ESAT Paper 4 Chemistry Worked Solutions

Correct Answer: D. 2, 8, 8, 2

Fastest Approach (🚀):
Ca = 20 electrons. 2 + 8 + 8 + 2 = 20. Option D.

Step-by-Step Breakdown:

1. Identify the number of electrons

Calcium (Ca) has an atomic number of 20. A neutral atom of calcium therefore has 20 electrons.

2. Fill the electron shells

Electrons fill shells starting from the lowest energy level closest to the nucleus:
1st shell holds a maximum of 2 electrons.
2nd shell holds a maximum of 8 electrons.

• 3rd shell holds a maximum of 8 electrons (before the 4th shell starts filling).

Total so far: $2 + 8 + 8 = 18$ electrons. This leaves 2 electrons.

• 4th shell holds the remaining 2 electrons.

Therefore, the configuration is 2, 8, 8, 2.

Common Mistake (⚠️):
Confusing calcium with a different element, or packing too many electrons into the 3rd shell (e.g., 2, 8, 10).

Correct Answer: E. 1 and 3

Fastest Approach (🚀):
Cl goes from 0 to -1 and +5. That's disproportionation (1). Disproportionation is a subset of redox (3). No solids are formed, so not precipitation (2). Thus 1 and 3.

Step-by-Step Breakdown:

1. Analyze for Redox and Disproportionation

Let's determine the oxidation states of chlorine:

• Reactant: In $\text{Cl}_2$, chlorine is in its elemental form, so its oxidation state is 0.
• Product 1: In $\text{Cl}^-$, the oxidation state is -1. (Chlorine is reduced).
• Product 2: In the chlorate(V) ion, $\text{ClO}_3^-$, oxygen is -2. So, $\text{Cl} + 3(-2) = -1 \implies \text{Cl} - 6 = -1 \implies \text{Cl} = \text{+5}$. (Chlorine is oxidised).

Because the same element (chlorine) is simultaneously oxidised and reduced, this is a disproportionation reaction (Statement 1 is true). Because oxidation states change, it is inherently a redox reaction (Statement 3 is true).

2. Analyze for Precipitation

A precipitation reaction requires aqueous reactants to form an insoluble solid product. All products here ($\text{Cl}^-$, $\text{ClO}_3^-$, and $\text{H}_2\text{O}$) are soluble ions or liquids. No solid is formed. (Statement 2 is false).

Therefore, 1 and 3 are correct.

Common Mistake (⚠️):
Failing to realize that disproportionation is just a specific type of redox reaction, so if 1 is true, 3 must also be true.

Correct Answer: B. increase the pressure and decrease the temperature

Fastest Approach (🚀):
Left side has 2 moles gas, right has 3. To favor left (NOCl), increase pressure. Decomposition is endothermic, so forming NOCl is exothermic. To favor exothermic, decrease temperature. Increase P, decrease T.

Step-by-Step Breakdown:

1. Determine the goal

The question asks to maximise the concentration of $\text{NOCl}$, which is a reactant. This means we want to shift the equilibrium to the left.

2. Effect of Pressure

Count the moles of gas on each side:

• Left side: 2 moles (from $2\text{NOCl}$)
• Right side: 3 moles (from $2\text{NO} + 1\text{Cl}_2$)

Le Chatelier's principle states that increasing pressure shifts the equilibrium to the side with fewer moles of gas. Since we want to shift left (2 moles), we must increase the pressure.

3. Effect of Temperature

The prompt says "Upon being heated... NOCl decomposes". This implies the forward decomposition reaction requires heat, making it endothermic. Therefore, the reverse reaction (forming NOCl) is exothermic.
To shift equilibrium in the exothermic direction, we must remove heat by decreasing the temperature.

Therefore, we should increase pressure and decrease temperature to maximise NOCl.

Common Mistake (⚠️):
Misidentifying the decomposition as exothermic, or trying to shift the equilibrium to the right (to the products) by mistake.

Correct Answer: C. $\text{C}_4\text{H}_{10}\text{O}$

Fastest Approach (🚀):
Mass of C per mol = $0.65 \times 74 = 48.1\text{ g}$. Number of C atoms = $48.1 / 12 \approx 4$. Option C is the only one with 4 carbon atoms ($\text{C}_4\text{H}_{10}\text{O}$). Check its mass: $(4\times 12) + (10\times 1) + 16 = 74$. Matches perfectly.

Step-by-Step Breakdown:

1. Direct Method using Molecular Weight

Since we are given the exact molecular weight of $74\text{ g/mol}$, we can find the exact mass of each element present in one mole of the compound.

• Carbon: $65\%$ of $74\text{ g} = 0.65 \times 74 = 48.1\text{ g}$
• Hydrogen: $14\%$ of $74\text{ g} = 0.14 \times 74 = 10.36\text{ g}$
• Oxygen: $21\%$ of $74\text{ g} = 0.21 \times 74 = 15.54\text{ g}$

2. Find the number of atoms

Divide the mass of each element by its relative atomic mass ($A_r$ for $\text{C}=12$, $\text{H}=1$, $\text{O}=16$):

• C atoms: $48.1 / 12 \approx 4.01 \rightarrow 4$
• H atoms: $10.36 / 1 \approx 10.36 \rightarrow 10$
• O atoms: $15.54 / 16 \approx 0.97 \rightarrow 1$

The molecular formula is $\text{C}_4\text{H}_{10}\text{O}$.

3. Verification

Let's check the molecular weight of $\text{C}_4\text{H}_{10}\text{O}$:
$(4 \times 12) + (10 \times 1) + (1 \times 16) = 48 + 10 + 16 = 74\text{ g/mol}$.
This matches the given molecular weight exactly.

Common Mistake (⚠️):
Making rounding errors while calculating the empirical formula from percentages (e.g. dividing by smallest moles) which can be tedious without a calculator.

Correct Answer: A. 1.54 g

Fastest Approach (🚀):
Theo AgCl = $1.82 / 0.856 = 2.126\text{g}$. Moles AgCl = $2.126 / 143.4 = 0.0148$. Stoichiometry: 1 BaCl2 per 2 AgCl. Moles BaCl2 = $0.0148 / 2 = 0.00741$. Mass BaCl2 = $0.00741 \times 208.3 = 1.54\text{g}$.

