A 5.00 g mixture of Na₂CO₃ and NaHCO₃ is heated until only Na₂CO₃ remains. Its mass decreases by 0.620 g. What was the mass of NaHCO₃ in the original mixture? Use Na = 23, C = 12, O = 16, H = 1.
A. 0.840 g
B. 1.68 g
C. 2.48 g
D. 3.36 g
Answer: B
On heating, 2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O. Every 168 g of NaHCO₃ loses 62 g as gases. Therefore the mass of NaHCO₃ is 0.620 × 168/62 = 1.68 g. The original carbonate does not contribute to this mass loss.
An oxide contains 70.0% iron and 30.0% oxygen by mass. With Fe = 56 and O = 16, its empirical formula is:
A. FeO
B. Fe₂O₃
C. Fe₃O₄
D. FeO₂
Answer: B
Take 100 g of oxide. The amounts are 70/56 = 1.25 mol Fe and 30/16 = 1.875 mol O. Dividing by 0.625 gives Fe:O = 2:3, so the empirical formula is Fe₂O₃.
For 2Al + 3Cl₂ → 2AlCl₃, 0.400 mol Al is mixed with 0.450 mol Cl₂. Assuming complete reaction, how much Al remains?
A. 0 mol
B. 0.050 mol
C. 0.100 mol
D. 0.150 mol
Answer: C
The available chlorine consumes (2/3) × 0.450 = 0.300 mol aluminium. Aluminium initially present is 0.400 mol. Thus 0.100 mol remains; chlorine is the limiting reagent.
An electron in a hydrogen atom falls from n = 4 to n = 2. What is the magnitude of the emitted photon energy? Use Eₙ = −13.6/n² eV.
A. 0.850 eV
B. 1.89 eV
C. 2.55 eV
D. 3.40 eV
Answer: C
E₄ = −0.850 eV and E₂ = −3.40 eV. The atom loses 2.55 eV, which is the photon energy. Use the magnitude of the energy difference, not the energy of either level alone.
An electron and a proton have the same non-zero kinetic energy and move non-relativistically. The ratio λₑ/λₚ of their de Broglie wavelengths is:
A. mₑ/mₚ
B. mₚ/mₑ
C. √(mₑ/mₚ)
D. √(mₚ/mₑ)
Answer: D
Since K = p²/(2m), momentum p = √(2mK). Therefore λ = h/√(2mK). At equal kinetic energy, wavelength is inversely proportional to √m, giving λₑ/λₚ = √(mₚ/mₑ).
What is the maximum number of electrons in one atom with n = 3 and mₗ = 0?
A. 2
B. 6
C. 10
D. 18
Answer: B
For n = 3, the allowed l values are 0, 1 and 2. Each of these subshells has exactly one orbital with mₗ = 0. Three such orbitals can hold two electrons each, giving six electrons.
Which gas-phase molecule has a permanent dipole moment?
A. CO₂
B. BF₃
C. SF₄
D. XeF₄
Answer: C
CO₂ is linear, BF₃ is trigonal planar and XeF₄ is square planar; their bond dipoles cancel by symmetry. SF₄ has a seesaw molecular shape because of one equatorial lone pair, so its dipoles do not cancel.
Using the usual molecular orbital description, which species has the highest bond order?
A. O₂
B. O₂⁺
C. O₂⁻
D. O₂²⁻
Answer: B
O₂ has bond order 2. Removing one electron from an antibonding π* orbital gives O₂⁺ a bond order of 2.5. Adding one or two antibonding electrons gives O₂⁻ and O₂²⁻ bond orders of 1.5 and 1 respectively.
In the usual molecular orbital model, which species is diamagnetic?
A. O₂
B. O₂⁺
C. N₂
D. NO
Answer: C
N₂ has all electrons paired and is diamagnetic. O₂ has two unpaired electrons. O₂⁺ and NO have an odd number of electrons and each has an unpaired electron.
Among the following ground-state gaseous atoms, which has the lowest first ionisation energy?
