Chemistry · Some Basic Concepts of Chemistry · Practice set

Advanced Stoichiometry and Concentration · Practice with Solutions

Twenty-four applied problems on the mole concept topics that decide the physical chemistry paper: back titration, double titration, oleum, volume strength and eudiometry, each fully solved. Free to read or download.

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In short

Stoichiometry questions in JEE are rarely about the mole itself. They are about applications: back titration, double titration with an indicator change, the strength of oleum, the volume strength of hydrogen peroxide, and eudiometry. This set works through twenty-four such problems, with the full solution shown for each.

Contents

·How to Use This Set
·Concentration Terms
·Applied Stoichiometry
1Answer Key
2Solutions: Concentration Terms
3Solutions: Applied Stoichiometry

How to Use This Set

Section A is concentration terms: 8 single-correct and 2 numerical.
Section B is applied stoichiometry: 8 single-correct, 2 multiple-correct and 4 numerical.
Take molar volume at STP as 22.4 L and use the atomic masses given in each question.
Every number here was computed and checked before printing, including a sanity check that no purity came out above 100%.

Think it through

The four ideas that carry this whole topic

Mole ratio, not mass ratio. Every stoichiometry error traces back to comparing masses.
Oxygen by difference. In combustion analysis you never measure oxygen directly.
Indirect measurement. Back titration and double titration both find a quantity by measuring what is left over, not what reacted.
Volume or mass? Any concentration term defined with a volume drifts with temperature. Terms built from masses do not.

Trap alert

Three sanity checks that catch most mistakes

A purity or a yield above 100% means a coefficient was dropped.
A diluted solution can never be stronger than the stock it came from.
An average molar mass must lie between the molar masses of the components.
Run these before you move on. They cost five seconds and save four marks.

SECTION A

Concentration Terms

Molarity, molality, mole fraction, normality, ppm · Q1 to Q10

A solution is 98% H₂SO₄ by mass and its density is 1.84 g/mL. Its molarity is:

(A)18.4 M

(B)9.2 M

(D)36.8 M

A 1.00 molar aqueous solution has density 1.04 g/mL. The solute has molar mass 60 g/mol. The molality is:

(A)1.00 m

(B)0.96 m

(D)1.04 m

The molarity of pure water at 25 °C, taking its density as 1.00 g/mL, is:

(A)1.00 M

(B)55.6 M

(D)100 M

200 mL of 0.50 M NaCl is mixed with 300 mL of 0.20 M NaCl. The molarity of the final solution is:

(A)0.35 M

(B)0.25 M

(D)0.32 M

A 0.50 molal aqueous solution has a solute mole fraction of about:

(A)0.50

(B)0.0089

(D)0.0050

The normality of a 0.10 M H₂SO₄ solution used in complete neutralisation is:

(A)0.05 N

(B)0.10 N

(D)0.20 N

Which concentration term is independent of temperature?

(A)Molarity

(B)Normality

(C)Molality

(D)Formality

A sample of water contains 0.001 M MgSO₄. Its hardness expressed in ppm of CaCO₃ is:

(A)100 ppm

(B)10 ppm

(D)120 ppm

Numerical Answer Type

Calculate the mass of glucose (molar mass 180 g/mol) needed to prepare 250 mL of a 0.20 M aqueous solution. Give the answer in gram.

NUMERICAL ANSWER · Round off to TWO decimal places.

36.0 g of water is mixed with 46.0 g of ethanol (molar mass 46 g/mol). Find the mole fraction of ethanol.

NUMERICAL ANSWER · Round off to TWO decimal places.