Step-by-Step Breakdown:

1. Calculate Theoretical Yield of AgCl

The problem gives the actual yield ($1.82\text{g}$) and percentage yield ($85.6\%$).
$$\text{Theoretical Yield} = \frac{\text{Actual Yield}}{\text{Percentage Yield}} = \frac{1.82}{0.856} = 2.126\text{ g}$$

2. Convert Theoretical Yield to Moles

Calculate the molar mass of $\text{AgCl}$:
$M_r(\text{AgCl}) = 107.9 + 35.5 = 143.4\text{ g/mol}$
$$\text{Moles of AgCl} = \frac{\text{Mass}}{M_r} = \frac{2.126}{143.4} = 0.0148\text{ moles}$$

3. Use Stoichiometry to find Moles of BaCl$_2$

From the balanced equation: $1\text{ mole of BaCl}_2 \rightarrow 2\text{ moles of AgCl}$.
Therefore, the moles of $\text{BaCl}_2$ required is half the moles of $\text{AgCl}$.
$$\text{Moles of BaCl}_2 = \frac{0.0148}{2} = 0.0074\text{ moles}$$

4. Calculate Mass of BaCl$_2$

Calculate the molar mass of $\text{BaCl}_2$:
$M_r(\text{BaCl}_2) = 137.3 + (2 \times 35.5) = 137.3 + 71.0 = 208.3\text{ g/mol}$
$$\text{Mass of BaCl}_2 = \text{Moles} \times M_r = 0.0074 \times 208.3 = 1.54\text{ g}$$

Common Mistake (⚠️):
Forgetting to divide the actual yield by the percentage yield, or forgetting the 1:2 molar ratio between BaCl2 and AgCl.

Correct Answer: A. 1 only

Fastest Approach (🚀):
$3.6 / 9 = 0.4$. So 1 is true. 3 spots doesn't mean exactly 3 compounds (could be overlap, could be invisible spots). Separation is by solubility/affinity, not purely molecular weight (isomers can separate). Only 1 is true.

Step-by-Step Breakdown:

1. Evaluate Statement 1 ($R_f$ value)

The retention factor ($R_f$) is calculated as:
$R_f = \frac{\text{distance travelled by compound}}{\text{distance travelled by solvent front}}$
For the spot that moved $3.6\text{ cm}$, the $R_f$ value is $3.6 / 9 = 0.4$. Therefore, statement 1 is true.

2. Evaluate Statement 2 (Number of compounds)

Seeing three spots only guarantees there are at least three compounds. There could be multiple compounds with identical $R_f$ values in this specific solvent, causing their spots to overlap entirely. There could also be compounds in the mixture that are invisible without a locating agent or UV light. Therefore, we cannot say it 'must' contain exactly three. (Statement 2 is not necessarily true).

3. Evaluate Statement 3 (Molecular weights)

Paper chromatography separates compounds based primarily on their relative solubilities in the mobile phase and their affinities for the stationary phase (polarity). It does not separate them strictly by molecular weight. Two completely different compounds with different polarities can have the exact same molecular weight (e.g., isomers) and still separate into different spots. Therefore, they do not 'must' have different molecular weights. (Statement 3 is not necessarily true).

Only Statement 1 must be true.

Common Mistake (⚠️):
Assuming that one spot perfectly corresponds to exactly one compound, or confusing paper chromatography with gel filtration (which separates by size).

Correct Answer: F. 1, 2 and 3

Fastest Approach (🚀):
1: True, H2SO4 has mobile ions, ethanol is covalent. 2: True, formula is H2SO4 (2 protons). 3: True, 1 mole H2SO4 gives 2 moles H+, ethanoic gives 1 mole H+, so H2SO4 needs more base.

Step-by-Step Breakdown:

1. Evaluate Statement 1

Sulfuric acid is a strong acid that dissociates in water to form mobile ions ($2\text{H}^+$ and $\text{SO}_4^{2-}$). These mobile ions carry electrical charge, making the solution a good conductor. Ethanol is a covalent molecule that dissolves in water but does not dissociate into ions, so its solution is a very poor conductor. (Statement 1 is true).

2. Evaluate Statement 2

$\text{H}_2\text{SO}_4$ has two ionizable hydrogen atoms per molecule. Acids that can donate two protons ($\text{H}^+$ ions) per molecule are called diprotic acids. (Statement 2 is true).

3. Evaluate Statement 3

Because sulfuric acid is diprotic, 1 mole of $\text{H}_2\text{SO}_4$ provides 2 moles of $\text{H}^+$ ions. Neutralizing it requires 2 moles of a monoprotic base (like $\text{NaOH}$). Ethanoic acid ($\text{CH}_3\text{COOH}$) is monoprotic, so 1 mole of it provides only 1 mole of $\text{H}^+$ ions, requiring only 1 mole of $\text{NaOH}$ to neutralize. Therefore, a greater amount of base is required for sulfuric acid. (Statement 3 is true).

All three statements are true.

Common Mistake (⚠️):
Misreading 'ethanoic acid' as something diprotic, or assuming ethanol conducts electricity.

Correct Answer: C. $4\text{Ag}^+(aq) + 4\text{e}^- \rightarrow 4\text{Ag}(s)$

Fastest Approach (🚀):
Cathode = reduction (electrons on the left). Ions present: $\text{Ag}^+$, $\text{H}^+$. Ag is less reactive, so $\text{Ag}^+$ is reduced. Equation: $\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag}$. Option C is just this multiplied by 4.

Step-by-Step Breakdown:

1. Identify the ions attracted to the cathode

The cathode is the negatively charged electrode. It attracts cations (positive ions) from the solution. Aqueous silver nitrate contains $\text{Ag}^+$ and $\text{H}^+$ (from the dissociation of water).

2. Determine which ion is discharged

At the cathode, reduction (gain of electrons) occurs. The rule is that the less reactive element is discharged. Comparing silver and hydrogen on the reactivity series, silver is significantly lower. Therefore, silver ions ($\text{Ag}^+$) will be preferentially discharged.