A. B
B. C
C. N
D. O
Answer: A
Across this part of Period 2, increasing effective nuclear charge generally raises first ionisation energy. Although O lies below N because pairing makes removal easier, boron has the lowest first ionisation energy among these four atoms.
Equal masses of He and O₂ are placed in a rigid vessel at one temperature. Assuming ideal behaviour, the ratio of the partial pressure of He to that of O₂ is:
A. 1:8
B. 1:1
C. 4:1
D. 8:1
Answer: D
For each gas pᵢ = nᵢRT/V. With equal mass m, n(He) = m/4 and n(O₂) = m/32. Their ratio, and hence their partial-pressure ratio, is 8:1.
An ideal gas mixture contains 2.0 mol N₂ and 1.0 mol CO₂. Its total pressure is 6.0 bar. The partial pressure of CO₂ is:
A. 1.0 bar
B. 2.0 bar
C. 3.0 bar
D. 4.0 bar
Answer: B
CO₂ has mole fraction 1/(2 + 1) = 1/3. Dalton's law gives p(CO₂) = (1/3) × 6.0 = 2.0 bar.
For a reaction, ΔH° = +40.0 kJ mol⁻¹ and ΔS° = +100 J K⁻¹ mol⁻¹. Treat both as temperature independent. Under standard conditions the reaction is spontaneous when:
A. T < 400 K
B. T > 400 K
C. At every positive temperature
D. At no positive temperature
Answer: B
Convert entropy to 0.100 kJ K⁻¹ mol⁻¹. Then ΔG° = 40.0 − 0.100T. Spontaneity requires ΔG° < 0, hence T > 400 K. At exactly 400 K, ΔG° = 0.
A system absorbs 500 J of heat and does 200 J of work on its surroundings. Its change in internal energy is:
A. +700 J
B. +300 J
C. −300 J
D. −700 J
Answer: B
In the chemistry sign convention, ΔU = q + w. Heat absorbed gives q = +500 J. Work done by the system gives w = −200 J. Therefore ΔU = +300 J.
At 298 K, for A(g) → 2B(g), ΔU = 10.00 kJ mol⁻¹ of reaction. Assuming ideal gases, what is ΔH? Use R = 8.314 J mol⁻¹ K⁻¹.
A. 7.52 kJ mol⁻¹
B. 10.00 kJ mol⁻¹
C. 12.48 kJ mol⁻¹
D. 14.96 kJ mol⁻¹
Answer: C
For ideal gases, ΔH = ΔU + Δn(gas)RT. Here Δn = 2 − 1 = 1. The correction is 8.314 × 298/1000 = 2.48 kJ mol⁻¹, so ΔH = 12.48 kJ mol⁻¹.
For A ⇌ B, Kc = 4.0. Initially a vessel contains 1.0 mol A and no B. If volume is constant, the equilibrium amount of B is:
A. 0.20 mol
B. 0.50 mol
C. 0.75 mol
D. 0.80 mol
Answer: D
Let x mol A react. Then n(B) = x and n(A) = 1 − x. The common volume cancels in Kc = x/(1 − x) = 4. Solving gives x = 0.80 mol.
For N₂O₄(g) ⇌ 2NO₂(g), one mole of pure N₂O₄ is allowed to dissociate. The equilibrium total pressure is P and the degree of dissociation is α. Using partial pressures, Kp is:
A. α²P/(1 − α²)
B. 4α²P/(1 − α²)
C. 2αP/(1 − α)
D. 4α²/(P(1 − α²))
Answer: B
At equilibrium the amounts are 1 − α and 2α, totalling 1 + α. Hence p(NO₂) = 2αP/(1 + α) and p(N₂O₄) = (1 − α)P/(1 + α). Substitute into p(NO₂)²/p(N₂O₄) to get 4α²P/(1 − α²). This uses the pressure-based school convention.
A buffer contains 0.100 mol CH₃COOH and 0.100 mol CH₃COO⁻. After adding 0.010 mol HCl, what is its approximate pH? Use pKa = 4.76 and log₁₀(9/11) = −0.087.
