ESAT Chemistry Mock Module 1 Worked Solutions

Correct Answer: D. $74$

Fastest Approach (🚀):
$\text{OH} = 17$, doubled $= 34$.
$40+34 = 74$.

Matches Option D.

Step-by-Step Breakdown:

1. Read the formula

$\text{Ca(OH)}_2$ contains one calcium, and two hydroxide groups — so two oxygens and two hydrogens.

2. Add up

$\text{Ca}: 1\times 40 = 40$
$\text{O}: 2\times 16 = 32$
$\text{H}: 2\times 1 = 2$

$M_r = 40+32+2 = 74$

3. The bracket is the whole question

Applying the $2$ to only the hydrogen gives $40+16+2 = 58$, which is Option C and the commonest wrong answer. The subscript multiplies every atom inside the bracket.

4. Why $M_r$ matters

It converts between grams and moles, and every quantitative chemistry calculation begins there. An error here propagates through the whole of the rest of a multi-step question, so it is worth the two seconds to check the bracket.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $57$ — Bracket Misapplied
  Using one $\text{OH}$ and one extra $\text{H}$.
• B. $114$ — Bracket Over-applied
  Doubling the calcium as well.
• C. $58$ — Bracket Misapplied
  Applying the subscript to hydrogen only.
• E. $34$ — Term Omitted
  The mass of the two hydroxide groups alone.

Common Mistake (⚠️):
Applying the subscript to the last atom in the bracket only, giving $58$. The bracket exists precisely to group the atoms the subscript applies to.

Takeaway (📌):
$M_r$ is the sum of every $A_r$ in the formula, with bracket subscripts multiplying everything inside.

Correct Answer: C. mass $\tfrac{1}{1836}$, charge $-1$

Fastest Approach (🚀):
Electron: negligible mass, charge $-1$.

Matches Option C.

Step-by-Step Breakdown:

1. The three particles

| particle | relative mass | relative charge | location |
|---|---|---|---|
| proton | $1$ | $+1$ | nucleus |
| neutron | $1$ | $0$ | nucleus |
| electron | $\tfrac{1}{1836}$ | $-1$ | shells around the nucleus |

2. Why almost all the mass is nuclear

An electron is roughly two thousand times lighter than a proton, so the nucleus holds essentially the whole mass of the atom while occupying a minute fraction of its volume.

3. Why the atom is neutral

An atom has equal numbers of protons and electrons, so the $+1$ and $-1$ charges cancel exactly. Gaining or losing electrons produces an ion; changing the proton count would change the element entirely.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. mass $1$, charge $-1$ — Mass Wrong
  The charge is right, but the electron is far lighter than a proton.
• B. mass $1$, charge $0$ — Wrong Particle
  Describes a neutron's mass and charge.
• D. mass $\tfrac{1}{1836}$, charge $+1$ — Charge Wrong
  Right mass, but electrons are negative.
• E. mass $0$, charge $+1$ — Both Wrong
  Describes neither particle correctly.

Common Mistake (⚠️):
Giving the electron a relative mass of $1$, or a positive charge. Only the proton is both massive and positive.

Takeaway (📌):
Proton $1$ and $+1$; neutron $1$ and $0$; electron negligible and $-1$. Mass is nuclear, charge is balanced.

Correct Answer: B. $\text{Mg}$, because it loses electrons

Fastest Approach (🚀):
$\text{Mg}: 0\to 2+$, so electrons lost $\Rightarrow$ oxidised.

Matches Option B.

Step-by-Step Breakdown:

1. Track the charges

$\text{Mg}$ starts as a neutral atom, charge $0$, and ends as $\text{Mg}^{2+}$. Its charge has risen by two, so it has lost two electrons.

$\text{Cu}^{2+}$ starts at $2+$ and ends as neutral $\text{Cu}$. Its charge has fallen by two, so it has gained two electrons.

2. Apply OIL RIG

Oxidation Is Loss of electrons; Reduction Is Gain.

Magnesium loses, so magnesium is oxidised. Copper gains, so the copper ion is reduced.

3. Half-equations

$\text{Mg}\rightarrow\text{Mg}^{2+}+2\text{e}^{-}$ (oxidation)
$\text{Cu}^{2+}+2\text{e}^{-}\rightarrow\text{Cu}$ (reduction)

The two electrons released are exactly the two consumed, so adding the halves recovers the full equation.

4. Why no oxygen is needed

The oxygen definition is the older and narrower one. Defining oxidation as electron loss covers every reaction the oxygen definition does, and many more besides — including this one, where no oxygen appears at all. Option E applies the narrow definition and misses the reaction entirely.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\text{Cu}^{2+}$, because it gains electrons — Definition Reversed
  Gaining electrons is reduction, not oxidation.
• C. $\text{Mg}^{2+}$, because it has a positive charge — Product Not Reactant
  $\text{Mg}^{2+}$ is the product of the oxidation, not the species oxidised.
• D. $\text{Cu}$, because it is a metal — Product Not Reactant
  $\text{Cu}$ is the product of reduction.
• E. Neither, because no oxygen is involved — Narrow Definition
  The electron definition covers reactions with no oxygen.

Common Mistake (⚠️):
Assuming redox requires oxygen. The electron definition is the general one, and most redox reactions in this module involve no oxygen at all.

Takeaway (📌):
OIL RIG. Track each species' charge across the arrow: rising means oxidised, falling means reduced.

Correct Answer: C. Air is a mixture, and its components can be separated by physical means

Fastest Approach (🚀):
Not chemically bonded $\Rightarrow$ mixture.
Mixtures separate physically.

Matches Option C.

Step-by-Step Breakdown:

1. The three categories

Element — one type of atom only. Oxygen, copper, argon.
Compound — two or more elements chemically bonded in a fixed ratio, with properties unlike its constituents.
Mixture — two or more substances physically together but not bonded, each keeping its own properties, in any proportion.

2. Which is air

Air contains nitrogen, oxygen, argon and carbon dioxide, all as separate molecules with no bonds between them. Each keeps its own properties — oxygen still supports combustion in air. The proportions also vary slightly with place and time, which a compound's could not.

So air is a mixture.

3. How it is separated

By fractional distillation of liquid air: cool it until it liquefies, then warm it slowly so the components boil off at their different boiling points. That is a physical process exploiting a physical difference — no bonds are broken because none exist between the components.

Option E gets the classification right and the separation wrong.