3. Formulate the half-equation

Silver ions gain one electron to form solid silver atoms:
$\text{Ag}^+(aq) + \text{e}^- \rightarrow \text{Ag}(s)$

Looking at the options, Option C shows this exact reaction, simply multiplied by a coefficient of 4:
$4\text{Ag}^+(aq) + 4\text{e}^- \rightarrow 4\text{Ag}(s)$
This is a perfectly valid and balanced half-equation for the cathode reaction.

Common Mistake (⚠️):
Choosing A (which is the anode reaction for water) or D (which is oxidation, happening at the anode if silver electrodes were used).

Correct Answer: B. $4.184 \times 10^7\text{ J}$

Fastest Approach (🚀):
$500\text{ L} = 500,000\text{ mL} = 500,000\text{ g}$. $\Delta T = 27 - 7 = 20$. $q = 500,000 \times 4.184 \times 20 = 10,000,000 \times 4.184 = 4.184 \times 10^7$.

Step-by-Step Breakdown:

1. Determine the mass of the water ($m$)

The volume is $500\text{ litres}$. We need to convert this to millilitres (cm$^3$) to use the density.
$500\text{ L} = 500 \times 1000\text{ mL} = 500,000\text{ mL}$.
Since the density is $1\text{ g/mL}$, the mass is:
$m = 500,000\text{ g} = 5 \times 10^5\text{ g}$.

2. Determine the change in temperature ($\Delta T$)

$\Delta T = \text{Final Temperature} - \text{Initial Temperature}$
$\Delta T = 27 - 7 = 20^\circ\text{C}$.

3. Calculate the energy ($q$)

Use the specific heat capacity formula: $q = mc\Delta T$
$q = (500,000\text{ g}) \times (4.184\text{ J/g}^\circ\text{C}) \times (20^\circ\text{C})$
$q = 500,000 \times 20 \times 4.184$
$q = 10,000,000 \times 4.184$
$q = 41,840,000\text{ J}$

In standard form, this is $4.184 \times 10^7\text{ J}$.

Common Mistake (⚠️):
Forgetting to convert litres to millilitres/grams, resulting in a magnitude error (e.g. calculating $4.184 \times 10^4$).

Correct Answer: C. 0.09 M

Fastest Approach (🚀):
Since volumes are in mL, you can use millimoles (mmol).
Sol 1: $80 \times 0.15 = 12\text{ mmol}$.
Sol 2: $120 \times 0.05 = 6\text{ mmol}$.
Total mmol = $12 + 6 = 18\text{ mmol}$.
Total volume = $80 + 120 = 200\text{ mL}$.
Conc = $18 / 200 = 0.09\text{ M}$.

Step-by-Step Breakdown:

1. Calculate moles in the first solution

$\text{Concentration} = 0.15\text{ mol/L}$
$\text{Volume} = 80\text{ mL} = 0.080\text{ L}$
$\text{Moles}_1 = C \times V = 0.15 \times 0.080 = 0.012\text{ moles}$

2. Calculate moles in the second solution

$\text{Concentration} = 0.05\text{ mol/L}$
$\text{Volume} = 120\text{ mL} = 0.120\text{ L}$
$\text{Moles}_2 = C \times V = 0.05 \times 0.120 = 0.006\text{ moles}$

3. Calculate final concentration

$\text{Total Moles} = 0.012 + 0.006 = 0.018\text{ moles}$
$\text{Total Volume} = 0.080 + 0.120 = 0.200\text{ L}$

$\text{Final Concentration} = \frac{\text{Total Moles}}{\text{Total Volume}} = \frac{0.018}{0.200} = 0.09\text{ M}$

Common Mistake (⚠️):
Simply taking the average of $0.15$ and $0.05$ (which is $0.10$), ignoring the fact that the volumes are different.

Correct Answer: A. 0.08 M

Fastest Approach (🚀):
Moles HCl = $0.040\text{ L} \times 1.25\text{ M} = 0.050\text{ mol}$. Moles Ba(OH)2 = $0.050 / 2 = 0.025\text{ mol}$. Conc Ba(OH)2 = $0.025\text{ mol} / 0.300\text{ L} = 0.0833\text{ M}$. Rounds to $0.08\text{ M}$. Option A.

Step-by-Step Breakdown:

1. Calculate Moles of HCl used

$C(\text{HCl}) = 1.25\text{ mol/L}$
$V(\text{HCl}) = 40\text{ mL} = 0.040\text{ L}$
$\text{Moles of HCl} = C \times V = 1.25 \times 0.040 = 0.050\text{ moles}$

2. Calculate Moles of Ba(OH)$_2$ reacted

The balanced equation is $2\text{HCl} + 1\text{Ba(OH)}_2 \rightarrow \text{BaCl}_2 + 2\text{H}_2\text{O}$.
The molar ratio of $\text{HCl}$ to $\text{Ba(OH)}_2$ is $2 : 1$.
Therefore, the moles of $\text{Ba(OH)}_2$ is half the moles of $\text{HCl}$:
$\text{Moles of Ba(OH)}_2 = 0.050 / 2 = 0.025\text{ moles}$

3. Calculate Concentration of Ba(OH)$_2$

$\text{Moles} = 0.025\text{ moles}$
$\text{Volume} = 300\text{ mL} = 0.300\text{ L}$
$\text{Concentration} = \text{Moles} / \text{Volume} = 0.025 / 0.300 = 0.08333...\text{ M}$

This is closest to $0.08\text{ M}$.

Common Mistake (⚠️):
Forgetting the 2:1 stoichiometric ratio, which would result in a concentration of 0.16 M (not an option) or dividing the wrong way to get 0.33 M.

Correct Answer: E. Bromoethane

Fastest Approach (🚀):
Cream precipitate = Silver Bromide (AgBr). Therefore the original compound contained bromine. Bromoethane is the only option with bromine. Option E.

Step-by-Step Breakdown:

1. Understand the chemical test

The sequence of warming an organic compound with $\text{NaOH}(aq)$ followed by adding $\text{AgNO}_3(aq)$ is the standard qualitative test for identifying halogenoalkanes (alkyl halides).

• Step 1: $\text{NaOH}$ hydrolyses the C-X bond, releasing the halide ion $X^-$ into the aqueous solution.
• Step 2: $\text{Ag}^+$ ions from the silver nitrate react with the halide ions to form an insoluble silver halide precipitate ($\text{AgX}$).

