ESAT Chemistry Mock 1

During the electrolysis of dilute $\displaystyle \ce{Na2SO4(aq)}$ :

- At the cathode (negative electrode), water is reduced in preference to $\displaystyle \ce{Na+}$ ions: $\displaystyle \ce{2H2O(l) + 2e- -> H2(g) + 2OH-(aq)}$ - At the anode (positive electrode), water is oxidised in preference to $\displaystyle \ce{SO4^{2-}}$ ions: $\displaystyle \ce{2H2O(l) -> O2(g) + 4H+(aq) + 4e-}$ Overall reaction: $\displaystyle \ce{2H2O(l) -> 2H2(g) + O2(g)}$ .

For every $\displaystyle 4\text{ mol}$ of electrons transferred, $\displaystyle 2\text{ mol}$ of $\displaystyle \ce{H2(g)}$ and $\displaystyle 1\text{ mol}$ of $\displaystyle \ce{O2(g)}$ are produced. According to Avogadro's law, the volume of $\displaystyle \ce{H2}$ collected is twice the volume of $\displaystyle \ce{O2}$ at any given time.

From the graph, the rate of production of Gas 1 is double that of Gas 2. Therefore, Gas 1 is $\displaystyle \ce{H2}$ and Gas 2 is $\displaystyle \ce{O2}$ .

- Statement 1 is correct: $\displaystyle \ce{H2}$ (Gas 1) is produced at the cathode (negative electrode).
- Statement 2 is correct: Oxidation occurs at the anode (where Gas 2 is produced), releasing electrons. Reduction occurs at the cathode (where Gas 1 is produced), accepting electrons. In the external circuit, electrons flow from the anode (producing Gas 2) to the cathode (producing Gas 1).
- Statement 3 is incorrect: Only water is decomposed and removed from the solution as gases. The amount of $\displaystyle \ce{Na2SO4}$ remains constant while the volume of water decreases, so the concentration of $\displaystyle \ce{Na2SO4}$ in the solution increases.

Thus, statements 1 and 2 only are correct.

Oxygen occupies $\displaystyle 0.2095 \times 200 = 41.9\text{ cm}^3$ , so $\displaystyle 200 - 41.9 = 158.1\text{ cm}^3$ remains. The closest option is $\displaystyle 158\text{ cm}^3$ .

The retention factor $\displaystyle R_\text{f}$ is defined as: $\displaystyle R_\text{f} = \dfrac{\text{distance moved by compound}}{\text{distance moved by solvent front}}$ For Solvent 1 (vertical axis), the solvent front is at $\displaystyle 12.0\text{ cm}$ .
For Solvent 2 (horizontal axis), the solvent front is at $\displaystyle 15.0\text{ cm}$ .

Let us calculate the $\displaystyle R_\text{f}$ values for compound $\displaystyle \text{X}$ :
- In Solvent 1: $\displaystyle R_{\text{f}, 1} = \dfrac{4.8\text{ cm}}{12.0\text{ cm}} = 0.40$ - In Solvent 2: $\displaystyle R_{\text{f}, 2} = \dfrac{6.0\text{ cm}}{15.0\text{ cm}} = 0.40$ Both $\displaystyle R_\text{f}$ values are equal to $\displaystyle 0.40$ , making statement C correct.

Evaluating the other options:
- A is incorrect: In Solvent 1 alone, $\displaystyle \text{V}$ and $\displaystyle \text{W}$ both travel $\displaystyle 9.0\text{ cm}$ , while $\displaystyle \text{X}$ and $\displaystyle \text{Y}$ both travel $\displaystyle 4.8\text{ cm}$ . Only 3 distinct spots would be observed.
- B is incorrect: In Solvent 2 alone, $\displaystyle \text{W}$ and $\displaystyle \text{X}$ both travel $\displaystyle 6.0\text{ cm}$ , giving only 4 distinct spots.
- D is incorrect: For $\displaystyle \text{Z}$ , $\displaystyle R_{\text{f}, 1} = \dfrac{2.4}{12.0} = 0.20$ and $\displaystyle R_{\text{f}, 2} = \dfrac{12.0}{15.0} = 0.80$ , so $\displaystyle R_{\text{f}, 1} < R_{\text{f}, 2}$ .
- E is incorrect: For $\displaystyle \text{Y}$ , $\displaystyle R_{\text{f}, 1} = \dfrac{4.8}{12.0} = 0.40$ and $\displaystyle R_{\text{f}, 2} = \dfrac{10.5}{15.0} = 0.70$ (the values in the statement are inverted).