4. The contrast

Water is a compound. Splitting it into hydrogen and oxygen requires electrolysis — a chemical process — because covalent bonds must be broken. No amount of distillation will do it.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Air is a compound, because it contains more than one element — Definition Confused
  Containing several elements does not make something a compound.
• B. Air is an element, because it behaves as a single gas — Factually Wrong
  Air contains several different substances.
• D. Air is a compound, because its composition is fixed — Definition Confused
  Air's composition varies slightly; a compound's is fixed.
• E. Air is a mixture, but separating it requires a chemical reaction — Separation Wrong
  Right classification, but mixtures separate physically.

Common Mistake (⚠️):
Calling anything with several elements in it a compound. The test is whether the elements are chemically bonded, not whether more than one is present.

Takeaway (📌):
Mixture: not bonded, variable proportions, separated physically. Compound: bonded, fixed ratio, separated only by chemical reaction.

Correct Answer: A. Elements are arranged in order of increasing atomic number, with Groups as vertical columns and Periods as horizontal rows

Fastest Approach (🚀):
Order by atomic number. Groups down, Periods across.

Matches Option A.

Step-by-Step Breakdown:

1. The ordering

Elements are placed in order of increasing atomic number — the proton count — not relative atomic mass. The two orders almost always agree, but not quite: argon ($Z=18$, $A_r=40$) comes before potassium ($Z=19$, $A_r=39$). Ordering by mass would swap them and put argon among the reactive metals, which is how the principle was settled.

2. The two directions

Groups are the vertical columns. Elements in a Group share the same number of outer electrons, and therefore similar chemical properties.

Periods are the horizontal rows. Elements in a Period have the same number of occupied shells.

3. The mnemonic worth having

Groups go down, like a group photo lined up in columns; Periods go across, like a sentence ending in a full stop. Whatever the aid, fix it once — the two terms are interchanged constantly under time pressure.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. Elements are arranged in order of increasing relative atomic mass, in vertical Periods — Two Errors
  Mass ordering, and Periods are horizontal not vertical.
• C. Groups are horizontal rows containing elements with the same number of shells — Terms Swapped
  Describes Periods, not Groups.
• D. Elements are arranged alphabetically within each Group — Implausible
  Alphabetical order would depend on the language used.
• E. Periods are vertical columns of elements with similar chemical properties — Terms Swapped
  Describes Groups, not Periods.

Common Mistake (⚠️):
Ordering by relative atomic mass. It usually gives the same sequence, and the exceptions are precisely what the atomic-number ordering was introduced to fix.

Takeaway (📌):
Increasing atomic number. Groups are vertical and share outer electrons; Periods are horizontal and share shell count.

Correct Answer: E. $\text{Al}_2\text{O}_3$

Fastest Approach (🚀):
Swap the charges: $\text{Al}^{3+}$ and $\text{O}^{2-} \Rightarrow \text{Al}_2\text{O}_3$.

Matches Option E.

Step-by-Step Breakdown:

1. The compound must be neutral

Total positive charge must cancel total negative charge exactly. An ionic compound with a net charge cannot exist as a bulk solid.

2. Find the smallest whole-number ratio

The lowest common multiple of $3$ and $2$ is $6$:
$2\times(+3) = +6$ and $3\times(-2) = -6$

So two aluminium ions to three oxide ions: $\text{Al}_2\text{O}_3$.

3. The swap shortcut

Take each ion's charge number and write it as the subscript of the other ion: $\text{Al}^{3+}$ and $\text{O}^{2-}$ give $\text{Al}_2\text{O}_3$. Then cancel any common factor — $\text{Mg}^{2+}$ with $\text{O}^{2-}$ would give $\text{Mg}_2\text{O}_2$, which simplifies to $\text{MgO}$. Option E is the uncancelled form of a different pairing.

4. What the formula does and does not describe

An ionic formula gives the ratio, not a molecule. Aluminium oxide is a giant lattice of alternating ions; there is no discrete $\text{Al}_2\text{O}_3$ particle in it.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\text{AlO}$ — Charge Unbalanced
  Net charge $+1$ per unit, so not neutral.
• B. $\text{Al}_3\text{O}_3$ — Not Simplified
  Not the smallest ratio, and not charge-balanced.
• C. $\text{Al}_3\text{O}_2$ — Swapped
  The subscripts are the wrong way round.
• D. $\text{AlO}_3$ — Charge Unbalanced
  Net charge $-3$ per unit.

Common Mistake (⚠️):
Writing $\text{AlO}$ by pairing one of each. That would leave a net charge of $+1$ per unit, which is impossible for a neutral compound.

Takeaway (📌):
Balance the charges to zero and reduce to the smallest whole-number ratio. Swapping charge numbers is the fast route.

Correct Answer: D. $0.5$

Fastest Approach (🚀):
$M_r = 12+4 = 16$.
$n = \dfrac{8}{16} = 0.5\ \text{mol}$.

Matches Option D.

Step-by-Step Breakdown:

1. Find the relative molar mass

$\text{CH}_4$: $12+(4\times 1) = 16$

2. Convert

$n = \dfrac{\text{mass}}{M_r} = \dfrac{8.0}{16} = 0.5\ \text{mol}$

3. Check the direction

$8\ \text{g}$ is half of $16\ \text{g}$, so it must be half a mole. Any answer greater than one is inverted — which rules out Options B, C and E without arithmetic.

4. What half a mole actually is

One mole contains Avogadro's number of particles, $6.02\times 10^{23}$. Half a mole of methane is therefore about $3.01\times 10^{23}$ molecules — and five times that many atoms, since each molecule contains five.

5. Why moles rather than grams

A balanced equation gives a ratio of particles. Two substances with the same mass contain quite different numbers of molecules, so masses cannot be compared directly. Converting to moles is what makes the equation's ratio usable.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $0.125$ — Mr Error
  Dividing by $64$, or using $M_r = 64$.
• B. $2.0$ — Inverted
  The fraction inverted.
• C. $8.0$ — No Conversion
  Quoting the mass in grams as a number of moles.
• E. $128$ — Operation Error
  Multiplying rather than dividing.

Common Mistake (⚠️):
Dividing $M_r$ by the mass, giving $2$. Compare the mass with $M_r$ first: less than $M_r$ always means less than one mole.

Takeaway (📌):
$n = \dfrac{m}{M_r}$. One mole is $M_r$ in grams and contains $6.02\times 10^{23}$ particles.

Correct Answer: A. $26$ protons, $30$ neutrons, $23$ electrons

Fastest Approach (🚀):
Protons $= 26$; neutrons $= 56-26 = 30$.
$3+$ means $3$ electrons lost: $26-3 = 23$.

Matches Option A.

Step-by-Step Breakdown:

1. Read the two numbers

The lower number is the atomic number, $26$ — the proton count, which defines the element as iron.
The upper number is the mass number, $56$ — protons plus neutrons.

