ESAT Paper 2 Chemistry Sample Questions

Correct Answer: E. 130

Step-by-Step Breakdown:

1. Turn the electron count into the atomic number

Charge is the count of electrons missing or spare. This ion has gained two electrons relative to the neutral atom, so it carries two electrons more than protons:

$Z=54-2=52$

That proton count is the atomic number, the figure written at the bottom left.

2. Add the neutrons to reach the mass number

The mass number totals protons and neutrons and nothing else, because an electron is far too light to register in it:

$A=Z+n=52+78=130$

Sanity check: read the notation back. From ${}_{52}^{130}X$ the neutrons are $130-52=78$, and an ion of charge $2-$ built on $52$ protons holds $52+2=54$ electrons. Both match what was measured.

The key is $130$.

Why the Other Options Are Wrong (❌):

• A. 78 · Neutron count read as the mass number
  Stops at the neutrons. The mass number counts protons and neutrons together, so $78$ omits every one of the $52$ protons; neutrons are never printed on their own in ${}_Z^{A}X$, they are only ever the difference $A-Z$.
• B. 132 · Ion treated as a neutral atom
  Reads the $54$ electrons straight off as the proton count, which is true only of a neutral atom. This ion has gained two electrons, so the protons come to $54-2=52$ and the mass number to $52+78=130$.
• C. 134 · Charge applied in the wrong direction
  Moves the electron count the wrong way. A negative ion has gained two electrons, so it holds two electrons more than protons and the proton count is $54-2=52$, not $56$.
• D. 184 · Electrons counted into the mass number
  Adds the $54$ electrons to the nucleons as well. An electron is far too light to register in a mass number, which totals the $52$ protons and the $78$ neutrons and nothing else.
• F. 128 · Charge applied a second time
  Uses the charge twice. It turns the electron count into the proton count once, $54-2=52$, and this option applies the same correction again to the finished total: $130-2$. A mass number counts the $52$ protons and the $78$ neutrons and nothing else, so once the charge has delivered the protons its work is done and $A=52+78=130$.

Common Mistake (⚠️):
Treating the ion as though it were neutral. Only in a neutral atom does the electron count double as the proton count; this ion has gained two electrons, so its $54$ electrons and its $52$ protons differ by $2$, and using $54$ as the atomic number shifts the mass number by that same $2$.

Takeaway (📌):
Charge first, then add. The charge turns the electron count into the proton count, $54-2=52$, and the mass number is that proton count plus the neutrons. Electrons never enter a mass number at all.

Correct Answer: C. 0.68

Step-by-Step Breakdown:

1. Recover the total volume of oxygen

A mean rate is the total change divided by the time it took, so the total is the rate multiplied by the interval. Only the oxygen reaches the syringe, so its reading follows that one substance:

$$V=2.4\times 100=240\mathrm{c}{\mathrm{m}}^3$$$$=0.24\mathrm{d}{\mathrm{m}}^3$$

2. Convert to moles of oxygen, then to moles of hydrogen peroxide

$$n({\mathrm{O}}_2)=\frac{0.24}{24}=0.01\mathrm{m}\mathrm{o}\mathrm{l}$$

The balanced equation pairs $2{\mathrm{H}}_2{\mathrm{O}}_2$ with $1{\mathrm{O}}_2$, so the amount of hydrogen peroxide is $\frac{2}{1}$ times the amount of oxygen:

$$n({\mathrm{H}}_2{\mathrm{O}}_2)=\frac{2}{1}\times 0.01$$$$=0.02\mathrm{m}\mathrm{o}\mathrm{l}$$

3. Convert the amount to a mass

$$m=n\times M_r=0.02\times 34=0.68g$$

Check the units before committing. Dividing $240\mathrm{c}{\mathrm{m}}^3$ by a molar volume quoted in $\mathrm{d}{\mathrm{m}}^3$ would have given $10\mathrm{m}\mathrm{o}\mathrm{l}$ of oxygen and a final mass of $680g$, which a syringe holding $240\mathrm{c}{\mathrm{m}}^3$ cannot account for.

