Physical Chemistry & Stoichiometry — Part 1: 40 MCQs with Explanations

Each formula unit of Al₂(SO₄)₃ contains 12 oxygen atoms (4×3). So 0.2 mol compound contains 0.2 × 12 = 2.4 mol oxygen atoms.

Molar mass of O₂ = 32 g/mol. So 32 g = 1 mole. At STP, 1 mole of any gas occupies 22.4 L.

H₂SO₄ is a strong diprotic acid but second dissociation is incomplete. For dilute solutions, we consider complete first dissociation: [H⁺] ≈ 0.002 M (since each H₂SO₄ gives 2H⁺ approximately). pH = -log(0.002) = 2.7 ≈ 3? Wait, more accurately: First dissociation complete: [H⁺] = 0.001 M from first H⁺, plus partial second dissociation. But for MDCAT level, often approximated as [H⁺] = 0.002 M → pH = -log(0.002) = 2.7 ≈ 3. However, standard approximation for 0.001 M H₂SO₄ is pH ≈ 3. Actually, precise calculation: Ka₂ = 0.01, so [H⁺] = 0.001 + [SO₄²⁻]. Solving quadratic gives [H⁺] ≈ 0.0019 M, pH ≈ 2.72 ≈ 3. But many textbooks approximate as pH = 3 for 0.001 M H₂SO₄. However, option 2 (pH=2) would be for 0.01 M strong monoprotic acid. Let me recalculate: For 0.001 M H₂SO₄, considering only first dissociation complete: [H⁺] = 0.001 M → pH = 3. But second dissociation adds more H⁺. Standard answer in MDCAT is pH ≈ 3 for 0.001 M H₂SO₄? Actually, careful calculation shows pH ≈ 2.7-2.8, which rounds to 3. But option 2 is pH=2 which would be for 0.01 M strong acid. I think the expected answer is 3 (option 3). Wait the options are: 1,2,3,4 corresponding to pH values 1,2,3,4. For 0.001 M H₂SO₄, pH is approximately 3 (more precisely 2.7-2.8). So answer should be option 3 (pH=3).

Volume depends on the amount of substance (extensive property). Density, temperature, and concentration are intensive properties (independent of amount).

Moles of H₂SO₄ = 4.9/98 = 0.05 mol. Volume = 250 mL = 0.25 L. Molarity = 0.05/0.25 = 0.2 M.

Density is independent of the amount of substance (intensive property). Mass, volume, and enthalpy depend on amount (extensive properties).

Surface tension depends on nature of liquid, not just concentration of solute particles. Colligative properties depend only on number of solute particles.

In pure water at 25°C, [H⁺] = [OH⁻] = √(10⁻¹⁴) = 10⁻⁷ M, so pH = -log(10⁻⁷) = 7.

n-factor = 2 (two replaceable H⁺ ions). Equivalent weight = 98/2 = 49 g/equiv.

[OH⁻] ≈ √(Kb·C) = √(1.8×10⁻⁵ × 0.1) = √(1.8×10⁻⁶) ≈ 1.34×10⁻³ M. pOH = 2.87, pH = 14 - 2.87 = 11.13 ≈ 11.

By convention, ΔHf° of elements in their standard states is zero. O₂(g) is the standard state of oxygen.

Henderson-Hasselbalch: pH = pKa + log([salt]/[acid]) = 4.74 + log(0.1/0.2) = 4.74 + log(0.5) = 4.74 - 0.30 = 4.44.

Rate ∝ 1/√M and rate ∝ 1/√d (at same T and P), where M is molar mass and d is density.

[OH⁻] = 0.01 M = 10⁻² M. pOH = -log(10⁻²) = 2.

Molar mass H₂ = 2 g/mol. 4 g H₂ = 2 mol. From stoichiometry, 2 mol H₂ produces 2 mol H₂O.

In aqueous solution, (CH₃)₂NH is strongest base among methylamines due to combined inductive and solvation effects.

In neutral/weakly alkaline medium: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. n-factor = 3. Eq. wt. = 158/3 ≈ 52.67 g/equiv.

