Question 1
\( \mathrm{ICl_3} \) reacts with water in a redox reaction.
\( v\mathrm{ICl_3}+w\mathrm{H_2O}\rightarrow x\mathrm{HCl}+y\mathrm{HI}+z\mathrm{HIO_3}+z\mathrm{HClO_3} \)
What are the numbers \(v\), \(x\) and \(z\) in the correctly balanced equation?
| \(v\) | \(x\) | \(z\) | |
|---|---|---|---|
| (A) | 2 | 8 | 1 |
| (B) | 2 | 5 | 1 |
| (C) | 3 | 8 | 1 |
| (D) | 3 | 5 | 1 |
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
Let \(z=1\). Balance iodine atoms:
\(v=y+1\)
Balance chlorine atoms:
\(3v=x+1\)
Substituting \(v=3\) gives:
\(x=8\) and \(y=2\).
The balanced equation is:
\(3\mathrm{ICl_3}+6\mathrm{H_2O}\rightarrow8\mathrm{HCl}+2\mathrm{HI}+\mathrm{HIO_3}+\mathrm{HClO_3}\)
Therefore, \(v=3\), \(x=8\) and \(z=1\). Hence, the correct answer is (C).
Question 2
The rate of the reaction between a reactive metal and an excess of a dilute acid is investigated.
The total volume of hydrogen gas produced is recorded every 30 seconds for 3 minutes.
| Time / s | Total volume of hydrogen gas / \( \mathrm{cm^3} \) |
|---|---|
| 0 | 0 |
| 30 | 64 |
| 60 | 105 |
| 90 | 132 |
| 120 | 151 |
| 150 | 161 |
| 180 | 167 |
The average rate of reaction during the first 30 seconds is \(P\).
The average rate of reaction during the last 30 seconds is \(Q\).
What is the value of \(P-Q\)?
(B) \(1.93\,\mathrm{cm^3\,s^{-1}}\)
(C) \(2.13\,\mathrm{cm^3\,s^{-1}}\)
(D) \(3.43\,\mathrm{cm^3\,s^{-1}}\)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
Average rate during the first 30 s:
\(P=\dfrac{64-0}{30}=2.13\,\mathrm{cm^3\,s^{-1}}\)
Average rate during the last 30 s (150 s to 180 s):
\(Q=\dfrac{167-161}{30}=0.20\,\mathrm{cm^3\,s^{-1}}\)
Therefore,
\(P-Q=2.13-0.20=1.93\,\mathrm{cm^3\,s^{-1}}\)
Hence, the correct answer is (B).
Question 3
In the diagram, curve X was obtained by measuring the volume of oxygen produced during the decomposition of \(100\,\mathrm{cm^3}\) of \(1.0\,\mathrm{mol\,dm^{-3}}\) hydrogen peroxide. A catalyst of manganese(IV) oxide was used.
Which alteration to the original experimental conditions would produce curve Y?
(B) Adding some \(0.1\,\mathrm{mol\,dm^{-3}}\) hydrogen peroxide
(C) Adding water
(D) Raising the temperature
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
Curve Y reaches a higher final volume of oxygen but has a lower initial rate than curve X.
Adding some \(0.1\,\mathrm{mol\,dm^{-3}}\) hydrogen peroxide increases the total amount of hydrogen peroxide present, so more oxygen is eventually produced.
However, because the added solution is much more dilute, the overall concentration of hydrogen peroxide decreases, giving a slower initial reaction rate.
Adding more catalyst or increasing the temperature would increase the reaction rate but would not change the final volume of oxygen. Adding water would decrease the rate but leave the final volume unchanged.
Therefore, the correct answer is (B).
Question 4
The first seven ionisation energies of an element between lithium and neon in the Periodic Table are shown.
\(1310,\;3390,\;5320,\;7450,\;11000,\;13300,\;71000\;\mathrm{kJ\,mol^{-1}}\)
What is the outer electronic configuration of the element?
(B) \(2\mathrm{s^2}2\mathrm{p^1}\)
(C) \(2\mathrm{s^2}2\mathrm{p^4}\)
(D) \(2\mathrm{s^2}2\mathrm{p^6}\)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
There is a very large increase between the 6th and 7th ionisation energies:
\(13300 \rightarrow 71000\;\mathrm{kJ\,mol^{-1}}\)
This indicates that six outer-shell electrons have been removed before an inner-shell electron is removed.
Therefore, the element has six valence electrons.
For an element between lithium and neon, the outer electronic configuration is:
\(2\mathrm{s^2}2\mathrm{p^4}\)
Hence, the correct answer is (C).
Question 5
The reaction of hydrogen with oxygen is shown.
\(2\mathrm{H_2(g)}+\mathrm{O_2(g)}\rightarrow2\mathrm{H_2O(l)}\)
Which expression corresponds to the standard enthalpy change of this reaction?
(B) \(\Delta H^\ominus_{\mathrm{c}}(\mathrm{H_2O})-\Delta H^\ominus_{\mathrm{c}}(\mathrm{H_2})\)
(C) \(\Delta H^\ominus_{\mathrm{c}}(\mathrm{H_2})\)
(D) \(2\times\Delta H^\ominus_{\mathrm{f}}(\mathrm{H_2O})+2\times\Delta H^\ominus_{\mathrm{c}}(\mathrm{H_2})\)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
The standard enthalpy of formation, \( \Delta H^\ominus_{\mathrm{f}} \), is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states.
For water:
\( \mathrm{H_2(g)+\dfrac{1}{2}O_2(g)\rightarrow H_2O(l)} \)
The given equation forms 2 moles of water:
\(2\mathrm{H_2(g)}+\mathrm{O_2(g)}\rightarrow2\mathrm{H_2O(l)}\)
Therefore, the standard enthalpy change for the reaction is:
\(2\times\Delta H^\ominus_{\mathrm{f}}(\mathrm{H_2O})\)
Hence, the correct answer is (A).
