ESAT Paper 3 Chemistry Worked Solutions

Correct Answer: B. Lowers the yield and raises the rate of getting to equilibrium

Fastest Approach (🚀):
Split the question into two independent sub-questions. (1) Yield: the reaction is exothermic, so raising temperature pushes the equilibrium toward the endothermic (reverse) direction, lowering the yield. (2) Rate: raising temperature always increases the rate of both forward and reverse reactions, regardless of $\Delta H$'s sign, so the rate of reaching equilibrium always increases. Combine: lower yield, faster rate.

Step-by-Step Breakdown:

1. Effect of Temperature on Equilibrium Yield

The reaction is described as an equilibrium exothermic reaction ($\Delta H$ is negative). According to Le Chatelier's principle, if you increase the temperature of a system at equilibrium, the system shifts to oppose the change, which means it favours the endothermic direction (the reverse reaction). Shifting the equilibrium to the left means the yield of product is lowered.

2. Effect of Temperature on Reaction Rate

Increasing the temperature means a higher proportion of molecules have energy greater than or equal to the activation energy. This increases the frequency of successful collisions. Therefore, the rate of both the forward and reverse reactions increases, so the rate of getting to equilibrium is raised — this is true regardless of whether the reaction is exothermic or endothermic.

3. Combine the Two Effects

Lower yield + raised rate of getting to equilibrium matches option B.

Why the Other Options Are Wrong (❌):

• A. Raises the yield and raises the rate of getting to equilibrium — Le Chatelier's Principle Error
  Gets the yield wrong: for an exothermic reaction, raising the temperature favours the endothermic (reverse) direction, so the yield falls rather than rises. The rate reasoning (raised) is correct, but pairing it with a raised yield mixes up which direction Le Chatelier's principle predicts.
• C. Lowers the yield and lowers the rate of getting to equilibrium — Collision Theory Error
  Gets the rate wrong: raising temperature always increases the rate of both the forward and reverse reactions, because more particles exceed the activation energy in both directions. The rate of reaching equilibrium cannot fall with increasing temperature.
• D. Raises the yield and lowers the rate of getting to equilibrium — Conceptual Misunderstanding
  Gets both effects backwards: incorrectly treats heating an exothermic reaction as favouring the forward (exothermic) direction (raising yield), and incorrectly assumes rate falls with rising temperature.
• E. No change in the yield and raises the rate of getting to equilibrium — Le Chatelier's Principle Error
  Assumes temperature has no effect on equilibrium position at all, ignoring Le Chatelier's principle. Any temperature change shifts $K_c$ (and hence the yield) whenever $\Delta H \neq 0$; only the rate-increases part of this option is correct.

Common Mistake (⚠️):
Assuming that because the reaction is exothermic, raising the temperature must also slow down the approach to equilibrium — this confuses the equilibrium position shift (Le Chatelier) with the rate of reaching equilibrium (collision theory). Increasing temperature always speeds up the approach to equilibrium, for both exothermic and endothermic reactions, because more particles in both directions exceed the activation energy.

Takeaway (📌):
Never let the direction of a Le Chatelier yield shift tell you anything about the rate of reaching equilibrium — raising temperature always increases that rate, regardless of $\Delta H$'s sign. Only a catalyst or concentration/pressure change (not temperature) can shift yield without necessarily being tied to a rate increase in the same way.

Correct Answer: C. $+30\ \text{kJmol}^{-1}$

Fastest Approach (🚀):
Picture the reaction profile: a transition-state peak $130\ \text{kJmol}^{-1}$ above the reactants, with products sitting $100\ \text{kJmol}^{-1}$ above the reactants (endothermic, $\Delta H=+100$). The reverse activation energy is just the drop from the peak down to the products: $130-100=30\ \text{kJmol}^{-1}$.

Step-by-Step Breakdown:

1. Identify the Given Quantities

The heat change of the reaction is $\Delta H = +100\ \text{kJmol}^{-1}$. Since this is positive, the reaction is endothermic: the products sit $100\ \text{kJmol}^{-1}$ higher in energy than the reactants.

The activation energy of the forward reaction is $E_{a(\text{fwd})} = +130\ \text{kJmol}^{-1}$ — the energy needed to climb from the reactants up to the transition state (the peak of the energy barrier).

2. Locate the Transition State Relative to the Products

The transition state sits $130\ \text{kJmol}^{-1}$ above the reactants. Since the products sit $100\ \text{kJmol}^{-1}$ above the reactants, the transition state sits only
$$130 - 100 = 30\ \text{kJmol}^{-1}$$
above the products.

3. Read Off the Reverse Activation Energy

The activation energy of the reverse reaction is the energy needed to climb from the products up to that same transition state:
$$E_{a(\text{rev})} = E_{a(\text{fwd})} - \Delta H = 130 - 100 = +30\ \text{kJmol}^{-1}$$

The correct answer is C.

Why the Other Options Are Wrong (❌):

• A. $+130\ \text{kJmol}^{-1}$ — Conceptual Misunderstanding
  Simply repeats the forward activation energy unchanged, ignoring that the reverse reaction starts from the products, which already sit $100\ \text{kJmol}^{-1}$ higher up the profile than the reactants.
• B. $+50\ \text{kJmol}^{-1}$ — Calculation Error
  Does not correspond to $E_{a(\text{fwd})}-\Delta H$ or any other consistent combination of the given values; likely from a halving or averaging slip rather than the correct subtraction $130-100$.
• D. $+100\ \text{kJmol}^{-1}$ — Conceptual Misunderstanding
  Confuses the reverse activation energy with the enthalpy change itself ($\Delta H=+100$), rather than the energy gap between the transition state and the products.
• E. $+230\ \text{kJmol}^{-1}$ — Sign Error
  Adds the enthalpy change to the forward activation energy ($130+100$) instead of subtracting it — the sign error that would arise from applying $E_{a(\text{fwd})}=E_{a(\text{rev})}-\Delta H$ backwards.

Common Mistake (⚠️):
Adding the enthalpy change to the forward activation energy ($130+100=230$, option E) instead of subtracting it, or simply re-reporting the forward activation energy unchanged (option A) without accounting for the products already sitting higher up the profile.

Takeaway (📌):
For any energy profile, $E_{a(\text{fwd})} - E_{a(\text{rev})} = \Delta H$. Given any two of forward $E_a$, reverse $E_a$, and $\Delta H$ (with the correct sign), the third follows immediately without needing to redraw the whole diagram.

Correct Answer: B. Statement or choice matching Option B as derived in the step-by-step solution.

Step-by-Step Breakdown:

1. Write the Balanced Equation

Sodium reacts with water to form sodium hydroxide and hydrogen gas:
$2Na_{(s)} + 2H_2O_{(l)} \rightarrow 2NaOH_{(aq)} + H_{2(g)}$

2. Calculate Moles of Sodium

Moles ($n$) = Mass / Relative Atomic Mass ($A_r$)
$n(Na) = \frac{0.46\text{ g}}{23\text{ gmol}^{-1}} = 0.02\text{ mol}$

3. Calculate Moles of Hydrogen Gas

From the balanced equation, the molar ratio of $Na$ to $H_2$ is 2:1.
Therefore, moles of $H_2 = \frac{0.02}{2} = 0.01\text{ mol}$.

