A-Level Chemistry Practice Test 11

60 original multiple-choice questions with answers & explanations — covering the full A-Level Chemistry syllabus (AQA, Edexcel & OCR), AS & A2.

Note: These are original practice questions written in A-Level multiple-choice style — not reproduced exam questions. Use them to test your readiness for the real A-Level Chemistry exams.

📱 How to use: Try each question first, then click “Show Answer” to check. Covers the full A-Level course: atomic structure, amount of substance, bonding, energetics & thermodynamics, kinetics, equilibria, redox & electrode potentials, acids & buffers, inorganic chemistry, and organic chemistry with spectroscopy.

⚛ Atomic Structure & Periodicity

Q1An element X has two isotopes, X-63 with 69% abundance and X-65 with 31% abundance. What is the relative atomic mass of X?

(A) 63.0
(B) 64.4
(C) 64.0
(D) 63.6
Show Answer

Correct: (D)

Ar = (63 × 0.69) + (65 × 0.31) = 43.47 + 20.15 = 63.62 ≈ 63.6.

Q2The first four ionisation energies of an element are 590, 1145, 4912 and 6474 kJ mol^-1. The large increase between the second and third values shows that this element is in

(A) Group 1
(B) Group 2
(C) Group 3
(D) Group 13
Show Answer

Correct: (B)

Two electrons are removed from the outer shell before an inner shell is broken into, which is characteristic of a Group 2 element.

Q3What is the electronic configuration of an Fe^2+ ion? (Fe: atomic number 26)

(A) 1s^2 2s^2 2p^6 3s^2 3p^6 3d^4 4s^2
(B) 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1
(C) 1s^2 2s^2 2p^6 3s^2 3p^6 3d^8
(D) 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6
Show Answer

Correct: (D)

Fe is [Ar] 3d^6 4s^2. Transition metal ions lose the 4s electrons first, so Fe^2+ is [Ar] 3d^6.

Q4In the mass spectrum of chlorine gas, Cl2, peaks are seen at m/z 70, 72 and 74 in an approximate ratio of 9:6:1. This pattern arises because

(A) chlorine has isotopes of mass 70, 72 and 74
(B) Cl2 molecules contain different combinations of Cl-35 and Cl-37 atoms
(C) the molecules fragment into single atoms in the spectrometer
(D) chlorine has three isotopes in equal abundance
Show Answer

Correct: (B)

The peaks are Cl-35/Cl-35 (70), Cl-35/Cl-37 (72) and Cl-37/Cl-37 (74); with abundances of about 3:1, the statistical ratio is 9:6:1.

Q5Which statement about the elements Na to Ar (Period 3) is correct?

(A) The melting point increases steadily across the period
(B) Argon has a higher melting point than sulfur
(C) Silicon has the highest melting point because of its giant covalent structure
(D) Aluminium has a lower melting point than sodium because it has more electrons
Show Answer

Correct: (C)

Silicon forms a giant covalent lattice with strong bonds throughout, giving it the highest melting point in the period.

⚖ Amount of Substance

Q6What volume, in dm^3, is occupied by 0.250 mol of a gas at 100 kPa pressure and 300 K? (R = 8.31 J K^-1 mol^-1)

(A) 6.23 dm^3
(B) 0.623 dm^3
(C) 62.3 dm^3
(D) 2.49 dm^3
Show Answer

Correct: (A)

V = nRT/p = (0.250 × 8.31 × 300) / 100 000 = 6.23 × 10^-3 m^3 = 6.23 dm^3.

Q7A 25.0 cm^3 sample of sodium hydroxide solution of concentration 0.100 mol dm^-3 required 20.0 cm^3 of sulfuric acid for complete neutralisation. What is the concentration of the sulfuric acid?

(A) 0.125 mol dm^-3
(B) 0.250 mol dm^-3
(C) 0.0625 mol dm^-3
(D) 0.0313 mol dm^-3
Show Answer

Correct: (C)

Moles NaOH = 0.100 × 0.0250 = 2.50 × 10^-3. H2SO4 reacts with 2 NaOH, so moles acid = 1.25 × 10^-3. Concentration = 1.25 × 10^-3 / 0.0200 = 0.0625 mol dm^-3.

Q8In a preparation, the theoretical yield of a product is 8.00 g but only 6.40 g of pure product is obtained. The percentage yield is

(A) 125%
(B) 80.0%
(C) 64.0%
(D) 20.0%
Show Answer

Correct: (B)

Percentage yield = (6.40 / 8.00) × 100 = 80.0%.