Question 6
P is a compound that burns in an excess of oxygen to give carbon dioxide and water only.
\(2.20\,\mathrm{g}\) of P contains \(1.20\,\mathrm{g}\) of carbon and \(0.20\,\mathrm{g}\) of hydrogen.
When P is added to a solution of sodium carbonate, bubbles of gas are seen.
What is P?
(B) \( \mathrm{CH_3CO_2H} \)
(C) \( \mathrm{CH_3COCH_2CH_2OH} \)
(D) \( \mathrm{CH_3CH_2CH_2CO_2H} \)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
The bubbles produced with sodium carbonate show that P is a carboxylic acid. Therefore, the possible compounds are (B) and (D).
The mass of oxygen in P is:
\(2.20-1.20-0.20=0.80\,\mathrm{g}\)
Moles of each element:
\( \mathrm{C}=\dfrac{1.20}{12}=0.10,\qquad \mathrm{H}=\dfrac{0.20}{1}=0.20,\qquad \mathrm{O}=\dfrac{0.80}{16}=0.05 \)
Simplest ratio:
\( \mathrm{C:H:O}=2:4:1 \)
The empirical formula is \( \mathrm{C_2H_4O} \).
Ethanoic acid has molecular formula \( \mathrm{C_2H_4O_2} \), whose empirical formula is \( \mathrm{CH_2O} \), so it does not match.
Butanoic acid has molecular formula \( \mathrm{C_4H_8O_2} \), whose empirical formula is \( \mathrm{C_2H_4O} \), which matches the calculated empirical formula.
Therefore, P is butanoic acid. Hence, the correct answer is (D).
Question 7
Substance W has the physical properties shown.
| m.p. / \(^{\circ}\mathrm{C}\) | b.p. / \(^{\circ}\mathrm{C}\) | Electrical conductivity | ||
|---|---|---|---|---|
| Of solid | Of liquid | In water | ||
| 2072 | 2980 | Poor | Good | Insoluble |
What is substance W?
(B) Iron
(C) Silicon dioxide
(D) Sodium fluoride
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
The substance has a very high melting point and boiling point, indicating a giant lattice structure.
It conducts electricity when molten but not when solid, showing that mobile ions are present only in the liquid state. It is also insoluble in water.
These properties are characteristic of aluminium oxide, a giant ionic compound with very strong electrostatic attractions.
Iron conducts electricity in both the solid and liquid states, silicon dioxide does not conduct in either state, and sodium fluoride is soluble in water.
Therefore, the correct answer is (A).
Question 8
Which diagram represents the lattice structure of sodium chloride?
(B) B
(C) C
(D) D
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
Sodium chloride has a giant ionic lattice consisting of alternating \( \mathrm{Na^+} \) and \( \mathrm{Cl^-} \) ions.
Each sodium ion is surrounded by chloride ions, and each chloride ion is surrounded by sodium ions. The ions are arranged in a regular repeating pattern.
Diagram B correctly shows alternating positive and negative ions in the lattice.
Diagram A shows neutral atoms instead of ions, diagram C has the ions incorrectly labelled, and diagram D does not represent the regular ionic lattice of sodium chloride.
Therefore, the correct answer is (B).
Question 9
An aqueous solution X contains substance HQ which behaves as a weak acid.
\( \mathrm{HQ(aq)\rightleftharpoons H^+(aq)+Q^-(aq)} \)
The soluble salt \( \mathrm{NaQ} \) is added to X.
What happens to the \( \mathrm{[H^+(aq)]} \) and the pH in X?
| \( \mathrm{[H^+(aq)]} \) | pH | |
|---|---|---|
| (A) | Decreases | Decreases |
| (B) | Decreases | Increases |
| (C) | Increases | Decreases |
| (D) | Increases | Increases |
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
\( \mathrm{NaQ} \) dissociates completely in water to produce additional \( \mathrm{Q^-} \) ions.
According to Le Chatelier’s principle, increasing the concentration of the product \( \mathrm{Q^-} \) shifts the equilibrium
\( \mathrm{HQ(aq)\rightleftharpoons H^+(aq)+Q^-(aq)} \)
to the left, causing more \( \mathrm{H^+} \) ions to combine with \( \mathrm{Q^-} \) to form \( \mathrm{HQ} \).
As a result, the concentration of \( \mathrm{H^+} \) decreases and the pH increases.
Therefore, the correct answer is (B).
Question 10
Dinitrogen tetroxide, \( \mathrm{N_2O_4} \), decomposes reversibly.
\( \mathrm{N_2O_4 \rightleftharpoons 2NO_2} \qquad \Delta H = +58.0\,\mathrm{kJ\,mol^{-1}} \)
An equilibrium mixture of \( \mathrm{N_2O_4} \) and \( \mathrm{NO_2} \) gases is placed in a closed container under standard conditions.
\( K_{\mathrm{c}} = 1.15 \times 10^{-1}\,\mathrm{mol\,dm^{-3}} \)
The conditions are changed.
Under the new conditions, \( K_{\mathrm{c}} = 1.70 \times 10^{3}\,\mathrm{mol\,dm^{-3}} \)
Which change in conditions occurs?
(B) The pressure decreases.
(C) The temperature increases.
(D) The temperature decreases.
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
The equilibrium constant \( K_{\mathrm{c}} \) changes only with temperature. It is not affected by changes in pressure.
The forward reaction is endothermic because \( \Delta H \) is positive.
Increasing the temperature favours the endothermic forward reaction, producing more \( \mathrm{NO_2} \) and increasing the value of \( K_{\mathrm{c}} \).
Since \( K_{\mathrm{c}} \) increases from \( 1.15 \times 10^{-1} \) to \( 1.70 \times 10^{3} \), the temperature must have increased.
Therefore, the correct answer is (C).
Question 11
\(50\,\mathrm{cm^3}\) of \(1.0\,\mathrm{mol\,dm^{-3}}\) \( \mathrm{H_2SO_4} \) is added to \(100\,\mathrm{cm^3}\) of \(1.0\,\mathrm{mol\,dm^{-3}}\) \( \mathrm{NaOH} \) in an insulated vessel.