4. Calculate Volume of Gas

Volume = moles $\times$ molar volume
$\text{Volume} = 0.01\text{ mol} \times 24\text{ dm}^3\text{mol}^{-1} = 0.24\text{ dm}^3$

The correct answer is B.

Correct Answer: E. Statement or choice matching Option E as derived in the step-by-step solution.

Step-by-Step Breakdown:

1. analyse Reaction 1 (Na metal with water)

$2Na + 2H_2O \rightarrow 2NaOH + H_2$

• Water ($H_2O$) is a covalent molecule. The O-H covalent bonds in water are broken to form $H_2$ gas and $OH^-$ ions.
• This involves breaking covalent bonds.

2. analyse Reaction 2 (Electrolysis of molten titanium oxide)

• Titanium oxide (e.g., $TiO_2$) in a molten state consists of ions ($Ti^{4+}$ and $O^{2-}$). Electrolysis involves the movement and discharge of these ions.

The forces overcome are the strong ionic bonds (electrostatic attraction between oppositely charged ions) in the molten state.
This does NOT primarily involve breaking covalent bonds.

3. analyse Reaction 3 (Burning sodium metal in oxygen)

$4Na + O_2 \rightarrow 2Na_2O$

• Oxygen ($O_2$) is a diatomic molecule with a double covalent bond ($O=O$).
• For the reaction to occur, this O=O covalent bond must be broken.
• This involves breaking covalent bonds.

Since reactions 1 and 3 involve breaking covalent bonds, the correct answer is E.

Correct Answer: D. Statement or choice matching Option D as derived in the step-by-step solution.

Step-by-Step Breakdown:

1. Determine the Atomic Number (Protons)

• The $Cu^{2+}$ ion has a $2+$ charge, meaning it has lost 2 electrons compared to the neutral atom.
• If $Cu^{2+}$ has 27 electrons, the neutral Copper ($Cu$) atom has $27 + 2 = 29$ electrons.
• Therefore, the atomic number (number of protons) is $Z = 29$.

2. Calculate Mass Numbers of the Isotopes

• Mass Number = Protons + Neutrons
• Isotope 1: $29 + 34 = 63$. (Abundance = 70%)
• Isotope 2: $29 + 36 = 65$. (Abundance = 30%)

3. Calculate Relative Atomic Mass ($A_r$)

$A_r = \frac{(\text{Mass}_1 \times \%_1) + (\text{Mass}_2 \times \%_2)}{100}$
$A_r = \frac{(63 \times 70) + (65 \times 30)}{100}$
$A_r = \frac{4410 + 1950}{100}$
$A_r = \frac{6360}{100} = 63.6$

4. Round to the Nearest Whole Number

$63.6$ rounded to the nearest whole number is 64.

The correct answer is D.

Correct Answer: C. 3 and 4

Fastest Approach (🚀):
Scan for elements that are free (uncombined, oxidation state 0) turning into combined ions or vice versa — that's the fastest visual cue for a redox change. $H_2+F_2\rightarrow 2HF$ and $Mg+FeSO_4\rightarrow MgSO_4+Fe$ both show elements starting at oxidation state 0, immediately flagging them as redox; the other two involve no free elements at all.

Step-by-Step Breakdown:

1. Define a Redox Reaction

A redox reaction is one in which oxidation (loss of electrons / increase in oxidation state) and reduction (gain of electrons / decrease in oxidation state) occur together — i.e. electrons are transferred between species.

2. Test Each Reaction for a Change in Oxidation State

• Reaction 1: $Br + e^- \rightarrow Br^-$. This is only a half-equation (reduction of Br to Br⁻); it isn't a complete reaction showing both an oxidation and a reduction, so it cannot itself be classed as 'a redox reaction' in the sense the question means.
• Reaction 2: $HCl + AgNO_3 \rightarrow AgCl + HNO_3$. Checking oxidation states: H(+1), Cl(–1), Ag(+1), N(+5), O(–2) — identical on both sides. This is a double (precipitation) reaction with no oxidation state changes, so it is not redox.
• Reaction 3: $H_2 + F_2 \rightarrow 2HF$. Hydrogen goes from $0$ (in $H_2$) to $+1$ (in HF) — oxidised. Fluorine goes from $0$ (in $F_2$) to $-1$ (in HF) — reduced. This is a redox reaction.
• Reaction 4: $Mg + FeSO_4 \rightarrow MgSO_4 + Fe$. Magnesium goes from $0$ to $+2$ — oxidised. Iron goes from $+2$ to $0$ — reduced. This is a redox reaction (a displacement reaction).

3. Select the Redox Pair

Reactions 3 and 4 both show a genuine change in oxidation state on both sides, so they are the two redox reactions.

The correct answer is C.

Why the Other Options Are Wrong (❌):

• A. 1 and 2 — Conceptual Misunderstanding
  Reaction 1 is only a reduction half-equation (not a full redox reaction on its own), and reaction 2 is a double decomposition with no oxidation state changes at all — neither belongs in the redox pair.
• B. 2 and 3 — Oxidation State Error
  Reaction 3 is correctly identified as redox, but reaction 2 involves no oxidation state changes (H, Cl, Ag, N, and O all keep the same oxidation state) — it is a precipitation reaction, not redox.
• D. 1 and 4 — Conceptual Misunderstanding
  Reaction 4 is correctly identified as redox, but reaction 1 is only a reduction half-equation, not a complete redox reaction — it has no paired oxidation to balance it.
• E. 1 and 3 — Conceptual Misunderstanding
  Reaction 3 is correctly identified as redox, but reaction 1 is only a half-equation showing reduction alone, not a full redox reaction with both oxidation and reduction present.

Common Mistake (⚠️):
Assuming reaction 1 counts as 'a redox reaction' just because it shows electron transfer — it's only a reduction half-equation, not a balanced overall reaction, and needs a paired oxidation half-equation to be complete. Also assuming reaction 2 must be redox simply because several compounds are involved, without actually checking that no oxidation states change (it's a simple precipitation reaction).

Takeaway (📌):
Always assign oxidation states explicitly rather than guessing from reaction 'shape' — precipitation and simple double-decomposition reactions (like reaction 2) are essentially never redox, while any reaction featuring a free element on one side (like reactions 3 and 4) almost always is.

Correct Answer: D. $CO = -90 + 3y + z - 2x\ \text{kJmol}^{-1}$

Fastest Approach (🚀):
Write bonds broken minus bonds formed using only the three given variables ($x$, $y$, $z$) plus the unknown $CO$ — since every answer option contains only $x,y,z$, that confirms no separate C–O single-bond term is needed, and you can go straight to $-90=(CO+2x)-(3y+z)$.