SECTION B

Applied Stoichiometry

Limiting reagent, purity, combustion, titrations, oleum, eudiometry · Q11 to Q24

5.60 g of iron is heated with 4.00 g of sulfur to form FeS. Which is correct? (Fe = 56, S = 32)

(A)Sulfur is limiting; 11.0 g FeS forms

(B)Iron is limiting; 0.80 g FeS forms

(C)Both are fully consumed; 9.60 g FeS forms

(D)Iron is limiting; 8.80 g FeS forms

10.0 g of limestone that is 80% CaCO₃ by mass is heated. The reaction gives 85% of the theoretical yield of CO₂. The volume of CO₂ at STP is:

(A)2.24 L

(B)1.52 L

(D)1.90 L

0.240 g of an organic compound on complete combustion gives 0.352 g CO₂ and 0.144 g H₂O. Its molar mass is 60 g/mol. Its molecular formula is:

(A)CH₂O

(B)C₃H₈O

(C)C₂H₆O

(D)C₂H₄O₂

2.00 g of impure CaCO₃ is dissolved in 50.0 mL of 1.00 M HCl. The unreacted acid needs 20.0 mL of 1.00 M NaOH to neutralise. The purity of the sample is:

(A)50%

(B)85%

(D)60%

A 25.0 mL mixture of NaOH and Na₂CO₃ is titrated with 0.100 M HCl. The phenolphthalein end point comes at 15.0 mL, and methyl orange needs a further 5.0 mL. The mass of Na₂CO₃ present is:

(A)0.053 g

(B)0.106 g

(D)0.021 g

A sample of H₂O₂ is labelled '20 volume'. Its molarity is:

(A)2.00 M

(B)1.79 M

(D)20.0 M

Oleum is labelled 109%. The percentage of free SO₃ in it is about:

(A)9%

(B)40%

(D)18%

10 mL of a gaseous hydrocarbon is burnt in excess oxygen. 40 mL of CO₂ forms, and cooling the products to room temperature causes a contraction of 50 mL due to water. The hydrocarbon is:

(A)C₄H₈

(B)C₄H₁₀

(C)C₂H₆

(D)C₅H₁₀

Multiple Correct Answer Type

A gaseous mixture contains CH₄ and C₂H₆. Which statements are correct? One or more options may be right.

(A) If mixed 40:60 by moles, the average molar mass is 24.4
(B) If mixed 40:60 by mass, the average molar mass is 22.2
(C) Mixing by mass and by moles in the same ratio gives the same average molar mass
(D) The average molar mass always lies between 16 and 30

ONE OR MORE OPTIONS MAY BE CORRECT

For a 1.0 M aqueous solution of a solute, which statements are correct? One or more options may be right.

(A) Its molarity falls if the solution is warmed
(B) Its molality falls if the solution is warmed
(C) Its molality can be found from the molarity only if the density is known
(D) Its mole fraction is unaffected by temperature

ONE OR MORE OPTIONS MAY BE CORRECT

Numerical Answer Type

100.0 g of KClO₃ is heated until it decomposes completely to KCl and O₂. Find the volume of O₂ released at STP, in litre. (K = 39, Cl = 35.5, O = 16)

NUMERICAL ANSWER · Round off to TWO decimal places.

A 9.20 g mixture of CaCO₃ and MgCO₃ is heated strongly until decomposition is complete. It gives 2.24 L of CO₂ at STP. Find the mass percentage of CaCO₃ in the mixture. (Ca = 40, Mg = 24, C = 12, O = 16)

NUMERICAL ANSWER · Round off to ONE decimal place.

What volume, in millilitre, of 18.0 M concentrated H₂SO₄ is needed to prepare 500 mL of 0.500 M solution?

NUMERICAL ANSWER · Round off to TWO decimal places.

A '20 volume' sample of H₂O₂ is diluted so that 100 mL of it becomes 500 mL. Find the volume strength of the diluted solution.

NUMERICAL ANSWER · Round off to the nearest integer.

Answer Key

Mark your paper first, then read every solution, including the ones you got right.

99.00

100.33

11D

12B

13D

14C

15A

16A

17C

18B

19A, B, D

20A, C, D

2127.43

2254.3

2313.89

244

Solutions: Concentration Terms

Each solution names the idea, shows the working, then names the exact mistake behind each wrong option.

1Answer AMedium · Concentration interconversion

Core principleTake exactly 1 litre. Find its mass from the density, then the mass of solute from the percentage.

