in the solution

• K_H = Henry’s Law constant (depends on nature of the gas and temperature)

Graphical Representation:

If we plot p vs x, we get a straight line with slope K_H 

Interpretation:

• A high K_H value indicates low solubility of the gas.

• A low K_H value means high solubility of the gas.

K_H  increases with temperature, so gas solubility decreases as temperature rises.

Effect of Temperature on Gas Solubility

The dissolution of gases in liquids is typically an exothermic process:

Gas (g) → Gas (in solution) + Heat

Hence, by Le Chatelier’s Principle:

When temperature increases, the equilibrium shifts backwards (i.e., towards the gaseous state), reducing gas solubility.

This is why:

• Cold water contains more dissolved oxygen than warm water.

• Aquatic animals thrive better in cold water due to higher oxygen content.

Applications of Henry’s Law

🔹 Carbonated Beverages:

• CO₂ is dissolved under high pressure in soda bottles.

• When opened, pressure is released, solubility drops, and CO₂ escapes as bubbles.

🔹 Scuba Diving and Bends:

• At high underwater pressures, more N₂ dissolves in blood.

• Upon sudden ascent, pressure drops quickly, and nitrogen comes out as bubbles in blood.

• This causes "bends" (decompression sickness), which can be fatal.

• Helium, a less soluble gas, is used in diving tanks along with oxygen.

🔹 High Altitude Anoxia:

• At high altitudes, pO₂ (partial pressure of oxygen) is low.

• Less oxygen dissolves in blood, leading to anoxia – a condition marked by weakness, confusion, and difficulty in thinking clearly.

Conclusion

• Solubility of gases is governed primarily by Henry’s Law, pressure, and temperature.

• Higher pressure = higher gas solubility

• Higher temperature = lower gas solubility

• Understanding this is crucial for real-world applications like scuba diving, soda bottling, and managing oxygen supply at high altitudes.

Key Formula:

p = K_H . x

Where:

• p = partial pressure of the gas

• x = mole fraction of the gas in the solution

• K_H = Henry's Law constant

Important Terms:

Term	Definition
Solubility	Maximum amount of a substance that can dissolve in a solvent at given T and p
Dynamic Equilibrium	Equal rate of dissolution and escape of gas
Henry’s Law	Solubility ∝ partial pressure of the gas
KH	Henry’s law constant; varies with gas and temperature