A. 4.67
B. 4.76
C. 4.85
D. 5.76
Answer: A
Strong acid first reacts with acetate: H⁺ + CH₃COO⁻ → CH₃COOH. The new amounts are 0.090 mol base and 0.110 mol acid. Henderson–Hasselbalch gives pH = 4.76 + log(0.090/0.110) = 4.673.
At 25 °C, a 0.0100 M solution of a weak monoprotic acid has Ka = 1.00 × 10⁻⁶. Its approximate pH is:
A. 2.00
B. 3.00
C. 4.00
D. 6.00
Answer: C
For small dissociation, [H⁺] ≈ √(KaC) = √(10⁻⁶ × 10⁻²) = 10⁻⁴ M, giving pH ≈ 4.00. The estimated dissociation is 1%, so the approximation is reasonable.
For AgCl, Ksp = 1.0 × 10⁻¹⁰. Ignoring complex formation and activity effects, its approximate molar solubility in 0.010 M NaCl is:
A. 1.0 × 10⁻⁵ M
B. 1.0 × 10⁻⁸ M
C. 1.0 × 10⁻¹⁰ M
D. 1.0 × 10⁻¹² M
Answer: B
Write Ksp = [Ag⁺][Cl⁻]. The NaCl supplies approximately 0.010 M chloride, far more than dissolving AgCl contributes. Thus s = 10⁻¹⁰/10⁻² = 10⁻⁸ M. The common ion suppresses solubility.
A nonvolatile nonelectrolyte has mole fraction 0.020 in an ideal solution. The relative lowering of the solvent vapour pressure is:
A. 0.010
B. 0.020
C. 0.040
D. 0.980
Answer: B
Raoult's law gives p = x(solvent)p°. Therefore (p° − p)/p° = 1 − x(solvent) = x(solute) = 0.020.
A 0.010 molal aqueous CaCl₂ solution dissociates completely. In the ideal dilute limit, what is its freezing-point depression? Kf(water) = 1.86 K kg mol⁻¹.
A. 0.0186 K
B. 0.0372 K
C. 0.0558 K
D. 0.186 K
Answer: C
CaCl₂ produces one Ca²⁺ and two Cl⁻ ions, so the ideal van't Hoff factor is 3. ΔTf = iKfm = 3 × 1.86 × 0.010 = 0.0558 K.
A solute dimerises in a dilute solution. If 60% of its original molecules become part of dimers, what is the van't Hoff factor?
A. 0.40
B. 0.60
C. 0.70
D. 1.60
Answer: C
Start with 1 mol of monomer units. After association, 0.40 mol monomers remain and 0.60/2 = 0.30 mol dimers form. The total number of solute particles is 0.70 mol, so i = 0.70.
For Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), E°cell = 1.10 V. At 298 K, [Zn²⁺] = 1.0 M and [Cu²⁺] = 0.010 M. Use 0.0592 V for 2.303RT/F. Ecell is approximately:
A. 0.982 V
B. 1.041 V
C. 1.100 V
D. 1.159 V
Answer: B
Two electrons are transferred and Q = [Zn²⁺]/[Cu²⁺] = 100. E = E° − (0.0592/2)log Q = 1.10 − 0.0592 = 1.0408 V. Pure solids do not enter Q.
How much copper is deposited when 0.100 mol electrons pass through CuSO₄ solution with 100% current efficiency? Use Cu = 63.5.
A. 3.175 g
B. 6.350 g
C. 12.70 g
D. 0.635 g
Answer: A
The cathode reaction is Cu²⁺ + 2e⁻ → Cu. Thus 0.100 mol electrons deposit 0.0500 mol Cu. Its mass is 0.0500 × 63.5 = 3.175 g.
For a galvanic reaction with n = 2 and E°cell = 0.500 V, ΔG° is approximately: F = 96485 C mol⁻¹.
A. −96.5 kJ mol⁻¹
B. +96.5 kJ mol⁻¹
C. −48.2 kJ mol⁻¹
D. +48.2 kJ mol⁻¹
Answer: A
Use ΔG° = −nFE°. This gives −2 × 96485 × 0.500 = −96485 J mol⁻¹, or about −96.5 kJ mol⁻¹. A positive standard cell potential corresponds to a negative standard Gibbs energy change.
