Problems & Solutions
Practice problems with solutions.
1 & 2. Mixtures: Definition and Types
Problem 1: Classify the following as homogeneous or heterogeneous mixtures: (a) Salt dissolved in water (b) Oil and water (c) Air (d) Sand and iron filings
Answer:
- Salt in water → Homogeneous (uniform composition)
- Oil and water → Heterogeneous (visible layers)
- Air → Homogeneous (uniform gases)
- Sand + iron → Heterogeneous (different components visible)
Explanation: Homogeneous mixtures have the same composition throughout, while heterogeneous mixtures have visibly different parts.
Problem 2: Why is brass considered a homogeneous mixture?
Answer: Brass is an alloy of copper and zinc. Its composition is uniform at the microscopic level.
Explanation: Alloys are homogeneous because the metals are evenly mixed, even if they are different elements.
3. Properties of Elements
Problem 1: Sodium reacts vigorously with water, but gold does not. Explain why.
Answer:
- Sodium → highly reactive alkali metal, easily loses one electron → reacts violently with water.
- Gold → inert metal, does not react easily due to stable electron configuration.
Explanation: Reactivity depends on how easily an element loses or gains electrons.
Problem 2: Describe physical vs chemical properties of chlorine.
Answer:
- Physical: Greenish-yellow gas, strong odor, density greater than air.
- Chemical: Reacts with Na → NaCl, reacts with H₂ → HCl gas.
Explanation: Physical properties are observed without changing composition; chemical properties describe reactivity.
4. Isotopes: Atomic Mass Calculations
Problem 1: Chlorine has two isotopes: Cl-35 (mass = 34.97, abundance = 75.76%) and Cl-37 (mass = 36.97, abundance = 24.24%). Calculate average atomic mass.
Answer: Average Mass = (34.97 × 0.7576) + (36.97 × 0.2424) = 35.46 amu
Explanation: Weighted average accounts for both mass and natural abundance of isotopes.
Problem 2: An element has isotopes X-10 (20%) and X-11 (80%). Calculate average atomic mass if masses are 10.01 and 11.01 amu.
Answer: Average Mass = (10.01 × 0.2) + (11.01 × 0.8) = 10.81 amu
Explanation: Use fractional abundances to weight each isotope.
5. Percentage & Mass Calculation
Problem 1: Calculate % of H in H₂O and mass of H in 100 g water.
Answer:
- Molar mass H₂O = 2×1.01 + 16 = 18.02 g/mol
- %H = (2.02 / 18.02) × 100 = 11.2%
- Mass of H in 100 g water = 11.2 g
Explanation: Percent composition shows the mass fraction of each element in a compound.
Problem 2: Calculate % O in CO₂.
Answer:
- Molar mass CO₂ = 12 + 2×16 = 44 g/mol
- %O = (32 / 44) × 100 = 72.73%
Explanation: Same method, use atomic masses and total molar mass.
6. Chemical Formula Names
Problem 1: Name the following compounds: (a) NaCl (b) CO₂ (c) MgF₂ (d) N₂O₅
Answer:
- NaCl → Sodium chloride
- CO₂ → Carbon dioxide
- MgF₂ → Magnesium fluoride
- N₂O₅ → Dinitrogen pentoxide
Explanation: Ionic compounds → cation first, anion with -ide. Covalent → use prefixes for number of atoms.
7, 9, 10. Equilibrium
Problem 1: For the reaction N₂ + 3H₂ ⇌ 2NH₃, what happens if H₂ is added?
Answer: Shift right (toward products).
Explanation: According to Le Châtelier's principle, adding reactant favors formation of products.
Problem 2: Kc vs Kp relation:
Kp = Kc(RT)^Δn Δn = moles gaseous products - moles gaseous reactants
Example: If Δn = -2, R = 0.0821 L·atm/mol·K, T = 298 K, Kc = 0.5
Kp = 0.5 × (0.0821 × 298)^-2
Explanation: Kp uses pressures, Kc uses concentrations; Δn accounts for gas moles difference.
