NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions
NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions provides step-by-step explanations of concentration terms, solubility of gases in liquids (Henry's Law), Raoult's Law for ideal and non-ideal solutions, colligative properties, and abnormal molar masses (van 't Hoff factor).
Overview of Topics & Subtopics (Click to Jump)
1. Types of Solutions & Concentration Expressing Units
Short Note:
• Solution: A homogeneous mixture of two or more chemically non-reacting substances. Binary solution contains one solute and one solvent.
• Key Concentration Terms:
- Mass Percentage (w/w): $(\text{Mass of solute} / \text{Total mass of solution}) \times 100$
- Mole Fraction ($x_A$): $x_A = n_A / (n_A + n_B)$. Sum of mole fractions of all components is always $1$.
- Molarity ($M$): Moles of solute dissolved per liter of solution ($M = n / V_{\text{in L}}$). Temperature dependent.
- Molality ($m$): Moles of solute per kilogram of solvent ($m = n / W_{\text{solvent in kg}}$). Independent of temperature.
- Parts per Million (ppm): $(\text{Mass of solute} / \text{Total mass of solution}) \times 10^6$
2. Solubility, Henry's Law & Raoult's Law
Short Note:
• Henry's Law: At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid surface: $p = K_H \cdot x$ (where $K_H$ is Henry's law constant; higher $K_H$ implies lower solubility).
• Raoult's Law: For a solution of volatile liquids, the partial vapor pressure of each component is directly proportional to its mole fraction in solution: $p_A = p_A^0 \cdot x_A$.
• Ideal Solutions: Obey Raoult's law at all concentrations ($\Delta H_{\text{mix}} = 0$, $\Delta V_{\text{mix}} = 0$, $A-B$ interactions are equal to $A-A$ and $B-B$ interactions). Example: $n$-hexane and $n$-heptane.
• Non-Ideal Solutions: Do not obey Raoult's law.
- Positive Deviation: $A-B$ interaction < $A-A$ interaction ($\Delta H_{\text{mix}} > 0$, $\Delta V_{\text{mix}} > 0$). Forms minimum boiling azeotropes (e.g., Ethanol + Acetone).
- Negative Deviation: $A-B$ interaction > $A-A$ interaction ($\Delta H_{\text{mix}} < 0$, $\Delta V_{\text{mix}} < 0$). Forms maximum boiling azeotropes (e.g., Chloroform + Acetone).
3. Colligative Properties & van 't Hoff Factor
Short Note:
Colligative properties depend only on the number of solute particles, not on their chemical identity.
1. Relative Lowering of Vapor Pressure: $(p_1^0 - p_1) / p_1^0 = i \cdot x_2$
2. Elevation of Boiling Point: $\Delta T_b = T_b - T_b^0 = i \cdot K_b \cdot m$ ($K_b$ = Ebullioscopic constant)
3. Depression of Freezing Point: $\Delta T_f = T_f^0 - T_f = i \cdot K_f \cdot m$ ($K_f$ = Cryoscopic constant)
4. Osmotic Pressure ($\pi$): $\pi = i \cdot C R T = i \cdot (n_2 / V) R T$
• van 't Hoff Factor ($i$): Account for dissociation or association of solute particles.
$i = \text{Observed Colligative Property} / \text{Calculated Colligative Property} = \text{Normal Molar Mass} / \text{Abnormal Molar Mass}$
- For Dissociation (e.g., $\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-$): $i > 1$
- For Association (e.g., Benzoic acid in benzene dimerizes): $i < 1$
- For Non-electrolytes (e.g., Glucose, Urea): $i = 1$
Intext & Conceptual Questions
• According to Henry's Law ($p = K_H \cdot x$), the Henry's law constant $K_H$ increases with an increase in temperature.
• Since solubility ($x$) is inversely proportional to $K_H$, the solubility of oxygen gas in water decreases as temperature rises.
• Cold water has a lower temperature and higher dissolved oxygen concentration compared to warm water. Thus, aquatic species breathe more comfortably in cold water.
| Feature | Ideal Solution | Non-Ideal Solution |
|---|---|---|
| Raoult's Law | Obeys Raoult's law over the entire range of concentrations. | Does not obey Raoult's law over the entire concentration range. |
| Enthalpy of Mixing ($\Delta H_{\text{mix}}$) | $\Delta H_{\text{mix}} = 0$ (No heat evolved or absorbed). | $\Delta H_{\text{mix}} \neq 0$ (Heat is absorbed or evolved). |
| Volume of Mixing ($\Delta V_{\text{mix}}$) | $\Delta V_{\text{mix}} = 0$ (Total volume equals sum of individual volumes). | $\Delta V_{\text{mix}} \neq 0$ (Volume changes upon mixing). |
| Intermolecular Forces | Solute-Solvent ($A-B$) forces equal $A-A$ and $B-B$ forces. | $A-B$ forces are stronger or weaker than $A-A$ and $B-B$ forces. |
Textbook Chapter End Exercise Questions
Given:
• Mass of Benzene ($m_{\text{solute}}$) = $22\text{ g}$
• Mass of $\text{CCl}_4$ ($m_{\text{solvent}}$) = $122\text{ g}$
Step 1: Total Mass of Solution
Total Mass = 22 g + 122 g = 144 g
Step 2: Mass % of Benzene
Mass % of C₆H₆ = (Mass of C₆H₆ / Total Mass) × 100
Mass % of C₆H₆ = (22 / 144) × 100 ≈ 15.28%
Step 3: Mass % of Carbon Tetrachloride
Mass % of CCl₄ = 100 - 15.28 = 84.72%
• Mass % of Benzene = 15.28%
• Mass % of CCl₄ = 84.72%
Given:
• Initial boiling point of water ($T_b^0$) = $99.63^\circ\text{C}$
• Desired boiling point of solution ($T_b$) = $100.00^\circ\text{C}$
• Elevation in boiling point ($\Delta T_b$) = $100.00 - 99.63 = 0.37^\circ\text{C} = 0.37\text{ K}$
• Mass of solvent water ($w_1$) = $500\text{ g} = 0.5\text{ kg}$
• Molar mass of Sucrose $\text{C}_{12}\text{H}_{22}\text{O}_{11}$ ($M_2$) = $12(12) + 22(1) + 11(16) = 342\text{ g/mol}$
• $K_b = 0.52\text{ K kg mol}^{-1}$
Step 1: Use Elevation of Boiling Point Formula
$\Delta T_b = (K_b \times w_2 \times 1000) / (M_2 \times w_1)$
Step 2: Rearrange to Solve for Mass of Sucrose ($w_2$)
$w_2 = (\Delta T_b \times M_2 \times w_1) / (K_b \times 1000)$
$w_2 = (0.37 \times 342 \times 500) / (0.52 \times 1000)$
$w_2 = 63.27 / 0.52 \approx 121.67\text{ g}$
Mass of sucrose to be added = 121.67 g.
Read More & Chapter Resources
These short notes are designed for quick revision. Read the main points once, then use the related material below to practise the same chapter. Revising a concept and checking it with questions can make your preparation more focused.