2. Identify the precipitate

The colours of the silver halide precipitates are:

• $\text{AgCl}$: White precipitate
• $\text{AgBr}$: Cream precipitate
• $\text{AgI}$: Yellow precipitate

Because a cream precipitate forms, the halide ion must be bromide ($\text{Br}^-$).

3. Identify the compound

The original organic compound must contain bromine. Looking at the options:
A. Propanoic acid (No halogen)
B. 2-iodohexane (Contains iodine $\rightarrow$ yellow ppt)

• C. 1-chloropentane (Contains chlorine $\rightarrow$ white ppt)
• D. Butan-2-ol (No halogen)
• E. Bromoethane (Contains bromine $\rightarrow$ cream ppt)

Therefore, the compound is Bromoethane.

Common Mistake (⚠️):
Confusing the colours of the silver halide precipitates (e.g., thinking iodine is cream).

Correct Answer: D. $^{39}_{19}\text{K}^+$

Fastest Approach (🚀):
$^{40}_{20}\text{Ca}^{2+}$ has $40-20=20$ neutrons and $20-2=18$ electrons. We need $20\text{n}, 18\text{e}$. Option D: $^{39}_{19}\text{K}^+$ has $39-19=20$ neutrons and $19-1=18$ electrons. Perfect match.

Step-by-Step Breakdown:

1. Determine the particles in the target ion ($^{40}_{20}\text{Ca}^{2+}$)

• Protons: Atomic number (bottom number) = $20$
• Neutrons: Mass number - Atomic number = $40 - 20 = 20$
• Electrons: Protons - Charge = $20 - (+2) = 18$

Target to match: 20 neutrons, 18 electrons.

2. Evaluate the options

• A. $^{35}_{17}\text{Cl}^-$:
• Neutrons: $35 - 17 = 18$ (Incorrect)
• B. $^{37}_{17}\text{Cl}$:
• Neutrons: $37 - 17 = 20$
• Electrons: $17 - 0 = 17$ (Incorrect)
• C. $^{40}_{18}\text{Ar}$:
• Neutrons: $40 - 18 = 22$ (Incorrect)
• D. $^{39}_{19}\text{K}^+$:
• Neutrons: $39 - 19 = 20$
• Electrons: $19 - (+1) = 18$ (Correct match!)
• E. $^{39}_{19}\text{K}$:
• Neutrons: $39 - 19 = 20$
• Electrons: $19 - 0 = 19$ (Incorrect)

Therefore, $^{39}_{19}\text{K}^+$ is the correct ion.

Common Mistake (⚠️):
Adding the charge instead of subtracting it to find electrons, or confusing atomic number and mass number.

Correct Answer: D. 3 only

Fastest Approach (🚀):
Ti atoms are transition metals; neutral atoms don't have noble gas configurations (1 is false). Molten ionic compounds conduct via mobile ions, not delocalised electrons (2 is false). Moles Ti = $7.2/48 = 0.15$. Moles O2 = $3.6/24 = 0.15$. So 0.15 mol Ti and 0.30 mol O atoms. Ratio is 1:2. Empirical formula is TiO2 (3 is true).

Step-by-Step Breakdown:

1. Evaluate Statement 1 (Electron configuration)

Titanium is a transition metal. While a titanium ion (like $\text{Ti}^{4+}$) might achieve a noble gas configuration (Argon core), the neutral titanium atoms produced after electrolysis have the configuration $[\text{Ar}] 4s^2 3d^2$. This is not a noble gas configuration. (Statement 1 is false).

2. Evaluate Statement 2 (Conductivity mechanism)

Titanium oxide is an ionic compound. When solid, its ions are locked in a lattice. When molten, its ions become free to move and carry electrical charge. The conductivity of molten ionic compounds is due to mobile ions, not delocalised electrons (which is the mechanism for metals and graphite). (Statement 2 is false).

3. Evaluate Statement 3 (Empirical formula)

Let's calculate the moles of the products formed during electrolysis:

• Moles of Ti: $n = \frac{m}{A_r} = \frac{7.2\text{ g}}{48\text{ g/mol}} = 0.15\text{ moles}$
• Moles of O$_2$ gas: $n = \frac{V}{V_m} = \frac{3.6\text{ dm}^3}{24\text{ dm}^3/\text{mol}} = 0.15\text{ moles}$

Since 1 mole of $\text{O}_2$ gas contains 2 moles of Oxygen atoms, $0.15\text{ moles}$ of $\text{O}_2$ contains $0.30\text{ moles}$ of Oxygen atoms.
Ratio of Ti atoms : O atoms = $0.15 : 0.30 = 1 : 2$.
The empirical formula is therefore $\text{TiO}_2$. (Statement 3 is true).

Only statement 3 is correct.

Common Mistake (⚠️):
Believing that molten ionic compounds conduct via delocalised electrons (a very common misconception tested in exams).

Correct Answer: C. 2 only

Fastest Approach (🚀):
1: Na(metal) + O2(covalent) -> Na2O(ionic). Covalent broken, but none formed (ionic formed). 2: NaCl(ionic) + H2O(covalent) -> NaOH(ionic) + H2(covalent) + Cl2(covalent). Covalent broken and formed. 3: Fe2O3(ionic) + Al(metal) -> Al2O3(ionic) + Fe(metal). No covalent bonds involved. Only 2.

Step-by-Step Breakdown:

1. Analyze Reaction 1 (Burning sodium)

Reaction: $4\text{Na}(s) + \text{O}_2(g) \rightarrow 2\text{Na}_2\text{O}(s)$

• Bonds broken: Metallic bonds in $\text{Na}$. Covalent double bond in $\text{O}_2$.
• Bonds formed: Ionic bonds in $\text{Na}_2\text{O}$.

Covalent bonds are broken, but no covalent bonds are formed. (Statement 1 is false).

2. Analyze Reaction 2 (Electrolysis of aq NaCl)

Reaction: $2\text{NaCl}(aq) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g) + \text{Cl}_2(g)$

• Bonds broken: Ionic bonds in $\text{NaCl}$. Covalent bonds in $\text{H}_2\text{O}$.
• Bonds formed: Ionic bonds in $\text{NaOH}$. Covalent bonds in $\text{H}_2$ and $\text{Cl}_2$.