Step-by-Step Breakdown:

1. Set Up the Bond-Energy Equation

Using $\Delta H = \Sigma(\text{bonds broken}) - \Sigma(\text{bonds formed})$ for $CO + 2H_2 \rightarrow CH_3OH$:

Bonds broken (reactants):

• $2$ mol of H–H bonds: $2x$
• The C≡O multiple bond in carbon monoxide: call its total energy $CO$ (the value the question asks for)

Bonds formed (products, methanol $CH_3OH$):

• $3$ mol of C–H bonds: $3y$
• $1$ mol of O–H bond: $z$

2. Account for the C–O Single Bond

Methanol's carbon–oxygen link is a single bond, carried through from the C–O σ-bond that was already part of the C≡O triple bond in carbon monoxide — it isn't newly broken and reformed. Only the extra ($\pi$) bonding that makes up the 'multiple' character of $CO$ needs to be broken; the underlying single-bond energy is shared by both the reactant's multiple bond and the product's C–O bond, so it cancels out of the calculation. This is exactly why the question asks specifically for the multiple bonding energy of $CO$ rather than 'the C–O bond energy' — the wording signals that only the extra multiple-bond contribution is unknown, and no separate C–O single-bond value is needed (or given) to answer it.

3. Solve for CO

$$\Delta H = \big(CO + 2x\big) - \big(3y + z\big)$$
$$-90 = CO + 2x - 3y - z$$
$$CO = -90 + 3y + z - 2x\ \text{kJmol}^{-1}$$

This matches option D.

The correct answer is D.

Why the Other Options Are Wrong (❌):

• A. $CO = -90 - 3y + z - 2x\ \text{kJmol}^{-1}$ — Sign Error
  Uses the wrong sign for the $3y$ (C-H) term — bonds formed should be subtracted as $+3y$ inside the bracket, not added with a flipped sign, corresponding to swapping the bonds-broken/bonds-formed roles for the C-H term.
• B. $CO = -90 - 3y - z - 2x\ \text{kJmol}^{-1}$ — Sign Error
  Flips the sign of both product bond terms ($3y$ and $z$), effectively treating the bonds formed in methanol as if they were also bonds broken.
• C. $CO = -90 + 2y + z - 2x\ \text{kJmol}^{-1}$ — Counting Error
  Miscounts the C-H bonds formed in methanol as $2$ instead of $3$ — methanol, $CH_3OH$, has three C-H bonds (plus one O-H bond), not two.
• E. $CO = -90 + 3y + z - x\ \text{kJmol}^{-1}$ — Counting Error
  Miscounts the H-H bonds broken as $1x$ instead of $2x$ — the equation $CO+2H_2\rightarrow CH_3OH$ breaks two moles of H-H bonds, not one.

Common Mistake (⚠️):
Trying to also include a separate 'C–O single bond formed' term on the products side — since no such bond energy is given among $x,y,z$, this looks like missing data, but the question deliberately asks only for the multiple bonding contribution of CO, which sidesteps needing that value at all.

Takeaway (📌):
When a question asks for a 'multiple bond' energy specifically (rather than simply 'the bond energy'), that phrasing is often a signal that only the excess energy beyond a single bond is being isolated — check whether an assumed single-bond contribution would cancel out of the equation before assuming data is missing.

Correct Answer: C. Statement or choice matching Option C as derived in the step-by-step solution.

Step-by-Step Breakdown:

1. Write the Balanced Equation

Potassium bicarbonate ($KHCO_3$) neutralises sulfuric acid ($H_2SO_4$):
$2KHCO_3 + H_2SO_4 \rightarrow K_2SO_4 + 2H_2O + 2CO_2$

2. Calculate Moles of $KHCO_3$

$\text{Moles} = \text{concentration} \times \text{volume (in dm}^3\text{)}$
$n(KHCO_3) = 0.1 \text{ mol dm}^{-3} \times \frac{50}{1000} \text{ dm}^3 = 0.005 \text{ mol}$

3. Calculate Moles of $H_2SO_4$

From the balanced equation, the molar ratio of $KHCO_3$ to $H_2SO_4$ is 2:1.
$n(H_2SO_4) = \frac{0.005}{2} = 0.0025 \text{ mol}$

4. Calculate Concentration of $H_2SO_4$

The volume of acid is $100 \text{ cm}^3 = 0.1 \text{ dm}^3$.
Concentration in $\text{mol dm}^{-3} = \frac{0.0025 \text{ mol}}{0.1 \text{ dm}^3} = 0.025 \text{ mol dm}^{-3}$

5. Convert to $\text{g dm}^{-3}$

Concentration ($\text{g dm}^{-3}$) = Concentration ($\text{mol dm}^{-3}$) $\times M_r$
Concentration = $0.025 \times 98 = 2.45 \text{ g dm}^{-3}$

The correct answer is C.

Correct Answer: B. $[CrCl_2(H_2O)_4]^+$

Fastest Approach (🚀):
Set up one linear equation per option (metal oxidation number + sum of ligand contributions = overall ion charge) and solve for the metal directly — there's no need to know anything else about the chemistry of each ion.

Step-by-Step Breakdown:

1. Calculate the Oxidation Number of the Metal in Each Ion

Use the rule that oxidation numbers sum to the overall charge of the ion, with O almost always $-2$, Cl in a complex almost always $-1$, neutral ligands (like $H_2O$) contributing $0$, and cyanide $CN^-$ contributing $-1$.

• A. $MnO_4^{2-}$: $x + 4(-2) = -2 \Rightarrow x = +6$
• B. $[CrCl_2(H_2O)_4]^+$: $x + 2(-1) + 4(0) = +1 \Rightarrow x = +3$
• C. $VO_2^+$: $x + 2(-2) = +1 \Rightarrow x = +5$
• D. $[Fe(CN)_6]^{4-}$: $x + 6(-1) = -4 \Rightarrow x = +2$

2. Identify the +3 Ion

Only chromium in option B has an oxidation number of $+3$.

The correct answer is B.

Why the Other Options Are Wrong (❌):

• A. $MnO_4^{2-}$ — Oxidation State Error
  $MnO_4^{2-}$ gives manganese an oxidation number of $x+4(-2)=-2 \Rightarrow x=+6$, not $+3$ — this is the manganate(VI) ion.
• C. $VO_2^+$ — Oxidation State Error
  $VO_2^+$ gives vanadium an oxidation number of $x+2(-2)=+1 \Rightarrow x=+5$, not $+3$.
• D. $[Fe(CN)_6]^{4-}$ — Oxidation State Error
  $[Fe(CN)_6]^{4-}$ gives iron an oxidation number of $x+6(-1)=-4 \Rightarrow x=+2$, not $+3$ — mistaking cyanide's contribution as $+1$ instead of $-1$ is the usual source of this error.

Common Mistake (⚠️):
Forgetting that neutral ligands like water ($H_2O$) contribute $0$ to the charge balance (sometimes mistakenly treated as contributing $-2$ from the oxygen, as if it were behaving like an oxide ion), which would throw off the oxidation number calculated for chromium in option B.

Takeaway (📌):
Whenever a metal sits inside a complex ion, isolate its oxidation number the same way every time: metal + (sum of ligand oxidation numbers × how many of each) = overall ion charge. This works regardless of how exotic the ligand or ion looks.

Correct Answer: A. $4z > 2x+y$

Fastest Approach (🚀):
Count bonds broken (2 mol H-H + 1 mol O=O) and bonds formed (4 mol O-H, since 2 mol water × 2 O-H bonds each) exactly as for an enthalpy calculation, then flip the usual '=' for '<0' and rearrange.

Step-by-Step Breakdown:

1. Set Up the Bond-Energy Equation

For $2H_2+O_2\rightarrow 2H_2O$: $\Delta H = \Sigma(\text{bonds broken})-\Sigma(\text{bonds formed})$.