Q9Calcium carbonate is heated to produce calcium oxide: CaCO3 → CaO + CO2. (Mr: CaCO3 = 100, CaO = 56, CO2 = 44.) The atom economy for the production of CaO is

(A) 44%
(B) 100%
(C) 56%
(D) 36%
Show Answer

Correct: (C)

Atom economy = (56 / 100) × 100 = 56%.

Q10A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its empirical formula is

(A) C2H4O2
(B) CHO
(C) CH3O
(D) CH2O
Show Answer

Correct: (D)

Moles: C = 40.0/12 = 3.33, H = 6.7/1 = 6.7, O = 53.3/16 = 3.33. Ratio 1:2:1, so the empirical formula is CH2O.

Q11What mass of sodium chloride is needed to make 250 cm^3 of a 0.200 mol dm^-3 solution? (Mr NaCl = 58.5)

(A) 11.7 g
(B) 2.93 g
(C) 1.17 g
(D) 29.3 g
Show Answer

Correct: (B)

Moles = 0.200 × 0.250 = 0.0500 mol. Mass = 0.0500 × 58.5 = 2.93 g.

🔗 Bonding & Structure

Q12What is the bond angle in an NH4^+ ion?

(A) 107°
(B) 120°
(C) 180°
(D) 109.5°
Show Answer

Correct: (D)

NH4^+ has four bonding pairs and no lone pairs arranged tetrahedrally, giving a bond angle of 109.5°.

Q13Which substance has the highest boiling point?

(A) CH4
(B) CH3CH3
(C) CH3OH
(D) CH3Cl
Show Answer

Correct: (C)

Methanol molecules are held together by hydrogen bonds, which are much stronger than the van der Waals forces in the other substances listed.

Q14Given: ΔHf(NaCl) = −411 kJ mol^-1; atomisation of Na = +107 kJ mol^-1; first ionisation energy of Na = +496 kJ mol^-1; atomisation of Cl (per mol of Cl atoms) = +121 kJ mol^-1; electron affinity of Cl = −349 kJ mol^-1. The lattice enthalpy of formation of NaCl is

(A) −376 kJ mol^-1
(B) −411 kJ mol^-1
(C) −1572 kJ mol^-1
(D) −786 kJ mol^-1
Show Answer

Correct: (D)

By a Born-Haber cycle: ΔHlatt = −411 − (107 + 496 + 121 − 349) = −411 − 375 = −786 kJ mol^-1.

Q15Which molecule is polar overall?

(A) CCl4
(B) CHCl3
(C) CO2
(D) BF3
Show Answer

Correct: (B)

CHCl3 has polar C−Cl bonds arranged asymmetrically, so the dipoles do not cancel. In CCl4, CO2 and BF3 the bond dipoles cancel by symmetry.

Q16Which of these ions has the largest ionic radius? (All are isoelectronic.)

(A) Ca^2+
(B) K^+
(C) Cl^-
(D) S^2-
Show Answer

Correct: (D)

For isoelectronic ions, radius decreases as the nuclear charge increases. S^2- has the smallest nuclear charge (16) of the four, so its electron cloud is held least tightly and is largest.

Q17Graphite conducts electricity whereas diamond does not, because in graphite

(A) the atoms are arranged in a giant ionic lattice
(B) each carbon atom has one delocalised electron that can move along the layers
(C) the covalent bonds are weaker than in diamond
(D) there are no covalent bonds between the layers
Show Answer

Correct: (B)

Each carbon in graphite forms three bonds, leaving one p-electron per atom delocalised above and below the layers, able to carry charge.

🔥 Energetics & Thermodynamics

Q18Burning 0.0200 mol of a fuel raised the temperature of 50.0 g of water by 12.0 K. (Specific heat capacity of water = 4.18 J g^-1 K^-1.) The enthalpy change of combustion of the fuel is approximately

(A) −2508 J mol^-1
(B) −50.2 kJ mol^-1
(C) −125 kJ mol^-1
(D) −12.5 kJ mol^-1
Show Answer

Correct: (C)

q = mcΔT = 50.0 × 4.18 × 12.0 = 2508 J. Per mole: 2508 / 0.0200 = 125 400 J mol^-1, so ΔH ≈ −125 kJ mol^-1.