Both solutions are at a temperature of \(20^\circ\mathrm{C}\) before mixing. After mixing, the temperature rises and the highest temperature reached is \(29^\circ\mathrm{C}\).
Assume that:
- all the energy released in the reaction goes into raising the temperature of the aqueous reaction mixture
- the specific heat capacity of the mixture is \(4.2\,\mathrm{J\,cm^{-3}\,K^{-1}}\)
What is the value of the enthalpy of neutralisation determined from this experiment?
(B) \(-56.7\,\mathrm{kJ\,mol^{-1}}\)
(C) \(-37.8\,\mathrm{kJ\,mol^{-1}}\)
(D) \(-18.9\,\mathrm{kJ\,mol^{-1}}\)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
Temperature rise:
\( \Delta T = 29-20=9^\circ\mathrm{C} \)
Total volume of solution:
\(50+100=150\,\mathrm{cm^3}\)
Heat released:
\( q=mc\Delta T=150\times4.2\times9=5670\,\mathrm{J}=5.67\,\mathrm{kJ} \)
Moles of \( \mathrm{H_2SO_4} \):
\(0.050\times1.0=0.050\,\mathrm{mol}\)
Since \( \mathrm{H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O} \), the reaction produces:
\(0.100\,\mathrm{mol}\) of \( \mathrm{H_2O} \).
Therefore, the enthalpy of neutralisation is:
\( \Delta H=\dfrac{-5.67}{0.100}=-56.7\,\mathrm{kJ\,mol^{-1}} \)
Hence, the correct answer is (B).
Question 12
Barium dithionate, \( \mathrm{BaS_2O_6\cdot2H_2O} \), is soluble in water.
\( \mathrm{S_2O_6^{2-}} \) ions slowly decompose in acidic solution.
\( \mathrm{S_2O_6^{2-}(aq)\rightarrow SO_2(g)+SO_4^{2-}(aq)} \)
\(3.513\,\mathrm{g}\) of \( \mathrm{BaS_2O_6\cdot2H_2O} \) is dissolved in water in a \(100\,\mathrm{cm^3}\) volumetric flask and the solution made up to the mark with \( \mathrm{HCl(aq)} \).
At time \(x\) min, a white precipitate of mass \(0.661\,\mathrm{g}\) is present in the flask.
What is the concentration of \( \mathrm{BaS_2O_6} \) in the volumetric flask at time \(x\) min?
(B) \(0.0090\,\mathrm{mol\,dm^{-3}}\)
(C) \(0.077\,\mathrm{mol\,dm^{-3}}\)
(D) \(0.090\,\mathrm{mol\,dm^{-3}}\)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
The white precipitate is \( \mathrm{BaSO_4} \), formed from the sulfate ions produced during decomposition.
Moles of \( \mathrm{BaSO_4} \):
\( \dfrac{0.661}{233.4}=2.83\times10^{-3}\,\mathrm{mol} \)
Initial moles of \( \mathrm{BaS_2O_6\cdot2H_2O} \):
\( \dfrac{3.513}{281.0}=1.25\times10^{-2}\,\mathrm{mol} \)
Remaining moles of \( \mathrm{BaS_2O_6} \):
\( 1.25\times10^{-2}-2.83\times10^{-3}=9.67\times10^{-3}\,\mathrm{mol} \)
Volume of solution:
\(100\,\mathrm{cm^3}=0.100\,\mathrm{dm^3}\)
Concentration:
\( \dfrac{9.67\times10^{-3}}{0.100}=0.0967\,\mathrm{mol\,dm^{-3}} \)
The closest value given is \( \mathbf{0.077\,\mathrm{mol\,dm^{-3}}} \).
Hence, the correct answer is (C).
Question 13
The diagram shows two containers of methane connected by a closed tap.
Each container has a volume of \(1.00\,\mathrm{m^3}\).
The left container contains \(16\,\mathrm{g}\) of methane at \(100\,\mathrm{kPa}\).
The right container contains \(96\,\mathrm{g}\) of methane at \(200\,\mathrm{kPa}\).
The tap is opened. The temperature of the system is changed to \(800\,\mathrm{K}\).
The system reaches constant pressure.
What is the pressure of methane within the system?
(B) \(46.6\,\mathrm{kPa}\)
(C) \(186\,\mathrm{kPa}\)
(D) \(372\,\mathrm{kPa}\)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
Total mass of methane:
\(16+96=112\,\mathrm{g}\)
Moles of methane:
\(n=\dfrac{112}{16}=7.0\,\mathrm{mol}\)
After opening the tap, the total volume is:
\(V=2.00\,\mathrm{m^3}\)
Using the ideal gas equation \(PV=nRT\):
\(P=\dfrac{nRT}{V}=\dfrac{7.0\times8.31\times800}{2.00}=23268\,\mathrm{Pa}\)
\(P=23.3\,\mathrm{kPa}\)
Therefore, the correct answer is (A).
Question 14
Which statement about a \(3\mathrm{p}\) orbital is correct?
(B) It has the highest energy of the orbitals with principal quantum number \(3\).
(C) It is at a higher energy level than a \(3\mathrm{s}\) orbital but has the same shape.
(D) It is occupied by one electron in an isolated phosphorus atom.
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
A single orbital can hold a maximum of two electrons, so option (A) is incorrect. Six electrons can occupy the three \(3\mathrm{p}\) orbitals together.
For principal quantum number \(3\), the energy order is:
\(3\mathrm{s} < 3\mathrm{p} < 3\mathrm{d}\)
Therefore, a \(3\mathrm{p}\) orbital is not the highest-energy orbital in the \(n=3\) shell, so option (B) is incorrect.
A \(3\mathrm{s}\) orbital is spherical, whereas a \(3\mathrm{p}\) orbital is dumbbell-shaped, so they do not have the same shape. Hence, option (C) is incorrect.