Bonds broken (reactants): $2$ mol H–H ($2x$) $+$ $1$ mol O=O ($y$) $= 2x+y$

Bonds formed (products): each $H_2O$ has $2$ O–H bonds, and there are $2$ mol of $H_2O$, giving $2\times 2=4$ mol of O–H bonds $=4z$

2. Apply the Exothermic Condition

A reaction is exothermic when $\Delta H<0$, i.e. more energy is released forming bonds than is used breaking them:
$$(2x+y)-4z<0$$
$$2x+y<4z$$
$$4z>2x+y$$

This matches option A.

The correct answer is A.

Why the Other Options Are Wrong (❌):

• B. $4z > x+y$ — Counting Error
  Drops the factor of $2$ on the H–H term — there are $2$ mol of H–H bonds broken (from $2H_2$), not $1$.
• C. $2z>2x+y$ — Counting Error
  Only counts $2$ mol of O–H bonds formed instead of $4$ — each of the $2$ mol of $H_2O$ actually contributes $2$ O–H bonds, giving $4z$, not $2z$.
• D. $z>2x+2y$ — Counting Error
  Miscounts both product and reactant bonds: only $1z$ (instead of $4z$) for the O–H bonds formed, and $2y$ (instead of $y$) for the O=O bond broken.
• E. $2z>x+y$ — Counting Error
  Combines both errors above — undercounts the O–H bonds formed as $2z$ instead of $4z$, and undercounts the H-H bonds broken as $x$ instead of $2x$.

Common Mistake (⚠️):
Forgetting the factor of $2$ on the O-H bonds formed (each water molecule has two O–H bonds, and there are two moles of water, so $4z$ total, not $2z$) — this is the difference between the correct option A and the too-small option C ($2z>2x+y$).

Takeaway (📌):
Always count TOTAL bonds of each type across ALL molecules in the balanced equation (moles of molecule × bonds per molecule), not just bonds per single molecule — a very common source of a factor-of-2 error in these problems.

Correct Answer: E. Statement or choice matching Option E as derived in the step-by-step solution.

Step-by-Step Breakdown:

1. Determine the Molar Masses

• Aluminium ($Al$): 27
• Aluminium oxide ($Al_2O_3$): $2(27) + 3(16) = 54 + 48 = 102$
• Hydrated aluminium oxide ($Al_2O_3 \cdot 2H_2O$): $102 + 2(18) = 138$

2. Calculate Mass of Al in Hydrated Aluminium Oxide

One mole of hydrated aluminium oxide (138 g) contains 2 moles of aluminium atoms.
Mass of Al in 1 mole = $2 \times 27 = 54\text{ g}$.

3. Calculate Required Mass of Hydrated Aluminium Oxide

• We need $108\text{ tonnes}$ of Aluminium.
• Mass of hydrated aluminium oxide needed = $\text{Mass of Al} \times \frac{\text{Molar mass of Hydrated } Al_2O_3}{\text{Mass of Al in one mole}}$
• Mass = $108 \times \frac{138}{54} = 2 \times 138 = 276\text{ tonnes}$.

4. Calculate Required Mass of Bauxite

• The bauxite is 75.0% hydrated aluminium oxide by mass.
• $0.75 \times \text{Mass of Bauxite} = 276\text{ tonnes}$
• Mass of Bauxite = $\frac{276}{0.75} = \frac{276}{(3/4)} = 368\text{ tonnes}$.

The correct answer is E.

Correct Answer: C. Statement or choice matching Option C as derived in the step-by-step solution.

Step-by-Step Breakdown:

1. analyse the Reaction and Options

The reaction is: $S_{(g)} + 2T_{(g)} \rightarrow W_{(g)} + 2X_{(g)}$ with $\Delta H = -12\text{ kJ mol}^{-1}$.
A. A solid catalyst would slow down the rate of reaction. False. Catalysts increase the rate of reaction by providing an alternative pathway with a lower activation energy.
B. If the chemical 'T' was a powder the reaction would be faster. False. The equation states T is a gas ($T_{(g)}$). A gas cannot be a powder.
C. A high activation energy would give a slower rate than a lower activation energy. True. Higher activation energy means fewer particles have sufficient energy to react upon collision, leading to a slower rate.
D. Increasing the temperature would decrease the rate. False. Increasing temperature always increases the rate of reaction.

• E. The rate of reaction can be monitored by measuring the change in gas volume. False. There are 3 moles of gas reactants ($1 S + 2 T$) and 3 moles of gas products ($1\ \text{W} + 2 X$). The total volume of gas does not change, assuming constant temperature and pressure.

The correct answer is C.

Correct Answer: B. Statement or choice matching Option B as derived in the step-by-step solution.

Step-by-Step Breakdown:

1. Determine the Number of Protons

The number of protons is equal to the atomic number.
$\text{Protons} = \frac{x-1}{2}$.

2. Determine the Number of Neutrons

The number of neutrons is the mass number minus the atomic number.
$\text{Neutrons} = x - \frac{x-1}{2} = \frac{2x - (x - 1)}{2} = \frac{x + 1}{2}$.

3. Determine the Number of Electrons

• The charge of the ion is -1, meaning it has gained one electron compared to its neutral state.
• $\text{Electrons} = \text{Protons} - \text{Charge} = \frac{x-1}{2} - (-1) = \frac{x-1}{2} + 1 = \frac{x-1+2}{2} = \frac{x+1}{2}$.

Comparing these values to the options:

• Protons = $\frac{x-1}{2}$
• Neutrons = $\frac{x+1}{2}$
• Electrons = $\frac{x+1}{2}$

This matches option B.

Correct Answer: F. Mn is reduced, Fe is oxidised

Step-by-Step Breakdown:

1. Determine the Oxidation States

• In $MnO_4^-$, Mn is +7. In the products, it is $Mn^{2+}$ (+2). Mn goes from +7 to +2, so it is reduced.
• Fe goes from $Fe^{2+}$ to $Fe^{3+}$. Fe goes from +2 to +3, so it is oxidised.

2. Define Disproportionation

Disproportionation is a specific type of redox reaction where the same element is simultaneously oxidised and reduced.
In this reaction, manganese is reduced and iron is oxidised. These are different elements.

• Therefore, this is a standard redox reaction, but NOT a disproportionation reaction.

The statement that best describes this is: "No, but it is a redox reaction."

The correct answer is F.

Correct Answer: E. Statement or choice matching Option E as derived in the step-by-step solution.

Step-by-Step Breakdown:

1. Understand Silicon's Structure

Silicon is in Group 14, directly below carbon. Like diamond (a form of carbon), silicon forms a giant covalent lattice (or macromolecular structure).
In this structure, each silicon atom is covalently bonded to four other silicon atoms in a tetrahedral arrangement.

2. Evaluate the Options

• A. It forms ions with a charge of 4+: Silicon generally forms covalent bonds, not ionic.

B. Its structure is a giant ionic lattice: Silicon is a metalloid, it forms a giant covalent lattice, not ionic.
C. It has a strong attraction between positive nuclei and delocalised electrons: This describes metallic bonding. Silicon is not a metal.

