Which concentration term is independent of temperature?

All three are independent of temperature

Mole Concept, Stoichiometry and Concentration Terms

Every numerical in physical chemistry — thermodynamics, equilibrium, electrochemistry, colligative properties — eventually reduces to counting moles. That makes this the highest-leverage chapter in the whole Chemistry syllabus: a candidate who is fluent here converts arithmetic into marks everywhere else, while one who is shaky loses marks in four separate units without ever realising the mole concept was the culprit. The core idea is deceptively simple. A balanced equation is a recipe in particles, not in grams, so the only legitimate currency for comparing reactants is the mole. Everything else in this chapter — limiting reagent, percentage yield, empirical formula, molarity versus molality — is a consequence of taking that one sentence seriously.

The Mole as a Counting Unit

A mole is 6.022 × 10²³ particles — nothing more mysterious than a dozen, only vastly larger. Its usefulness is that the mass of one mole in grams is numerically equal to the atomic or molecular mass in atomic mass units, so a balance in the laboratory becomes a particle counter. From 2019 the mole is defined as exactly 6.02214076 × 10²³ entities rather than by reference to 12 g of carbon-12, but for exam arithmetic 6.022 × 10²³ is what you use.

The four bridges to and from moles
You are given	Divide or multiply by	Worked example
Mass of a substance	Molar mass M — n = m/M	9 g of H₂O → 9/18 = 0.5 mol
Number of particles	Avogadro number Nₐ — n = N/Nₐ	3.011 × 10²³ atoms → 0.5 mol
Volume of a gas at STP	22.4 L per mole (22.7 L at 1 bar, 273 K)	11.2 L of O₂ at STP → 0.5 mol
Volume and molarity of a solution	n = M × V(litres)	250 mL of 2 M NaOH → 0.5 mol

⚠️ One mole of a molecule is not one mole of atoms

Read the question for what is being counted. One mole of H₂SO₄ contains 1 mol of sulphur atoms, 2 mol of hydrogen atoms, 4 mol of oxygen atoms and therefore 7 mol of atoms in total — that is 7 × 6.022 × 10²³ atoms, not one Avogadro number. Likewise 1 mol of CO₂ has 3 mol of atoms, and 1 mol of glucose C₆H₁₂O₆ has 24 mol of atoms. Examiners set the trap by asking for "the number of atoms in 1 mole of ..." and offering 6.022 × 10²³ as the first option, which is the answer to a question they did not ask.

Two derived ideas follow immediately. Empirical formula is the simplest whole-number ratio of atoms, obtained by dividing each percentage by the atomic mass and then by the smallest quotient. Molecular formula is the empirical formula multiplied by n = molecular mass / empirical formula mass. Acetic acid CH₃COOH has empirical formula CH₂O (mass 30) and molecular mass 60, so n = 2 — the same empirical formula as formaldehyde HCHO and as glucose, which is precisely why an empirical formula alone can never identify a compound.

Limiting Reagent and Percentage Yield

When two reactants are both given by mass, one of them runs out first and stops the reaction. That one is the limiting reagent, and the entire yield is calculated from it; the other is in excess and part of it survives untouched. The procedure never varies: convert both masses to moles, divide each mole count by its stoichiometric coefficient, and the smaller quotient identifies the limiting reagent.

Balance the equation first — an unbalanced equation gives wrong coefficients and therefore the wrong limiting reagent.
Convert every given mass or volume to moles.
Divide each mole count by its coefficient in the balanced equation.
The smallest ratio is the limiting reagent. Compute product moles from it alone.
Convert product moles back to grams or litres as the question demands.

🎯 Why divide by the coefficient rather than compare moles directly

Consider N₂ + 3H₂ → 2NH₃ with 1 mol N₂ and 2 mol H₂. Hydrogen has more moles, so a careless candidate calls nitrogen limiting. But the recipe demands three hydrogens per nitrogen: 1/1 = 1 for N₂ against 2/3 = 0.67 for H₂, so hydrogen is limiting and only 1.33 mol of NH₃ can form. Dividing by the coefficient converts raw moles into "how many complete reaction packets can I assemble", which is the only fair comparison. Comparing moles directly is the single most common error in this chapter.

The mass calculated from the limiting reagent is the theoretical yield. What is actually isolated is the actual yield, and percentage yield = actual/theoretical × 100. Note the asymmetry that questions exploit: percentage yield can never exceed 100 % for a correctly identified limiting reagent, so an answer above 100 % is a signal that you picked the wrong limiting reagent or used an unbalanced equation.

🧠 Gay-Lussac shortcut for gases at the same T and P

For reactions between gases measured at the same temperature and pressure, volume ratios equal mole ratios, so you can work directly in litres and skip the conversion entirely. In 2CO + O₂ → 2CO₂, mixing 10 L of CO with 10 L of O₂ consumes only 5 L of O₂, leaves 5 L of O₂ unreacted and produces 10 L of CO₂ — total volume falls from 20 L to 15 L. This saves genuine seconds in the exam, but it applies only to gases at identical T and P; the moment a liquid or solid enters the equation you must return to moles.