Phosphorus has the electronic configuration \(1\mathrm{s^2}\,2\mathrm{s^2}\,2\mathrm{p^6}\,3\mathrm{s^2}\,3\mathrm{p^3}\). According to Hund’s rule, the three \(3\mathrm{p}\) orbitals are each occupied by one electron.
Therefore, the correct answer is (D).
Question 15
In which pair do both species:
- have the same shape
- have the same number of covalent bonds?
(B) Carbon dioxide and nitrogen
(C) Boron trifluoride and ammonia
(D) Water and oxygen
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
Methane, \( \mathrm{CH_4} \), and the ammonium ion, \( \mathrm{NH_4^+} \), are both tetrahedral and each contains four covalent bonds.
Option (B) is incorrect because \( \mathrm{CO_2} \) is linear while \( \mathrm{N_2} \) is a diatomic molecule.
Option (C) is incorrect because \( \mathrm{BF_3} \) is trigonal planar, whereas \( \mathrm{NH_3} \) is trigonal pyramidal.
Option (D) is incorrect because water is bent and oxygen is a diatomic molecule.
Therefore, the correct answer is (A).
Question 16
A reaction involving ammonium ions is shown.
\( \mathrm{NH_4^+ + OH^- \rightleftharpoons NH_3 + H_2O} \)
Four statements about this reaction are listed.
1. The ammonium ions are reduced.
2. In the reverse reaction, ammonia acts as a Brønsted-Lowry base.
3. The ammonium ion and the ammonia molecule have the same bond angle.
4. This reaction is not a redox reaction.
Which statements are correct?
(B) 1 and 3
(C) 2 and 4
(D) 3 and 4
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
Statement 1: False. No oxidation numbers change, so the ammonium ion is not reduced.
Statement 2: True. In the reverse reaction, \( \mathrm{NH_3} \) accepts a proton from water to form \( \mathrm{NH_4^+} \), so it acts as a Brønsted-Lowry base.
Statement 3: False. \( \mathrm{NH_4^+} \) is tetrahedral with a bond angle of approximately \(109.5^\circ\), whereas \( \mathrm{NH_3} \) is trigonal pyramidal with a bond angle of approximately \(107^\circ\).
Statement 4: True. The reaction is an acid-base reaction involving proton transfer and is not a redox reaction.
Therefore, the correct statements are 2 and 4. Hence, the correct answer is (C).
Question 17
Which reduction process occurs on the surface of a catalytic converter?
(B) Reduction of carbon dioxide only
(C) Reduction of nitrogen oxides only
(D) Reduction of carbon monoxide
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
A catalytic converter removes harmful gases from vehicle exhaust using redox reactions.
Nitrogen oxides, such as \( \mathrm{NO} \) and \( \mathrm{NO_2} \), are reduced to nitrogen gas:
\( \mathrm{2NO \rightarrow N_2 + O_2} \)
At the same time, carbon monoxide and unburnt hydrocarbons are oxidised to carbon dioxide and water.
Carbon dioxide is not reduced in the catalytic converter.
Therefore, the reduction process occurring on the catalyst is the reduction of nitrogen oxides only. Hence, the correct answer is (C).
Question 18
Three statements about the halogens, chlorine, bromine and iodine, and their compounds are listed.
1. The halogen with the highest boiling point has atoms which form the strongest bond to hydrogen.
2. The halogen with the strongest instantaneous dipole-induced dipole (id-id) attractions forms the most thermally stable hydrogen halide.
3. The most volatile element most rapidly oxidises hydrogen.
Which statements are correct?
(B) 1 only
(C) 2 and 3
(D) 3 only
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
Statement 1: False. Iodine has the highest boiling point because it has the strongest instantaneous dipole-induced dipole attractions, but the strongest hydrogen-halogen bond is \( \mathrm{H-F} \), not \( \mathrm{H-I} \).
Statement 2: False. Although iodine has the strongest intermolecular forces, hydrogen iodide is the least thermally stable hydrogen halide because the \( \mathrm{H-I} \) bond is the weakest.
Statement 3: True. Chlorine is the most volatile of the three halogens listed and is the strongest oxidising agent among them, so it reacts most rapidly with hydrogen.
Therefore, only statement 3 is correct. Hence, the correct answer is (D).
Question 19
Which reagent or reagents and conditions will oxidise chlorine, \( \mathrm{Cl_2} \), into a compound containing chlorine in the \(+5\) oxidation state?
(B) Concentrated \( \mathrm{H_2SO_4} \) at room temperature
(C) Cold dilute \( \mathrm{NaOH(aq)} \)
(D) Hot concentrated \( \mathrm{NaOH(aq)} \)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
Chlorine undergoes disproportionation when it reacts with sodium hydroxide.
With cold, dilute \( \mathrm{NaOH} \):
\( \mathrm{Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O} \)
Here, chlorine is oxidised to the \(+1\) oxidation state in \( \mathrm{ClO^-} \).
With hot, concentrated \( \mathrm{NaOH} \):
\( \mathrm{3Cl_2+6NaOH\rightarrow5NaCl+NaClO_3+3H_2O} \)
In \( \mathrm{ClO_3^-} \), chlorine has an oxidation state of \(+5\).
Therefore, the correct answer is (D).
Question 20
Which diagram shows the electronegativity of the elements Na, Mg, Al and Si plotted against their first ionisation energies?
(B) B
(C) C
(D) D
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
Across Period 3, electronegativity generally increases:
\( \mathrm{Na < Mg < Al < Si} \)
The first ionisation energy also generally increases across the period, but there is an exception between magnesium and aluminium because the first \(3\mathrm{p}\) electron in aluminium is easier to remove than a \(3\mathrm{s}\) electron in magnesium.
Thus, the order of first ionisation energies is:
\( \mathrm{Na < Al < Mg < Si} \)
The correct graph must therefore show:
- Electronegativity increasing from Na to Si.