The key is $0.68$.

Why the Other Options Are Wrong (❌):

• A. 0.02 · Amount in moles reported as a mass
  The amount in moles was written down as though it were already a mass: $n({\mathrm{H}}_2{\mathrm{O}}_2)=0.02\mathrm{m}\mathrm{o}\mathrm{l}$ was quoted as $0.02g$. The last conversion is still owed, and it is the one the stem hands over: multiplying by $M_r({\mathrm{H}}_2{\mathrm{O}}_2)=34$ gives $0.68$.
• B. 0.34 · Stoichiometric ratio ignored
  The equation's coefficients were skipped and the amount of hydrogen peroxide was set equal to the amount of oxygen, $0.01\mathrm{m}\mathrm{o}\mathrm{l}$ rather than $0.02\mathrm{m}\mathrm{o}\mathrm{l}$, before multiplying by $34$. The equation pairs $2{\mathrm{H}}_2{\mathrm{O}}_2$ with $1{\mathrm{O}}_2$, so the amount of hydrogen peroxide is $\frac{2}{1}$ times the amount of oxygen, and this option is the key $0.68$ scaled by $\frac{1}{2}$.
• D. 0.36 · Wrong relative formula mass used
  The right amount, $0.02\mathrm{m}\mathrm{o}\mathrm{l}$, was multiplied by the relative formula mass of water instead: $18$ rather than the $34$ the stem supplies. $M_r({\mathrm{H}}_2\mathrm{O})=18$ is genuine, but it belongs to another substance in the equation, and the question asks for the mass of hydrogen peroxide.
• E. 680 · Molar volume units mismatched
  The volume was left in $\mathrm{c}{\mathrm{m}}^3$ and divided by a molar volume quoted in $\mathrm{d}{\mathrm{m}}^3$: $240÷24=10\mathrm{m}\mathrm{o}\mathrm{l}$ of oxygen instead of $0.01\mathrm{m}\mathrm{o}\mathrm{l}$. One mole of gas occupies $24\mathrm{d}{\mathrm{m}}^3$, which is $24000\mathrm{c}{\mathrm{m}}^3$, so this option is the key $0.68$ multiplied by a thousand.
• F. 0.17 · Mole ratio turned upside down
  The equation's coefficients were read the wrong way up. The $0.01\mathrm{m}\mathrm{o}\mathrm{l}$ of oxygen was scaled by $\frac{1}{2}$ instead of $\frac{2}{1}$ before the relative formula mass was applied, so this option is the key $0.68$ divided by ${\left(\frac{2}{1}\right)}^2$. The equation pairs $2{\mathrm{H}}_2{\mathrm{O}}_2$ with $1{\mathrm{O}}_2$, and it is the hydrogen peroxide that is being counted, so the amount wanted is $0.02\mathrm{m}\mathrm{o}\mathrm{l}$ and the mass is $0.68$.
• G. 16.32 · Molar volume never divided in
  The volume was converted to $\mathrm{d}{\mathrm{m}}^3$ and then treated as though it were already an amount: $0.24$ went straight into the $2$ to $1$ scaling and then into the multiplication by $34$, with no division by the molar volume anywhere. A volume becomes an amount only by dividing by $24\mathrm{d}{\mathrm{m}}^3\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$, which turns $0.24\mathrm{d}{\mathrm{m}}^3$ into $0.01\mathrm{m}\mathrm{o}\mathrm{l}$, so this option is the key $0.68$ multiplied by $24$.
• H. 1.36 · Equation coefficient counted a second time
  The $2$ to $1$ from the equation was used twice: once correctly, turning $0.01\mathrm{m}\mathrm{o}\mathrm{l}$ of oxygen into $0.02\mathrm{m}\mathrm{o}\mathrm{l}$ of hydrogen peroxide, and again at the mass step, as though the coefficient scaled the relative formula mass as well. It does not: $M_r({\mathrm{H}}_2{\mathrm{O}}_2)=34$ is the mass of one mole of hydrogen peroxide whatever the equation's coefficients are, so the mass is $0.02\times 34=0.68$ and the ratio is spent once.