2. Neutrons

$\text{neutrons} = \text{mass number}-\text{atomic number} = 56-26 = 30$

3. Electrons

A neutral iron atom would have $26$ electrons. The $3+$ charge means it has lost three:
$26-3 = 23$

4. Why positive means fewer

Electrons carry the negative charge, so removing them leaves a net positive charge. A $3+$ ion has three fewer electrons than protons, not three more — which is the trap Option E is built on.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $26$ protons, $30$ neutrons, $26$ electrons — Charge Ignored
  Ignores the charge, giving the neutral atom.
• C. $26$ protons, $56$ neutrons, $23$ electrons — Neutron Count
  Uses the mass number as the neutron count.
• D. $29$ protons, $30$ neutrons, $26$ electrons — Protons Altered
  Changes the proton count, which would change the element.
• E. $26$ protons, $30$ neutrons, $29$ electrons — Charge Reversed
  Adds electrons for a positive charge.

Common Mistake (⚠️):
Adding electrons for a positive charge. Positive means electrons were removed; the proton count never changes in ion formation.

Takeaway (📌):
Protons from the bottom number, neutrons from the difference, electrons from protons minus the charge.

Correct Answer: C. $+7$

Fastest Approach (🚀):
$x-8 = -1 \implies x = +7$.

Matches Option C.

Step-by-Step Breakdown:

1. Fix the known states

Oxygen is $-2$ in almost every compound, and there are four oxygens:
$4\times(-2) = -8$

2. Set the total equal to the ion's charge

The oxidation states in an ion sum to the charge on the ion, which is $-1$:
$x+(-8) = -1$

3. Solve

$x = -1+8 = +7$

Manganese is in oxidation state $+7$ — its highest, which is why permanganate is such a powerful oxidising agent: it has a long way to fall.

4. Watch the sign

Writing $x-8 = -1$ and then $x = -9$ or $x = -7$ is a sign slip, and $-7$ sits at Option B waiting for it. A transition metal in a compound with oxygen is essentially always positive, which is a useful plausibility check.

5. The rules worth carrying

Uncombined element: $0$.
Simple ion: the ionic charge.
Oxygen: $-2$ (except in peroxides, $-1$).
Hydrogen: $+1$ (except in metal hydrides, $-1$).
Sum over a neutral compound: $0$. Over an ion: the charge.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $+2$ — Guessed From Memory
  A common manganese state, but not in this ion.
• B. $-7$ — Sign Error
  Sign error in solving the equation.
• D. $+8$ — Charge Ignored
  Summing to zero rather than to the ion's charge.
• E. $-1$ — Wrong Quantity
  Quoting the ion's charge rather than manganese's oxidation state.

Common Mistake (⚠️):
Setting the sum to zero for an ion. A neutral compound sums to zero; an ion sums to its charge, and here that difference changes the answer by one.

Takeaway (📌):
Assign the known states, sum to the overall charge, solve for the unknown. Oxygen $-2$ and hydrogen $+1$ carry most questions.

Correct Answer: B. Sodium transfers one electron to chlorine, forming $\text{Na}^{+}$ and $\text{Cl}^{-}$ held by electrostatic attraction in a giant lattice

Fastest Approach (🚀):
Metal loses, non-metal gains.
Ions in a giant lattice, not molecules.

Matches Option B.

Step-by-Step Breakdown:

1. Which way the electron goes

Sodium has one outer electron; losing it leaves the configuration $2,8$ — the same as neon. Chlorine has seven; gaining one gives $2,8,8$ — the same as argon.

So sodium loses and chlorine gains. Both reach a noble gas configuration, which is why the reaction is so favourable.

2. The ions formed

$\text{Na}\rightarrow\text{Na}^{+}+\text{e}^{-}$
$\text{Cl}+\text{e}^{-}\rightarrow\text{Cl}^{-}$

3. What holds them together

The strong electrostatic attraction between oppositely charged ions. That attraction acts in every direction equally, so each ion surrounds itself with as many oppositely charged neighbours as will fit — six, in sodium chloride.

4. Why there is no NaCl molecule

The result is a giant ionic lattice extending through the whole crystal, not discrete pairs. The formula $\text{NaCl}$ states the $1:1$ ratio, not a molecule. Option D gets the transfer right and the structure wrong, and that structure is what explains the high melting point.

5. Why the metal always loses

Metals have few outer electrons and low ionisation energies; non-metals are a few short and attract electrons strongly. Option A reverses this and would require sodium to gain seven electrons, which is energetically hopeless.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Sodium gains an electron and chlorine loses one, forming a covalent bond — Direction Reversed
  Transfer reversed, and transfer gives ionic not covalent bonding.
• C. Both atoms share a pair of electrons to complete their outer shells — Bond Type
  Sharing is covalent bonding, between non-metals.
• D. Sodium transfers one electron to chlorine, forming discrete $\text{NaCl}$ molecules — Structure Wrong
  Correct transfer, but ionic compounds form lattices not molecules.
• E. Chlorine transfers one electron to sodium, forming a metallic bond — Two Errors
  Direction reversed, and metallic bonding needs two metals.

Common Mistake (⚠️):
Describing sodium chloride as molecules. Ionic compounds form giant lattices, and that difference explains every one of their physical properties.

Takeaway (📌):
Ionic bonding: metal loses, non-metal gains, oppositely charged ions in a giant lattice held by electrostatic attraction.

Correct Answer: E. Alkaline earth metals — Group 3

Fastest Approach (🚀):
Alkaline earth metals are Group $2$.

Matches Option E.

Step-by-Step Breakdown:

1. The families and their columns

Group 1 — alkali metals: lithium, sodium, potassium. One outer electron, forming $1+$ ions.
Group 2 — alkaline earth metals: magnesium, calcium. Two outer electrons, forming $2+$ ions.
Group 17 — halogens: fluorine, chlorine, bromine, iodine. Seven outer electrons, forming $1-$ ions.
Group 18 — noble gases: helium, neon, argon. Full outer shells, hence almost entirely unreactive.
Transition metals — the central block between Groups 2 and 13.

2. Spot the error

The alkaline earth metals sit in Group $2$, immediately beside the alkali metals. Group $3$ is part of the transition block, not a main-group family.

3. Why the position predicts the charge

For main-group metals, the Group number gives the outer electrons and so the positive charge of the ion. For the non-metals near the right, the charge is the Group number minus $18$. That single rule generates most ionic formulae without any memorisation.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. Alkali metals — Group 1 — Correct Pairing
  Correct.
• B. Halogens — Group 17 — Correct Pairing
  Correct.
• C. Noble gases — Group 18 — Correct Pairing
  Correct.
• D. Transition metals — the central block between Groups 2 and 13 — Correct Pairing
  Correct.