The starting material is 3-methylpent-2-ene, $\displaystyle \ce{CH3-CH=C(CH3)-CH2-CH3}$ .

In an electrophilic addition reaction, $\displaystyle \ce{HBr}$ adds across the carbon–carbon double bond between C2 and C3:

- If $\displaystyle \ce{H+}$ adds to C2 and $\displaystyle \ce{Br-}$ adds to C3 (via the more stable tertiary carbocation), the product is 3-bromo-3-methylpentane: $\displaystyle \ce{CH3CH2CBr(CH3)CH2CH3}$ (Product 1, major product).
- If $\displaystyle \ce{H+}$ adds to C3 and $\displaystyle \ce{Br-}$ adds to C2 (via the secondary carbocation), the product is 2-bromo-3-methylpentane: $\displaystyle \ce{CH3CH2CH(CH3)CHBrCH3}$ (Product 2, minor product).

Product 3 (1-bromo-3-methylpentane, $\displaystyle \ce{CH3CH2CH(CH3)CH2CH2Br}$ ) requires bromine to attach to C1, which cannot happen because the double bond is between C2 and C3.

Therefore, only products 1 and 2 are formed.

1. Identify the element: An atomic number of $\displaystyle 4$ corresponds to the electron configuration $\displaystyle 1s^2 2s^2$ . This places element $\displaystyle \ce{X}$ (Beryllium) in Group 2.

2. Determine the formula: Group 2 elements have $\displaystyle 2$ valence electrons and form ions/compounds with a valency of $\displaystyle 2$ . Reacting with chlorine (valency $\displaystyle 1$ ) gives the formula $\displaystyle \ce{XCl2}$ .

3. Determine the structure: Although Group 2 chlorides are typically ionic, element $\displaystyle \ce{X}$ is at the top of the group. The $\displaystyle \ce{X^2+}$ ion is very small with a high charge density, making it highly polarising. This results in bonds with significant covalent character.

4. Reason about polymerization: As a simple monomer, $\displaystyle \ce{XCl2}$ would be linear with only $\displaystyle 4$ electrons in the valence shell of $\displaystyle \ce{X}$ (electron deficient). To resolve this in the solid state, the molecules polymerise: chlorine atoms donate lone pairs to form coordinate (dative) bonds with adjacent $\displaystyle \ce{X}$ atoms, creating polymeric chains.

Melting changes particle arrangement and freedom of movement, not chemical identity. Particles in a liquid remain close together, so statement 3 is incorrect.

We must evaluate the products at both electrodes based on electrode potentials and concentration.

At the cathode (reduction):
Species present are the metal ion ( $\displaystyle M^{n+}$ ) and water.
- Metals more reactive than hydrogen (e.g., Na, Mg, K) are not deposited from aqueous solution; water is reduced to hydrogen gas ( $\displaystyle 2\mathrm{H}_2\mathrm{O} + 2\mathrm{e}^- \rightarrow \mathrm{H}_2 + 2\mathrm{OH}^-$ ).
- Metals less reactive than hydrogen (e.g., Cu, Ag) are deposited as the metal ( $\displaystyle \mathrm{M}^{n+} + n\mathrm{e}^- \rightarrow \mathrm{M}$ ).
- Therefore, only $\displaystyle \mathrm{AgF}$ and $\displaystyle \mathrm{CuCl}_2$ can produce the metal.