A first-order reaction is 75% complete after 20 minutes. Its half-life is:
A. 5 min
B. 10 min
C. 15 min
D. 20 min
Answer: B
75% completion leaves 25% of the initial reactant, which is (1/2)². Two half-lives have elapsed in 20 minutes, so the half-life is 10 minutes.
For a reaction, doubling [A] at fixed [B] multiplies the initial rate by 4. Tripling [B] at fixed [A] multiplies it by 3. The rate law is:
A. k[A][B]
B. k[A]²[B]
C. k[A][B]²
D. k[A]²[B]²
Answer: B
If rate ∝ [A]ᵐ, then 2ᵐ = 4 and m = 2. Likewise 3ⁿ = 3 gives n = 1. Therefore rate = k[A]²[B]. Reaction orders must be inferred from rate data, not simply from a balanced overall equation.
A reaction is zero-order in A over the stated concentration range. Which graph is a straight line with negative slope?
A. ln[A] versus time
B. 1/[A] versus time
C. [A] versus time
D. Rate versus time with slope −k
Answer: C
The zero-order integrated law is [A] = [A]₀ − kt, so [A] versus time is linear with slope −k. Its rate is constant within the applicable range, so the rate–time slope is zero.
An ideal face-centred cubic unit cell contains how many atoms in total?
A. 1
B. 2
C. 4
D. 6
Answer: C
Eight corner atoms each contribute 1/8, giving one atom. Six face-centred atoms each contribute 1/2, giving three atoms. The total is four atoms per unit cell.
In an ideal NaCl crystal, the coordination numbers of Na⁺ and Cl⁻ are respectively:
A. 4 and 4
B. 6 and 6
C. 8 and 8
D. 6 and 3
Answer: B
The rock-salt structure places each Na⁺ in an octahedral environment of six Cl⁻ ions, and each Cl⁻ likewise has six nearest Na⁺ neighbours. Both coordination numbers are six.
The oxidation state and d-electron count of cobalt in [Co(NH₃)₅Cl]Cl₂ are:
A. +2, d⁷
B. +3, d⁶
C. +3, d⁷
D. +1, d⁸
Answer: B
Two external Cl⁻ ions require a complex charge of +2. NH₃ is neutral and the coordinated Cl has charge −1, so x − 1 = +2 and Co is +3. Co is [Ar]3d⁷4s²; removing three electrons gives d⁶.
In the usual crystal-field model, [Fe(CN)₆]⁴⁻ has how many unpaired electrons?
A. 0
B. 2
C. 4
D. 5
Answer: A
The iron oxidation state is +2, so it is d⁶. CN⁻ is a strong-field ligand and this octahedral complex is low spin. Its six d electrons occupy t₂g⁶, all paired.
An octahedral d⁸ metal ion has two unpaired electrons. Its spin-only magnetic moment is closest to:
A. 1.00 BM
B. 1.73 BM
C. 2.83 BM
D. 4.90 BM
Answer: C
Use μ = √[n(n + 2)] BM, where n is the number of unpaired electrons. For n = 2, μ = √8 = 2.83 BM. This is the spin-only estimate.
Which pair consists of linkage isomers?
A. [Co(NH₃)₅(NO₂)]²⁺ with N-bound NO₂ and the same complex with O-bound ONO
B. cis- and trans-[Pt(NH₃)₂Cl₂]
C. Two mirror images of [Co(en)₃]³⁺
D. [Co(NH₃)₆]³⁺ and [Co(NH₃)₆]²⁺
Answer: A
The ambidentate nitrite ligand can bind through N or through O, giving linkage isomerism. The platinum pair shows geometrical isomerism, the en pair optical isomerism, and the last pair differs in oxidation state.
Which species is the most acidic in water under ordinary conditions?
A. Ethanol
B. Phenol
C. Ethanoic acid
D. Ethane
Answer: C
Ethanoic acid forms a carboxylate ion with negative charge delocalised over two electronegative oxygen atoms. Its pKa is much lower than those of phenol, ethanol and ethane, so it is the strongest acid of this set.