• D. It has strong intermolecular forces: Giant covalent structures don't have molecules, they are one large continuous network of bonds.
• E. Covalent bonds are broken on melting: Because the entire structure is held together by strong covalent bonds, these bonds must be broken to melt the solid. This requires a very large amount of energy, leading to a high melting point.

The correct answer is E.

Correct Answer: C. Statement or choice matching Option C as derived in the step-by-step solution.

Step-by-Step Breakdown:

1. analyse Statement A

• "During chromatography, a mobile phase moves in a definite direction over a stationary phase." This is a correct description of the general mechanism of chromatography.

2. analyse Statement B

• "Chromatography relies on the ability of different components to have different affinities for a static phase and a mobile phase." This is correct; separation occurs because components partition differently between the two phases.

3. analyse Statement C

• "Absorption is the process by which a solid holds molecules... as a thin film on the surface..." This definition describes adsorption, not absorption. Absorption is when a substance permeates into the bulk of a solid or liquid. Therefore, this statement is incorrect due to the use of the wrong term.

4. analyse Statement D

• "In thin layer chromatography, the stationary phase is a solid and the mobile phase is a liquid..." This is correct. TLC uses a solid plate (like silica) and a liquid solvent. Gas chromatography uses a gas mobile phase and a solid/liquid stationary phase. This statement is correct.

Since statement C is the incorrect one, it is the correct answer.

Correct Answer: D. Statement or choice matching Option D as derived in the step-by-step solution.

Step-by-Step Breakdown:

1. Evaluate Option A

• Electrolysis of aqueous rubidium chloride would produce hydrogen gas at the cathode, not rubidium metal, because rubidium is much more reactive than hydrogen.

2. Evaluate Option B

Melting and boiling points of alkali metals decrease* down the group. Rubidium is below sodium, so it has lower melting and boiling points.

3. Evaluate Option C

Reactivity increases* down the alkali metal group. Rubidium reacts much more vigorously and quickly with water than sodium does.

4. Evaluate Option D

• Because rubidium is highly reactive with oxygen and moisture in the air, it must be stored under oil or in an inert atmosphere to prevent it from reacting. This statement is correct.

5. Evaluate Option E

• Rubidium is in Group 1 and forms a +1 ion ($Rb^+$). Sulfate is a -2 ion ($SO_4^{2-}$). The correct formula is $Rb_2SO_4$, not $RbSO_4$.

The correct answer is D.

Correct Answer: D. Statement or choice matching Option D as derived in the step-by-step solution.

Step-by-Step Breakdown:

1. Calculate Energy Gained by Water

• The formula is $q = m c \Delta T$.
• Mass of water ($m$) = $150\text{ g}$.
• Specific heat capacity ($c$) = $4\text{ J g}^{-1} {^\circ C}^{-1}$.
• Temperature rise ($\Delta T$) = $60^\circ C$.
• $q = 150 \times 4 \times 60 = 36,000\text{ J} = 36\text{ kJ}$.

2. Calculate Total Energy Released by Methanol

• Only 80% of the energy released is transferred to the water.
• $\text{Total Energy} \times 0.8 = 36\text{ kJ}$.
• $\text{Total Energy} = \frac{36}{0.8} = 45\text{ kJ}$.

3. Calculate Moles of Methanol Required

• Burning 1 mole releases $720\text{ kJ}$.
• Moles needed = $\frac{45\text{ kJ}}{720\text{ kJmol}^{-1}} = 0.0625\text{ mol}$.

4. Calculate Mass of Methanol

• $\text{Mass} = \text{Moles} \times M_r$.
• $\text{Mass} = 0.0625 \times 32 = 2.00\text{ g}$.

The correct answer is D.

Correct Answer: F. 2, 3, and 4 only

Fastest Approach (🚀):
Add up $2+8+3=13$ to identify aluminium immediately, then go statement-by-statement: statement 4 (atomic number 13) and, via its $+3$ valency, statements 2 and 3 (forming $X_2O_3$ and $XBr_3$) fall out directly from 'Al forms $Al^{3+}$', while 1 and 5 are simply mismatched group/classification facts to be recognised as false.

Step-by-Step Breakdown:

1. Identify Element X

The electronic structure $2,8,3$ gives a total of $2+8+3=13$ electrons, so the atomic number is $13$ — this is aluminium (Al).

2. Evaluate Each Statement

• 1. Group 12, period 3: False. Al has $3$ electrons in its outer shell, placing it in Group 13 (historically 'Group III'), not Group 12 (a transition-metal group).
• 2. Reacts with oxygen to form $X_2O_3$: True. Al forms $Al^{3+}$ and oxygen forms $O^{2-}$, combining as $Al_2O_3$.
• 3. Reacts with bromine to form $XBr_3$: True. Al forms $Al^{3+}$ and bromine forms $Br^-$, combining as $AlBr_3$.
• 4. Atomic number is 13: True — directly from the electron count above.
• 5. Alkali metal: False. Alkali metals are Group 1 (1 outer electron); aluminium is a Group 13 post-transition metal.

3. Select the Correct Combination

Statements 2, 3, and 4 are true; statements 1 and 5 are false. This matches option F, '2, 3, and 4 only'.

The correct answer is F.

Why the Other Options Are Wrong (❌):

• A. 1 and 5 only — Conceptual Misunderstanding
  Both statements here are false: aluminium is in Group 13, not Group 12 (statement 1), and it is a post-transition metal, not an alkali metal (statement 5).
• B. 2 and 3 only — Incomplete Selection
  Statements 2 and 3 are indeed true, but this option wrongly omits statement 4 (atomic number 13), which is also true — the complete correct set needs all three of 2, 3, and 4.
• C. 2 and 5 only — Conceptual Misunderstanding
  Statement 2 is true, but statement 5 is false — aluminium is a Group 13 post-transition metal, not an alkali metal.
• D. 3 and 4 only — Incomplete Selection
  Statements 3 and 4 are indeed true, but this option wrongly omits statement 2 ($Al_2O_3$ formation), which is also true.
• E. 1, 4 and 5 only — Conceptual Misunderstanding
  Statement 4 is true, but statements 1 and 5 are both false (wrong group, and aluminium is not an alkali metal).

Common Mistake (⚠️):
Misreading '3 outer electrons' as meaning Group 3 or Group 12 (where the transition metals sit) instead of Group 13 — the historical 'Group III' naming for the boron/aluminium group causes exactly this kind of confusion with the transition-metal group numbering.

Takeaway (📌):
An element's number of outer-shell electrons gives its main group number directly (in the 1–8 / 13–18 convention) and its ionic charge in simple compounds — use that one fact to check both formula statements (like $X_2O_3$, $XBr_3$) and group/classification statements in a single pass.

Correct Answer: D. 1 and 2 only

Fastest Approach (🚀):
Statements 1 and 2 are essentially restating the definition of 'isotope' (same element, same protons, same chemistry) — treat these as automatically true for any set of isotopes. Only statement 3 needs an actual calculation: neutrons = mass number − 28, checked against each of the four given mass numbers.

Step-by-Step Breakdown:

1. Identify the Common Feature of the Isotopes

Nickel has atomic number $28$, so every isotope of nickel — regardless of mass number — has exactly $28$ protons (and, as a neutral atom, $28$ electrons). The four isotopes (mass numbers $58,60,61,62$) differ only in neutron number.