Covalent bonds are both broken and formed. (Statement 2 is true).

3. Analyze Reaction 3 (Thermite reaction)

Reaction: $\text{Fe}_2\text{O}_3(s) + 2\text{Al}(s) \rightarrow \text{Al}_2\text{O}_3(s) + 2\text{Fe}(s)$

• Bonds broken: Ionic bonds in $\text{Fe}_2\text{O}_3$. Metallic bonds in $\text{Al}$.
• Bonds formed: Ionic bonds in $\text{Al}_2\text{O}_3$. Metallic bonds in $\text{Fe}$.

No covalent bonds are broken or formed. (Statement 3 is false).

Only reaction 2 involves both the breaking and forming of covalent bonds.

Common Mistake (⚠️):
Assuming metal oxides like Na2O or Al2O3 contain covalent bonds.

Correct Answer: C. $\text{P} + 2\text{Q} \rightleftharpoons 2\text{R} + \text{S}$, $\Delta H$ is +ve

Fastest Approach (🚀):
Temperature increases yield $\implies$ forward reaction is endothermic $\implies \Delta H$ is +ve. Eliminates D, E, F (all -ve).
Pressure doesn't affect yield $\implies$ equal total moles of gas on both sides.
Check remaining options A, B, C:
A: left=4, right=5 (unequal) -- eliminated.
B: left=1+3=4, right=1+3=4 (equal) -- survives.
C: left=1+2=3, right=2+1=3 (equal) -- survives.

Both B and C satisfy every stated constraint. The question as posed does not contain enough information to distinguish them (see editorial note). Answer key currently gives Option C, but this is not uniquely derivable.

Step-by-Step Breakdown:

1. Use the temperature/yield clue

An increase in temperature increases the equilibrium yield, which by Le Chatelier's principle means the forward reaction is endothermic ($\Delta H$ is positive). This eliminates any option with $\Delta H$ negative: D, E, and F are ruled out.

2. Use the pressure/yield clue

An increase in pressure increases the rate but does NOT shift the equilibrium yield. By Le Chatelier's principle, pressure only affects equilibrium position when the two sides have different numbers of gas moles -- so the surviving option must have EQUAL total moles of gas on both sides.

3. Check the remaining options (A, B, C) for equal moles

• A: $3P+Q \rightleftharpoons 2R+3S$ -- left $=4$, right $=5$. Not equal. Eliminated.
• B: $P+3Q \rightleftharpoons R+3S$ -- left $=4$, right $=4$. Equal.
• C: $P+2Q \rightleftharpoons 2R+S$ -- left $=3$, right $=3$. Equal.

4. The problem: two options survive

Both B and C have $\Delta H$ positive AND equal total gas moles on both sides -- both are fully consistent with everything stated in the question. Nothing in the given text distinguishes them further. The answer key marks C as correct, but this cannot be derived uniquely from the information given; either the question is missing a discriminating detail (e.g. a specific mole ratio, a diagram, or additional data), or the intended answer requires context not present in this transcription. This should be reviewed against the original source before being presented to students as having a single correct answer.

Common Mistake (⚠️):
Stopping as soon as one option (Option C, a 'nice' 1+2=2+1 pattern) satisfies both derived constraints, without checking whether any OTHER option also satisfies them. Here, Option B independently satisfies both constraints too, which the original solution material did not acknowledge.

Correct Answer: D. $+50\text{ kJ mol}^{-1}$

Fastest Approach (🚀):
$E_{a\text{(reverse)}} = E_{a\text{(forward)}} - \Delta H = 150 - 100 = 50$. Activation energy is always positive. So $+50\text{ kJ mol}^{-1}$.

Step-by-Step Breakdown:

1. Visualize the reaction profile

• $\Delta H = +100\text{ kJ mol}^{-1}$. This means the reaction is endothermic. The products are $100\text{ kJ}$ higher in energy than the reactants.
• $E_{a\text{(forward)}} = +150\text{ kJ mol}^{-1}$. This is the energy required to go from the reactants up to the peak (transition state).

2. Calculate the reverse activation energy

The reverse activation energy is the energy required to go from the products up to the peak.
Since the peak is $150\text{ kJ}$ above the reactants, and the products are already $100\text{ kJ}$ above the reactants, the peak is only $50\text{ kJ}$ above the products.
$E_{a\text{(reverse)}} = E_{a\text{(forward)}} - \Delta H$
$E_{a\text{(reverse)}} = 150 - 100 = 50\text{ kJ mol}^{-1}$.

3. Note on signs

Activation energy represents a barrier that must be overcome by absorbing energy. Therefore, activation energy is always represented as a positive value. Thus, it is $+50\text{ kJ mol}^{-1}$.

Common Mistake (⚠️):
Adding the values to get 250, or choosing a negative value thinking that reverse reactions must have negative activation energies.

Correct Answer: F. 1 and 3 only

Fastest Approach (🚀):
1: Rate depends on [H+]. Strong acid reacts faster (True). 2: Neutralization depends on total moles of acid. Both are 1 M monoprotic, so they need the same volume of NaOH (False). 3: Conductivity depends on free ions. Strong acid has more free ions (True). Tests 1 and 3 work.

Step-by-Step Breakdown:

1. Evaluate Test 1 (Reaction rate with Mg)

The rate of reaction with a metal depends on the concentration of free $\text{H}^+$ ions in the solution. Because HX is a strong acid, it fully dissociates, providing a high $[\text{H}^+]$. HY partially dissociates, providing a low $[\text{H}^+]$. Therefore, HX will react with the magnesium much faster, showing it is the stronger acid.

2. Evaluate Test 2 (Neutralization volume)

Neutralization measures the total amount of acid present. Both HX and HY have a concentration of $1\text{ mol dm}^{-3}$ and are monoprotic. Even though HY is weak, as the small amount of free $\text{H}^+$ is neutralized by $\text{NaOH}$, Le Chatelier's principle causes more HY to dissociate until all of it has reacted. Therefore, both $20\ \text{cm}^3$ samples will require exactly $20\ \text{cm}^3$ of $1\text{ mol dm}^{-3}$ $\text{NaOH}$ to neutralize. This test will not show a difference.