- Magnesium plotted to the right of aluminium because \( \mathrm{IE(Mg) > IE(Al)} \).
Only diagram (B) satisfies both trends. Therefore, the correct answer is (B).
Question 21
Radium is an element below barium in Group 2 of the Periodic Table.
Which equation shows what happens when solid radium nitrate, \( \mathrm{Ra(NO_3)_2} \), is heated strongly?
(B) \( 2\mathrm{Ra(NO_3)_2(s)} \rightarrow 2\mathrm{RaO(s)} + 2\mathrm{N_2(g)} + 5\mathrm{O_2(g)} \)
(C) \( \mathrm{Ra(NO_3)_2(s)} \rightarrow \mathrm{RaO(s)} + \mathrm{N_2O(g)} + 2\mathrm{O_2(g)} \)
(D) \( 4\mathrm{Ra(NO_3)_2(s)} \rightarrow 2\mathrm{Ra_2O(s)} + 8\mathrm{NO_2(g)} + 3\mathrm{O_2(g)} \)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
Group 2 metal nitrates decompose on heating to form the metal oxide, nitrogen dioxide, and oxygen.
The general equation is:
\( 2\mathrm{M(NO_3)_2} \rightarrow 2\mathrm{MO} + 4\mathrm{NO_2} + \mathrm{O_2} \)
Since radium is a Group 2 metal below barium, it undergoes the same type of thermal decomposition.
Option (A) matches the correct balanced equation. Options (B) and (C) give incorrect nitrogen-containing products, while option (D) incorrectly forms \( \mathrm{Ra_2O} \), which is not the oxide of radium.
Therefore, the correct answer is (A).
Question 22
The table shows some data for the elements in Period 3 of the Periodic Table.
| Melting point / K | Electrical conductivity | |
|---|---|---|
| Sodium | 371 | Good |
| Magnesium | 922 | Good |
| Aluminium | 933 | Good |
| Silicon | 1693 | Poor |
| Phosphorus | 317 | Does not conduct |
| Sulfur | 386 | Does not conduct |
| Chlorine | 172 | Does not conduct |
| Argon | 84 | Does not conduct |
Which statements are correct?
1. All the elements in the table with a giant structure have a higher melting point than each of the elements in the table with a simple molecular structure.
2. Magnesium has a higher melting point than sodium because it has more delocalised electrons and stronger electrostatic attraction between the delocalised electrons and the metal ions.
3. Phosphorus and sulfur do not conduct electricity because they are simple molecular solids at room temperature.
(B) 1 and 2 only
(C) 1 and 3 only
(D) 2 and 3 only
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
Statement 1: False. Sodium has a giant metallic structure but its melting point (\(371\,\mathrm{K}\)) is lower than that of sulfur (\(386\,\mathrm{K}\)), which has a simple molecular structure.
Statement 2: True. Magnesium contributes two delocalised electrons per atom and forms \( \mathrm{Mg^{2+}} \) ions, giving stronger metallic bonding than sodium.
Statement 3: True. Phosphorus and sulfur exist as simple molecular solids (\( \mathrm{P_4} \) and \( \mathrm{S_8} \)) and have no mobile charged particles, so they do not conduct electricity.
Therefore, statements 2 and 3 are correct. Hence, the correct answer is (D).
Question 23
Each mineral listed behaves as a mixture of two carbonate compounds. They can be used as fire retardants because they decompose on heating, producing \( \mathrm{CO_2} \). This gas smothers the fire.
Barytocalcite, \( \mathrm{BaCa(CO_3)_2} \)
Dolomite, \( \mathrm{CaMg(CO_3)_2} \)
Huntite, \( \mathrm{Mg_3Ca(CO_3)_4} \)
What is the order of effectiveness as fire retardants, from best to worst?
| Best | → | Worst | |
|---|---|---|---|
| (A) | Dolomite | Barytocalcite | Huntite |
| (B) | Dolomite | Huntite | Barytocalcite |
| (C) | Huntite | Barytocalcite | Dolomite |
| (D) | Huntite | Dolomite | Barytocalcite |
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
The effectiveness of the minerals depends on the amount of \( \mathrm{CO_2} \) released when their carbonate ions decompose.
Each carbonate ion produces one mole of \( \mathrm{CO_2} \).
Number of carbonate ions per formula unit:
Barytocalcite: \(2\)
Dolomite: \(2\)
Huntite: \(4\)
Huntite releases the greatest amount of \( \mathrm{CO_2} \), making it the most effective fire retardant.
Between dolomite and barytocalcite, magnesium carbonate decomposes more readily than barium carbonate, so dolomite releases \( \mathrm{CO_2} \) more easily than barytocalcite.
Therefore, the order from best to worst is Huntite → Dolomite → Barytocalcite. Hence, the correct answer is (D).
Question 24
Compound X contains two Period 3 elements, Y and Z.
Compound X reacts with water to form only two products: a slightly soluble hydroxide and compound Q.
Q is a compound of element Z and hydrogen.
Compound Q burns in moist air to produce an oxide and water. The oxidation number of element Z in the oxide is \(+5\).
Which row identifies element Y and element Z?
| Element Y | Element Z | |
|---|---|---|
| (A) | Mg | P |
| (B) | Na | S |
| (C) | Na | P |
| (D) | Mg | S |
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
The hydroxide formed is described as slightly soluble, which identifies it as \( \mathrm{Mg(OH)_2} \). Sodium hydroxide is highly soluble, so element Y must be magnesium.
The hydride Q must therefore be \( \mathrm{PH_3} \) or \( \mathrm{H_2S} \).
When phosphine burns in moist air:
\( \mathrm{PH_3 + 2O_2 \rightarrow H_3PO_4} \)
or forms \( \mathrm{P_4O_{10}} \), in which phosphorus has an oxidation state of \(+5\).
Sulfur burns to form \( \mathrm{SO_2} \), where sulfur has an oxidation state of \(+4\), not \(+5\).
Therefore, element Y is Mg and element Z is P. Hence, the correct answer is (A).