Concentration Terms — and Which One Is Temperature-Safe

The concentration terms the paper actually uses
Term	Definition	Depends on temperature?
Molarity M	moles of solute per litre of solution	Yes — volume expands on heating, so M falls
Molality m	moles of solute per kilogram of solvent	No — mass does not change with T
Mole fraction x	moles of one component / total moles	No — a pure ratio of mole counts
Mass percentage	mass of solute / mass of solution × 100	No
ppm	mass of solute / mass of solution × 10⁶	No
Normality N	gram equivalents per litre = M × n-factor	Yes, for the same reason as molarity

🎯 Why colligative properties are defined through molality

Elevation of boiling point and depression of freezing point are measured by heating or cooling the solution — precisely the operations that change its volume. If they were defined through molarity, the concentration would drift during the very measurement being made. Molality uses the solvent mass, which no amount of heating alters, so the number stays fixed while the thermometer moves. This is also why molality is the natural unit whenever temperature is a variable, and molarity the natural unit for titrations, which are done at a fixed room temperature with volumetric glassware.

Two relations are worth memorising because they appear as one-step questions. Dilution: M₁V₁ = M₂V₂, since adding solvent changes volume but not the number of moles of solute. Mixing two solutions of the same solute: M_final = (M₁V₁ + M₂V₂)/(V₁ + V₂) — a weighted average, so the answer must always lie between M₁ and M₂. If your mixed molarity comes out above the higher of the two, you have added where you should have averaged.

To convert molality to molarity you need the density of the solution, because the bridge between "per kg of solvent" and "per litre of solution" is mass ÷ density. Take 1 kg of solvent containing m moles of solute of molar mass M_s: the solution mass is (1000 + m·M_s) g, its volume is that divided by the density d in g/mL, and molarity is m divided by that volume in litres. Questions that give you density are almost always asking for this conversion.

Equivalent Concept and Redox Balancing

The equivalent concept exists so that titrations can be solved without balancing an equation. At the end point, gram equivalents of acid = gram equivalents of base, that is N₁V₁ = N₂V₂. The link to molarity is N = M × n-factor, where the n-factor counts how many units of reactivity one formula unit supplies.

Finding the n-factor
Species type	n-factor is	Example
Acid	Number of replaceable H⁺	H₂SO₄ → 2, HCl → 1, H₃PO₃ → 2 (only two are acidic)
Base	Number of replaceable OH⁻	NaOH → 1, Ca(OH)₂ → 2
Salt in a precipitation	Total charge on cation or anion	Na₂CO₃ → 2, AlCl₃ → 3
Oxidising or reducing agent	Electrons gained or lost per formula unit	KMnO₄ in acid → 5, in neutral → 3, in base → 1

⚠️ The n-factor of KMnO₄ depends on the medium, not on KMnO₄

This is the most frequently examined subtlety in the equivalent concept. In acidic medium Mn goes from +7 to +2, a gain of 5 electrons. In neutral or faintly alkaline medium it stops at MnO₂, Mn +7 → +4, a gain of 3. In strongly alkaline medium it becomes manganate MnO₄²⁻, Mn +7 → +6, a gain of just 1. So the equivalent mass of the same crystal of KMnO₄ is 158/5 = 31.6 in acid but 158 in strong base. Any question that mentions the medium is telling you which n-factor to use, and any candidate who memorises "KMnO₄ = 5" without the condition will get one in three of these wrong.

For balancing redox equations the ion-electron (half-reaction) method is reliable and mechanical. Split the reaction into oxidation and reduction halves; balance all atoms except O and H; balance O by adding H₂O; balance H by adding H⁺ (in acid) or by adding H₂O to the side deficient in H and OH⁻ to the other (in base); balance charge with electrons; multiply the halves so the electrons cancel; add. The final check is always the same two conservations — atoms balance and total charge balances — and if either fails, the answer is wrong regardless of how plausible it looks.

ℹ️ Oxidation number rules that settle most cases

Free elements are 0; oxygen is −2 except in peroxides (−1), superoxides (−½) and OF₂ (+2); hydrogen is +1 except in metal hydrides (−1); fluorine is always −1; the sum over a neutral molecule is 0 and over an ion equals its charge. Fractional values are legitimate and simply mean the atoms are not all equivalent — in Fe₃O₄ the average is +8/3 because the solid contains one Fe²⁺ and two Fe³⁺, and in C₃O₂ carbon averages +4/3. An answer of −½ or +8/3 is not an arithmetic slip.