Q19Using standard enthalpies of formation (kJ mol^-1): CH4 = −75, CO2 = −394, H2O = −286, the enthalpy change for CH4 + 2O2 → CO2 + 2H2O is

(A) −605 kJ mol^-1
(B) −1041 kJ mol^-1
(C) −891 kJ mol^-1
(D) −819 kJ mol^-1
Show Answer

Correct: (C)

ΔH = [−394 + 2(−286)] − [−75] = −966 + 75 = −891 kJ mol^-1.

Q20Using bond enthalpies (kJ mol^-1): H−H = 436, Cl−Cl = 242, H−Cl = 431, the enthalpy change for H2 + Cl2 → 2HCl is

(A) −184 kJ mol^-1
(B) +184 kJ mol^-1
(C) −247 kJ mol^-1
(D) −862 kJ mol^-1
Show Answer

Correct: (A)

Bonds broken = 436 + 242 = 678; bonds formed = 2 × 431 = 862. ΔH = 678 − 862 = −184 kJ mol^-1.

Q21Which process is accompanied by an increase in entropy?

(A) CaCO3(s) → CaO(s) + CO2(g)
(B) H2O(l) → H2O(s)
(C) N2(g) + 3H2(g) → 2NH3(g)
(D) H2O(g) → H2O(l)
Show Answer

Correct: (A)

A gas is produced from a solid, so the disorder (entropy) of the system increases.

Q22A reaction has ΔH = −100 kJ mol^-1 and ΔS = −200 J K^-1 mol^-1. The value of ΔG at 298 K is

(A) −159.6 kJ mol^-1
(B) −40.4 kJ mol^-1
(C) +40.4 kJ mol^-1
(D) −99.4 kJ mol^-1
Show Answer

Correct: (B)

ΔG = ΔH − TΔS = −100 − 298(−0.200) = −100 + 59.6 = −40.4 kJ mol^-1, so the reaction is feasible.

Q23Using standard entropies (J K^-1 mol^-1): N2 = 192, H2 = 131, NH3 = 193, the entropy change for N2 + 3H2 → 2NH3 is

(A) −199 J K^-1 mol^-1
(B) +199 J K^-1 mol^-1
(C) −1 J K^-1 mol^-1
(D) −587 J K^-1 mol^-1
Show Answer

Correct: (A)

ΔS = 2(193) − [192 + 3(131)] = 386 − 585 = −199 J K^-1 mol^-1.

⏱ Kinetics & Rate Equations

Q24In a series of experiments, doubling the concentration of A (with B constant) increased the rate by a factor of 4, and doubling the concentration of B (with A constant) doubled the rate. The rate equation is

(A) rate = k[A][B]
(B) rate = k[A]^2[B]^2
(C) rate = k[A][B]^2
(D) rate = k[A]^2[B]
Show Answer

Correct: (D)

Rate ∝ [A]^2 (second order in A) and rate ∝ [B] (first order in B), so rate = k[A]^2[B].

Q25For the reaction in Q24, when [A] = 0.100 mol dm^-3 and [B] = 0.200 mol dm^-3 the rate is 2.00 × 10^-3 mol dm^-3 s^-1. The value of the rate constant k is

(A) 1.00 mol^-2 dm^6 s^-1
(B) 0.100 mol^-2 dm^6 s^-1
(C) 10.0 mol^-2 dm^6 s^-1
(D) 1.00 mol^-1 dm^3 s^-1
Show Answer

Correct: (A)

k = rate / ([A]^2[B]) = 2.00 × 10^-3 / (0.0100 × 0.200) = 2.00 × 10^-3 / 2.00 × 10^-3 = 1.00, with units mol^-2 dm^6 s^-1 for an overall third-order reaction.

Q26A first-order reaction has a half-life of 693 s. Its rate constant is

(A) 1.44 × 10^-3 s^-1
(B) 1.00 × 10^-3 s^-1
(C) 0.693 s^-1
(D) 9.63 × 10^-6 s^-1
Show Answer

Correct: (B)

For a first-order reaction k = ln 2 / t1/2 = 0.693 / 693 = 1.00 × 10^-3 s^-1.

Q27Raising the temperature of a reaction mixture increases the rate mainly because

(A) the activation energy decreases
(B) the particles collide less often but with more energy
(C) a greater proportion of collisions have energy at least equal to the activation energy
(D) the rate constant decreases
Show Answer

Correct: (C)

The Maxwell-Boltzmann distribution shifts so that many more particles exceed the activation energy, and collisions are also slightly more frequent.

Q28A graph of initial rate against concentration of a reactant is a straight line passing through the origin. The reaction is which order with respect to this reactant?