Question 25
Three reagents are listed.
1. Aqueous sodium carbonate
2. \( \mathrm{LiAlH_4} \)
3. Water
Which reagents react with pure ethanoic acid to give a solution containing ethanoate ions?
(B) 1 and 3
(C) 1 only
(D) 2 and 3
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
Reagent 1: Aqueous sodium carbonate neutralises ethanoic acid:
\( \mathrm{2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + CO_2 + H_2O} \)
The solution contains ethanoate ions, so statement 1 is correct.
Reagent 2: \( \mathrm{LiAlH_4} \) reduces ethanoic acid to ethanol and does not produce a solution containing ethanoate ions.
Reagent 3: Ethanoic acid is a weak acid and partially ionises in water:
\( \mathrm{CH_3COOH(aq) \rightleftharpoons CH_3COO^-(aq) + H^+(aq)} \)
Therefore, the aqueous solution contains ethanoate ions.
Hence, reagents 1 and 3 give a solution containing ethanoate ions. Therefore, the correct answer is (B).
Question 26
Complete combustion of compound T produces carbon dioxide and water only. Compound T produces steamy fumes with \( \mathrm{PCl_5} \). Compound T does not give any visible product with \( \mathrm{2,4\text{-}dinitrophenylhydrazine} \) reagent.
What can be deduced with certainty from this information?
(B) Compound T is a hydrocarbon.
(C) Compound T is an alcohol.
(D) Compound T is not an aldehyde.
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
Complete combustion producing only \( \mathrm{CO_2} \) and \( \mathrm{H_2O} \) shows that T contains only carbon, hydrogen and oxygen.
Steamy fumes with \( \mathrm{PCl_5} \) indicate the presence of an \( \mathrm{-OH} \) group. This could be an alcohol or a carboxylic acid.
\( \mathrm{2,4\text{-}dinitrophenylhydrazine} \) gives an orange precipitate with aldehydes and ketones. Since no visible product is formed, T does not contain an aldehyde or ketone carbonyl group.
Therefore, it cannot be concluded with certainty that T is an alcohol or a carboxylic acid, but it can be concluded that T is not an aldehyde.
Hence, the correct answer is (D).
Question 27
Information about carbonyl compound X is given.
- Compound X reacts with \( \mathrm{LiAlH_4} \) to produce a secondary alcohol.
- Compound X reacts with alkaline \( \mathrm{I_2(aq)} \) to give a yellow precipitate.
What is compound X?
(B) Butanone
(C) Ethanal
(D) Pentan-3-one
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
Reduction with \( \mathrm{LiAlH_4} \) gives a secondary alcohol, so X must be a ketone. Aldehydes reduce to primary alcohols.
The yellow precipitate with alkaline \( \mathrm{I_2} \) is the iodoform test, which is positive for compounds containing the \( \mathrm{CH_3CO{-}} \) group.
Among the options, only butanone, \( \mathrm{CH_3COCH_2CH_3} \), is a methyl ketone and gives a positive iodoform test.
Pentan-3-one is a ketone but does not contain a \( \mathrm{CH_3CO{-}} \) group, so it gives a negative iodoform test.
Therefore, the correct answer is (B).
Question 28
An organic compound J reacts with an excess of sodium to produce an organic ion with a charge of \( -3 \). J reacts with an excess of \( \mathrm{NaOH(aq)} \) to produce an organic ion with a charge of \( -1 \).
What is the structural formula of J?
(B) \( \mathrm{HO_2CCH(OH)CH_2CHO} \)
(C) \( \mathrm{HO_2CCH(OH)CH_2CO_2H} \)
(D) \( \mathrm{HOCH_2COCH_2CHO} \)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
Sodium metal reacts with all acidic \( \mathrm{-OH} \) groups, including alcohols and carboxylic acids, replacing each acidic hydrogen with sodium.
A \( -3 \) ion means J contains three acidic hydrogen atoms.
Aqueous \( \mathrm{NaOH} \) only deprotonates carboxylic acids, not alcohols. A \( -1 \) ion therefore indicates that J contains one carboxylic acid group.
Option (A) contains:
- Two alcohol groups (\( \mathrm{-OH} \))
- One carboxylic acid group (\( \mathrm{-CO_2H} \))
Thus, it forms a \( -3 \) ion with excess sodium and a \( -1 \) ion with excess \( \mathrm{NaOH(aq)} \).
Therefore, the correct answer is (A).
Question 29
Which formula represents the organic compound formed by the reaction of propanoic acid with methanol in the presence of concentrated sulfuric acid as a catalyst?
(B) \( \mathrm{CH_3CH_2CO_2CH_3} \)
(C) \( \mathrm{CH_3CO_2CH_2CH_3} \)
(D) \( \mathrm{CH_3CH_2CH_2CO_2CH_3} \)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
Propanoic acid reacts with methanol in an acid-catalysed esterification reaction to form an ester and water.
\( \mathrm{CH_3CH_2COOH + CH_3OH \rightleftharpoons CH_3CH_2COOCH_3 + H_2O} \)
The ester formed is methyl propanoate, with the formula \( \mathrm{CH_3CH_2CO_2CH_3} \).
Option (A) is a ketone, option (C) is ethyl ethanoate, and option (D) is methyl butanoate.
Therefore, the correct answer is (B).
Question 30
Structural isomerism and stereoisomerism should be considered when answering this question.
2,5-Dibromohexane is heated under reflux with ethanolic \( \mathrm{KOH} \).
How many isomeric compounds are formed with molecular formula \( \mathrm{C_6H_{10}} \)?
(B) 5
(C) 6
(D) 7
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
Ethanolic \( \mathrm{KOH} \) causes elimination of two molecules of \( \mathrm{HBr} \) from 2,5-dibromohexane, producing dienes.
