Practice Questions With Full Worked Solutions

Solution

Moles of H₂SO₄ = 0.050 L × 0.2 mol/L = 0.01 mol. Since H₂SO₄ is diprotic, this supplies 0.02 mol of H⁺.

Moles of NaOH = 0.050 L × 0.1 mol/L = 0.005 mol, supplying 0.005 mol of OH⁻.

Neutralization consumes H⁺ and OH⁻ in a 1:1 ratio, leaving excess H⁺ = 0.02 − 0.005 = 0.015 mol. Total volume = 100 mL = 0.1 L, so [H⁺] = 0.015/0.1 = 0.15 M.

pH = −log(0.15) = 1 − log(1.5) ≈ 1 − 0.18 = 0.82.

Answer: the resulting solution is strongly acidic, with pH ≈ 0.82. The step students most often skip is remembering that each mole of H₂SO₄ gives two moles of H⁺.

Solution

Let α = 0.4 be the degree of dissociation. Starting from 2 mol of PCl₅ in 1 L (so concentration = mol here):

At equilibrium: [PCl₅] = 2(1−α) = 2(0.6) = 1.2 M; [PCl₃] = [Cl₂] = 2α = 0.8 M each (both formed in equal amounts from the same dissociation).

K_c = [PCl₃][Cl₂] / [PCl₅] = (0.8 × 0.8) / 1.2 = 0.64/1.2 ≈ 0.53 mol/L.

Answer: K_c ≈ 0.53 mol/L. Setting up an ICE (initial–change–equilibrium) table before substituting is what prevents sign errors in this type of question.

Solution

S_N1 reactions proceed through a carbocation intermediate, so the rate tracks how stable that carbocation is — more substituted carbocations are stabilised by hyperconjugation and the electron-donating (+I) effect of alkyl groups.

tert-Butyl bromide ionises to a tertiary carbocation (three alkyl groups stabilising the positive charge) — most stable, forms fastest. Isopropyl bromide gives a secondary carbocation — less stabilised. Ethyl bromide would have to form a primary carbocation, which is so unstable that it barely reacts by this pathway at all; in practice, ethyl bromide reacts mainly by S_N2 instead.

Answer: tert-butyl bromide > isopropyl bromide > ethyl bromide. Remember: S_N1 favours more-substituted substrates; S_N2 favours less-substituted ones — the two mechanisms pull in opposite directions.

Solution

A dihybrid cross between two double heterozygotes (TtRr × TtRr) follows Mendel's law of independent assortment, producing the classic 9:3:3:1 phenotypic ratio: Tall Round : Tall wrinkled : Dwarf round : Dwarf wrinkled.

Alternatively, treat each trait separately: P(dwarf) = 1/4 (from Tt × Tt) and P(round) = 3/4 (from Rr × Rr). Since the two genes assort independently, P(dwarf and round) = 1/4 × 3/4 = 3/16.

Answer: 3/16 of the offspring are expected to be dwarf with round seeds. Breaking a dihybrid cross into two independent monohybrid crosses and multiplying probabilities is usually faster than drawing the full 16-box Punnett square.

Solution

At rest, the neuron's interior is negative relative to the outside (resting potential, about −70 mV), maintained by the Na⁺/K⁺ pump and the membrane's differential permeability.

When a stimulus brings the membrane to threshold, voltage-gated Na⁺ channels open and Na⁺ rushes in, driving the potential rapidly positive — this is depolarization. The Na⁺ channels then inactivate, and voltage-gated K⁺ channels open, letting K⁺ flow out and bringing the potential back down — this is repolarization, which often briefly overshoots to more negative than the resting value (hyperpolarization) before the Na⁺/K⁺ pump restores the original ionic gradients.

Answer: while the Na⁺ channels are open and then inactivated (roughly the depolarization and early repolarization phase), the neuron cannot fire another action potential no matter how strong the stimulus — this is the absolute refractory period. During the later hyperpolarization phase, only an unusually strong stimulus can trigger a new impulse (relative refractory period).

Solution

N_t = N₀e^rt = 100 × e^(0.2)(5) = 100 × e¹.

e¹ ≈ 2.718, so N_t ≈ 100 × 2.718 = 271.8.

Answer: approximately 272 bacteria after 5 hours. This exponential model (N_t = N₀e^rt) assumes unlimited resources; it's the same growth pattern discussed alongside the logistic (S-shaped) model for populations that eventually hit a carrying capacity.