(A) Zero order
(B) Second order
(C) Third order
(D) First order
Show Answer

Correct: (D)

Rate directly proportional to concentration means first order in that reactant.

⚖ Chemical Equilibria (Kc & Kp)

Q29At equilibrium in a 1.00 dm^3 vessel for N2O4(g) ⇌ 2NO2(g), there are 0.200 mol of N2O4 and 0.400 mol of NO2. The value of Kc is

(A) 0.800 mol dm^-3
(B) 0.400 mol dm^-3
(C) 2.00 mol dm^-3
(D) 0.200 mol dm^-3
Show Answer

Correct: (A)

Kc = [NO2]^2 / [N2O4] = (0.400)^2 / 0.200 = 0.160 / 0.200 = 0.800 mol dm^-3.

Q30For the equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g), which expression gives Kp?

(A) Kp = p(SO3)^2 / (p(SO2)^2 × p(O2))
(B) Kp = p(SO2)^2 × p(O2) / p(SO3)^2
(C) Kp = p(SO3) / (p(SO2) × p(O2))
(D) Kp = p(SO3)^2 / (p(SO2) × p(O2))
Show Answer

Correct: (A)

Kp is the product of the partial pressures of the products, each raised to its stoichiometric power, divided by the same for the reactants.

Q31For an exothermic equilibrium, increasing the temperature will

(A) increase the yield of products
(B) have no effect on the position of equilibrium
(C) increase the value of Kc
(D) decrease the yield of products
Show Answer

Correct: (D)

The system opposes the change by moving in the endothermic (reverse) direction, so the yield of products falls and Kc decreases.

Q32Hydrogen and iodine react: H2(g) + I2(g) ⇌ 2HI(g). Starting with 1.00 mol of each reactant in a 1.00 dm^3 vessel, 1.60 mol of HI is present at equilibrium. Kc is

(A) 2.56
(B) 64
(C) 16
(D) 6.4
Show Answer

Correct: (B)

1.60 mol HI formed consumes 0.80 mol of each reactant, leaving [H2] = [I2] = 0.200. Kc = (1.60)^2 / (0.200 × 0.200) = 2.56 / 0.0400 = 64 (no units).

Q33Adding a catalyst to a system at equilibrium will

(A) increase the value of Kc
(B) shift the position of equilibrium to the right
(C) increase the yield of products
(D) have no effect on the position of equilibrium or on Kc
Show Answer

Correct: (D)

A catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached faster but its position and Kc are unchanged.




🔋 Redox & Electrode Potentials

Q34What is the (average) oxidation state of sulfur in Na2S2O3?

(A) +2
(B) +4
(C) +6
(D) −2
Show Answer

Correct: (A)

2(+1) + 2(S) + 3(−2) = 0, so 2(S) = +4 and the average oxidation state of S is +2.

Q35Standard electrode potentials: Zn^2+/Zn = −0.76 V, Fe^2+/Fe = −0.44 V, Cu^2+/Cu = +0.34 V, Ag^+/Ag = +0.80 V. The strongest reducing agent is

(A) Ag
(B) Cu
(C) Fe
(D) Zn
Show Answer

Correct: (D)

The species with the most negative electrode potential is oxidised most readily, so zinc metal is the strongest reducing agent.

Q36The EMF of a cell combining the Zn^2+/Zn and Cu^2+/Cu half-cells is

(A) +1.10 V
(B) −1.10 V
(C) +0.42 V
(D) +1.56 V
Show Answer

Correct: (A)

EMF = E(cathode) − E(anode) = +0.34 − (−0.76) = +1.10 V.

Q37When aqueous copper(II) sulfate is electrolysed using inert graphite electrodes, the products are

(A) copper at the cathode and hydrogen at the anode
(B) hydrogen at the cathode and oxygen at the anode
(C) copper at the cathode and oxygen at the anode
(D) copper at both electrodes
Show Answer

Correct: (C)

Cu^2+ ions are discharged at the cathode as copper; at the anode, hydroxide ions from water are discharged, giving oxygen.

Q38Given Fe^3+/Fe^2+ = +0.77 V and Cl2/Cl^- = +1.36 V, which statement is correct?

(A) Chlorine cannot oxidise Fe^2+ ions
(B) Chlorine will oxidise Fe^2+ ions to Fe^3+, with EMF +0.59 V
(C) Fe^3+ ions will oxidise Cl^- ions to chlorine
(D) The reaction of chlorine with Fe^2+ has EMF −0.59 V
Show Answer

Correct: (B)

EMF = 1.36 − 0.77 = +0.59 V. A positive EMF means the reaction (Cl2 oxidising Fe^2+) is feasible.