At the anode (oxidation):
Species present are the halide ion ( $\displaystyle X^-$ ) and water.
- For concentrated solutions of $\displaystyle \mathrm{Cl}^-$ , $\displaystyle \mathrm{Br}^-$ , and $\displaystyle \mathrm{I}^-$ , the halogen is produced ( $\displaystyle 2\mathrm{X}^- \rightarrow \mathrm{X}_2 + 2\mathrm{e}^-$ ). Although oxidation of water is often thermodynamically favoured, the high overpotential for oxygen evolution favours the halogen.
- For $\displaystyle \mathrm{F}^-$ , the oxidation potential is extremely high (standard potential +2.87 V). It is not energetically feasible to oxidise $\displaystyle \mathrm{F}^-$ in the presence of water; water is oxidised to oxygen ( $\displaystyle 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{O}_2 + 4\mathrm{H}^+ + 4\mathrm{e}^-$ ).
- Therefore, $\displaystyle \mathrm{AgF}$ produces oxygen, not fluorine.

Conclusion:
- $\displaystyle \mathrm{AgF}$ produces Ag and $\displaystyle \mathrm{O}_2$ .
- $\displaystyle \mathrm{CuCl}_2$ produces Cu and $\displaystyle \mathrm{Cl}_2$ .

Only $\displaystyle \mathrm{CuCl}_2$ produces both the metal and the halogen.

1. **Determine the Group of element $\displaystyle \ce{Q}$ :**
Looking at the graph of $\displaystyle \log_{10}(\text{IE})$ against the number of electrons removed, the values increase gradually for the first 4 electrons (from $\displaystyle \approx 2.90$ to $\displaystyle \approx 3.64$ ). A prominent, sharp jump occurs between the 4th and 5th electron removed (jumping to $\displaystyle \approx 4.21$ ). This large increase indicates that the 5th electron is removed from a lower, completely filled inner principal quantum shell ( $\displaystyle n=2$ ). Therefore, element $\displaystyle \ce{Q}$ has 4 valence electrons in its outer shell ( $\displaystyle n=3$ ), placing it in Group 14 of the Periodic Table (element $\displaystyle \ce{Q}$ is silicon, $\displaystyle Z=14$ ).

2. Determine the formula of the chloride:
With 4 valence electrons, element $\displaystyle \ce{Q}$ forms a stable tetrachloride with chlorine: ** $\displaystyle \ce{QCl4}$ **.

3. Determine the number of unpaired electrons in the ground state:
The full ground-state electronic configuration of neutral atom $\displaystyle \ce{Q}$ ( $\displaystyle Z=14$ ) is $\displaystyle 1s^2 2s^2 2p^6 3s^2 3p^2$ .
- The $\displaystyle 3s$ orbital is completely filled with 2 paired electrons.
- By Hund's rule of maximum multiplicity, the 2 electrons in the $\displaystyle 3p$ subshell occupy separate degenerate orbitals with parallel spins ( $\displaystyle 3p_x^1 3p_y^1 3p_z^0$ ).
- Thus, there are 2 unpaired electrons in the ground-state neutral atom.

Therefore, row D is correct.

Statement 1 is correct: The Period 2 hydrides of Groups 15, 16, and 17 ( $\displaystyle \ce{NH3}$ , $\displaystyle \ce{H2O}$ , and $\displaystyle \ce{HF}$ ) exhibit hydrogen bonding between molecules. $\displaystyle \ce{H2O}$ can form an average of two hydrogen bonds per molecule, giving it the highest boiling point ( $\displaystyle 373\text{ K}$ ) among all the hydrides shown. Thus, Curve $\displaystyle \text{W}$ corresponds to Group 16.

Statement 2 is incorrect: When simple molecular substances boil, intermolecular forces are overcome, not intramolecular covalent bonds. The increase in boiling point from Period 3 to Period 5 is due to an increasing number of electrons per molecule, which increases polarisability and leads to stronger London (dispersion) forces. In fact, the intramolecular covalent bond strengths decrease down each group.