Common Mistake (⚠️):
Starting the mole calculation from the rate itself. A figure in $\mathrm{c}{\mathrm{m}}^3{\mathrm{s}}^{-1}$ is not a volume, so it has to be multiplied by the $100s$ interval before the molar gas volume can be used; feeding $2.4$ straight into the division gives an amount $100$ times too small.

Takeaway (📌):
Rate times time gives the amount of oxygen; the molar gas volume, the $2$ to $1$ from the balanced equation and $M_r({\mathrm{H}}_2{\mathrm{O}}_2)=34$ then run the chain backwards to the hydrogen peroxide. Following a product is only ever a way of following the reactant that made it.

Correct Answer: E. 42

Step-by-Step Breakdown:

1. Find the amount of copper displaced

The manganese is in excess, so every copper ion reacts and the copper(II) sulfate is the limiting reagent. The volume in cubic decimetres is $0.25{\text{dm}}^3$, so

$$n(\mathrm{C}\mathrm{u}\mathrm{S}{\mathrm{O}}_4)=c\times V=0.6\times 0.25$$$$=0.15\text{mol}$$

The reaction is $\mathrm{M}\mathrm{n}+\mathrm{C}{\mathrm{u}}^{2+}\to \mathrm{M}{\mathrm{n}}^{2+}+\mathrm{C}\mathrm{u}$, one copper atom for each copper ion, so $0.15\text{mol}$ of copper is displaced.

2. Convert that amount into a total heat

$$Q=0.15\times 294=44.1\text{kJ}$$$$=44100\text{J}$$

3. Turn the heat into a temperature rise

What warms up is the $250{\text{cm}}^3$ of solution, and at a density of $1{\text{g cm}}^{-3}$ that is $250\text{g}$. The metal's mass is ignored and the cup loses no heat, so all $44100\text{J}$ goes into the solution. Rearranging $Q=mc\Delta T$,

$$\Delta T=\frac{Q}{mc}=\frac{44100}{250\times 4.2}$$$$=\frac{44100}{1050}=42{}^{\circ }\text{C}$$

Sanity check: $1050\text{J}$ raises this solution by one degree, and $44100$ is $42$ times $1050$, so $42$ degrees. Notice that the volume never reaches the answer: it fixes the amount reacting and the mass warmed in the same proportion, so it cancels, and the rise depends only on the concentration, the energy per mole and the specific heat capacity.

The temperature rise is $42{}^{\circ }\text{C}$.

Why the Other Options Are Wrong (❌):

• A. 4.2 · Volume conversion off by ten
  The volume was converted the wrong way, $250{\text{cm}}^3$ taken as $0.025{\text{dm}}^3$, giving $n=0.015\text{mol}$ and $Q=4.41\text{kJ}$, then $4410÷1050=4.2$. A ${\text{dm}}^3$ is a thousand ${\text{cm}}^3$, so the volume is $0.25{\text{dm}}^3$ and the amount reacting is ten times larger than this.
• B. 21 · Ionic charge used as a mole ratio
  The $2+$ charge on the copper ion was used as a stoichiometric factor and the amount divided by $2$ to $0.075\text{mol}$: $0.075\times 294=22.05\text{kJ}$ and $22050÷1050=21$. The charge counts the electrons transferred, not the atoms: each copper ion gives one copper atom, so the amount stays at $0.15\text{mol}$.
• C. 176.4 · Specific heat capacity omitted
  The specific heat capacity was dropped from the denominator: $44100÷250=176.4$. Dividing joules by a mass alone leaves ${\text{J g}}^{-1}$, an energy per gram and not a temperature at all; it is the $4.2$ that turns joules per gram into degrees.
• D. 280 · Scaling by the amount omitted
  The per-mole figure was treated as the total heat released: $294\text{kJ}=294000\text{J}$ and $294000÷1050=280$. Only $0.15\text{mol}$ of copper is displaced, so only that fraction of the $294\text{kJ}$ is ever given out, and $280$ is the rise a whole mole would produce.
• F. 0.042 · Kilojoules divided by a joules-per-degree
  The amount and the heat are both right, $0.15\text{mol}$ giving $44.1\text{kJ}$, but that heat was divided by $1050$ while still in kilojoules. A specific heat capacity of $4.2{\text{J g}}^{-1}{}^{\circ }{\text{C}}^{-1}$ prices a degree in JOULES, so the heat has to be rewritten as $44100\text{J}$ before it meets $1050$, and $44100÷1050=42$, a thousand times this.
• G. 168 · Concentration read off as the amount
  A concentration is a count of moles per cubic decimetre, not a count of moles. Taking $0.6{\text{mol dm}}^{-3}$ as $0.6\text{mol}$ of copper leaves the volume out of the calculation altogether, and the rise then comes out as $42÷0.25$. Only $0.25{\text{dm}}^3$ was poured, so the amount is $0.6\times 0.25=0.15\text{mol}$ and the rise is $42{}^{\circ }\text{C}$.
• H. 44.1 · The heat reported instead of the rise
  This is where the calculation stands after two steps and not at the end: $0.15\times 294=44.1\text{kJ}$ is the heat the displacement gives out, in kilojoules, and the question asks how much warmer the solution gets, in ${}^{\circ }\text{C}$. Warming $250\text{g}$ of solution by one degree costs $250\times 4.2=1050\text{J}$, so $44100\text{J}$ buys $42{}^{\circ }\text{C}$.