Math1 L weighs 1840 g. Solute = 98% of that = 1803 g. Moles = 1803/98 = 18.4. So M = 18.4 M. The shortcut is M = 1000d×w/(100×M_w).

TrapUsing 98 g of acid per 100 g of solution but forgetting to scale to 1 litre gives 1.0 M. The density is what converts volume to mass, so it can never be left out.

2Answer CHard · Molarity to molality

Core principleMolarity is per litre of solution. Molality is per kilogram of solvent. So you must subtract the solute mass.

Math1 L of solution = 1040 g. It holds 1 mol = 60 g of solute. So solvent = 980 g = 0.980 kg. Molality = 1.00/0.980 = 1.02 m.

Trap(A) assumes the two are equal, which is only true for very dilute solutions. Molality is always larger than molarity when the solute is denser than the solvent, and density is the only bridge between them.

3Answer BMedium · Concentration terms

Core principleTreat water as both the solvent and the solute. One litre of water is 1000 g.

MathMoles in 1 L = 1000/18 = 55.6. So the molarity is 55.6 M.

TrapAnswering 18 M inverts the calculation. This value is worth memorising, because it is the reason the solvent concentration is treated as effectively constant in dilute-solution equilibria.

4Answer DEasy · Mixing solutions

Core principleMoles add. Volumes add. Divide the total moles by the total volume.

MathMoles = (0.200)(0.50) + (0.300)(0.20) = 0.100 + 0.060 = 0.160. Volume = 0.500 L. M = 0.160/0.500 = 0.32 M.

Trap(A) is the simple average of 0.50 and 0.20, which ignores that the volumes are unequal. Averaging concentrations is only valid when equal volumes are mixed.

5Answer BHard · Molality to mole fraction

Core principleMolality fixes the solvent at exactly 1 kg. Convert that to moles of solvent, then form the fraction.

Math1 kg of water = 1000/18 = 55.6 mol. x_solute = 0.50/(0.50 + 55.6) = 0.0089.

Trap(B) treats molality as if it were already a fraction. Note how small the mole fraction is: even a 0.5 molal solution is overwhelmingly solvent by particle count.

6Answer DEasy · Normality and n-factor

Core principleNormality = Molarity × n-factor. For an acid the n-factor is the number of replaceable H⁺ ions.

MathH₂SO₄ gives 2 H⁺, so n = 2 and N = 0.10 × 2 = 0.20 N.

Trap(A) divides instead of multiplying. Normality is never smaller than molarity, because the n-factor is at least 1.

7Answer CMedium · Temperature dependence

Core principleAny term defined using a volume changes with temperature, because the volume expands on heating.

MathMolality is moles per kilogram of solvent. Both are masses, and mass does not change with temperature. So molality is temperature independent, as are mole fraction and mass percentage.

TrapMolarity, normality and formality all use litres of solution, so all three drift as the temperature changes. This is exactly why molality is used in colligative property work.

8Answer AHard · Hardness of water

Core principleHardness is always reported as the equivalent mass of CaCO₃, whatever salt is actually present.

Math0.001 mol per litre × 100 g/mol (CaCO₃) = 0.1 g/L = 100 mg/L = 100 ppm.

Trap(D) uses the molar mass of MgSO₄ (120). The whole convention is that you convert to CaCO₃, so the molar mass of the salt actually present never enters the final number.

9Answer 9.00Easy · Molarity

Core principleMoles = molarity × volume in litres. Then convert moles to mass.

Mathn = 0.20 × 0.250 = 0.050 mol. Mass = 0.050 × 180 = 9.00 g.

TrapUsing 250 instead of 0.250 gives 9000 g. Always convert millilitres to litres before multiplying by molarity.

10Answer 0.33Medium · Mole fraction

Core principleConvert both masses to moles first. Mole fraction compares particle counts, never masses.

Mathn(water) = 36/18 = 2.00. n(ethanol) = 46/46 = 1.00. x(ethanol) = 1.00/3.00 = 0.33.