Statement 3 is correct: Curve $\displaystyle \text{Z}$ represents Group 14 hydrides ( $\displaystyle \ce{CH4}$ , $\displaystyle \ce{SiH4}$ , $\displaystyle \ce{GeH4}$ , $\displaystyle \ce{SnH4}$ ). Carbon is not sufficiently electronegative, and $\displaystyle \ce{CH4}$ has no lone pairs of electrons, so it cannot form hydrogen bonds. As a result, its boiling point trend increases monotonically with molecular size and shows no Period 2 anomaly.

Therefore, statements 1 and 3 only are correct.

Step 1: Separation by density (Separating Funnel)
Liquids P and Q are miscible, so they form a single phase. Liquid R is immiscible with P (and the P/Q mixture) and has a significantly different density ( $\displaystyle 1.00$ vs $\displaystyle \approx 0.80$ ), so it forms a separate layer.
The separating funnel allows Liquid R to be drained off and isolated pure.

Step 2: Separation by boiling point (Distillation)
The remaining phase contains a mixture of P and Q. Fractional distillation relies on differences in boiling point to separate components. Since P and Q have the same boiling point ( $\displaystyle 78$ $\displaystyle ^\circ$ C), they cannot be separated by this method.

Therefore, only Liquid R is obtained in a pure form.

A burning splint gives a squeaky pop with $\displaystyle \ce{H2}$ ; a glowing splint relights in $\displaystyle \ce{O2}$ ; limewater turns cloudy with $\displaystyle \ce{CO2}$ ; damp blue litmus turns red then bleaches with $\displaystyle \ce{Cl2}$ .

Statement 1 is correct: Hydrochloric acid is a strong monoprotic acid that fully dissociates in aqueous solution, giving $\displaystyle [\ce{H+}] = \pu{0.10 mol dm-3}$ and $\displaystyle \text{pH} = -\log_{10}(0.10) = 1.0$ . Citric acid is a weak acid and only partially ionises in water ( $\displaystyle K_{a1} \ll 1$ ). Consequently, in a $\displaystyle \pu{0.10 mol dm-3}$ solution of citric acid, $\displaystyle [\ce{H+}] < \pu{0.10 mol dm-3}$ , which means its pH is greater than $\displaystyle 1.0$ .

Statement 2 is correct: The dihydrogen citrate ion, $\displaystyle \ce{H2A-}$ , can donate a proton (acting as an acid to form $\displaystyle \ce{HA^{2-}}$ ) or accept a proton (acting as a base to form $\displaystyle \ce{H3A}$ ). Thus, it is amphiprotic.

Statement 3 is correct: Citric acid possesses three carboxylic acid groups ( $\displaystyle -\ce{COOH}$ ) and one tertiary alcohol group ( $\displaystyle -\ce{OH}$ ). The alcohol group does not react with aqueous metal hydroxides, so citric acid behaves as a tribasic (triprotic) acid ( $\displaystyle \ce{H3A}$ ). Each formula unit of barium hydroxide, $\displaystyle \ce{Ba(OH)2}$ , provides two hydroxide ions ( $\displaystyle \ce{OH-}$ ). The balanced equation for complete neutralisation is: $\displaystyle \ce{2H3A + 3Ba(OH)2 -> Ba3A2 + 6H2O}$ Moles of citric acid $\displaystyle = 0.0200\text{ dm}^3 \times 0.10\text{ mol dm}^{-3} = 2.0 \times 10^{-3}\text{ mol}$ .
Moles of $\displaystyle \ce{Ba(OH)2}$ required $\displaystyle = \dfrac{3}{2} \times (2.0 \times 10^{-3}\text{ mol}) = 3.0 \times 10^{-3}\text{ mol}$ .
Volume of $\displaystyle \pu{0.10 mol dm-3}$ $\displaystyle \ce{Ba(OH)2}$ required $\displaystyle = \dfrac{3.0 \times 10^{-3}\text{ mol}}{0.10\text{ mol dm}^{-3}} = 0.0300\text{ dm}^3 = 30.0\text{ cm}^3$ .

Therefore, all three statements (1, 2, and 3) are correct.

For an overhead cable, low mass is important while good conductivity and corrosion resistance are also needed. Metal P best combines these properties.