3. Evaluate Test 3 (Electrical conductance)

Electrical conductivity depends on the concentration of mobile ions. The strong acid HX fully dissociates into $\text{H}^+$ and $\text{X}^-$ ions, making it a strong electrolyte and an excellent conductor. The weak acid HY only partially dissociates, meaning mostly neutral $\text{HY}$ molecules are present, resulting in poor conductivity. This test will show a difference.

Therefore, tests 1 and 3 would show the difference.

Common Mistake (⚠️):
Believing that a weak acid requires less base to neutralize it because it has a lower pH (lower concentration of H+). Total titratable acidity is independent of acid strength.

Correct Answer: A. 0.44 g

Fastest Approach (🚀):
Moles gas = $240 / 24000 = 0.01\text{ mol}$. Na2CO3 + 2HCl -> 2NaCl + H2O + CO2. Moles Na2CO3 = $0.01\text{ mol}$. Molar mass Na2CO3 = $(2\times23) + 12 + (3\times16) = 106$. Pure mass = $0.01 \times 106 = 1.06\text{g}$. Impurity = $1.50 - 1.06 = 0.44\text{g}$. Option A.

Step-by-Step Breakdown:

1. Calculate moles of gas produced

The gas produced when a carbonate reacts with acid is carbon dioxide ($\text{CO}_2$).
$\text{Moles of CO}_2 = \frac{\text{Volume}}{\text{Molar Gas Volume}} = \frac{240\text{ cm}^3}{24,000\text{ cm}^3/\text{mol}} = 0.01\text{ moles}$

2. Calculate moles of pure sodium carbonate

The balanced chemical equation is:
$\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$
From the equation, 1 mole of $\text{Na}_2\text{CO}_3$ produces 1 mole of $\text{CO}_2$. Therefore, $0.01\text{ moles}$ of $\text{Na}_2\text{CO}_3$ must have reacted.

3. Calculate the mass of pure sodium carbonate

Calculate the molar mass ($M_r$) of $\text{Na}_2\text{CO}_3$:
$M_r = (2 \times 23) + 12 + (3 \times 16) = 46 + 12 + 48 = 106\text{ g/mol}$
$\text{Mass of pure Na}_2\text{CO}_3 = \text{Moles} \times M_r = 0.01 \times 106 = 1.06\text{ g}$

4. Calculate the mass of the impurity

The total mass of the sample was $1.50\text{g}$.
$\text{Mass of impurity} = \text{Total mass} - \text{Mass of pure substance}$
$\text{Mass of impurity} = 1.50\text{ g} - 1.06\text{ g} = 0.44\text{ g}$

Common Mistake (⚠️):
Stopping at the mass of the pure substance (1.06 g) and choosing option F, forgetting that the question specifically asks for the mass of the impurity.

Correct Answer: F. 600 cm³

Fastest Approach (🚀):
Equation: $2\text{Li} + 2\text{H}_2\text{O} \rightarrow 2\text{LiOH} + \text{H}_2$.
Moles Li = $0.35 / 7 = 0.05\text{ mol}$.
Moles H2 = $0.05 / 2 = 0.025\text{ mol}$.
Volume H2 = $0.025 \times 24.0\text{ dm}^3 = 0.60\text{ dm}^3$.
Convert to cm3: $0.60 \times 1000 = 600\text{ cm}^3$. Option F.

Step-by-Step Breakdown:

1. Write the balanced equation

Alkali metals react with water to form a metal hydroxide and hydrogen gas.
$2\text{Li}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{LiOH}(aq) + \text{H}_2(g)$

2. Calculate moles of Lithium

$\text{Moles of Li} = \frac{\text{Mass}}{A_r} = \frac{0.35\text{ g}}{7\text{ g/mol}} = 0.05\text{ moles}$

3. Calculate moles of Hydrogen gas

From the equation, 2 moles of Li produce 1 mole of $\text{H}_2$. Therefore, the moles of $\text{H}_2$ is half the moles of Li.
$\text{Moles of H}_2 = \frac{0.05}{2} = 0.025\text{ moles}$

4. Calculate the volume of Hydrogen gas

The molar volume is given as $24.0\text{ dm}^3$.
$\text{Volume of H}_2 = \text{Moles} \times \text{Molar Volume} = 0.025 \times 24.0\text{ dm}^3 = 0.60\text{ dm}^3$

5. Convert to cm$^3$

The options are in cm$^3$. $1\text{ dm}^3 = 1000\text{ cm}^3$.
$\text{Volume in cm}^3 = 0.60 \times 1000 = 600\text{ cm}^3$.

Common Mistake (⚠️):
Forgetting the 2:1 ratio (giving $1200\text{ cm}^3$) or failing to convert dm$^3$ to cm$^3$ (leading to a search for a $0.6$ option in cm$^3$).

Correct Answer: C. 2 only

Fastest Approach (🚀):
Balance electrons: Cathode needs $\times2$. So $4\text{H}_2\text{O} + 4\text{e}^- \rightarrow 2\text{H}_2 + 4\text{OH}^-$. Ratio of $\text{H}_2$ to $\text{O}_2$ is 2:1 (1 is false). Water is consumed, so $\text{Na}_2\text{SO}_4$ concentration increases (2 is true). $4\text{H}^+$ from anode and $4\text{OH}^-$ from cathode neutralize back to water, so pH is unchanged (3 is false). Only 2 is true.

Step-by-Step Breakdown:

1. Evaluate Statement 1 (Mole ratio)

To find the overall ratio of gases produced, we must balance the electrons transferred in both half-equations. The anode releases 4 electrons, but the cathode equation only accepts 2.
We multiply the cathode equation by 2:
$4\text{H}_2\text{O}(l) + 4\text{e}^- \rightarrow 2\text{H}_2(g) + 4\text{OH}^-(aq)$
Comparing this to the anode:
$2\text{H}_2\text{O}(l) \rightarrow 1\text{O}_2(g) + 4\text{H}^+(aq) + 4\text{e}^-$
For every 4 electrons transferred, 2 moles of hydrogen and 1 mole of oxygen are produced. The ratio is 2:1, not 1:1. (Statement 1 is false).

