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Mole Concept and Stoichiometry — ICSE Class 10 Chemistry Important Questions

13 ICSE Class 10 Chemistry practice questions on Mole Concept and Stoichiometry, each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

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Reviewed by Classmate AI Team · 18 October 2026

Last updated 25 October 2026

13
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5
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4
Key concepts
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Quick answer

ICSE Mole Concept questions use 111 mole =6.022×1023=6.022\times10^{23}=6.022×10²³ particles === molar mass in grams =22.4 dm3=22.4\ \text{dm}^3=22.4 dm³ of any gas at STP. Common numericals convert between mass, moles, volume and number of molecules, find percentage composition, and derive empirical and molecular formulae using vapour density (M=2×V.D.M=2\times\text{V.D.}M=2× V.D.).

Syllabus reference (2026–27): ICSE 2027 Chemistry syllabus (PDF) · CISCE (Council for the Indian School Certificate Examinations)

About Mole Concept and Stoichiometry

In this ICSE Class 10 Chemistry chapter Mole Concept and Stoichiometry you use Avogadro's number 6.022×10236.022\times10^{23}6.022×10²³ and the molar volume 22.4 dm322.4\ \text{dm}^322.4 dm³ at STP to interconvert mass, moles, volume and number of particles, calculate percentage composition, and determine empirical and molecular formulae from experimental data. These calculations underpin all quantitative chemistry in the ICSE syllabus.

Mole and Avogadro's numberGay-Lussac's law and Avogadro's lawMolar volume and gas calculations at STPPercentage composition of compoundsEmpirical and molecular formula (vapour density)

Key concepts & formulas

The mole

One mole of any substance contains 6.022×10236.022\times10^{23}6.022×10²³ particles (Avogadro's number) and has a mass equal to its relative atomic/molecular mass in grams.

Molar volume

At STP, 111 mole of any gas occupies 22.4 dm322.4\ \text{dm}^322.4 dm³. So moles =volume at STP22.4=\dfrac{\text{volume at STP}}{22.4}=(volume at STP)/22.4 and volume =moles×22.4 dm3=\text{moles}\times22.4\ \text{dm}^3=moles×22.4 dm³.

Mole relationships

moles=massmolar mass=number of particles6.022×1023=volume at STP22.4 dm3\text{moles}=\dfrac{\text{mass}}{\text{molar mass}}=\dfrac{\text{number of particles}}{6.022\times10^{23}}=\dfrac{\text{volume at STP}}{22.4\ \text{dm}^3}moles=mass/(molar mass)=number of particles6.022×10²³=(volume at STP)/(22.4 dm³).

Vapour density

For a gas, molecular mass =2×=2\times=2× vapour density. Molecular formula =(empirical formula)n=(\text{empirical formula})_n=(empirical formula)_n where n=molecular massempirical formula massn=\dfrac{\text{molecular mass}}{\text{empirical formula mass}}n=(molecular mass)/(empirical formula mass).

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Important questions with answers

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Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The number of molecules present in 0.50.50.5 mole of carbon dioxide is:

  1. (a)

    3.011×10233.011\times10^{23}3.011×10²³

  2. (b)

    6.022×10236.022\times10^{23}6.022×10²³

  3. (c)

    1.204×10241.204\times10^{24}1.204×10²⁴

  4. (d)

    1.5×10231.5\times10^{23}1.5×10²³

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Answer: (a) 3.011×10233.011\times10^{23}3.011×10²³.

Number of molecules === moles ×6.022×1023=0.5×6.022×1023=3.011×1023\times6.022\times10^{23}=0.5\times6.022\times10^{23}=3.011\times10^{23}×6.022×10²³=0.5×6.022×10²³=3.011×10²³.

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Q2MCQModerate1 mark

The volume occupied by 8 g8\ \text{g}8 g of oxygen gas at STP is: (O=16O=16O=16)

  1. (a)

    2.8 dm32.8\ \text{dm}^32.8 dm³

  2. (b)

    5.6 dm35.6\ \text{dm}^35.6 dm³

  3. (c)

    11.2 dm311.2\ \text{dm}^311.2 dm³

  4. (d)

    22.4 dm322.4\ \text{dm}^322.4 dm³

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Answer: (b) 5.6 dm35.6\ \text{dm}^35.6 dm³.

Moles of O2=832=0.25O_2=\dfrac{8}{32}=0.25O₂=8/32=0.25. Volume at STP =0.25×22.4=5.6 dm3=0.25\times22.4=5.6\ \text{dm}^3=0.25×22.4=5.6 dm³.

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Q3MCQModerate1 mark

A compound contains 40%40\%40% carbon, 6.7%6.7\%6.7% hydrogen and 53.3%53.3\%53.3% oxygen by mass. Its empirical formula is: (C=12, H=1, O=16C=12,\ H=1,\ O=16C=12, H=1, O=16)

  1. (a)

    CH2OCH_2OCH₂O

  2. (b)

    CH3OCH_3OCH₃O

  3. (c)

    C2H4OC_2H_4OC₂H₄O

  4. (d)

    CHOCHOCHO

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Answer: (a) CH2OCH_2OCH₂O.

Mole ratio C:H:O=4012:6.71:53.316=3.33:6.7:3.33=1:2:1C:H:O=\dfrac{40}{12}:\dfrac{6.7}{1}:\dfrac{53.3}{16}=3.33:6.7:3.33=1:2:1C:H:O=40/12:6.7/1:53.3/16=3.33:6.7:3.33=1:2:1, giving CH2OCH_2OCH₂O.

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Q4MCQHOTS1 mark

Which of the following samples contains the greatest number of molecules?