Moles HCl = 0.01, moles NaOH = 0.02. After neutralization, excess NaOH = 0.01 mol in 200 mL. [OH⁻] = 0.01/0.2 = 0.05 M. pOH = 1.3, pH = 12.7.

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. So 2 moles propane produce 6 moles CO₂.

NH₄⁺ is acidic: Ka = Kw/Kb = 10⁻¹⁴/1.8×10⁻⁵ = 5.56×10⁻¹⁰. [H⁺] = √(Ka·C) = √(5.56×10⁻¹¹) = 7.46×10⁻⁶ M. pH = 5.13.

Normal rainwater has pH ~5.6 due to dissolved CO₂. Acid rain has pH < 5.6 due to dissolved SO₂ and NOₓ.

Sodium benzoate, common salt (NaCl), and vinegar (acetic acid solution) are all used as food preservatives.

Molar mass of O in CO₂ = 32 g/mol (2 atoms × 16 g/mol). 8.0 g oxygen = 8/32 = 0.25 mol of O atoms. Since each CO₂ molecule contains 2 O atoms, moles of CO₂ = 0.25/2 = 0.125 mol. Wait, let's recalculate: 8g O atoms = 8/16 = 0.5 mol O atoms. Each CO₂ has 2 O atoms, so moles of CO₂ = 0.5/2 = 0.25 mol.

HCl is a strong acid that completely dissociates: [H⁺] = 0.01 M = 10⁻² M. pH = -log[H⁺] = -log(10⁻²) = 2.

Work is a path function, not a state function. State functions (internal energy, enthalpy, entropy) depend only on the initial and final states, not on the path taken.

1 mole of any substance contains Avogadro's number (6.022 × 10²³) of particles. For helium (monatomic), this equals the number of atoms.

Equivalent weight = Molar mass / n-factor. For H₂SO₄ acting as diprotic acid, n-factor = 2. Equivalent weight = 98/2 = 49 g/equiv.

For weak acid: [H⁺] ≈ √(Ka·C) = √(1.8×10⁻⁵ × 0.1) = √(1.8×10⁻⁶) ≈ 1.34×10⁻³ M. pH = -log(1.34×10⁻³) ≈ 2.87 ≈ 3.

By convention, the standard enthalpy of formation (ΔHf°) of elements in their most stable form at 1 bar and specified temperature is zero.

Henderson-Hasselbalch equation: pH = pKa + log([salt]/[acid]) = -log(1.8×10⁻⁵) + log(0.1/0.1) = 4.74 + 0 = 4.74.

Graham's law: Rate ∝ 1/√M, where M is molar mass. Rate is inversely proportional to the square root of molecular mass (or density at same T and P).

[OH⁻] = 0.001 M = 10⁻³ M. pOH = 3. pH = 14 - pOH = 11.

From stoichiometry: 2 mol KClO₃ produces 3 mol O₂. So for 1.5 mol O₂: (2/3)×1.5 = 1.0 mol KClO₃ required.

In aqueous solution, (CH₃)₂NH is the strongest base among methylamines due to combined inductive effect and solvation effects (order: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃).

pH = -log(2.5×10⁻⁴) = 4 - log(2.5) = 4 - 0.40 = 3.60.

Heat of combustion of graphite: C(graphite) + O₂ → CO₂; ΔH = -393.5 kJ/mol. This is also the standard enthalpy of formation of CO₂ since graphite and O₂ are elements in standard states.

In acidic medium: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. n-factor = 5. Equivalent weight = 158/5 = 31.6 g/equiv.

Equal moles of strong acid and strong base neutralize completely: HCl + NaOH → NaCl + H₂O. Resulting solution is neutral salt in water, pH = 7 at 25°C.

Combustion reaction: C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O. So 1 mole butane produces 4 moles CO₂.

NH₄⁺ is acidic: Ka = Kw/Kb = 10⁻¹⁴/1.8×10⁻⁵ = 5.56×10⁻¹⁰. [H⁺] ≈ √(Ka·C) = √(5.56×10⁻¹⁰ × 0.1) = √(5.56×10⁻¹¹) ≈ 7.46×10⁻⁶ M. pH = -log(7.46×10⁻⁶) ≈ 5.13.