2. Evaluate Statement 2 (Concentration)

The net reaction is the electrolysis of water ($2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2$). The sodium and sulfate ions are spectator ions and do not react. Because water (the solvent) is being consumed and lost as gas, while the moles of $\text{Na}_2\text{SO}_4$ solute remain constant, the concentration of the sodium sulfate solution increases. (Statement 2 is true).

3. Evaluate Statement 3 (Acidity)

While $4\text{H}^+$ ions are produced at the anode, $4\text{OH}^-$ ions are simultaneously produced at the cathode. As these mix in the bulk solution, they will neutralise each other to reform water ($4\text{H}^+ + 4\text{OH}^- \rightarrow 4\text{H}_2\text{O}$). Therefore, the overall solution will remain neutral, not acidic. (Statement 3 is false).

Only statement 2 can be deduced.

Common Mistake (⚠️):
Failing to balance the electrons between the half-equations, leading to the false conclusion that H2 and O2 are produced 1:1.

Correct Answer: F. 1 and 3 only

Fastest Approach (🚀):
O2- has 10e. Mg2+ has 10e. Same config (1 is true).
S2- neutrons = 32-16 = 16. O-18 neutrons = 18-8 = 10. 16 is not double 10 (2 is false).
O2- (10e) + O-18 (8e) = 18e. S2- has 16+2 = 18e. Sum is equal (3 is true). 1 and 3 are correct.

Step-by-Step Breakdown:

1. Prepare the data for each species

• $^{24}_{12}\text{Mg}^{2+}$: Protons = 12. Neutrons = $24 - 12 = 12$. Electrons = $12 - 2 = 10$.
• $^{16}_{8}\text{O}^{2-}$: Protons = 8. Neutrons = $16 - 8 = 8$. Electrons = $8 + 2 = 10$.
• $^{18}_{8}\text{O}$: Protons = 8. Neutrons = $18 - 8 = 10$. Electrons = $8 - 0 = 8$.
• $^{32}_{16}\text{S}^{2-}$: Protons = 16. Neutrons = $32 - 16 = 16$. Electrons = $16 + 2 = 18$.

2. Evaluate Statement 1

Both $^{16}_{8}\text{O}^{2-}$ and $^{24}_{12}\text{Mg}^{2+}$ have 10 electrons. Therefore, they are isoelectronic and share the same electron configuration (2, 8). (Statement 1 is true).

3. Evaluate Statement 2

$^{32}_{16}\text{S}^{2-}$ has 16 neutrons. $^{18}_{8}\text{O}$ has 10 neutrons. 16 is not double 10. (Statement 2 is false).

4. Evaluate Statement 3

Electrons in $^{16}_{8}\text{O}^{2-}$ (10) + electrons in $^{18}_{8}\text{O}$ (8) = 18.
Electrons in $^{32}_{16}\text{S}^{2-}$ = 18.
The sum is equal. (Statement 3 is true).

Statements 1 and 3 are correct.

Common Mistake (⚠️):
Calculating the neutrons of standard Oxygen-16 (8) instead of the Oxygen-18 isotope provided in statement 2, which would make 16 double 8 and lead to selecting H.

Correct Answer: D. 1, 3 and 4 only

Fastest Approach (🚀):
Eq: $2\text{Li} + 2\text{H}_2\text{O} \rightarrow 2\text{LiOH} + \text{H}_2$. 1: Li goes $0 \rightarrow +1$, H goes $+1 \rightarrow 0$. Redox is true. 2: $7\text{g}$ Li is $1\text{ mol}$. Produces $0.5\text{ mol}$ $\text{H}_2$ ($1\text{g}$), not $2\text{g}$. False. 3: Forms LiOH (strong base), pH > 7. True. 4: $14\text{g}$ Li is $2\text{ mol}$. Reacts with $2\text{ mol}$ water ($36\text{g}$). True. So 1, 3, and 4.

Step-by-Step Breakdown:

1. Evaluate Statement 1 (Redox)

The reaction is $2\text{Li} + 2\text{H}_2\text{O} \rightarrow 2\text{LiOH} + \text{H}_2$.

• Lithium goes from an oxidation state of 0 to +1 (Oxidation).
• Hydrogen in water goes from +1 to 0 in $\text{H}_2$ gas (Reduction).

Since oxidation and reduction both occur, it is a redox reaction. (Statement 1 is true).

2. Evaluate Statement 2 (Hydrogen mass)

$7\text{g}$ of Lithium is $1\text{ mole}$ ($7/7 = 1$).
According to the 2:1 stoichiometric ratio ($2\text{Li} \rightarrow 1\text{H}_2$), $1\text{ mole}$ of Li produces $0.5\text{ moles}$ of $\text{H}_2$ gas.
The mass of $0.5\text{ moles}$ of $\text{H}_2$ is $0.5 \times 2 = 1\text{g}$. It does not produce $2\text{g}$. (Statement 2 is false).

3. Evaluate Statement 3 (pH)

The reaction produces Lithium Hydroxide ($\text{LiOH}$), which is a strong alkali. Alkalis dissolve in water to produce solutions with a pH greater than 7 (the pH of neutral water). (Statement 3 is true).

4. Evaluate Statement 4 (Water mass)

$14\text{g}$ of Lithium is $2\text{ moles}$ ($14/7 = 2$).
According to the 2:2 stoichiometric ratio ($2\text{Li} + 2\text{H}_2\text{O}$), $2\text{ moles}$ of Li react exactly with $2\text{ moles}$ of water.
The mass of $2\text{ moles}$ of $\text{H}_2\text{O}$ is $2 \times (2 + 16) = 2 \times 18 = 36\text{g}$. (Statement 4 is true).

Statements 1, 3, and 4 are correct.

Common Mistake (⚠️):
Thinking 1 mole of Li produces 1 mole of H2 (which would make statement 2 true and result in a wrong answer).

Correct Answer: E. C₄F₆

Fastest Approach (🚀):
Mass C = $81 - 57 = 24\text{g}$. Moles C = $24/12 = 2$. Moles F = $57/19 = 3$. Empirical formula = $\text{C}_2\text{F}_3$. Molecular mass is twice this, so molecular formula is $\text{C}_4\text{F}_6$. Option E.