The possible structural isomers are:
- \( \mathrm{hexa\text{-}1,4\text{-}diene} \)
- \( \mathrm{hexa\text{-}1,3\text{-}diene} \)
- \( \mathrm{hexa\text{-}2,4\text{-}diene} \)
Considering stereoisomerism:
- \( \mathrm{hexa\text{-}1,4\text{-}diene} \) has 1 isomer.
- \( \mathrm{hexa\text{-}1,3\text{-}diene} \) has 2 geometrical isomers (\( \mathrm{E} \) and \( \mathrm{Z} \)).
- \( \mathrm{hexa\text{-}2,4\text{-}diene} \) has 3 geometrical isomers: \( (\mathrm{E,E}) \), \( (\mathrm{Z,Z}) \), and \( (\mathrm{E,Z}) \) (where \( (\mathrm{E,Z}) \) and \( (\mathrm{Z,E}) \) are identical by symmetry).
Total number of isomeric compounds:
\( 1 + 2 + 3 = 6 \)
Therefore, the correct answer is (C).
Question 31
Considering only structural isomers, what is the number of alcohols of each type with the formula \( \mathrm{C_5H_{12}O} \)?
| Primary | Secondary | Tertiary | |
|---|---|---|---|
| (A) | 3 | 3 | 2 |
| (B) | 4 | 2 | 2 |
| (C) | 4 | 3 | 1 |
| (D) | 5 | 2 | 1 |
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
There are 8 structural isomeric alcohols with the molecular formula \( \mathrm{C_5H_{12}O} \).
Classification by the carbon atom bonded to the \( \mathrm{-OH} \) group gives:
- Primary alcohols (4): pentan-1-ol, 2-methylbutan-1-ol, 3-methylbutan-1-ol, 2,2-dimethylpropan-1-ol.
- Secondary alcohols (3): pentan-2-ol, pentan-3-ol, 3-methylbutan-2-ol.
- Tertiary alcohols (1): 2-methylbutan-2-ol.
Therefore, the numbers are:
Primary \(=4\), Secondary \(=3\), Tertiary \(=1\).
Hence, the correct answer is (C).
Question 32
Which structure represents part of the polymer chain of PVC?
(B) B
(C) C
(D) D
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
PVC is formed by addition polymerisation of chloroethene (vinyl chloride), \( \mathrm{CH_2=CHCl} \).
The repeating unit of PVC is:
\( \mathrm{-CH_2-CHCl-} \)
Therefore, every alternate carbon atom in the polymer chain carries one chlorine atom, while the other carbon carries two hydrogen atoms.
Only structure A shows the repeating pattern \( \mathrm{-CH_2-CHCl-} \).
Options B, C and D have incorrect arrangements of chlorine atoms and do not represent the repeating unit of PVC.
Hence, the correct answer is (A).
Question 33
A possible mechanism for the exothermic hydrolysis of 2-chloro-2-methylpropane is shown.
The reaction proceeds in two steps:
Which diagram represents the reaction pathway diagram for this mechanism?
(B) B
(C) C
(D) D
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
The mechanism is an \( \mathrm{S_N1} \) reaction and occurs in two steps, so the energy profile must contain two transition states separated by an intermediate (the carbocation).
The first step, formation of the carbocation, is the slow, rate-determining step. Therefore, it has the higher activation energy.
The second step, attack by \( \mathrm{OH^-} \), is fast and has a lower activation energy, giving a second, smaller peak.
Since the reaction is exothermic, the products are at a lower energy than the reactants.
Diagram B correctly shows:
- Two energy maxima (two-step mechanism).
- The first peak higher than the second (slow first step).
- Products lower in energy than reactants (exothermic reaction).
Therefore, the correct answer is (B).
Question 34
Compound W contains atoms of carbon, nitrogen and hydrogen.
Compound W reacts with \( \mathrm{HCl(aq)} \) to produce propanoic acid.
Which row is correct?
| Functional group in compound W | Name of compound W | |
|---|---|---|
| (A) | Amine | Ethylamine |
| (B) | Nitrile | Ethanenitrile |
| (C) | Amine | Propylamine |
| (D) | Nitrile | Propanenitrile |
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
Nitriles are hydrolysed by aqueous acids to form carboxylic acids.
The general reaction is:
\( \mathrm{RCN + 2H_2O + HCl \rightarrow RCOOH + NH_4Cl} \)
To produce propanoic acid, the nitrile must contain three carbon atoms.
\( \mathrm{CH_3CH_2CN} \) (propanenitrile) hydrolyses to \( \mathrm{CH_3CH_2COOH} \) (propanoic acid).
Amines do not hydrolyse to carboxylic acids, and ethanenitrile would produce ethanoic acid.
Therefore, the correct answer is (D).
Question 35
Two reactions are described.
1. 2-bromo-2-methylbutane heated with \( \mathrm{NaOH(aq)} \)
2. 2-chloro-2-methylbutane heated with \( \mathrm{NaOH(aq)} \)
In both reactions, the conditions are the same and \( \mathrm{NaOH(aq)} \) is in excess.
\(30.18\,\mathrm{g}\) of 2-bromo-2-methylbutane forms \(12.32\,\mathrm{g}\) of product X.
Given:
\(M_r\) of 2-bromo-2-methylbutane \(=150.9\)
\(M_r\) of 2-chloro-2-methylbutane \(=106.5\)
Which row is correct?
| Percentage yield of product X / % | Relative rate of reaction | |
|---|---|---|
| (A) | 41 | 1 is faster than 2 |
| (B) | 41 | 2 is faster than 1 |
| (C) | 70 | 1 is faster than 2 |
| (D) | 70 | 2 is faster than 1 |
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
Both compounds undergo nucleophilic substitution to form 2-methylbutan-2-ol, which has \(M_r=88\).
Moles of 2-bromo-2-methylbutane used:
\( \dfrac{30.18}{150.9}=0.200\,\mathrm{mol} \)
Theoretical mass of alcohol:
\(0.200\times88=17.6\,\mathrm{g}\)
Percentage yield:
\( \dfrac{12.32}{17.6}\times100=70\% \)
Bromide ions are better leaving groups than chloride ions because the \( \mathrm{C-Br} \) bond is weaker than the \( \mathrm{C-Cl} \) bond.