2. Evaluate Each Statement

• 1. All of them have the same chemical properties. True — chemical properties are governed by electron configuration, which is identical for all isotopes of the same element (all have $28$ electrons arranged the same way).
• 2. All of them have nuclei containing 28 protons. True — protons define the element; every nickel isotope has $28$ protons by definition.
• 3. One of them has a nucleus that contains 62 neutrons. False — neutrons = mass number − protons. The heaviest isotope listed has mass number $62$, giving $62-28=34$ neutrons, not $62$. (Mistaking the mass number itself for a neutron count is the trap here.)

3. Select the Correct Combination

Statements 1 and 2 are true; statement 3 is false. This matches option D, '1 and 2 only'.

The correct answer is D.

Why the Other Options Are Wrong (❌):

• A. 1 only — Incomplete Selection
  Statement 1 is true, but this option wrongly excludes statement 2 (all nickel isotopes do have 28 protons — true by definition of being isotopes of the same element).
• B. 2 only — Incomplete Selection
  Statement 2 is true, but this option wrongly excludes statement 1 (isotopes of the same element do share identical chemical properties).
• C. 3 only — Conceptual Misunderstanding
  Statement 3 is false (the mass-62 isotope has $62-28=34$ neutrons, not $62$), so an option built around only statement 3 being correct cannot be right.
• E. 1 and 3 only — Conceptual Misunderstanding
  Statement 1 is true, but statement 3 is false — no nickel isotope in this list has 62 neutrons; the mass-62 isotope has 34 neutrons.
• F. 2 and 3 only — Conceptual Misunderstanding
  Statement 2 is true, but statement 3 is false for the same reason: 62 is the mass number, not the neutron count, of the heaviest isotope listed.
• G. 1, 2 and 3 — Conceptual Misunderstanding
  Statements 1 and 2 are true, but statement 3 is false, so all three cannot be correct together.
• H. None of them — Conceptual Misunderstanding
  Statements 1 and 2 are in fact both true (same protons, same chemistry, by definition of isotopes), so 'none of them' is too strong.

Common Mistake (⚠️):
Reading '62 neutrons' directly off the mass-62 isotope, instead of subtracting the $28$ protons first — the mass-62 isotope has $34$ neutrons, not $62$; the mass number counts protons AND neutrons together.

Takeaway (📌):
Whenever a question gives mass numbers and asks about neutrons, always compute neutrons = mass number − atomic number explicitly before answering — never assume the mass number itself is the neutron count.

Correct Answer: C. Aqueous potassium sulphate: Oxygen at the anode, Hydrogen at the cathode

Fastest Approach (🚀):
First check whether each electrolyte is aqueous or molten — that decides whether water can compete at all. For aqueous ones, apply the shortcuts (halide beats water at the anode; a metal less reactive than hydrogen beats water at the cathode); for molten ones, the ion present is always discharged directly, so just match cation → cathode, anion → anode with no exceptions.

Step-by-Step Breakdown:

1. Recall the Rules for Electrolysis with Inert Electrodes

• Anode (positive electrode): anions are attracted and oxidised. If a halide ion (Cl⁻, Br⁻, I⁻) is present in aqueous solution, it is discharged in preference to hydroxide/water, forming the halogen. If only an ion that is very hard to oxidise is present (e.g. $SO_4^{2-}$, $NO_3^-$), water is oxidised instead, releasing oxygen gas. In a molten (water-free) salt, whatever anion is present is discharged directly.
• Cathode (negative electrode): cations are attracted and reduced. If the metal is less reactive than hydrogen (e.g. $Cu^{2+}$), the metal is deposited. If the metal is more reactive than hydrogen (e.g. $Ca^{2+}$, $K^+$, $Na^+$) and water is present, water is reduced instead, releasing hydrogen gas. In a molten (water-free) salt, the metal cation is always reduced to the metal, however reactive it is.

2. Check Each Row

• A. Aqueous calcium bromide: anode — $Br^-$ is discharged in preference to water, giving bromine (correct). Cathode — calcium is more reactive than hydrogen, and water IS present (aqueous), so water is reduced to hydrogen, not calcium. Row A is wrong.
• B. Aqueous copper nitrate: anode — nitrate is very hard to oxidise, so water is oxidised to give oxygen, not nitrogen. Row B is wrong.
• C. Aqueous potassium sulphate: anode — sulphate is hard to oxidise, so water gives oxygen (correct). Cathode — potassium is more reactive than hydrogen, so water is reduced to hydrogen (correct). Row C is entirely correct.
• D. Molten aluminium oxide: no water is present, so the ions themselves are discharged directly: $O^{2-}$ at the anode gives oxygen, and $Al^{3+}$ at the cathode gives aluminium. The row has these swapped, so row D is wrong.
• E. Molten sodium chloride: no water is present, so $Cl^-$ at the anode gives chlorine (correct), but $Na^+$ at the cathode must be reduced directly to sodium metal (there is no water present to form hydrogen from). Row E is wrong.

3. Conclusion

Only row C has both electrode products correct.

The correct answer is C.

Why the Other Options Are Wrong (❌):

• A. Aqueous calcium bromide: Bromine at the anode, Calcium at the cathode — Reactivity Series Error
  Calcium is more reactive than hydrogen, so in this AQUEOUS solution, water — not calcium — is reduced at the cathode, giving hydrogen gas. The anode product (bromine) is correct, but the cathode product is wrong.
• B. Aqueous copper nitrate: Nitrogen at the anode, Copper at the cathode — Conceptual Misunderstanding
  Nitrate ions are very difficult to oxidise, so water is oxidised at the anode instead, giving oxygen — not nitrogen (nitrogen gas is not a plausible electrolysis product here at all).
• D. Molten aluminium oxide: Aluminium at the anode, Oxygen at the cathode — Electrode Assignment Error
  This row has the products swapped: in molten aluminium oxide, the negative $O^{2-}$ ions migrate to the anode to give oxygen, while the positive $Al^{3+}$ ions migrate to the cathode to give aluminium — the reverse of what the row states.
• E. Molten sodium chloride: Chlorine at the anode, Hydrogen at the cathode — Conceptual Misunderstanding
  Molten sodium chloride contains no water, so there is nothing to reduce to hydrogen at the cathode — the $Na^+$ ions themselves are reduced directly to sodium metal.

Common Mistake (⚠️):
Applying the 'water competes' rules to molten salts, where there is no water at all — this is exactly why row D has aluminium and oxygen swapped (expecting the metal to be 'too reactive' to be discharged, forgetting that rule only applies in aqueous solution) and why row E lists hydrogen instead of sodium at the cathode.

Takeaway (📌):
Always ask 'is water present?' before applying any competition rule in electrolysis. Aqueous electrolysis of a reactive-metal salt almost never gives the metal or a non-halogen anion product; molten electrolysis of the same salt always gives exactly the metal and the non-metal from the compound's own ions.

Correct Answer: D. $100\text{ cm}^3$

Fastest Approach (🚀):
Compare the volumes actually present (100:10, i.e. 10:1) to the required stoichiometric ratio (2:1 for X:Y). Since 10:1 is far more X-rich than the required 2:1, Y is limiting — work out everything from '10 cm³ of Y reacts completely'.