Step-by-Step Breakdown:

1. Determine the mass of Carbon

The compound is a fluorocarbon, so it contains only carbon and fluorine.
$\text{Total mass} = 81\text{ g}$
$\text{Mass of Fluorine} = 57\text{ g}$
$\text{Mass of Carbon} = 81 - 57 = 24\text{ g}$

2. Calculate moles of each element

• $\text{Moles of Carbon} = \frac{\text{Mass}}{A_r} = \frac{24\text{ g}}{12\text{ g/mol}} = 2\text{ moles}$
• $\text{Moles of Fluorine} = \frac{\text{Mass}}{A_r} = \frac{57\text{ g}}{19\text{ g/mol}} = 3\text{ moles}$

3. Determine empirical and molecular formula

The simplest whole number ratio of C : F is 2 : 3.
Therefore, the empirical formula is $\text{C}_2\text{F}_3$.

The prompt states the molecular mass is twice the empirical formula mass. This means the molecular formula contains twice as many atoms of each element.
$\text{Molecular Formula} = 2 \times (\text{C}_2\text{F}_3) = \mathbf{C_4F_6}$.

Common Mistake (⚠️):
Stopping at the empirical formula ($\text{C}_2\text{F}_3$) and selecting Option A, failing to read the first sentence of the prompt.

Correct Answer: F. Q: 3, R: 5

Fastest Approach (🚀):
Initial (P): Moles Mg=$0.05$, Moles Acid=$0.04$. Acid limits. Max $\text{H}_2$=$0.04$.
Q: Mg=$0.05$, Acid=$0.08$. Mg limits. Max $\text{H}_2$=$0.05$. Higher plateau than P. Acid is 2.0M (doubled), so steeper. Higher + Steeper = Line 3.
R: Mg=$0.05$, Acid=$0.02$. Acid limits. Max $\text{H}_2$=$0.02$. Half the plateau of P. Acid is 0.5M (halved), so less steep. Half plateau + less steep = Line 5. Option F.

Step-by-Step Breakdown:

1. Analyze the Baseline (Line P)

• $\text{Moles of Mg} = 1.2\text{g} / 24 = 0.05\text{ moles}$
• $\text{Moles of H}_2\text{SO}_4 = (40/1000)\text{ L} \times 1.0\text{ M} = 0.04\text{ moles}$

The ratio is 1:1. Because $0.04 < 0.05$, sulfuric acid is the limiting reactant.
The maximum amount of $\text{H}_2$ gas produced is proportional to $0.04\text{ moles}$. This is the height of Line P's plateau.

2. Analyze Experiment Q

• $\text{Moles of Mg} = 0.05\text{ moles}$
• $\text{Moles of H}_2\text{SO}_4 = (40/1000)\text{ L} \times 2.0\text{ M} = 0.08\text{ moles}$
• Plateau: Max $\text{H}_2$ is proportional to $0.05\text{ moles}$. Since $0.05 > 0.04$, the plateau must be higher than P.
• Rate (Slope): The acid concentration is doubled ($2.0\text{M}$ vs $1.0\text{M}$). This increases collision frequency, so the initial slope must be steeper than P.

Now, Mg is the limiting reactant ($0.05 < 0.08$).
Looking at graph Q, Line 3 is both steeper and reaches a higher plateau. Thus, Q = 3.

3. Analyze Experiment R

• $\text{Moles of Mg} = 0.05\text{ moles}$
• $\text{Moles of H}_2\text{SO}_4 = (40/1000)\text{ L} \times 0.5\text{ M} = 0.02\text{ moles}$
• Plateau: Max $\text{H}_2$ is proportional to $0.02\text{ moles}$. Since $0.02$ is exactly half of $0.04$, the plateau must be exactly half the height of P.
• Rate (Slope): The acid concentration is halved ($0.5\text{M}$ vs $1.0\text{M}$), so the initial slope must be less steep than P.

Sulfuric acid is the limiting reactant ($0.02 < 0.05$).
Looking at graph R, Line 5 is less steep and plateaus at half the height of P. Thus, R = 5.

Experiment Q is Line 3, Experiment R is Line 5.

Common Mistake (⚠️):
Assuming Mg is always the limiting reactant, leading to the false conclusion that Experiment Q will plateau at the same height as P but faster (Line 1).

Correct Answer: D. 69% ⁶³Cu and 31% ⁶⁵Cu

Fastest Approach (🚀):
$63.55$ is closer to $62.93$ than $64.93$, so $^{63}\text{Cu}$ must be the majority ($>50\%$). This eliminates A, C, E. Let's test D ($69\%$). $62.93(0.69) + 64.93(0.31) = 43.4217 + 20.1283 = 63.55$. It works perfectly.

Step-by-Step Breakdown:

1. Set up the equation

The relative atomic mass ($A_r$) is the weighted average of the isotopes.
Let $x$ be the fractional abundance of $^{63}\text{Cu}$.
Then $(1 - x)$ must be the fractional abundance of $^{65}\text{Cu}$.

$A_r = (\text{mass}_1 \times \text{abundance}_1) + (\text{mass}_2 \times \text{abundance}_2)$
$63.55 = (62.93 \times x) + (64.93 \times (1 - x))$

2. Solve for x

Expand the brackets:
$63.55 = 62.93x + 64.93 - 64.93x$
Combine the $x$ terms:
$63.55 = -2.00x + 64.93$
Rearrange to solve for $x$:
$2.00x = 64.93 - 63.55$
$2.00x = 1.38$
$x = 1.38 / 2.00 = 0.69$

3. Convert to percentages

$x = 0.69$, which corresponds to 69% for $^{63}\text{Cu}$.
$(1 - x) = 1 - 0.69 = 0.31$, which corresponds to 31% for $^{65}\text{Cu}$.

This matches Option D.

Common Mistake (⚠️):
Setting up the equation backwards or mixing up which percentage belongs to which isotope (e.g. choosing B or C). Always check: the $A_r$ (63.55) is closer to 63, so 63 must be the majority.

Correct Answer: .

Step-by-Step Breakdown:
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