Key takeaways

A mole is 6.022 × 10²³ particles; reach it from mass (÷M), particles (÷Nₐ), gas volume (÷22.4 L at STP) or M × V.
One mole of H₂SO₄ holds 7 mol of atoms — always read whether molecules, atoms or ions are being counted.
Limiting reagent: divide moles by coefficient, not by comparing moles. Smallest quotient limits the yield.
Percentage yield above 100 % means a wrong limiting reagent or an unbalanced equation — treat it as an error flag.
For gases at the same T and P, volume ratios are mole ratios, so you can work directly in litres.
Molarity and normality change with temperature; molality, mole fraction, mass percentage and ppm do not.
Mixing the same solute gives a weighted average, so M_final always lies between M₁ and M₂.
The n-factor of KMnO₄ is 5 in acid, 3 in neutral and 1 in strong base — the medium decides, not the reagent.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The total number of atoms present in 1 mole of H₂SO₄ is:
1. 6.022 × 10²³
2. 4 × 6.022 × 10²³
3. 7 × 6.022 × 10²³
4. 2 × 6.022 × 10²³
Show answer
Answer: C — 7 × 6.022 × 10²³
One formula unit contains 2 H, 1 S and 4 O — seven atoms in all — so one mole of the compound contains 7 mol of atoms, that is 7 × 6.022 × 10²³. The first option is offered because candidates reflexively answer with one Avogadro number whenever they read "1 mole", without checking whether the question counted molecules or atoms.
1 mol of N₂ is mixed with 2 mol of H₂ to form ammonia. The limiting reagent and the moles of NH₃ formed are:
1. N₂, 2 mol
2. H₂, 1.33 mol
3. H₂, 2 mol
4. N₂, 1.33 mol
Show answer
Answer: B — H₂, 1.33 mol
N₂ + 3H₂ → 2NH₃. Dividing by coefficients gives 1/1 = 1 for N₂ and 2/3 = 0.67 for H₂, so hydrogen is limiting. Two moles of H₂ produce 2 × (2/3) = 1.33 mol of NH₃, leaving 0.67 mol of N₂ unreacted. The tempting "N₂, 2 mol" comes from comparing raw mole counts, which wrongly makes the reagent with fewer moles look limiting even though the recipe needs three times as much hydrogen.
Which concentration term is independent of temperature?
1. Molarity
2. Normality
3. Molality
4. All three are independent of temperature
Show answer
Answer: C — Molality
Molality is moles of solute per kilogram of solvent, and mass does not change when the solution is heated, so the value is fixed. Molarity and normality are both defined per litre of solution, and volume expands on heating, so both fall as temperature rises. This is exactly why colligative properties — measured while heating or cooling — are defined through molality rather than molarity.
The n-factor of KMnO₄ acting as an oxidising agent in strongly alkaline medium is:
1. 5
2. 3
3. 1
4. 7
Show answer
Answer: C — 1
In strong base MnO₄⁻ is reduced only to manganate MnO₄²⁻, so Mn goes from +7 to +6 and gains a single electron, giving n = 1. The answer 5 belongs to acidic medium (+7 → +2) and 3 to neutral or faintly alkaline medium (+7 → MnO₂, +4). Since the same reagent has three different equivalent masses, the medium named in the question is never decorative.
The average oxidation number of Fe in Fe₃O₄ is:
1. +2
2. +3
3. +8/3
4. +4
Show answer
Answer: C — +8/3
Four oxygens contribute 4 × (−2) = −8, so three irons must total +8 and the average is +8/3. The fraction is not an error: Fe₃O₄ is a mixed oxide containing one Fe²⁺ and two Fe³⁺ ions, and the average of one +2 with two +3 is exactly 8/3. Options +2 and +3 are the individual values, correct for FeO and Fe₂O₃ separately but not for the average asked about here.
10 L of CO is mixed with 10 L of O₂ at the same T and P and burnt completely. The total final volume, measured at the same T and P, is:
1. 20 L
2. 15 L
3. 10 L
4. 5 L
Show answer
Answer: B — 15 L
2CO + O₂ → 2CO₂, so 10 L of CO needs only 5 L of O₂ and yields 10 L of CO₂, leaving 5 L of O₂ unused: total 15 L. The answer 10 L is the trap for candidates who forget that the leftover excess oxygen is still a gas occupying volume, and 20 L assumes volume is conserved, which holds only when the mole counts on the two sides happen to be equal.
A compound contains 40 % C, 6.7 % H and 53.3 % O by mass and has molecular mass 180. Its molecular formula is:
1. CH₂O
2. C₂H₄O₂
3. C₆H₁₂O₆
4. C₃H₆O₃
Show answer
Answer: C — C₆H₁₂O₆
Dividing by atomic masses gives 40/12 = 3.33, 6.7/1 = 6.7, 53.3/16 = 3.33; dividing by the smallest gives C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O with mass 30. Then n = 180/30 = 6 and the molecular formula is C₆H₁₂O₆. Option CH₂O is the empirical formula itself, which is the standard trap — the same empirical formula also fits formaldehyde and acetic acid, so it can never be the answer once a molecular mass is supplied.
100 mL of 0.5 M HCl is mixed with 400 mL of 0.1 M HCl. The molarity of the resulting solution is:
1. 0.60 M
2. 0.30 M
3. 0.18 M
4. 0.09 M
Show answer
Answer: C — 0.18 M
Total moles = 0.5 × 0.100 + 0.1 × 0.400 = 0.050 + 0.040 = 0.090 mol in 0.500 L, giving 0.18 M. Because mixing is a weighted average, the answer must lie between 0.1 M and 0.5 M — which immediately rules out 0.60 M (obtained by adding the molarities) and 0.09 M (the mole count mistaken for a molarity).