CN01 · Quiz — Basic Concepts of Chemistry

Question 1 of 20

The SI unit of amount of substance is the mole. One mole of any substance contains exactly:

A6.022 × 10²³ elementary entities (Avogadro's number, NA) B1.66 × 10⁻²⁷ particles C1 gram of the substance DThe same mass as one molecule of that substance

One mole contains exactly 6.02214076 × 10²³ elementary entities (Avogadro's number) — a fixed constant since the 2019 SI redefinition. 1 u (atomic mass unit) = 1/NA grams = 1.66054 × 10⁻²⁷ kg. The molar mass in g/mol is numerically equal to the relative atomic/molecular mass.

Question 2 of 20

Dalton's Law of Partial Pressures states that the total pressure of a gaseous mixture equals:

AThe pressure of the most abundant gas only BThe geometric mean of individual gas pressures CThe sum of the partial pressures of each individual gas DThe product of individual pressures divided by volume

Dalton's Law (1801): Ptotal = ΣPi = P1 + P2 + P3... where Pi = xiPtotal (xi = mole fraction). Applies to ideal gases that do not react with each other. Used in diving medicine (partial pressure of O2 and N2 at depth), respiratory physiology (pO2, pCO2), and gas collection over water (subtract water vapour pressure).

Question 3 of 20

The van der Waals equation (P + an²/V²)(V - nb) = nRT corrects the ideal gas law for:

AOnly temperature deviations in real gases BIntermolecular attractions and finite molecular volume in real gases COnly pressure deviations at very high temperatures DThe presence of charged particles in the gas

Van der Waals equation corrects for two real gas behaviours: (1) an²/V² = pressure correction for intermolecular attractions (cohesive forces pull molecules inward, reducing pressure on walls — add this back); (2) nb = volume correction for finite molecular volume (molecules occupy space — subtract from total volume). 'a' measures strength of attractions; 'b' is molecular volume per mole.

Question 4 of 20

Equivalent weight of H2SO4 in the neutralisation reaction H2SO4 + 2NaOH → Na2SO4 + 2H2O is:

A98 g/eq (same as molar mass) B49 g/eq (molar mass divided by 2 C32 g/eq D196 g/eq

Equivalent weight = molar mass / n-factor. For H2SO4 in neutralisation: n-factor = 2 (donates 2 H⁺ per molecule). Equivalent weight = 98/2 = 49 g/eq. For HCl: n-factor = 1, eq. wt = 36.5 g/eq. For H3PO4 in neutralisation (depends on reaction): n-factor = 1, 2, or 3. The n-factor changes with the specific reaction context.

Question 5 of 20

Raoult's Law states that for an ideal solution, the vapour pressure of a component is:

AEqual to the vapour pressure of the pure solvent at all concentrations BProportional to the mole fraction of that component: PA = xA × PA° CInversely proportional to the solute concentration DIndependent of temperature at constant composition

Raoult's Law: PA = xA × PA° (partial vapour pressure of component A = its mole fraction in solution × vapour pressure of pure A). Ideal solutions obey Raoult's Law (similar intermolecular forces between components). Positive deviations: interactions A-B weaker than A-A and B-B (e.g., ethanol-water). Negative deviations: A-B interactions stronger (e.g., acetone-chloroform).

Question 6 of 20

The normality (N) of a solution is related to molarity (M) by:

AN = M always (normality equals molarity for all solutes) BN = M × n-factor (n-factor = equivalents per mole) CN = M divided by the molar mass of the solute DN = M multiplied by the volume of solution in litres

N = M × n-factor. For H2SO4 (n=2): 1M H2SO4 = 2N H2SO4. For NaOH (n=1): 1M = 1N. For KMnO4 in acidic solution (n=5, gains 5e⁻): 1M KMnO4 = 5N. At equivalence point of a titration: N1V1 = N2V2 (normality × volume is equal). This is more convenient than the mole calculation for redox titrations.