Step-by-Step Breakdown:

1. Identify the Limiting Reagent

At constant temperature and pressure, gas volumes are directly proportional to moles, so the stoichiometric ratio $2X:1Y$ can be applied directly to volumes.

To react completely with all $10\text{ cm}^3$ of Y, we would need $2\times 10=20\text{ cm}^3$ of X. We have $100\text{ cm}^3$ of X available — far more than the $20\text{ cm}^3$ required — so Y is the limiting reagent and is fully used up; X is left in excess.

2. Calculate the Volumes Reacting and Remaining

• X reacted: $20\text{ cm}^3$ (leaving $100-20=80\text{ cm}^3$ of X unreacted)
• Y reacted: $10\text{ cm}^3$ (all of it; $0$ remaining)
• Z produced: since $2X$ makes $2Z$ (a 1:1 volume ratio between X consumed and Z produced), $20\text{ cm}^3$ of X reacting produces $20\text{ cm}^3$ of Z

3. Sum the Final Volume

$$\text{Final volume} = \underbrace{80}_{\text{X left over}} + \underbrace{20}_{\text{Z produced}} + \underbrace{0}_{\text{Y left over}} = 100\text{ cm}^3$$

The correct answer is D.

Why the Other Options Are Wrong (❌):

• A. $20\text{ cm}^3$ — Incomplete Calculation
  This is the volume of X that reacts (equivalently, the volume of Z produced) — not the total final volume in the syringe, which must also include the $80\text{ cm}^3$ of unreacted X.
• B. $55\text{ cm}^3$ — Calculation Error
  Does not correspond to any consistent stoichiometric calculation for this reaction — possibly from mishandling the mole ratio or averaging the two initial volumes rather than tracking what's consumed and produced.
• C. $80\text{ cm}^3$ — Incomplete Calculation
  This is only the leftover (unreacted) volume of X — it omits the $20\text{ cm}^3$ of gas Z produced by the reaction, which remains in the syringe.
• E. $110\text{ cm}^3$ — Conceptual Misunderstanding
  This is simply the sum of the two initial volumes ($100+10$), ignoring that the total number of gas moles decreases during the reaction (3 volumes of reactant become 2 volumes of product for every 'unit' of reaction).
• F. $120\text{ cm}^3$ — Conceptual Misunderstanding
  Overshoots by treating the reaction as if it increased the total gas volume — the total moles of gas can only decrease here, since $2X+Y\rightarrow 2Z$ converts 3 volumes of reactant into 2 volumes of product.

Common Mistake (⚠️):
Simply adding the two initial volumes and applying the equation's mole-ratio change to the WHOLE amount, as if both gases were present in exactly the stoichiometric ratio, instead of first identifying that X is in large excess and only part of it actually reacts.

Takeaway (📌):
In any 'excess reagent' gas-volume problem: (1) find the limiting reagent by comparing the given ratio to the stoichiometric ratio, (2) compute how much of the OTHER gas actually reacts and how much is left over, (3) add leftover reactant + product volume for the final total — never just apply the equation's volume change to the full initial volumes.

Correct Answer: D. $CrO_4^{2-} + 8H^+ + 3e^- \rightarrow Cr^{3+} + 4H_2O$

Fastest Approach (🚀):
Since the stem already tells you chromate ($CrO_4^{2-}$) forms $Cr^{3+}$, you only need to balance O with $4H_2O$, then H with $8H^+$, then check the charge gap ($+6$ on the left from $-2+8$, versus $+3$ on the right) to see it needs exactly $3e^-$ — this pins down option D without testing the other four.

Step-by-Step Breakdown:

1. Write the Skeleton Equation

• Chromate ion is $CrO_4^{2-}$. It forms $Cr^{3+}$.
• $CrO_4^{2-} \rightarrow Cr^{3+}$

2. Balance the Oxygen Atoms

• Add water ($H_2O$) to the right side to balance the 4 oxygen atoms.
• $CrO_4^{2-} \rightarrow Cr^{3+} + 4H_2O$

3. Balance the Hydrogen Atoms

• Add hydrogen ions ($H^+$) to the left side to balance the 8 hydrogen atoms.
• $CrO_4^{2-} + 8H^+ \rightarrow Cr^{3+} + 4H_2O$

4. Balance the Charges

• Total charge on left = $(-2) + (+8) = +6$.
• Total charge on right = $+3 + 0 = +3$.
• Add 3 electrons ($e^-$) to the left side to balance the charges.
• $CrO_4^{2-} + 8H^+ + 3e^- \rightarrow Cr^{3+} + 4H_2O$

This matches option D.

Why the Other Options Are Wrong (❌):

• A. $CrO_4^{2-} + 8H^+ + 2e^- \rightarrow Cr^{3+} + 4H_2O$ — Charge Balance Error
  Uses the correct product ion ($Cr^{3+}$) but the wrong number of electrons ($2e^-$ instead of $3e^-$): with only $2e^-$ the charges don't balance (left side totals $-2+8-2=+4$, right side totals $+3+0=+3$), so this half-equation is not properly balanced.
• B. $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow Cr^{3+} + 7H_2O$ — Conceptual Misunderstanding
  As printed, this represents the DICHROMATE ion's half-equation, not the chromate ion ($CrO_4^{2-}$) asked for in the stem — dichromate is $Cr_2O_7^{2-}$. (Note the two chromium atoms on the left would also need '$2Cr^{3+}$' on the right to balance atoms, not a single $Cr^{3+}$.)
• C. $Cr_2O_4^{2-} + 8H^+ + 5e^- \rightarrow Cr^{3+} + 4H_2O$ — Conceptual Misunderstanding
  $Cr_2O_4^{2-}$ is not a real ion (it mixes the two-chromium subscript of dichromate with the four-oxygen count of chromate); the electron count (5) is also wrong for a genuine $+6\rightarrow+3$ change.
• E. $CrO_4^{2-} + 14H^+ + 6e^- \rightarrow Cr^{3+} + 7H_2O$ — Conceptual Misunderstanding
  Uses the dichromate ion's stoichiometry ($14H^+$, $6e^-$, $7H_2O$) applied incorrectly to the chromate formula ($CrO_4^{2-}$) — mixing the two half-equations' quantities together.

Common Mistake (⚠️):
Getting the electron count right for the wrong species (or vice versa) — e.g. option A pairs the correct product ($Cr^{3+}$) with the dichromate-style electron count region but with the wrong number ($2e^-$ instead of $3e^-$), which silently breaks the charge balance ($+4$ left vs $+3$ right) unless you actually check it.

Takeaway (📌):
Never assume a half-equation is balanced just because the atoms match — always total the charge on both sides explicitly as a final check, since a plausible-looking electron count (like $2e^-$ in option A) can still leave the charges unequal.

Correct Answer: F. 392

Fastest Approach (🚀):
List each element's atom count once (N:2, H:20, Fe:1, S:2, O:14), multiply each by its given isotope mass, and add — grouping by element avoids missing any atoms from the water of crystallisation or from the polyatomic ions.