TrapComparing the masses directly gives 46/82 = 0.56. Mole fraction is about numbers of molecules, and water molecules are much lighter, so there are twice as many of them here.

Solutions: Applied Stoichiometry

11Answer DMedium · Limiting reagent

Core principleConvert both masses to moles, then compare against the 1:1 ratio in the equation.

Mathn(Fe) = 5.60/56 = 0.100. n(S) = 4.00/32 = 0.125. Fe is smaller, so Fe is limiting. FeS formed = 0.100 × 88 = 8.80 g, with 0.80 g of sulfur left over.

TrapComparing the raw masses suggests iron is in excess because 5.60 > 4.00. Limiting reagent is decided by moles against the stoichiometric ratio, never by mass.

12Answer BHard · Purity and yield

Core principleApply purity first to get the reacting mass, then stoichiometry, then the yield factor at the end.

MathPure CaCO₃ = 8.00 g = 0.0800 mol. Theoretical CO₂ = 0.0800 mol. Actual = 0.0800 × 0.85 = 0.0680 mol. Volume = 0.0680 × 22.4 = 1.52 L.

Trap(A) ignores both corrections. (B) applies only the purity. The two factors multiply, so 0.80 × 0.85 = 0.68 of the ideal. Applying only one of them is the most common single error in this topic.

13Answer DHard · Combustion analysis

Core principleGet carbon from the CO₂ and hydrogen from the H₂O. Find oxygen by difference, never by measurement.

MathC = 0.352 × 12/44 = 0.096 g. H = 0.144 × 2/18 = 0.016 g. O = 0.240 − 0.096 − 0.016 = 0.128 g. Mole ratio = 0.008 : 0.016 : 0.008 = 1 : 2 : 1, so the empirical formula is CH₂O (30). 60/30 = 2, giving C₂H₄O₂.

Trap(A) stops at the empirical formula. Oxygen must come from the mass difference, because the oxygen in the products comes partly from the burning air, not only from the compound.

14Answer CHard · Back titration

Core principleYou cannot titrate a solid directly. So use excess acid, then find how much acid survived, and subtract.

MathTotal HCl = 0.0500 mol. Excess = 0.0200 mol. Reacted with CaCO₃ = 0.0300 mol. Since CaCO₃ + 2HCl, n(CaCO₃) = 0.0150 mol = 1.50 g. Purity = 1.50/2.00 = 75%.

TrapForgetting the 1:2 ratio gives 0.0300 mol = 3.00 g, which exceeds the sample mass and should be an instant warning. Any purity above 100% means a stoichiometric coefficient was dropped.

15Answer AHard · Double titration

Core principlePhenolphthalein catches all the NaOH plus half the carbonate. Methyl orange then catches the second half of the carbonate alone. So the second volume measures carbonate directly.

MathV₂ = 5.0 mL corresponds to half the carbonate, and by symmetry the other half also took 5.0 mL. n(Na₂CO₃) = (5.0)(0.100)/1000 = 5.0 × 10⁻⁴ mol. Mass = 5.0 × 10⁻⁴ × 106 = 0.053 g.

Trap(A) uses the full 10 mL of carbonate titration and doubles the answer. The classic trap is forgetting that V₁ already contains half the carbonate, so NaOH = (V₁ − V₂), not V₁.

16Answer AHard · Volume strength

Core principle'20 volume' means 1 litre of the solution releases 20 litres of O₂ at STP on full decomposition.

Math2H₂O₂ gives O₂, so 20 L of O₂ needs 2(20/22.4) = 1.786 mol of H₂O₂ per litre. So M = 20/11.2 = 1.79 M.

Trap(D) reads the label as a concentration. The useful shortcuts are M = volume strength/11.2 and strength in g/L = volume strength × 34/11.2, which gives 60.7 g/L here.

17Answer CHard · Oleum labelling

Core principleThe label means that 109 g of oleum yields 109 g of H₂SO₄ when enough water is added. So the extra 9 g is the water absorbed.