Question 7 of 20

The empirical formula gives the simplest ratio of atoms. For a compound with 40% C, 6.67% H, and 53.3% O by mass, the empirical formula is:

AC2H4O2 BCH2O CC3H6O3 DCHO

Molar ratios: C: 40/12 = 3.33; H: 6.67/1 = 6.67; O: 53.3/16 = 3.33. Divide by smallest (3.33): C:H:O = 1:2:1 → empirical formula = CH2O. Molecular formula could be CH2O (formaldehyde, MW=30), C2H4O2 (acetic acid, MW=60), C6H12O6 (glucose, MW=180) etc., determined from molar mass.

Question 8 of 20

The limiting reagent in a chemical reaction is the reactant that:

AIs present in the smallest mass BIs present in the smallest number of moles CIs completely consumed first DHas the highest molar mass among the reactants

The limiting reagent (limiting reactant) is completely consumed when the reaction stops — it determines the theoretical yield. Example: N2 + 3H2 → 2NH3. If 1 mol N2 and 2 mol H2 are used: H2 requires 3 mol N2 per mol H2 used → H2 is limiting (only 2 mol available, need 3 for 1 mol N2). Theoretical yield = calculated assuming complete conversion of limiting reagent.

Question 9 of 20

Percentage yield of a reaction is calculated as:

A(Moles of product obtained / moles of reactant used) × 100 B(Actual yield / Theoretical yield) × 100 C(Mass of product / molar mass of product) × 100 D(Theoretical yield / Actual yield) × 100

% yield = (actual yield / theoretical yield) × 100. Actual yield < theoretical yield due to: incomplete reactions, side reactions, product losses during separation/purification. A 100% yield is ideal but rarely achieved. Industrial processes aim for high yields to be economically viable. In drug synthesis, overall yield over multiple steps becomes very important (each step's yield multiplies).

Question 10 of 20

Gay-Lussac's Law of Combining Volumes states that at the same T and P, gases react and form products in volumes that are in:

ARandom ratios depending on molecular size BSimple whole-number ratios CRatios equal to their molar masses DRatios equal to their densities

Gay-Lussac's Law (1808): volumes of gaseous reactants and products at the same T and P are in simple whole-number ratios equal to the stoichiometric coefficients. Example: H2(g) + Cl2(g) → 2HCl(g) — 1 vol H2 + 1 vol Cl2 → 2 vol HCl. This laid the groundwork for Avogadro's law and the concept of diatomic molecules.

Question 11 of 20

The molality (m) of a solution is defined as:

AMoles of solute per litre of solution (same as molarity) BMoles of solute per kilogram of solvent CGrams of solute per 100 mL of solution DMoles of solute per mole of solvent

Molality (m) = moles of solute / mass of solvent in kg. Unlike molarity (mol/L solution), molality does NOT change with temperature (solvent mass is temperature-independent). Used in colligative property calculations (boiling point elevation, freezing point depression): ΔTb = Kb × m; ΔTf = Kf × m. Molality is preferred for accurate thermodynamic work.

Question 12 of 20

The compressibility factor Z = PV/nRT for an ideal gas equals 1. For real gases at high pressure, Z is typically:

ALess than 1 — attractions dominate BGreater than 1 (Z > 1) CAlways exactly 1 regardless of conditions DEqual to the van der Waals constant 'a'

At high pressure: molecules are very close → repulsive forces dominate → actual volume > ideal volume → Z > 1. At moderate pressure (most real gases): attractive forces dominate → actual volume < ideal volume → Z < 1 (except for H2 and He which show Z > 1 even at moderate pressures due to very weak attractions). Z = 1 only for ideal gas.

Question 13 of 20

The law of conservation of mass and energy in chemical reactions is expressed through:

AThe Aufbau principle BHess's Law of constant heat summation CThe Pauli exclusion principle DGay-Lussac's Law of combining volumes

Hess's Law: the total enthalpy change (ΔH) for a chemical reaction is independent of the pathway — only the initial and final states matter (state function). This allows calculation of ΔH for reactions that cannot be measured directly (e.g., combustion of carbon to CO). ΔHreaction = ΣΔHf°(products) - ΣΔHf°(reactants).

Question 14 of 20

When 2 moles of hydrogen gas react with 1 mole of oxygen gas to form water: 2H2 + O2 → 2H2O, the volume of water vapour formed at STP is:

A22.4 L B44.8 L C11.2 L D33.6 L

At STP, 1 mole of any gas = 22.4 L. From the equation: 2 moles of H2 + 1 mole O2 → 2 moles H2O (gas). Volume of H2O(g) = 2 × 22.4 = 44.8 L. Note: at STP (0°C), water exists as vapour (water vapour). The molar volume of 22.4 L applies to any ideal gas at STP conditions (0°C, 1 atm).