Common Mistake (⚠️):
Placing the alkaline earth metals in Group $3$. They are Group $2$, which is what makes them form $2+$ ions.

Takeaway (📌):
Groups $1$, $2$, $17$ and $18$ are the four named main-group families, with the transition metals in the central block.

Correct Answer: B. $5$

Fastest Approach (🚀):
$3\ \text{C} \Rightarrow 3\text{CO}_2$; $8\ \text{H} \Rightarrow 4\text{H}_2\text{O}$.
O needed $= 6+4 = 10 \Rightarrow 5\text{O}_2$.

Matches Option B.

Step-by-Step Breakdown:

1. Balance carbon

Three carbons on the left, so three $\text{CO}_2$ on the right.

2. Balance hydrogen

Eight hydrogens on the left. Each water has two, so four $\text{H}_2\text{O}$.

3. Balance oxygen last

Oxygen atoms now required on the right:
$3\text{CO}_2$ gives $3\times 2 = 6$
$4\text{H}_2\text{O}$ gives $4\times 1 = 4$
total $= 10$

Oxygen arrives as $\text{O}_2$, so $\dfrac{10}{2} = 5$ molecules:

$\text{C}_3\text{H}_8(\text{g})+5\text{O}_2(\text{g})\rightarrow 3\text{CO}_2(\text{g})+4\text{H}_2\text{O}(\text{l})$

4. Why oxygen goes last

It appears in both products, so its requirement is not fixed until the others are settled. Balancing it first means redoing it. The general rule: the element in the most compounds is balanced last.

5. The rule that must not be broken

Only the numbers in front of formulae may be changed. Writing $\text{CO}_3$ to gain an oxygen would turn carbon dioxide into carbonate — a different substance, and a chemistry error rather than an arithmetic one.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $3$ — Partial Balance
  Balances the $\text{CO}_2$ oxygen only.
• C. $4$ — Partial Balance
  Balances the water oxygen only.
• D. $7$ — Count Error
  An arithmetic slip in totalling the oxygens.
• E. $10$ — Atoms Not Molecules
  The number of oxygen atoms, not $\text{O}_2$ molecules.

Common Mistake (⚠️):
Changing a subscript to balance an atom. Coefficients scale how many molecules there are; subscripts define what the molecule is.

Takeaway (📌):
Balance the fuel's carbon and hydrogen first, then oxygen. Never alter a formula, only its coefficient.

Correct Answer: A. $40.0\%$

Fastest Approach (🚀):
$M_r = 40+12+48 = 100$.
$\dfrac{40}{100} = 40\%$.

Matches Option A.

Step-by-Step Breakdown:

1. Find the formula mass

$\text{Ca}: 40$
$\text{C}: 12$
$\text{O}: 3\times 16 = 48$

$M_r = 40+12+48 = 100$

2. Take the fraction

$\%\text{Ca} = \dfrac{\text{mass of Ca in one formula unit}}{M_r}\times 100 = \dfrac{40}{100}\times 100 = 40\%$

3. Check the whole composition

$\%\text{C} = 12\%$ and $\%\text{O} = 48\%$. Together: $40+12+48 = 100\%$ ✓ — a complete check that costs nothing and catches any arithmetic slip.

4. Why Option A is the trap

$33.3\%$ is the fraction of atoms that are calcium — one atom in three... in fact one in five, since there are five atoms in the formula unit. Either way it is a count, not a mass, and the question asks by mass. Atoms differ in mass, so the two fractions rarely agree.

5. Where it is used

Percentage composition is how the metal content of an ore is quoted, and how a proposed formula is checked against experiment.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $33.3\%$ — Count Not Mass
  A fraction of atoms rather than of mass.
• C. $12.0\%$ — Wrong Element
  The percentage of carbon.
• D. $48.0\%$ — Wrong Element
  The percentage of oxygen.
• E. $25.0\%$ — Mr Error
  Using $M_r = 160$ or another mis-computed total.

Common Mistake (⚠️):
Counting atoms rather than weighing them. Percentage by mass needs each element's total mass, not how many of its atoms appear.

Takeaway (📌):
$\%$ by mass $= \dfrac{n\times A_r}{M_r}\times 100$. Check that all the percentages sum to $100$.

Correct Answer: D. $2,8,7$ — Group $17$ (Group $7$ of the main groups)

Fastest Approach (🚀):
$17 = 2+8+7$.
Seven outer electrons $\Rightarrow$ Group $17$.

Matches Option D.

Step-by-Step Breakdown:

1. Fill the shells

For the first twenty elements the shells hold $2$, then $8$, then $8$:
$17 = 2+8+7$

so the configuration is $2,8,7$.

2. Read the Group

The number of electrons in the outermost shell gives the Group. Seven outer electrons places it in Group $17$ under the modern numbering, which is Group $7$ in the older main-group scheme — the halogens. The element is chlorine.

3. Read the Period

The number of occupied shells gives the Period. Three shells means Period $3$.

4. Why the configuration predicts the chemistry

Seven outer electrons means one short of a full shell, so chlorine gains a single electron to form $\text{Cl}^{-}$. That single fact predicts its ionic charge, its reactivity and the formula of its compounds.

5. Why Option A is not quite right

The configuration is correct, but the Group label matters: under the current IUPAC numbering the halogens are Group $17$. Both conventions are in use, and Option B states the relationship between them.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $2,8,7$ — Group $7$ — Convention Only
  Configuration right; the Group numbering is the older convention without the link stated.
• B. $2,8,8,1$ — Group $1$ — Count Wrong
  That configuration has $19$ electrons, not $17$.
• C. $2,7,8$ — Group $8$ — Shell Capacity
  The second shell holds eight, not seven.
• E. $8,8,1$ — Group $1$ — Shell Omitted
  Omits the innermost shell entirely.

Common Mistake (⚠️):
Filling the second shell with $7$ before completing it with $8$. Shells fill in order and completely, for the first twenty elements.

Takeaway (📌):
Shells fill $2, 8, 8, 2$. Outer electrons give the Group; occupied shells give the Period.

Correct Answer: B. $\text{Cu}^{2+}$

Fastest Approach (🚀):
$\text{Cu}^{2+}$ is reduced $\Rightarrow$ it is the oxidising agent.

Matches Option B.

Step-by-Step Breakdown:

1. Identify what changes

$\text{Zn}: 0\to +2$ — oxidised, losing two electrons.
$\text{Cu}^{2+}: +2\to 0$ — reduced, gaining two electrons.
$\text{SO}_4^{2-}$ is unchanged throughout — a spectator ion, present on both sides.