Step-by-Step Breakdown:

1. Identify the Atoms in One Formula Unit

Formula: $(NH_4)_2Fe(SO_4)_2\cdot 6H_2O$

• Nitrogen: $2$
• Hydrogen: $(4\times 2)$ from the ammonium ions $+ (2\times 6)$ from the water of crystallisation $=8+12=20$
• Iron: $1$
• Sulfur: $2$
• Oxygen: $(4\times 2)$ from the sulfate ions $+6$ from the water $=8+6=14$

2. Multiply Each Atom Count by Its Isotopic Mass

Using $^1H=1$, $^{14}N=14$, $^{16}O=16$, $^{32}S=32$, $^{56}Fe=56$:

• N: $2\times 14=28$
• H: $20\times 1=20$
• Fe: $1\times 56=56$
• S: $2\times 32=64$
• O: $14\times 16=224$

3. Sum to Get $M_r$

$$M_r = 28+20+56+64+224=392$$

The correct answer is F.

(Note: adding up just the number of atoms — $2+20+1+2+14=39$ — answers a different question, 'how many atoms are in one formula unit', not the formula mass; don't stop at the atom count.)

Why the Other Options Are Wrong (❌):

• A. 144 — Incomplete Calculation
  Far too small for this formula's $M_r$ — consistent with only totalling one small fragment of the formula (for instance just the ammonium and water contributions) rather than every ion and every water molecule in $(NH_4)_2Fe(SO_4)_2\cdot 6H_2O$.
• B. 204 — Incomplete Calculation
  Also too small — consistent with omitting a major fragment of the formula (such as one full sulfate group or the iron) while still including the water of crystallisation.
• C. 284 — Omission Error
  Consistent with omitting the $6H_2O$ water of crystallisation entirely ($6\times 18=108$; $392-108=284$) while still correctly totalling the anhydrous salt $(NH_4)_2Fe(SO_4)_2$.
• D. 360 — Counting Error
  Consistent with undercounting the sulfur atoms as $1$ instead of $2$ (i.e. treating $Fe(SO_4)_2$ as if it contributed only one sulfur atom's mass while still correctly counting all 8 oxygens from the two sulfate groups): $392-32=360$.
• E. 374 — Counting Error
  Off from the correct $M_r$ by exactly one water molecule's mass ($H_2O=18$; $392-18=374$) — consistent with using $5H_2O$ instead of $6H_2O$ for the water of crystallisation.

Common Mistake (⚠️):
Stopping after counting atoms (giving $39$) instead of weighting each count by its atomic mass and summing to $392$ — the two calculations look superficially similar (both sum five numbers) but answer completely different questions.

Takeaway (📌):
Whenever isotope masses are given explicitly in a question, that is a strong signal the question wants a mass-weighted sum ($M_r$ or similar), not a plain atom count — a plain count would not need atomic mass data at all.

Correct Answer: E. Statement or choice matching Option E as derived in the step-by-step solution.

Step-by-Step Breakdown:

1. analyse the Solids and Tests

• We have $Na_2CO_3$ (carbonate), $NaHCO_3$ (bicarbonate), and $Na_2SO_4$ (sulfate).
• Test 1 (Dilute HCl): Carbonates and bicarbonates will react with acid to produce $CO_2$ gas (effervescence). Sulfates will not react. This test distinguishes the sulfate from the other two.
• Test 2 (Heating the solid): Bicarbonates decompose on gentle heating to release $CO_2$ gas and water vapour (e.g., $2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2$). Sodium carbonate and sodium sulfate are thermally stable at typical bunsen burner temperatures and will not evolve gas. This test distinguishes the bicarbonate from the other two.

2. Conclusion

By performing Test 1, we can identify sodium sulfate (no gas).
By performing Test 2, we can identify sodium bicarbonate (gas evolved).
The remaining solid is sodium carbonate.
Therefore, the combination of Test 1 and Test 2 (option e) is sufficient to distinguish all three solids.

The correct answer is E (1 and 2 only).

Correct Answer: E. $120\text{ cm}^3$

Fastest Approach (🚀):
Moles of acid $=C\times V=0.20\times 0.030=0.006\text{ mol}$; double it for a diprotic acid to get moles of NaOH needed ($0.012\text{ mol}$); divide by NaOH's concentration ($0.10\text{ mol dm}^{-3}$) to get the volume in dm³, then convert to cm³.

Step-by-Step Breakdown:

1. Write the Reaction Equation

• Let the diprotic acid be $H_2\ \text{A}$. Since it is diprotic, it requires two moles of NaOH per mole of acid to neutralise.
• $H_2\ \text{A} + 2NaOH \rightarrow Na_2\ \text{A} + 2H_2O$

2. Calculate Moles of Acid

• Moles ($n$) = Concentration ($C$) $\times$ Volume ($V$). (Volume must be in $dm^3$, so $30\text{ cm}^3 = 30 \times 10^{-3}\text{ dm}^3$)
• Moles of $H_2\ \text{A} = 0.20 \times (30 \times 10^{-3}) = 6 \times 10^{-3}\text{ mol}$.

3. Calculate Moles of NaOH Required

From the equation, 1 mole of acid requires 2 moles of NaOH.
Moles of $NaOH = 2 \times (6 \times 10^{-3}) = 12 \times 10^{-3}\text{ mol}$.

4. Calculate Volume of NaOH

• Volume ($V$) = Moles ($n$) / Concentration ($C$)
• Volume = $\frac{12 \times 10^{-3}\text{ mol}}{0.10\text{ mol dm}^{-3}} = 120 \times 10^{-3}\text{ dm}^3$
• Converting back to $cm^3$: $120 \times 10^{-3}\text{ dm}^3 = 120\text{ cm}^3$

The correct answer is E.

Why the Other Options Are Wrong (❌):

• A. $7.5\text{ cm}^3$ — Calculation Error
  Far too small — roughly 16 times below the correct volume; consistent with inverting which concentration belongs to which substance, compounded with the diprotic factor going the wrong way.
• B. $15\text{ cm}^3$ — Calculation Error
  Consistent with using a $1:1$ mole ratio (ignoring the diprotic nature of the acid) together with an inverted concentration ratio — two errors compounding to a much smaller volume than correct.
• C. $30\text{ cm}^3$ — Conceptual Misunderstanding
  Equal to the acid's own volume ($30\text{ cm}^3$) — consistent with assuming the neutralising volume must simply match the acid's volume, rather than actually computing moles from the concentrations given.
• D. $60\text{ cm}^3$ — Stoichiometry Error
  Consistent with treating the acid as monoprotic (using a $1:1$ acid:alkali mole ratio instead of $1:2$) — the single most common error for this question, giving exactly half the correct answer.

Common Mistake (⚠️):
Forgetting that the acid is diprotic and using a $1:1$ mole ratio instead of $1:2$ — this alone would give $60\text{ cm}^3$ (option D) instead of the correct $120\text{ cm}^3$.

Takeaway (📌):
Always check the acid's basicity (mono-, di-, triprotic) before setting up the mole ratio in a neutralisation calculation — the number of acidic hydrogens directly sets the stoichiometric ratio to the alkali, and missing it is one of the most common titration errors.

Correct Answer: A. Statement or choice matching Option A as derived in the step-by-step solution.

Step-by-Step Breakdown:
Note: The original image for this question was found to be missing from the dataset (a duplicate of Question 26 was provided instead). Unable to provide a worked solution.