Which substrate generally reacts fastest by an SN1 mechanism in a polar protic solvent?
A. CH₃Br
B. CH₃CH₂Br
C. (CH₃)₂CHBr
D. (CH₃)₃CBr
Answer: D
The rate-determining SN1 step forms a carbocation. The tertiary carbocation from tert-butyl bromide is stabilised by alkyl groups more than the secondary or primary alternatives. Methyl carbocations are especially unfavourable.
How many stereogenic carbon centres are present in CH₃CH(OH)CH₂CH₃?
A. 0
B. 1
C. 2
D. 3
Answer: B
The carbon carrying OH is attached to H, OH, CH₃ and CH₂CH₃: four different groups. It is stereogenic. None of the other carbons has four different substituents.
Reductive ozonolysis of an alkene produces only ethanal. Which alkene fits?
A. Ethene
B. Propene
C. But-1-ene
D. But-2-ene
Answer: D
Cut the C=C bond and replace each alkene carbon's double bond by C=O. CH₃CH=CHCH₃ gives two CH₃CHO molecules. Both cis- and trans-but-2-ene give this same constitutional product.
Addition of HBr to propene in the presence of a radical initiator under suitable peroxide conditions gives predominantly:
A. 1-bromopropane
B. 2-bromopropane
C. 1,2-dibromopropane
D. Propan-2-ol
Answer: A
In the radical pathway, Br· adds to the terminal carbon so that the intermediate carbon radical is secondary. That radical takes H from HBr, giving CH₃CH₂CH₂Br. This is the peroxide effect for HBr.
Which reagent converts a primary alcohol to an aldehyde while limiting further oxidation under suitable anhydrous conditions?
A. PCC in CH₂Cl₂
B. Hot aqueous acidified KMnO₄ in excess
C. NaBH₄
D. H₂/Ni
Answer: A
PCC under anhydrous conditions is a standard reagent for stopping primary-alcohol oxidation at the aldehyde. Strong aqueous oxidising conditions commonly continue to the acid. NaBH₄ and H₂/Ni are reducing reagents.
Which compound gives a positive iodoform test?
A. Methanol
B. Propan-1-ol
C. Propan-2-ol
D. Benzyl alcohol
Answer: C
Propan-2-ol is oxidised under the test conditions to propanone, a methyl ketone. Methyl ketones undergo haloform cleavage to yellow CHI₃. The other alcohols listed do not provide the required group.
An aldehyde undergoes a Cannizzaro reaction under suitable concentrated-base conditions when it has:
A. At least one α-hydrogen
B. No α-hydrogen
C. An ester group
D. A carbon–carbon triple bond
Answer: B
Aldehydes lacking an α-hydrogen cannot form an enolate for the usual aldol pathway. Under suitable concentrated-base conditions, one aldehyde molecule is oxidised to carboxylate and another is reduced to an alcohol.
Which bond directly joins amino-acid residues along the backbone of a polypeptide?
A. Glycosidic bond
B. Peptide bond
C. Phosphodiester bond
D. Hydrogen bond only
Answer: B
The amino group of one amino acid and the carboxyl group of another form an amide linkage, −CO−NH−, called a peptide bond. Hydrogen bonds help stabilise folding but do not form the covalent backbone.
At fixed temperature, an adsorption isotherm is described by θ = KP/(1 + KP), with K > 0. As P becomes very large, θ approaches:
A. 0
B. 1/2
C. 1
D. Infinity
Answer: C
Divide numerator and denominator by KP. Then θ = 1/(1 + 1/KP). As pressure becomes very large, 1/KP approaches zero, so θ approaches 1: saturation of the available sites.
In acidic solution, how many moles of Fe²⁺ are oxidised by 1 mol MnO₄⁻ when manganese ends as Mn²⁺?
A. 1
B. 3
C. 5
D. 7
Answer: C
Mn falls from +7 to +2, gaining five electrons. Each Fe²⁺ → Fe³⁺ supplies one electron, so five Fe²⁺ ions are needed per permanganate ion.