2. Apply the definition

An oxidising agent causes another species to be oxidised. To do that it must take the electrons, so it is itself reduced.

$\text{Cu}^{2+}$ takes the electrons from zinc, so $\text{Cu}^{2+}$ is the oxidising agent.

Correspondingly, $\text{Zn}$ is the reducing agent — it supplies the electrons and is itself oxidised.

3. Why the names feel backwards

The agent is named for what it does to the other species, not for what happens to it. Reading 'oxidising agent' as 'the one oxidised' inverts the answer, and that inversion is the entire point of the question.

4. The ionic equation

Stripping the spectator ion leaves
$\text{Zn}+\text{Cu}^{2+}\rightarrow\text{Zn}^{2+}+\text{Cu}$

which makes the electron transfer visible and shows the sulfate playing no part.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\text{Zn}$ — Agent Reversed
  Zinc is oxidised, so it is the reducing agent.
• C. $\text{SO}_4^{2-}$ — Spectator
  A spectator ion, unchanged throughout.
• D. $\text{Zn}^{2+}$ — Product Not Agent
  The product of the oxidation, not an agent.
• E. There is no oxidising agent, as no oxygen is transferred — Narrow Definition
  Redox is defined by electron transfer, not oxygen.

Common Mistake (⚠️):
Naming zinc as the oxidising agent because zinc is oxidised. Zinc is the reducing agent; the agent is named for its effect on the other species.

Takeaway (📌):
Oxidising agent: reduced itself. Reducing agent: oxidised itself. Spectator ions appear unchanged on both sides.

Correct Answer: B. $6$

Fastest Approach (🚀):
Triple bond $= 3$ pairs $= 6$ electrons.

Matches Option B.

Step-by-Step Breakdown:

1. What a covalent bond is

A shared pair of electrons, one contributed by each atom, held between the two nuclei by their mutual attraction.

2. Count for a triple bond

Three bonds means three shared pairs:
$3\times 2 = 6$ electrons shared.

3. Check against the atoms

Each nitrogen atom has five outer electrons and needs eight. Sharing three pairs means each atom counts $5+3 = 8$ in its outer shell — a full octet for both, achieved by sharing rather than transferring.

4. Why nitrogen gas is so unreactive

Three bonds must be broken to make nitrogen react, and the $\text{N}\equiv\text{N}$ bond enthalpy is very high — around $945\ \text{kJ mol}^{-1}$. That is why nitrogen makes up most of the atmosphere without taking part in much of anything, and why fixing it industrially needs the high temperatures and pressures of the Haber process.

5. Why Option E is wrong

$14$ is the total number of electrons in the molecule, counting both atoms' inner shells. The question asks only how many are shared.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $2$ — Single Bond
  The electrons in a single bond.
• C. $3$ — Bonds Not Electrons
  The number of bonds, not of electrons.
• D. $8$ — Wrong Quantity
  The number of outer electrons around each atom after sharing.
• E. $14$ — Wrong Quantity
  The total electrons in the molecule, not those shared.

Common Mistake (⚠️):
Answering $3$ — the number of bonds rather than of electrons. Each bond involves two electrons, and the question asks for electrons.

Takeaway (📌):
One covalent bond is one shared pair. Double the bond count for electrons, and check each atom reaches a full outer shell.

Correct Answer: C. Group $2$, Period $3$, forming $\text{X}^{2+}$

Fastest Approach (🚀):
Outer $= 2 \Rightarrow$ Group $2$. Shells $= 3 \Rightarrow$ Period $3$.
Metal loses $2$ electrons $\Rightarrow \text{X}^{2+}$.

Matches Option C.

Step-by-Step Breakdown:

1. Read the Group

The outermost shell holds $2$ electrons, so the element is in Group $2$ — the alkaline earth metals.

2. Read the Period

Three numbers means three occupied shells, so Period $3$.

3. Predict the ion

Atoms react to reach a noble gas configuration. With only two outer electrons it is far easier to lose them than to gain six, so the element forms $\text{X}^{2+}$ with the configuration $2,8$ — the same as neon.

The element is magnesium, $Z=12$.

4. Why metals lose and non-metals gain

Elements on the left have few outer electrons and lose them to reach the previous noble gas. Elements on the right are a few short and gain them to reach the next one. The dividing line runs diagonally down the table, and the Group number tells you which side of it an element sits on.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Group $3$, Period $2$, forming $\text{X}^{3+}$ — Group Wrong
  Three outer electrons would be needed for Group 3.
• B. Group $2$, Period $2$, forming $\text{X}^{2-}$ — Two Errors
  Two shells would be configuration 2,8; and Group 2 metals form positive ions.
• D. Group $12$, Period $3$, forming $\text{X}^{2+}$ — Block Confused
  Group 12 is in the transition block, not the main groups.
• E. Group $2$, Period $3$, forming $\text{X}^{2-}$ — Charge Reversed
  Right position, but a metal forms a positive ion.

Common Mistake (⚠️):
Giving a negative ion for a Group $2$ element. Losing two electrons is far easier than gaining six, so metals on the left always form positive ions.

Takeaway (📌):
Outer electrons give the Group, shell count gives the Period, and the easier route to a full shell gives the ion.

Correct Answer: A. The forward and reverse reactions continue at equal rates, so the concentrations stay constant

Fastest Approach (🚀):
Constant concentrations, not stopped reactions.
Equal rates, not equal amounts.

Matches Option A.

Step-by-Step Breakdown:

1. What 'dynamic' means

Both reactions continue indefinitely. Reactants keep forming products and products keep re-forming reactants. Nothing stops — which is why the word dynamic is attached to the term.

2. Why the concentrations hold steady

The forward and reverse rates are equal, so every product molecule formed is matched by one converting back. The concentrations therefore stay constant, even though individual molecules are constantly changing.

3. Constant is not equal

Option C is the most common error. The concentrations stop changing, but they are almost never equal to one another — the position of equilibrium may lie far to one side, and usually does.

4. Why the container must be closed

If a product escapes, it can never react back. The reverse reaction is starved, equilibrium is never established, and the forward reaction continues until the reactants are exhausted. That is why reversible reactions are studied in sealed vessels, and Option E is wrong.

