Revision Notes: Basic Mathematics for Chemistry | Chemistry Class 11 - NEET PDF Download

Basic Mathematics for Chemistry

Basic mathematics is essential for solving numerical problems in chemistry, including mole calculations, equilibrium constants, and thermodynamics. This section covers key mathematical tools used in Class 11 Chemistry, vital for NEET preparation.

Importance in Chemistry

• Enables precise calculations for stoichiometry, pH, energy changes, etc.
• Ensures accuracy in interpreting experimental data.

Scientific Notation

Definition

• Format: a × 10ⁿ (1 ≤ a < 10, n = integer).
• Purpose: Simplifies large/small numbers (e.g., Avogadro’s number: 6.022 × 10²³).

Operations

• Multiplication: Multiply coefficients, add exponents (e.g., (2 × 10³) × (3 × 10²) = 6 × 10⁵).
• Division: Divide coefficients, subtract exponents (e.g., (6 × 10⁵) / (2 × 10²) = 3 × 10³).

Significant Figures

Definition

• Digits contributing to the precision of a number.

Rules

• All non-zero digits are significant (e.g., 345 = 3 SF).
• Zeros between non-zeros are significant (e.g., 1005 = 4 SF).
• Leading zeros are not significant (e.g., 0.007 = 1 SF).
• Trailing zeros after decimal are significant (e.g., 4.50 = 3 SF).

Calculations

• Addition/Subtraction: Result has same decimal places as least precise number (e.g., 2.3 + 4.56 = 6.9).
• Multiplication/Division: Result has same SF as number with fewest SF (e.g., 2.5 × 3.20 = 8.0).

Logarithms

Definition

• Log: log_b(a) = x where bˣ = a.
• Common Log: log₁₀ (used in pH); ln (base e ≈ 2.718) in thermodynamics.

Properties

• log(ab) = log a + log b.
• log(a/b) = log a - log b.
• log(aⁿ) = n log a.
• log₁₀(10ⁿ) = n; log₁₀(1) = 0.

Applications

• pH: pH = -log[H⁺] (e.g., [H⁺] = 10⁻³ → pH = 3).
• Equilibrium: pK_a = -log K_a.

Percentage and Ratio Calculations

Percentage

• Formula: % = (Part / Whole) × 100.
• Example: %O in H₂O = [16/18] × 100 = 88.89%.

Ratio

• Simplifies proportions (e.g., H:O in H₂O = 2:1).
• Use: Empirical formula determination.

Mole Concept Calculations

Definition

• Mole: 6.022 × 10²³ particles (Avogadro’s number, N_A).

Formulas

• Moles (n): n = mass / molar mass.
• Particles: Number = n × N_A.
• Volume (gas): 1 mole = 22.4 L at STP (0°C, 1 atm).

Algebraic Equations

Linear Equations

• Form: ax + b = 0 (e.g., 2x + 5 = 0 → x = -2.5).
• Use: Stoichiometry balancing.

Quadratic Equations

• Form: ax² + bx + c = 0.
• Solution: x = [-b ± √(b² - 4ac)] / 2a.
• Example: Equilibrium problems (e.g., x² + 0.1x - 0.01 = 0).

Unit Conversions

Common Conversions

• 1 kg = 1000 g; 1 g = 1000 mg.
• 1 L = 1000 mL; 1 mL = 1 cm³.
• °C to K: K = °C + 273.

Dimensional Analysis

• Method: Use conversion factors (e.g., 2 L to mL: 2 × 1000 = 2000 mL).
• Use: Concentration (mol/L to g/L).

Points to Remember

• Scientific Notation: a × 10ⁿ for large/small values.
• Significant Figures: Reflect precision in measurements.
• Logarithms: pH = -log[H⁺]; pK_a = -log K_a.
• Mole: n = mass / molar mass; 22.4 L at STP.
• Equations: Linear (stoichiometry), Quadratic (equilibrium).
• Units: Convert consistently (e.g., g to kg).

Additional Key Concepts

• Antilog: 10ˣ reverses log (e.g., antilog(3) = 10³ = 1000).
• Molar mass: g/mol (e.g., H₂O = 18 g/mol).
• STP: 0°C, 1 atm; useful for gas laws.
• Percentage yield: (Actual/Theoretical) × 100.
• Log tables or calculators simplify complex problems.

General Points and Errors to Be Noted

• Do not ignore SF in final answers (e.g., 2.0 ≠ 2).
• Memorize log properties for pH/K_a calculations.
• Avoid unit mismatches (e.g., L vs. mL in molarity).
• Understand quadratic roots: Use positive/realistic value in context.

Example Problems

Example 1

Question: Express 0.000067 in scientific notation:

1. 6.7 × 10⁻⁵
2. 6.7 × 10⁻⁴
3. 67 × 10⁻⁶
4. 0.67 × 10⁻⁴

Answer: a) 6.7 × 10⁻⁵

Solution: Move decimal 5 places right: 0.000067 = 6.7 × 10⁻⁵.

Example 2

Question: pH of a solution with [H⁺] = 0.001 M is:

1. 1
2. 2
3. 3
4. 4

Answer: c) 3

Solution: pH = -log[10⁻³] = -(-3) = 3.

Example 3

Question: Moles in 4 g of He (molar mass = 4 g/mol) are:

1. 0.5
2. 1
3. 2
4. 4

Answer: b) 1

Solution: n = mass / molar mass = 4 / 4 = 1 mol.