1 is correct: The activation energy for step 1 ( $\displaystyle \ce{X -> Y}$ ) is $\displaystyle E_{\text{a,1}} = 110 - 40 = +70\text{ kJ mol}^{-1}$ . The activation energy for step 2 ( $\displaystyle \ce{Y -> Z}$ ) is $\displaystyle E_{\text{a,2}} = 80 - 20 = +60\text{ kJ mol}^{-1}$ . Since step 1 has the higher activation energy barrier, it is the slower, rate-determining step for the forward reaction.

2 is incorrect: The activation energy for the reverse of step 1 ( $\displaystyle \ce{Y -> X}$ ) is the energy difference between transition state 1 and intermediate $\displaystyle \ce{Y}$ : $\displaystyle E_{\text{a,rev,1}} = 110 - 20 = +90\text{ kJ mol}^{-1}$ . The value $\displaystyle +70\text{ kJ mol}^{-1}$ is the forward activation energy ( $\displaystyle 110 - 40$ ).

3 is correct: The overall enthalpy change is $\displaystyle \Delta H = H(\ce{Z}) - H(\ce{X}) = -50 - 40 = -90\text{ kJ mol}^{-1}$ .

Therefore, statements 1 and 3 only are correct.

The element with atomic number $\displaystyle Z=15$ is phosphorus ( $\displaystyle \text{P}$ ), which is in Group 15 (Period 3) of the periodic table. Its electronic configuration is $\displaystyle [2, 8, 5]$ .

To complete its octet, the atom needs to gain or share 3 electrons. Since hydrogen is a non-metal and phosphorus is a non-metal with a similar electronegativity (2.19 vs 2.20), they form covalent bonds. The element forms 3 single bonds with hydrogen atoms.

Therefore, the simplest hydride is $\displaystyle \text{PH}_3$ (or $\displaystyle \text{XH}_3$ ) and the bonding is covalent.

Balancing atoms gives $\displaystyle \ce{C5H10}$ , an alkene, so it decolourises bromine water.

First, calculate the number of moles of ions produced per mole of the ionic solid.This means that each formula unit of the solid must dissociate to produce 5 ions in solution.

Next, we examine each option to determine the number of ions it produces upon dissociation:

* A: $\displaystyle \ce{NaCl(s) -> Na+(aq) + Cl-(aq)}$ (produces 2 ions)
* B: $\displaystyle \ce{MgCl2(s) -> Mg^{2+}(aq) + 2Cl-(aq)}$ (produces 1 + 2 = 3 ions)
* C: $\displaystyle \ce{AlCl3(s) -> Al^{3+}(aq) + 3Cl-(aq)}$ (produces 1 + 3 = 4 ions)
* D: $\displaystyle \ce{Na3PO4(s) -> 3Na+(aq) + PO4^{3-}(aq)}$ (produces 3 + 1 = 4 ions)
* E: $\displaystyle \ce{Al2(SO4)3(s) -> 2Al^{3+}(aq) + 3SO4^{2-}(aq)}$ (produces 2 + 3 = 5 ions)
* F: $\displaystyle \ce{C6H12O6}$ (glucose) is a molecular compound. It dissolves but does not dissociate into ions.

Only $\displaystyle \ce{Al2(SO4)3}$ produces 5 ions per formula unit, which matches the calculated ratio. Therefore, this is the correct answer.

Write the general balanced equations for $\displaystyle 1.0\text{ mol}$ of $\displaystyle \ce{C_x H_y S_z}$ for each process:

1. Process 1 (complete combustion to $\displaystyle \ce{CO2}$ , $\displaystyle \ce{H2O}$ , and $\displaystyle \ce{SO2}$ ): $\displaystyle \ce{C_x H_y S_z + \left(x + \dfrac{y}{4} + z\right) O2 -> x CO2 + \dfrac{y}{2} H2O + z SO2}$ $\displaystyle \implies x + \dfrac{y}{4} + z = 7.0$ 2. Process 2 (combustion to $\displaystyle \ce{CO2}$ , $\displaystyle \ce{H2O}$ , and $\displaystyle \ce{S}$ ): $\displaystyle \ce{C_x H_y S_z + \left(x + \dfrac{y}{4}\right) O2 -> x CO2 + \dfrac{y}{2} H2O + z S}$ $\displaystyle \implies x + \dfrac{y}{4} = 6.0$ 3. Process 3 (combustion to $\displaystyle \ce{CO}$ , $\displaystyle \ce{H2O}$ , and $\displaystyle \ce{S}$ ): $\displaystyle \ce{C_x H_y S_z + \left(\dfrac{x}{2} + \dfrac{y}{4}\right) O2 -> x CO + \dfrac{y}{2} H2O + z S}$ $\displaystyle \implies \dfrac{x}{2} + \dfrac{y}{4} = 4.0$ Subtracting equation (2) from equation (1): $\displaystyle z = 7.0 - 6.0 = 1$ Subtracting equation (3) from equation (2): $\displaystyle \left(x + \dfrac{y}{4}\right) - \left(\dfrac{x}{2} + \dfrac{y}{4}\right) = \dfrac{x}{2} = 6.0 - 4.0 = 2.0 \implies x = 4$ Substituting $\displaystyle x = 4$ into equation (2): $\displaystyle 4 + \dfrac{y}{4} = 6.0 \implies \dfrac{y}{4} = 2.0 \implies y = 8$ Thus, the molecular formula is $\displaystyle \ce{C4H8S}$ .

The goal is to isolate organelle Q (2.5 μm).

1. Analyze the sizes: P (12.0 μm) > Q (2.5 μm) > R (0.05 μm).
2. Evaluate Option A: Centrifugation at 1000g pellets particles larger than 10 μm. Since P (12.0) > 10, P will pellet. Q (2.5) and R (0.05) will remain in the supernatant. This effectively removes the large contaminant P from Q.
3. Evaluate Option B: Centrifugation at 10 000g pellets particles larger than 2.0 μm. Both P (12.0) and Q (2.5) will pellet. The resulting pellet will contain a mixture of P and Q, making it difficult to separate them.
4. Evaluate Option C: Pellets everything (> 0.02 μm), resulting in no separation.
5. Evaluate Option D: Pellets P, but retaining the pellet discards the target Q (which is in the supernatant).

Therefore, the most effective initial step is to pellet the largest contaminant (P) and retain the supernatant containing the target (Q) for further purification (e.g., a subsequent faster spin).

1. **Atomic number of $\displaystyle \ce{X}$ ( $\displaystyle Z$ ):**
Each $\displaystyle \ce{^{16}_8O}$ atom has 8 protons. The total number of protons in $\displaystyle \ce{[XO4]^{3-}}$ is $\displaystyle Z + 4(8) = Z + 32$ .
Because the ion carries a charge of $\displaystyle -3$ , it has 3 more electrons than protons: $\displaystyle E = (Z + 32) + 3 = Z + 35 \implies Z = E - 35$ 2. **Neutrons in one atom of $\displaystyle \ce{X}$ :**
Each $\displaystyle \ce{^{16}_8O}$ atom contains $\displaystyle 16 - 8 = 8$ neutrons. The 4 oxygen atoms contribute $\displaystyle 4 \times 8 = 32$ neutrons.
The total number of neutrons in the ion is $\displaystyle N$ , so the number of neutrons in atom $\displaystyle \ce{X}$ is: $\displaystyle \text{neutrons in } \ce{X} = N - 32$ 3. **Mass number of $\displaystyle \ce{X}$ ( $\displaystyle A$ ):**
The mass number is the sum of protons and neutrons in one atom of $\displaystyle \ce{X}$ : $\displaystyle A = Z + (\text{neutrons in } \ce{X}) = (E - 35) + (N - 32) = N + E - 67$ Therefore, row C is correct.

$\displaystyle 0.25\text{ mol}$ $\displaystyle \ce{O2}$ contains $\displaystyle 0.50\text{ mol}$ O atoms, so the number of atoms is $\displaystyle 0.50 \times 6.0 \times 10^{23} = 3.0 \times 10^{23}$ .