5. Where this matters industrially

The Haber process runs as a reversible reaction that never goes to completion. Ammonia is removed as it forms, which prevents equilibrium and keeps the forward reaction running — an application of exactly the point Option E gets backwards.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. Both reactions have stopped, so the concentrations no longer change — Static Misconception
  Equilibrium is dynamic — the reactions continue.
• C. The concentrations of reactants and products are equal — Constant vs Equal
  Constant, but not generally equal.
• D. The forward reaction has gone to completion — Completion Assumed
  A reversible reaction does not go to completion.
• E. Equilibrium can be reached in an open container just as readily — Closed System Ignored
  An escaping product prevents the reverse reaction entirely.

Common Mistake (⚠️):
Reading 'concentrations no longer change' as 'the reactions have stopped'. They continue at equal and opposite rates, which is the whole content of the word dynamic.

Takeaway (📌):
Equilibrium: both reactions continue at equal rates, concentrations constant but not equal, and only in a closed system.

Correct Answer: E. $\text{CH}_3$

Fastest Approach (🚀):
C: $\dfrac{2.4}{12} = 0.2$. H: $\dfrac{0.6}{1} = 0.6$.
Divide by $0.2$: $1:3 \Rightarrow \text{CH}_3$.

Matches Option E.

Step-by-Step Breakdown:

1. Convert each mass to moles

$\text{C}: \dfrac{2.4}{12} = 0.2\ \text{mol}$
$\text{H}: \dfrac{0.6}{1} = 0.6\ \text{mol}$

2. Divide by the smallest

$\dfrac{0.2}{0.2} = 1$ and $\dfrac{0.6}{0.2} = 3$

3. Write the formula

The simplest whole-number ratio is $1:3$, so the empirical formula is $\text{CH}_3$.

4. Empirical against molecular

The empirical formula gives only the ratio. The actual molecule might be $\text{C}_2\text{H}_6$ — ethane — whose ratio is also $1:3$. Distinguishing them needs the relative molecular mass: if $M_r = 30$, then since the empirical unit weighs $15$, the molecular formula is twice the empirical one.

Option D is that molecular formula, and is wrong here only because the question asked for the empirical one.

5. Why the masses cannot be compared directly

There is four times as much carbon by mass, but carbon atoms are twelve times heavier — so there are actually three times as many hydrogen atoms. That inversion is what the mole conversion exists to handle.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\text{CH}_2$ — Ratio Error
  A ratio of $1:2$, from an arithmetic slip.
• B. $\text{C}_4\text{H}$ — Masses Compared
  The mass ratio used directly, without converting to moles.
• C. $\text{CH}_4$ — Ratio Error
  A ratio of $1:4$.
• D. $\text{C}_2\text{H}_6$ — Not Simplified
  A molecular formula with the same ratio, not the simplest one.

Common Mistake (⚠️):
Using the mass ratio $2.4:0.6 = 4:1$ and writing $\text{C}_4\text{H}$. Mass ratios must be divided by $A_r$ before they mean anything about atom counts.

Takeaway (📌):
Divide each mass by its $A_r$, then divide all results by the smallest. Empirical gives the ratio; $M_r$ is needed for the molecular formula.

Correct Answer: A. $35.5$

Fastest Approach (🚀):
$\dfrac{35(75)+37(25)}{100} = \dfrac{2625+925}{100} = 35.5$.

Matches Option A.

Step-by-Step Breakdown:

1. Set up the weighted mean

$A_r = \dfrac{\sum(\text{isotope mass}\times\text{percentage abundance})}{100}$

2. Substitute

$A_r = \dfrac{(35\times 75)+(37\times 25)}{100} = \dfrac{2625+925}{100} = \dfrac{3550}{100} = 35.5$

3. Check it is plausible before trusting it

The answer must fall between $35$ and $37$, and closer to $35$ because that isotope is three times as abundant. $35.5$ satisfies both. The unweighted average, $36.0$, is Option A and ignores the abundances entirely.

4. Why relative atomic masses are rarely whole numbers

A quoted $A_r$ is an average over the natural mixture of isotopes. Chlorine's $35.5$ does not describe any individual atom — every chlorine atom has a mass number of $35$ or $37$. That is exactly why the periodic table shows $35.5$ and no nuclide does.

5. What isotopes share

Same proton count, so the same element and the same electron configuration — and therefore identical chemistry. Only mass-dependent properties differ, such as density and rate of diffusion.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $36.0$ — Weights Ignored
  The unweighted average, ignoring the abundances.
• C. $35.0$ — Single Isotope
  The mass of the more abundant isotope alone.
• D. $37.0$ — Single Isotope
  The mass of the less abundant isotope alone.
• E. $72.0$ — Not Averaged
  The sum of the two masses, not an average.

Common Mistake (⚠️):
Taking the plain average of $35$ and $37$ to get $36$. The abundances are the weights, and ignoring them discards the whole point of the calculation.

Takeaway (📌):
$A_r$ is a weighted mean of the isotope masses. It always lies between them, nearer the more abundant one.

Correct Answer: E. Because chlorine is both oxidised, from $0$ to $+1$, and reduced, from $0$ to $-1$

Fastest Approach (🚀):
$\text{Cl}: 0 \to -1$ (in $\text{NaCl}$) and $0 \to +1$ (in $\text{NaClO}$).
Both directions, one element.

Matches Option E.

Step-by-Step Breakdown:

1. Assign the chlorine states

Reactant: $\text{Cl}_2$ is an uncombined element, so chlorine is $0$.

In $\text{NaCl}$: sodium is $+1$, and the compound is neutral, so chlorine is $-1$.

In $\text{NaClO}$: sodium is $+1$ and oxygen is $-2$, so
$(+1)+x+(-2) = 0 \implies x = +1$

2. Read both changes

$0\to -1$: chlorine has gained an electron — reduced.
$0\to +1$: chlorine has lost an electron — oxidised.

Both happen to the same element, in the same reaction, starting from the same species.

3. That is the definition

Disproportionation is a redox reaction in which a single element is simultaneously oxidised and reduced. It requires an element in an intermediate oxidation state that can move in both directions — which elemental chlorine at $0$ can, having accessible states at $-1$ and $+1$.

4. Why the other options fail

Forming two products is not sufficient — most reactions do that without any redox at all. Sodium stays at $+1$ throughout, so Option C is simply false. And Option D omits the oxidation half, which is the half that makes it disproportionation rather than ordinary reduction.

5. Where else it appears

The decomposition of hydrogen peroxide, $2\text{H}_2\text{O}_2\rightarrow 2\text{H}_2\text{O}+\text{O}_2$, is the other standard example: oxygen moves from $-1$ to both $-2$ and $0$.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. Because two products are formed from one reactant — Insufficient Test
  Two products is common and says nothing about oxidation states.
• B. Because the reaction is reversible — Irrelevant
  Reversibility is unrelated to disproportionation.
• C. Because sodium is oxidised while chlorine is reduced — Factually Wrong
  Sodium remains at +1 throughout.
• D. Because chlorine is reduced from $+1$ to $-1$ — Half Described
  Omits the oxidation half, which is what makes it disproportionation.