🧫 Acids, Bases & Buffers

Q39The pH of a 0.0500 mol dm^-3 solution of hydrochloric acid is

(A) 0.50
(B) 1.30
(C) 2.00
(D) 1.70
Show Answer

Correct: (B)

pH = −log(0.0500) = 1.30.

Q40The pH of a 0.0100 mol dm^-3 solution of sodium hydroxide is (Kw = 1.00 × 10^-14 at 298 K)

(A) 2.00
(B) 11.00
(C) 12.00
(D) 13.00
Show Answer

Correct: (C)

pOH = −log(0.0100) = 2.00, so pH = 14.00 − 2.00 = 12.00.

Q41An ethanoic acid solution has concentration 0.100 mol dm^-3 and Ka = 1.80 × 10^-5 mol dm^-3. Its pH is approximately

(A) 2.87
(B) 4.74
(C) 1.87
(D) 3.37
Show Answer

Correct: (A)

[H^+] = √(Ka × c) = √(1.80 × 10^-5 × 0.100) = √(1.80 × 10^-6) = 1.34 × 10^-3, so pH = 2.87.

Q42A buffer solution contains 0.200 mol dm^-3 ethanoic acid and 0.100 mol dm^-3 sodium ethanoate. (Ka of ethanoic acid = 1.74 × 10^-5, pKa = 4.76.) The pH of the buffer is

(A) 4.76
(B) 5.06
(C) 4.16
(D) 4.46
Show Answer

Correct: (D)

pH = pKa + log([salt]/[acid]) = 4.76 + log(0.100/0.200) = 4.76 − 0.30 = 4.46.

Q43At 50°C, Kw = 5.48 × 10^-14 mol^2 dm^-6. The pH of pure water at this temperature is

(A) 7.00
(B) 6.63
(C) 7.37
(D) 6.26
Show Answer

Correct: (B)

In pure water [H^+] = √Kw = √(5.48 × 10^-14) = 2.34 × 10^-7, so pH = 6.63. The water is still neutral because [H^+] = [OH^-].

⚗ Inorganic Chemistry

Q44Which statement about the Group 2 elements is correct?

(A) The reactivity with water increases down the group
(B) Barium reacts less vigorously with water than magnesium
(C) Magnesium hydroxide is more soluble than barium hydroxide
(D) Strontium does not react with cold water
Show Answer

Correct: (A)

Reactivity increases down Group 2 because the outer electrons are further from the nucleus and more easily lost; solubilities of the hydroxides also increase down the group.

Q45Chlorine gas is bubbled through aqueous potassium bromide. What is observed?

(A) No reaction occurs
(B) A yellow-green gas is given off
(C) The solution turns orange as bromine is formed
(D) A white precipitate forms
Show Answer

Correct: (C)

Chlorine is a stronger oxidising agent than bromine and displaces it: Cl2 + 2Br^- → 2Cl^- + Br2, giving an orange solution.

Q46When excess aqueous ammonia is added to aqueous copper(II) sulfate, the final colour of the solution is

(A) pale blue
(B) deep blue
(C) green
(D) colourless
Show Answer

Correct: (B)

NH3 ligands replace water ligands to form [Cu(NH3)4(H2O)2]^2+, which gives an intense deep blue solution.

Q47To test for sulfate ions in solution, the reagents used are

(A) silver nitrate solution and dilute nitric acid, giving a white precipitate
(B) barium chloride solution and dilute hydrochloric acid, giving a white precipitate
(C) sodium hydroxide solution, giving a white precipitate
(D) limewater, giving a cloudy solution
Show Answer

Correct: (B)

Ba^2+ ions form white insoluble BaSO4 with sulfate ions; the acid removes carbonate ions which would otherwise give a false positive.

Q48Which aqueous ion is colourless?

(A) Fe^3+(aq)
(B) Cu^2+(aq)
(C) Cr^3+(aq)
(D) Sc^3+(aq)
Show Answer

Correct: (D)

Sc^3+ has an empty 3d subshell (no d electrons), so no d-d transitions can occur and it is colourless.

🛢 Organic Chemistry I

Q49In the free-radical substitution of methane by chlorine, the propagation steps involve

(A) Cl• radicals only
(B) bond fission of C−H bonds in the initiation step
(C) Cl• and CH3• radicals reacting alternately with Cl2 and CH4
(D) ions formed by heterolytic fission
Show Answer

Correct: (C)

In propagation, Cl• abstracts H from CH4 giving CH3•, which then reacts with Cl2 regenerating Cl•, continuing the chain.