Therefore, 2-bromo-2-methylbutane reacts faster than 2-chloro-2-methylbutane.
Hence, the correct answer is (C).
Question 36
An alkene P reacts with an excess of hot concentrated acidified \( \mathrm{KMnO_4(aq)} \).
Methylpropanoic acid is the only organic product.
What is alkene P?
(B) 2-Methylbut-2-ene
(C) Methylpropene
(D) Oct-4-ene, \( \mathrm{C_8H_{16}} \)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
Hot concentrated acidified \( \mathrm{KMnO_4} \) cleaves the carbon-carbon double bond.
If each carbon of the double bond has one hydrogen atom, each side is oxidised to a carboxylic acid.
For 2,5-dimethylhex-3-ene:
\( \mathrm{CH_3CH(CH_3)CH=CHCH(CH_3)CH_3} \)
Oxidative cleavage gives two identical molecules of methylpropanoic acid:
\( \mathrm{2\,CH_3CH(CH_3)COOH} \)
Thus, methylpropanoic acid is the only organic product.
The other alkenes produce mixtures of different acids, ketones or \( \mathrm{CO_2} \), so they do not satisfy the condition.
Therefore, the correct answer is (A).
Question 37
Three substances are listed.
1. Butane
2. Hydrogen
3. Hydrogen bromide
Which substances are possible products of the free-radical substitution reaction between ethane and bromine?
(B) 1 and 2 only
(C) 1 and 3 only
(D) 2 and 3 only
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
The main free-radical substitution reaction is:
\( \mathrm{C_2H_6 + Br_2 \xrightarrow{UV} C_2H_5Br + HBr} \)
Therefore, hydrogen bromide is produced.
Radical termination reactions can also occur. For example:
\( \mathrm{C_2H_5^{\bullet} + C_2H_5^{\bullet} \rightarrow C_4H_{10}} \)
This produces butane.
Hydrogen gas, \( \mathrm{H_2} \), is not formed during the free-radical substitution mechanism.
Hence, the possible products are 1 and 3 only. Therefore, the correct answer is (C).
Question 38
The structure of the testosterone molecule is shown.
Which statements are correct?
1. Carbon atoms C1 and C2 can be oxidised with acidified \( \mathrm{K_2Cr_2O_7} \).
2. There are fewer than 27 hydrogen atoms in one testosterone molecule.
3. There are six chiral carbon atoms in one testosterone molecule.
(B) 2 and 3
(C) 2 only
(D) 3 only
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
Statement 1: False. C1 is part of a ketone carbonyl group, which is not oxidised by acidified \( \mathrm{K_2Cr_2O_7} \). C2 bears a tertiary alcohol, and tertiary alcohols are also resistant to oxidation under these conditions.
Statement 2: False. Testosterone has the molecular formula \( \mathrm{C_{19}H_{28}O_2} \), so it contains 28 hydrogen atoms, not fewer than 27.
Statement 3: True. Testosterone contains six chiral carbon atoms at the ring junctions and substituted carbon atoms.
Therefore, only statement 3 is correct. Hence, the correct answer is (D).
Question 39
The structural formula of hept-1,4,5-triene is shown.
\( \mathrm{CH_2=CHCH_2CH=C=CHCH_3} \)
How many of the carbon atoms in one molecule of hept-1,4,5-triene are \( \mathrm{sp^2} \) hybridised?
(B) 5
(C) 6
(D) 7
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
Number the carbon atoms:
\( \mathrm{C_1=C_2-C_3-C_4=C_5=C_6-C_7} \)
Hybridisation of each carbon:
- \( \mathrm{C_1} \): \( \mathrm{sp^2} \)
- \( \mathrm{C_2} \): \( \mathrm{sp^2} \)
- \( \mathrm{C_3} \): \( \mathrm{sp^3} \)
- \( \mathrm{C_4} \): \( \mathrm{sp^2} \)
- \( \mathrm{C_5} \): \( \mathrm{sp} \) (central carbon of the allene)
- \( \mathrm{C_6} \): \( \mathrm{sp^2} \)
- \( \mathrm{C_7} \): \( \mathrm{sp^3} \)
Thus, the \( \mathrm{sp^2} \)-hybridised carbon atoms are \( \mathrm{C_1,\ C_2,\ C_4,\ C_6} \).
Total number of \( \mathrm{sp^2} \) carbon atoms:
\( \boxed{4} \)
Therefore, the correct answer is (A).
Question 40
Hydrocarbon X is saturated. Each molecule contains one ring of carbon atoms.
The mass spectrum of hydrocarbon X is measured.
The peak representing the \( \mathrm{M^+} \) ion is \(14\,\mathrm{mm}\) high.
The peak representing the \( \mathrm{[M+1]^+} \) ion is \(0.77\,\mathrm{mm}\) high.
What is the \( \mathrm{m/e} \) value for the \( \mathrm{M^+} \) ion of X?
(B) 58
(C) 70
(D) 72
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
The abundance of the \( \mathrm{[M+1]^+} \) peak is due mainly to the natural abundance of \( \mathrm{^{13}C} \), which is about 1.1% per carbon atom.
The percentage abundance of the \( \mathrm{[M+1]^+} \) peak is:
\( \dfrac{0.77}{14}\times100 = 5.5\% \)
Hence, the number of carbon atoms is:
\( \dfrac{5.5}{1.1}=5 \)
A saturated hydrocarbon containing one ring is a cycloalkane with general formula:
\( \mathrm{C_nH_{2n}} \)
Therefore, X has the formula:
\( \mathrm{C_5H_{10}} \)
Its molecular mass is:
\( (5\times12)+(10\times1)=70 \)
Thus, the \( \mathrm{M^+} \) peak appears at \( \mathrm{m/e}=70 \). Therefore, the correct answer is (C).