Common Mistake (⚠️):
Calling any reaction with two products disproportionation. The test is on oxidation states — one element must move in both directions.

Takeaway (📌):
Disproportionation: one element simultaneously oxidised and reduced. Assign its state in every species and look for movement both ways.

Correct Answer: D. $28\ \text{g}$

Fastest Approach (🚀):
$n(\text{CaCO}_3) = \dfrac{50}{100} = 0.5$.
$1:1$, so $n(\text{CaO}) = 0.5$ and $m = 0.5\times 56 = 28\ \text{g}$.

Matches Option D.

Step-by-Step Breakdown:

1. Mass to moles

$n(\text{CaCO}_3) = \dfrac{50}{100} = 0.5\ \text{mol}$

2. Apply the ratio from the equation

The equation shows one $\text{CaCO}_3$ giving one $\text{CaO}$, a $1:1$ ratio:
$n(\text{CaO}) = 0.5\ \text{mol}$

3. Moles back to mass

$m = n\times M_r = 0.5\times 56 = 28\ \text{g}$

4. Why the mass falls

$50\ \text{g}$ in, $28\ \text{g}$ out. The missing $22\ \text{g}$ leaves as carbon dioxide, which is Option D and a check rather than an answer: $28+22 = 50\ \text{g}$, so mass is conserved exactly.

5. Why the ratio cannot be applied to grams

A $1:1$ ratio does not mean equal masses. It means equal numbers of particles, and $\text{CaCO}_3$ and $\text{CaO}$ particles have different masses. Option A applies the ratio to grams and is the standard error.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $50\ \text{g}$ — Ratio On Masses
  Applying the mole ratio to the masses directly.
• B. $22\ \text{g}$ — Wrong Product
  The mass of carbon dioxide released.
• C. $56\ \text{g}$ — One Mole Assumed
  Giving $M_r$ of $\text{CaO}$, which is the mass of one mole.
• E. $112\ \text{g}$ — Ratio Inverted
  Doubling instead of halving.

Common Mistake (⚠️):
Applying the $1:1$ ratio directly to the masses and answering $50\ \text{g}$. A balanced equation counts particles, so the ratio only works on moles.

Takeaway (📌):
Mass to moles, apply the equation's ratio, moles back to mass. Check that the masses conserve across the reaction.

Correct Answer: D. $2.0$

Fastest Approach (🚀):
$250\ \text{cm}^{3} = 0.25\ \text{dm}^{3}$.
$\dfrac{0.5}{0.25} = 2.0\ \text{mol dm}^{-3}$.

Matches Option D.

Step-by-Step Breakdown:

1. Convert the volume

$1\ \text{dm}^{3} = 1000\ \text{cm}^{3}$, so
$250\ \text{cm}^{3} = \dfrac{250}{1000} = 0.25\ \text{dm}^{3}$

2. Divide

$c = \dfrac{n}{V} = \dfrac{0.5}{0.25} = 2.0\ \text{mol dm}^{-3}$

3. Sanity-check

$0.5\ \text{mol}$ in a quarter of a litre is the same as $2\ \text{mol}$ in a full litre. Since the volume is less than $1\ \text{dm}^{3}$, the concentration must exceed the number of moles — so any answer below $0.5$ is wrong before it is computed.

4. Why the conversion is the whole difficulty

Dividing by $250$ instead of $0.25$ gives $0.002$, out by a factor of exactly $1000$. That is Option A, and it is by far the commonest error in solution calculations. Convert the volume before anything else touches it.

5. The other unit

Concentration in $\text{g dm}^{-3}$ is the mass rather than the moles per cubic decimetre. Multiply by $M_r$ to convert: here $2.0\times 58.5 = 117\ \text{g dm}^{-3}$.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $0.002$ — Volume Not Converted
  Dividing by $250$ rather than $0.25$.
• B. $125$ — Operation Error
  Multiplying by $250$.
• C. $0.5$ — No Division
  Quoting the number of moles as a concentration.
• E. $0.125$ — Inverted
  Multiplying by $0.25$ instead of dividing.

Common Mistake (⚠️):
Dividing by the volume in $\text{cm}^{3}$. The unit $\text{mol dm}^{-3}$ states the volume unit required, and ignoring it costs a factor of $1000$.

Takeaway (📌):
$c = \dfrac{n}{V}$ with $V$ in $\text{dm}^{3}$. Convert $\text{cm}^{3}$ by dividing by $1000$ before anything else.

Correct Answer: B. $2:1$, giving $2\text{Mg}+\text{O}_2\rightarrow 2\text{MgO}$

Fastest Approach (🚀):
$n(\text{Mg}) = \dfrac{4.8}{24} = 0.2$; $n(\text{O}_2) = \dfrac{3.2}{32} = 0.1$.
Ratio $2:1$.

Matches Option B.

Step-by-Step Breakdown:

1. Convert both masses to moles

$n(\text{Mg}) = \dfrac{4.8}{24} = 0.2\ \text{mol}$

Oxygen gas is $\text{O}_2$, so $M_r = 32$:
$n(\text{O}_2) = \dfrac{3.2}{32} = 0.1\ \text{mol}$

2. Simplify the ratio

$0.2 : 0.1 = 2 : 1$

3. Write the equation

$2\text{Mg}+\text{O}_2\rightarrow 2\text{MgO}$

Two magnesium atoms and two oxygen atoms on each side. ✓

4. Check against the mass data

$4.8+3.2 = 8.0\ \text{g}$ of magnesium oxide should form. From the equation, $0.2\ \text{mol}$ of $\text{MgO}$ with $M_r = 40$ gives $0.2\times 40 = 8.0\ \text{g}$ ✓ — mass conserved, which confirms the whole construction.

5. The diatomic trap

Using $A_r = 16$ for oxygen gives $0.2\ \text{mol}$ and a $1:1$ ratio, leading to the impossible formula $\text{MgO}_2$. Oxygen, hydrogen, nitrogen and the halogens all exist as diatomic molecules, and their $M_r$ is double the $A_r$.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $1:1$, giving $\text{Mg}+\text{O}_2\rightarrow\text{MgO}_2$ — Diatomic Ignored
  Treating oxygen as atomic, giving an impossible formula.
• C. $1:2$, giving $\text{Mg}+2\text{O}_2\rightarrow\text{MgO}_4$ — Ratio Inverted
  Ratio inverted and the formula impossible.
• D. $3:2$, giving $3\text{Mg}+2\text{O}_2\rightarrow 3\text{MgO}$ — Ratio Error
  Does not follow from the mole values.
• E. $2:1$, giving $2\text{Mg}+\text{O}_2\rightarrow\text{Mg}_2\text{O}$ — Formula Wrong
  Right ratio, but $\text{Mg}_2\text{O}$ is not charge-balanced.