Question 15 of 20

The mole fraction of a component in a mixture is defined as:

AMoles of component / mass of total mixture BMoles of component / total moles of all components in the mixture CMoles of component / volume of mixture in litres DMass of component / total mass of mixture

Mole fraction (xi) = ni / ntotal. Sum of all mole fractions in a mixture = 1 (Σxi = 1). Mole fraction is dimensionless. Used in Raoult's Law (vapour pressure), Dalton's Law (partial pressure: Pi = xi × Ptotal), and thermodynamic calculations. It is independent of temperature (unlike molarity).

Question 16 of 20

The percentage purity of a sample is important in stoichiometric calculations because:

APure substances have lower reactivity than impure ones BOnly the pure component reacts CImpurities increase the molar mass of the sample DPurity affects the density but not the number of moles

If a 10 g sample of NaOH is 95% pure: mass of pure NaOH = 9.5 g; moles = 9.5/40 = 0.2375 mol. Using 10 g directly gives 0.25 mol — overstating by ~5%. In pharmaceutical synthesis and industrial chemistry, purity affects dosage calculations and reaction yields critically. Always: mass of pure substance = total mass × (% purity / 100).

Question 17 of 20

The concept of 'theoretical yield' vs 'actual yield' arises because:

AChemists are not skilled enough to achieve theoretical yields BSide reactions, incomplete reactions, product losses during separation CChemical equations are never balanced correctly DLimiting reagent calculations contain inherent errors

Theoretical yield = maximum possible product if limiting reagent completely converts to product with no losses. Actual yield < theoretical due to: parallel side reactions producing unwanted products, reversible reactions not going to completion, losses during filtration/crystallisation/distillation, and physical losses. % yield = (actual/theoretical) × 100%.

Question 18 of 20

The critical temperature of a gas is:

AThe temperature at which the gas occupies zero volume BThe temperature above which the gas cannot be liquefied by compression alone CThe boiling point of the gas at standard pressure DThe temperature at which real and ideal gas behaviours coincide

Critical temperature (Tc): above Tc, the gas phase cannot be condensed to liquid no matter how high the pressure — kinetic energy exceeds intermolecular attractions. Tc = 8a/27Rb (from van der Waals). Examples: CO2 Tc = 31.1°C (dry ice used in fire extinguishers, above room temperature), N2 Tc = -147°C (must cool below -147°C to liquefy N2). Above Tc, only supercritical fluid exists.

Question 19 of 20

Ostwald's dilution law relates the degree of dissociation (α) of a weak electrolyte to its concentration (C) and dissociation constant (Ka):

AKa = Cα BKa ≈ Cα² (when α << 1, i.e. CKa = C²α DKa = α/C

For weak acid HA ⇌ H⁺ + A⁻: Ka = [H⁺][A⁻]/[HA] = (Cα)(Cα)/(C(1-α)) = Cα²/(1-α). When α << 1 (weak acid, not too dilute): Ka ≈ Cα². Therefore α ≈ √(Ka/C). As C decreases (dilution), α increases — more dissociation at lower concentration (Ostwald's dilution law). This explains why weak acids conduct electricity better in dilute solution.

Question 20 of 20

The pKa of a weak acid is related to Ka by:

ApKa = Ka × 10 BpKa = -log10 — smaller pKa CpKa = log10(Ka) DpKa = 1/Ka

pKa = -log10(Ka). Stronger acid: higher Ka → lower pKa. Examples: acetic acid (Ka = 1.8 × 10⁻⁵, pKa = 4.74), HF (Ka = 6.8 × 10⁻⁴, pKa = 3.17), HCN (Ka = 6.2 × 10⁻¹⁰, pKa = 9.21). HCl, HNO3 (strong acids): Ka >> 1, pKa < 0. At the half-equivalence point in a titration: pH = pKa (Henderson-Hasselbalch equation at equal concentrations of HA and A⁻).

CN01 — Basic Concepts of Chemistry