Q50Which halogenoalkane is hydrolysed most rapidly by aqueous sodium hydroxide?

(A) 1-chlorobutane
(B) 1-bromobutane
(C) 1-fluorobutane
(D) 1-iodobutane
Show Answer

Correct: (D)

The C−I bond is the weakest carbon-halogen bond, so it breaks most easily and hydrolysis is fastest.

Q51The major product of the reaction of propene with hydrogen bromide is

(A) 2-bromopropane
(B) 1-bromopropane
(C) 1,2-dibromopropane
(D) propyl hydrogen sulfate
Show Answer

Correct: (A)

Electrophilic addition follows Markovnikov's rule: H adds to the carbon already having more hydrogens, giving the more stable secondary carbocation and hence 2-bromopropane.

Q52When propan-1-ol is heated with acidified potassium dichromate(VI) under distillation conditions, the main organic product is

(A) propanal
(B) propanoic acid
(C) propene
(D) propanone
Show Answer

Correct: (A)

A primary alcohol is oxidised to an aldehyde, which distils off before it can be oxidised further to the carboxylic acid.

Q53Which alcohol cannot be oxidised by acidified potassium dichromate(VI)?

(A) butan-1-ol
(B) butan-2-ol
(C) propan-1-ol
(D) 2-methylpropan-2-ol
Show Answer

Correct: (D)

Tertiary alcohols have no hydrogen atom on the carbon bearing the −OH group, so they resist oxidation.

Q54Ethanol vapour passed over heated aluminium oxide, or ethanol heated with concentrated sulfuric acid at 170°C, gives mainly

(A) ethanoic acid
(B) ethene
(C) diethyl ether only
(D) ethane
Show Answer

Correct: (B)

The alcohol is dehydrated (elimination of water) to form the alkene ethene.

🔬 Organic Chemistry II & Analysis

Q55Tollens' reagent (ammoniacal silver nitrate) will give a silver mirror with

(A) propanone but not propanal
(B) both propanal and propanone
(C) propanal but not propanone
(D) neither propanal nor propanone
Show Answer

Correct: (C)

Aldehydes are oxidised by Tollens' reagent, reducing Ag^+ to silver; ketones are not oxidised.

Q56Propanone reacts with HCN, in the presence of a little alkali, to form

(A) 2-hydroxy-2-methylpropanenitrile
(B) 2-hydroxypropanenitrile
(C) propanenitrile
(D) 2-aminopropanoic acid
Show Answer

Correct: (A)

Nucleophilic addition of CN^- to the carbonyl carbon, followed by protonation, attaches a −CN group and an −OH group to the central carbon: (CH3)2C(OH)CN, 2-hydroxy-2-methylpropanenitrile.

Q57When ethanoic acid is warmed with ethanol in the presence of concentrated sulfuric acid, the ester formed is

(A) methyl ethanoate
(B) ethyl methanoate
(C) ethyl ethanoate
(D) propyl ethanoate
Show Answer

Correct: (C)

The ester takes its first name from the alcohol (ethyl) and its second from the acid (ethanoate): ethyl ethanoate.

Q58In the nitration of benzene, the electrophile that attacks the ring is

(A) NO2^- (nitrite ion)
(B) HNO3
(C) NO2^+ (nitronium ion)
(D) NO^+
Show Answer

Correct: (C)

Concentrated sulfuric acid protonates nitric acid and water is lost, generating the nitronium ion NO2^+, the electrophile in electrophilic substitution.

Q59Which amine is the strongest base?

(A) phenylamine
(B) ammonia
(C) ethylamine
(D) they are all equally basic
Show Answer

Correct: (C)

The ethyl group pushes electron density onto the nitrogen, making its lone pair more available to accept a proton. In phenylamine the lone pair is delocalised into the ring, making it a much weaker base.

Q60A spot on a TLC plate moves 3.2 cm while the solvent front moves 8.0 cm. The Rf value of the spot is

(A) 0.40
(B) 2.5
(C) 0.25
(D) 4.8
Show Answer

Correct: (A)

Rf = distance moved by spot / distance moved by solvent front = 3.2 / 8.0 = 0.40.

📥 Prefer pen and paper? Take this test offline.

Download Printable PDF – Test 11

Free printable version: all 60 questions + answer key. Explanations stay here on the website.