Common Mistake (⚠️):
Dividing the heat by the amount instead of multiplying, out of habit from the more familiar calorimetry question that asks for kilojoules per mole. Here the energy per mole is handed over, $294\text{kJ}$, and the temperature rise is what is wanted, so the amount $0.15\text{mol}$ multiplies up to a total heat of $44100\text{J}$ first, and only then does $Q=mc\Delta T$ get rearranged.

Takeaway (📌):
Work out what one degree costs before anything else. The product $mc$ is the price of a degree in joules, here $1050\text{J}$, and every calorimetry question is then either a heat divided by that price or that price multiplied by a rise.

Correct Answer: B. 43

Step-by-Step Breakdown:

1. Read the two numbers off the label, then take the difference

The label gives the nuclide as ${}_{34}^{77}\mathrm{S}\mathrm{e}$. The lower figure is the atomic number, $Z=34$, and it counts protons only: it is what makes the sample selenium rather than anything else. The upper figure is the mass number, $A=77$, and it counts protons and neutrons together, because the electrons are far too light to register in it. Neutrons appear in $A$ and nowhere else on the label, so removing the protons from the mass number leaves them:

$n=A-Z=77-34=43$

Sanity check: the neutron count cannot equal or exceed the mass number, since the protons take up part of it, so $77$ and $111$ are impossible before any arithmetic is attempted.

The key is $43$.

Why the Other Options Are Wrong (❌):

• A. 34 · Reads Z as the neutron count
  Quotes the atomic number instead of subtracting it. $Z=34$ is the proton count, and the neutrons are what is left of the mass number once those protons are removed, $77-34$, so $34$ answers a different question.
• C. 77 · Reads A as the neutron count
  Reads the mass number as a neutron count. $A=77$ is protons and neutrons together, so it already contains the $34$ protons and overstates the neutrons by exactly that many.
• D. 111 · Addition for subtraction
  Adds the two numbers, $77+34=111$, instead of subtracting them. That would make the nucleus heavier than the mass number printed on its own label.
• E. 9 · Subtracts Z twice
  Takes the protons off and then the electrons as well, $77-34-34=9$. The neutral atom does have $34$ electrons, but they sit outside the nucleus and were never counted in the mass number, so $Z$ comes off once only.

Common Mistake (⚠️):
Quoting the upper number as the neutron count. The mass number is protons and neutrons added together, not neutrons alone, so $77$ silently includes the $34$ protons and overstates the neutrons by exactly that many.

Takeaway (📌):
The bottom number identifies, the top number weighs. Protons come straight from the bottom number, neutrons only from the difference, and nothing on a nuclide label ever counts neutrons for you.