Common Mistake (⚠️):
Using $A_r = 16$ for oxygen gas instead of $M_r = 32$. Oxygen is diatomic, and treating it as atomic doubles its mole count and wrecks the ratio.

Takeaway (📌):
Convert every mass to moles using the right formula mass — diatomic for the elemental gases — then simplify and check mass is conserved.

Correct Answer: D. $6\ \text{dm}^{3}$

Fastest Approach (🚀):
$0.25\times 24 = 6\ \text{dm}^{3}$.

Matches Option D.

Step-by-Step Breakdown:

1. Apply the molar volume

$V = n\times V_m = 0.25\times 24 = 6\ \text{dm}^{3}$

2. Why the gas does not matter

Avogadro's law: equal volumes of any gases at the same temperature and pressure contain equal numbers of molecules. A mole of carbon dioxide, hydrogen or ammonia all occupy $24\ \text{dm}^{3}$ at room conditions, despite molar masses of $44$, $2$ and $17$.

That works because gas particles are so far apart that their own size is negligible; the volume is set by the spacing, not by the molecules.

3. Check the direction

A quarter of a mole must occupy a quarter of the molar volume. Any answer larger than $24\ \text{dm}^{3}$ is inverted, which eliminates Option A on sight.

4. The two-step version

Given a mass instead, convert to moles first. $11\ \text{g}$ of carbon dioxide is $\dfrac{11}{44} = 0.25\ \text{mol}$, and then the same multiplication applies.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $96\ \text{dm}^{3}$ — Inverted
  Dividing rather than multiplying.
• B. $0.01\ \text{dm}^{3}$ — Arithmetic Error
  Dividing by $24$ twice, or a decimal slip.
• C. $24\ \text{dm}^{3}$ — Quantity Ignored
  The volume of one mole, ignoring the quantity.
• E. $12\ \text{dm}^{3}$ — Constant Error
  Using a molar volume of 48, or halving incorrectly.

Common Mistake (⚠️):
Dividing the molar volume by the number of moles, giving $96\ \text{dm}^{3}$. Fewer moles must mean a smaller volume.

Takeaway (📌):
$V = n\times 24\ \text{dm}^{3}$ at room conditions, for any gas. Equal volumes hold equal numbers of molecules.

Correct Answer: B. $75\%$

Fastest Approach (🚀):
$\dfrac{21}{28} = \dfrac{3}{4} = 75\%$.

Matches Option B.

Step-by-Step Breakdown:

1. The definition

$\text{percentage yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}}\times 100$

2. Substitute

$\dfrac{21}{28}\times 100 = 0.75\times 100 = 75\%$

3. The bound

A percentage yield cannot exceed $100\%$ — that would mean creating matter. Option B is $\dfrac{28}{21}$, the fraction inverted, and can be rejected without arithmetic. A yield above $100\%$ in a real experiment means the product was impure or still wet.

4. Why yields fall short

Some product is lost in transferring between vessels or in filtering. Some reactions are reversible and never go to completion. Side reactions consume reactants into other products. None of these is a mistake — they are why industrial processes are judged on yield at all.

5. What must be in the same units

Both yields must be expressed the same way — both masses or both moles. Mixing a mass with a mole count gives a meaningless ratio, which is Option D's origin.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $133\%$ — Inverted
  The fraction inverted — impossible.
• C. $25\%$ — Complement
  The percentage lost rather than obtained.
• D. $7\%$ — Difference Not Ratio
  Computing $28-21$ as a percentage.
• E. $88\%$ — Arithmetic Error
  Using $\dfrac{21}{24}$ or another wrong denominator.

Common Mistake (⚠️):
Inverting the fraction to get $133\%$. Any yield above $100\%$ is impossible and should be caught before the arithmetic is finished.

Takeaway (📌):
Actual over theoretical, times $100$. Never above $100\%$, and both yields in the same units.

Correct Answer: C. $0.125\ \text{mol dm}^{-3}$

Fastest Approach (🚀):
$n = 0.025\times 0.1 = 0.0025\ \text{mol}$, and the ratio is $1:1$.
$c = \dfrac{0.0025}{0.020} = 0.125\ \text{mol dm}^{-3}$.

Matches Option C.

Step-by-Step Breakdown:

1. Start with the solution you know completely

Sodium hydroxide has both a volume and a concentration:
$n = c\times V = 0.100\times\dfrac{25.0}{1000} = 2.50\times 10^{-3}\ \text{mol}$

2. Apply the reacting ratio

The equation is $1:1$, so the acid also supplied $2.50\times 10^{-3}\ \text{mol}$.

3. Divide by the acid's volume

$c(\text{HCl}) = \dfrac{2.50\times 10^{-3}}{\dfrac{20.0}{1000}} = \dfrac{2.50\times 10^{-3}}{0.0200} = 0.125\ \text{mol dm}^{-3}$

4. Check it is plausible

The same number of moles was delivered in a smaller volume of acid, so the acid must be more concentrated than $0.100\ \text{mol dm}^{-3}$. That rules out Options A, C and E immediately, and is worth the two seconds.

5. When the ratio is not 1:1

With sulfuric acid, $\text{H}_2\text{SO}_4+2\text{NaOH}\rightarrow\text{Na}_2\text{SO}_4+2\text{H}_2\text{O}$, only half as many moles of acid are needed. Reading the ratio off the equation rather than assuming $1:1$ is what the balanced equation is given for.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $0.080\ \text{mol dm}^{-3}$ — Inverted
  The final division inverted.
• B. $0.100\ \text{mol dm}^{-3}$ — Volumes Ignored
  Assuming equal concentrations because the ratio is 1:1.
• D. $2.00\ \text{mol dm}^{-3}$ — Unit Error
  Volumes left in $\text{cm}^{3}$, out by a factor of 1000 somewhere.
• E. $0.050\ \text{mol dm}^{-3}$ — Ratio Error
  Halving, as though the ratio were 1:2.

Common Mistake (⚠️):
Inverting the final division, or assuming a $1:1$ ratio without reading the equation. Check whether the answer should be larger or smaller than the known concentration first.

Takeaway (📌):
Moles from the fully known solution, the equation's ratio, then divide by the other volume in $\text{dm}^{3}$.