A student prepares 250.0 mL of 0.1000 M Na₂CO₃ using anhydrous solid. What mass is required? M(Na₂CO₃) = 106.0 g mol⁻¹.
A. 0.2650 g
B. 1.060 g
C. 2.650 g
D. 26.50 g
Answer: C
Convert 250.0 mL to 0.2500 L. The required amount is CV = 0.1000 × 0.2500 = 0.02500 mol. Multiplying by 106.0 gives 2.650 g. Dissolve, then make the final solution volume up to the mark.
A burette has independent reading uncertainties of ±0.05 mL at each end. Using a worst-case sum, the uncertainty in a delivered volume of 20.00 mL is:
A. ±0.025 mL
B. ±0.05 mL
C. ±0.10 mL
D. ±1.00 mL
Answer: C
Delivered volume is final reading minus initial reading. In a worst-case bound, the absolute uncertainties add: 0.05 + 0.05 = 0.10 mL. This is 0.50% of 20.00 mL. A statistical root-sum-square estimate would be different; the question explicitly asks for a worst-case sum.
Which statements about molecular shape are correct in the usual VSEPR description?
A. NH₃ is trigonal pyramidal.
B. CO₂ is bent.
C. XeF₄ is square planar.
D. SF₆ is octahedral.
Answer: A, C, D
NH₃ has three bonding pairs and one lone pair, giving a trigonal-pyramidal molecular shape. CO₂ has two electron domains on carbon and is linear, so B is false. XeF₄ has four bonds and two opposite lone pairs in an octahedral electron-domain arrangement. SF₆ has six bonds and no lone pairs on sulfur.
Which statements are correct?
A. For an ideal gas, internal energy depends only on temperature.
B. For a cyclic process, ΔU = 0.
C. A spontaneous process must always have ΔH < 0.
D. At constant T and P, ΔG < 0 is the criterion for a spontaneous change in a closed system with only pressure–volume work.
Answer: A, B, D
An ideal gas has no intermolecular potential-energy contribution, so its internal energy depends only on temperature. Internal energy is a state function and returns to its initial value in a cycle. An endothermic change can still be spontaneous when the entropy contribution is favourable. Under the stated constraints, Gibbs energy decreases for a spontaneous change.
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at equilibrium, which statements are correct for ideal gases?
A. Adding a catalyst changes Kp.
B. Compressing the mixture at fixed temperature favours NH₃.
C. Adding an inert gas at fixed volume and temperature changes none of the reacting gases' partial pressures.
D. At equilibrium, forward and reverse reactions stop.
Answer: B, C
Compression raises the reacting partial pressures and favours the side with fewer gas molecules, here NH₃. An inert gas at fixed V and T does not change nᵢRT/V for any reacting species. A catalyst changes the speed of reaching equilibrium, not Kp. Equilibrium is dynamic: both reaction rates remain equal and non-zero.
For an operating galvanic cell, which statements are correct?
A. Oxidation occurs at the anode.
B. Reduction occurs at the cathode.
C. Electrons pass through the external circuit from anode to cathode.
D. A salt bridge primarily carries electrons between the half-cells.
Answer: A, B, C
Oxidation releases electrons at the anode. They travel through the external conductor to the cathode, where reduction consumes them. A salt bridge conducts ions and helps maintain electrical neutrality; electrons do not travel through it as the primary charge carriers.
For an elementary first-order decay obeying [A] = [A]₀e⁻ᵏᵗ, which statements are correct?
A. The half-life equals ln 2/k.
B. The units of k can be s⁻¹.
C. Doubling [A] at fixed temperature doubles the instantaneous rate.
D. The half-life is proportional to the initial concentration.
Answer: A, B, C
Setting [A]/[A]₀ = 1/2 gives t½ = ln 2/k. Because rate = k[A], k has inverse-time units and doubling [A] doubles the rate. The initial concentration cancels from the half-life expression, so D is false.
For octahedral [Co(en)₃]³⁺, where en is neutral ethane-1,2-diamine, which statements are correct?