Math9 g of water = 0.50 mol, which reacts with 0.50 mol of free SO₃ = 40 g. So free SO₃ = 40/109 = 36.7%.

Trap(A) reads the excess 9 as the percentage of SO₃ directly. The 9 g is the mass of water, and each mole of water consumes a whole mole of the much heavier SO₃.

18Answer BHard · Eudiometry

Core principleFor gases at the same temperature and pressure, volume ratios equal mole ratios. So read the subscripts straight off the volumes.

MathCO₂ is 40 mL from 10 mL, so x = 4. Water vapour is 50 mL from 10 mL, so y/2 = 5, giving y = 10. The hydrocarbon is C₄H₁₀.

Trap(A) forgets that the water volume gives y/2, not y, because each molecule of C_xH_y makes y/2 molecules of water. Halving the hydrogen count is the standard eudiometry slip.

19Answer A, B, DHard · Average molar mass

Core principleAverage molar mass is total mass divided by total moles. So the answer depends on whether the given ratio is a ratio of moles or of masses.

MathBy moles: M = 0.4(16) + 0.6(30) = 24.4. By mass, take 40 g and 60 g: n = 40/16 + 60/30 = 2.5 + 2.0 = 4.5 mol for 100 g, so M = 100/4.5 = 22.2. Any mixture must lie between the two pure values.

Trap(C) is the trap, and the two numbers above disprove it. A mass ratio favours the lighter component in particle count, so it always pulls the average molar mass down.

20Answer A, C, DHard · Temperature and concentration

Core principleAsk one question of each term: does its definition contain a volume? If yes, it changes with temperature. If it contains only masses or mole counts, it does not.

MathMolarity is per litre of solution, and warming expands the solution, so the molarity falls. Molality and mole fraction use only masses and mole counts, so both are unchanged. Converting between molarity and molality needs the mass of a known volume, which is the density.

Trap(B) is the trap. Warming does not evaporate or create solvent, so the kilogram of solvent in the molality definition is untouched. Only volume-based terms drift.

21Answer 27.43Medium · Stoichiometry, POAC

Core principleBalance the equation, or simply conserve oxygen atoms. Each KClO₃ carries three O atoms, and each O₂ takes two.

MathM(KClO₃) = 122.5, so n = 0.8163 mol. 2KClO₃ gives 3O₂, so n(O₂) = 1.5(0.8163) = 1.2245 mol. V = 1.2245 × 22.4 = 27.43 L.

TrapUsing a 1:1 ratio gives 18.29 L. Conserving atoms rather than trusting a remembered coefficient is the safer route, and it is what POAC formalises.

22Answer 54.3Hard · Mixture analysis

Core principleTwo unknowns need two equations. Use the total mass and the total moles of CO₂, since each carbonate gives exactly one CO₂.

MathLet x and y be the moles. x + y = 2.24/22.4 = 0.100, and 100x + 84y = 9.20. Solving gives x = y = 0.0500. Mass of CaCO₃ = 5.00 g, so the percentage is 5.00/9.20 = 54.3%.

TrapSetting up only one equation makes the problem unsolvable, which is the signal that a second relation is needed. Note the answer is asked as a mass percentage, not a mole percentage, which is 50%.

23Answer 13.89Medium · Dilution

Core principleDilution does not change the number of moles of solute. So M₁V₁ = M₂V₂.

Math18.0 × V = 0.500 × 500, so V = 250/18.0 = 13.89 mL.

TrapInverting the ratio gives 18000 mL, which is more than the final volume and is impossible. The concentrated stock is always the smaller volume.

24Answer 4Hard · Volume strength and dilution

Core principleVolume strength is directly proportional to concentration. So dilution scales it in exactly the same ratio as it scales molarity.

MathThe dilution factor is 500/100 = 5. So the volume strength becomes 20/5 = 4. As a check, M falls from 1.79 to 0.357, and 0.357 × 11.2 = 4.

TrapMultiplying by the dilution factor gives 100 volume, which would mean dilution made it stronger. Always sanity check the direction: adding water can only weaken a solution.

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