Problem 3: Predicting Equilibrium Shift Using Kp
Reaction: N2(g) + 3H2(g) ⇌ 2NH3(g)
Scenario 1: Products are less than required
Given:
- Equilibrium constant, Kp = 0.5
- Current pressures: PN2 = 1 atm, PH2 = 1 atm, PNH3 = 0.25 atm
Step 1: Calculate the reaction quotient (Qp) Qp = (PNH3^2) / (PN2 × PH2^3) = 0.0625
Step 2: Compare Qp with Kp
- Qp < Kp → Reaction shifts right (towards products)
Step 3: Predict changes in concentrations
| Species | Change |
|---|---|
| NH3 (product) | ↑ increases |
| N2 (reactant) | ↓ decreases |
| H2 (reactant) | ↓ decreases |
Scenario 2: Products are more than required
Given:
- Current pressures: PN2 = 1 atm, PH2 = 1 atm, PNH3 = 1 atm
Step 1: Calculate Qp = 1
Step 2: Compare Qp with Kp
- Qp > Kp → Reaction shifts left (towards reactants)
Step 3: Predict changes in concentrations
| Species | Change |
|---|---|
| NH3 (product) | ↓ decreases |
| N2 (reactant) | ↑ increases |
| H2 (reactant) | ↑ increases |
Quick Guide: Predicting Equilibrium Shift
| Reaction Quotient | Equilibrium Constant | Shift Direction | Effect on Species |
|---|---|---|---|
| Qc or Qp < Kc or Kp | - | Right (toward products) | Reactants ↓, Products ↑ |
| Qc or Qp > Kc or Kp | - | Left (toward reactants) | Reactants ↑, Products ↓ |
| Qc or Qp = Kc or Kp | - | No shift | No change |
8. Empirical & Molecular Formulas
Problem 1: 40%C, 6.7%H, 53.3%O → find empirical formula.
Answer:
- Assume 100 g → C=40 g, H=6.7 g, O=53.3 g
- Moles: C=3.33, H=6.7, O=3.33
- Divide by smallest → C₁H₂O₁ → CH₂O
Explanation: Convert % to moles, divide by smallest to get simplest ratio.
Problem 2: Empirical formula CH₂O, molar mass = 180 g/mol → molecular formula?
Answer:
- Empirical mass = 12+2+16=30
- n = 180 / 30 = 6
- Molecular formula = C₆H₁₂O₆
Explanation: Molecular formula = empirical × n
11. Acid-Base Theories
Problem 1: Classify HCl in water by three theories:
Answer:
- Arrhenius → acid (produces H⁺)
- Brønsted-Lowry → proton donor
- Lewis → electron pair acceptor
Explanation: Each theory defines acid-base differently but overlaps in behavior.
Problem 2: NH₃ in water acts as:
Answer:
- Arrhenius → base (produces OH⁻)
- Brønsted-Lowry → proton acceptor
- Lewis → electron pair donor
12. pH Calculations
Problem 1: [H⁺] = 1×10⁻³ M → pH?
Answer: pH = -log[H⁺] = 3
Problem 2: [OH⁻] = 1×10⁻⁴ M → pH?
Answer: pOH = -log[OH⁻] = 4 → pH = 10
13. Chapter 1 Review
Problem 1: Valence electrons of O, Cl, Na → 6, 7, 1
Problem 2: Ionization energy trend
- Across period → increases
- Down group → decreases
14. Elements & Electron Configuration
Problem 1: Fe (Z=26) → 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶ or [Ar] 4s² 3d⁶
15. Chemical Bonding
Problem 1: Bond order of O₂ → 2
Problem 2: EAN of Fe in K₄[Fe(CN)₆] → 36
16. Colloids and Colloidal Solutions
Problem 1: Identify type: Milk → Emulsion, Fog → Aerosol, Paint → Sol
Problem 2: Explain Tyndall effect → Light scattering by colloidal particles
Extra: How Soap Works (Micelle Formation)
Problem: Explain how soap cleans grease.
Answer:
- Soap has hydrophobic tail and hydrophilic head.
- Soap molecules form micelles → tails trap grease inside, heads interact with water → grease washed away.