A. The cobalt oxidation state is +3.
B. The coordination number is 6.
C. Each en ligand is bidentate.
D. The complex has three coordinated chloride ligands.
Answer: A, B, C
Neutral en contributes no charge, so cobalt is +3. Each of the three en ligands binds through two nitrogen atoms, giving coordination number six. There is no coordinated chloride in the formula. Octahedral tris-chelate complexes of this kind can show optical isomerism.
Which statements are correct?
A. Enantiomers are non-superimposable mirror images.
B. A meso compound can contain stereogenic centres but be achiral.
C. Every molecule containing a stereogenic carbon must be optically active as a bulk sample.
D. A racemic mixture has zero net optical rotation under ordinary measurement conditions.
Answer: A, B, D
Enantiomers are non-superimposable mirror images. Internal symmetry can make a molecule with stereogenic centres achiral, as in a meso compound. A racemic mixture contains equal amounts of enantiomers whose rotations cancel. Therefore the presence of a stereogenic carbon alone does not guarantee a rotating bulk sample.
Which statements hold for an ideal dilute solution of a nonvolatile nonelectrolyte?
A. Its vapour pressure is lower than that of the pure solvent at the same temperature.
B. Its freezing point is lower than that of the pure solvent.
C. Its boiling point at fixed external pressure is lower than that of the pure solvent.
D. Its osmotic pressure is approximately cRT.
Answer: A, B, D
A nonvolatile solute lowers solvent vapour pressure. Consequently the freezing point falls, while the boiling point at a fixed external pressure rises. In the dilute ideal limit, a nonelectrolyte has i = 1 and π = cRT. Thus C is the incorrect statement.
Which changes are oxidations?
A. Fe²⁺ → Fe³⁺
B. 2Cl⁻ → Cl₂
C. Cr₂O₇²⁻ → Cr³⁺ in acidic solution
D. CH₃CH₂OH → CH₃CHO
Answer: A, B, D
Fe increases from +2 to +3. Chloride increases from −1 to 0 in Cl₂. Dichromate chromium decreases from +6 to +3, so that is reduction. Converting ethanol to ethanal removes hydrogen and increases the oxidation state of the alcohol carbon, so it is oxidation.
Which sets (n, l, mₗ) are allowed for an electron in an atom?
A. (2, 2, 0)
B. (3, 3, 0)
C. (4, 2, −3)
D. (1, 0, 0)
Answer: D
The rules are n = 1, 2, …; l = 0 to n − 1; and mₗ = −l to +l. A and B both violate l ≤ n − 1. For C, mₗ cannot be −3 when l = 2. Only D obeys every restriction. A multiple-correct-format question can have just one correct choice.
For a titration of a measured strong acid with a standard strong base, which actions are appropriate?
A. Rinse the burette with a little of the titrant before filling it.
B. Remove air bubbles from the burette tip before recording the initial reading.
C. Read the bottom of a clear aqueous meniscus at eye level.
D. Rinse the receiving conical flask with the acid sample and leave that rinse in the flask in addition to the measured aliquot.
Answer: A, B, C
Conditioning the burette prevents residual water from diluting the titrant. A tip bubble can make the recorded delivered volume exceed the amount actually reaching the flask. Eye-level reading reduces parallax. Leaving extra acid rinse in the flask adds an unmeasured amount of analyte, invalidating the titre.
For a weak-acid/conjugate-base buffer in the range where Henderson–Hasselbalch applies, which statements are correct?
A. If acid and conjugate-base concentrations are equal, pH ≈ pKa.
B. Adding a small amount of strong acid converts some conjugate base into weak acid.
C. A buffer has a finite capacity to absorb added strong acid or base.
D. Dilution that preserves the base/acid ratio leaves the approximate pH unchanged, provided the buffer approximation remains valid.
Answer: A, B, C, D
pH = pKa + log([base]/[acid]), so equal concentrations give pH ≈ pKa. Added H⁺ consumes conjugate base. Buffer capacity is finite. Moderate dilution preserves the ratio and approximately the pH, but reduces capacity; at extreme dilution the approximation can fail. All four statements are correct, so all four must be selected.