  1. (a)

    1 g1\ \text{g}1 g of H2H_2H₂

  2. (b)

    1 g1\ \text{g}1 g of O2O_2O₂

  3. (c)

    1 g1\ \text{g}1 g of CO2CO_2CO₂

  4. (d)

    1 g1\ \text{g}1 g of N2N_2N₂

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Answer: (a) 1 g1\ \text{g}1 g of H2H_2H₂.

For a fixed mass, moles =1M=\dfrac{1}{M}=1/M is largest for the smallest molar mass. H2H_2H₂ has the least molar mass (222), giving 12=0.5\dfrac{1}{2}=0.51/2=0.5 mol, hence the most molecules.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): 22.4 dm322.4\ \text{dm}^322.4 dm³ of any gas at STP contains 6.022×10236.022\times10^{23}6.022×10²³ molecules.

Reason (R): Equal volumes of all gases, under the same conditions of temperature and pressure, contain equal numbers of molecules.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (a) By Avogadro's law equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules; since 111 mole occupies 22.4 dm322.4\ \text{dm}^322.4 dm³ at STP, it contains 6.022×10236.022\times10^{23}6.022×10²³ molecules. R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

(a) State Avogadro's law.

(b) What is the volume occupied by 111 mole of any gas at STP?

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(a) Avogadro's law: Equal volumes of all gases, measured under the same conditions of temperature and pressure, contain equal numbers of molecules.

(b) 111 mole of any gas occupies 22.4 dm322.4\ \text{dm}^322.4 dm³ (i.e. 22400 cm322400\ \text{cm}^322400 cm³) at STP.

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Q7Very ShortModerate2 marks

Calculate the number of moles and the number of molecules present in 11 g11\ \text{g}11 g of carbon dioxide. (C=12, O=16C=12,\ O=16C=12, O=16)

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Molar mass of CO2=12+2×16=44CO_2=12+2\times16=44CO₂=12+2×16=44.

Moles =1144=0.25 mol=\dfrac{11}{44}=0.25\ \text{mol}=11/44=0.25 mol.

Number of molecules =0.25×6.022×1023=1.51×1023=0.25\times6.022\times10^{23}=1.51\times10^{23}=0.25×6.022×10²³=1.51×10²³ molecules.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Calculate the percentage of nitrogen in ammonium nitrate, NH4NO3NH_4NO_3NH₄NO₃. (N=14, H=1, O=16N=14,\ H=1,\ O=16N=14, H=1, O=16)

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Molar mass of NH4NO3=(2×14)+(4×1)+(3×16)=28+4+48=80NH_4NO_3=(2\times14)+(4\times1)+(3\times16)=28+4+48=80NH₄NO₃=(2×14)+(4×1)+(3×16)=28+4+48=80.

Mass of nitrogen =2×14=28=2\times14=28=2×14=28.

% N=2880×100=35%\%\ \text{N}=\dfrac{28}{80}\times100=35\%% N=28/80×100=35%.

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Q9Short AnswerHOTS3 marks

A gaseous hydrocarbon contains 80%80\%80% carbon and 20%20\%20% hydrogen by mass. If its vapour density is 151515, find its molecular formula. (C=12, H=1C=12,\ H=1C=12, H=1)

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Mole ratio C:H=8012:201=6.67:20=1:3C:H=\dfrac{80}{12}:\dfrac{20}{1}=6.67:20=1:3C:H=80/12:20/1=6.67:20=1:3, so the empirical formula is CH3CH_3CH₃ (empirical mass =15=15=15).

Molecular mass =2×V.D.=2×15=30=2\times\text{V.D.}=2\times15=30=2× V.D.=2×15=30.

n=3015=2n=\dfrac{30}{15}=2n=30/15=2. Molecular formula =(CH3)2=C2H6=(CH_3)_2=C_2H_6=(CH₃)₂=C₂H₆ (ethane).

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Q10Short AnswerEasy3 marks

Calculate the mass of calcium oxide obtained when 50 g50\ \text{g}50 g of calcium carbonate is heated strongly. (Ca=40, C=12, O=16Ca=40,\ C=12,\ O=16Ca=40, C=12, O=16)

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CaCO3→ΔCaO+CO2↑CaCO_3 \xrightarrow{\Delta} CaO + CO_2\uparrowCaCO₃ → (Δ) CaO + CO₂↑

100 g100\ \text{g}100 g of CaCO3CaCO_3CaCO₃ (molar mass =100=100=100) gives 56 g56\ \text{g}56 g of CaOCaOCaO (molar mass =56=56=56).

50 g50\ \text{g}50 g of CaCO3CaCO_3CaCO₃ gives 56100×50=28 g\dfrac{56}{100}\times50=28\ \text{g}56/100×50=28 g of CaOCaOCaO.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

Zinc reacts with dilute sulphuric acid according to:

Zn+H2SO4→ZnSO4+H2↑Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2\uparrowZn + H₂SO₄ → ZnSO₄ + H₂↑

Calculate (i) the mass of zinc required to produce 2.24 dm32.24\ \text{dm}^32.24 dm³ of hydrogen at STP, and (ii) the mass of zinc sulphate formed. (Zn=65, S=32, O=16, H=1Zn=65,\ S=32,\ O=16,\ H=1Zn=65, S=32, O=16, H=1)

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Moles of H2H_2H₂: 2.2422.4=0.1 mol\dfrac{2.24}{22.4}=0.1\ \text{mol}2.24/22.4=0.1 mol.

From the equation, mole ratio Zn:H2:ZnSO4=1:1:1Zn:H_2:ZnSO_4=1:1:1Zn:H₂:ZnSO₄=1:1:1, so moles of Zn=0.1Zn=0.1Zn=0.1 and moles of ZnSO4=0.1ZnSO_4=0.1ZnSO₄=0.1.

(i) Mass of zinc =0.1×65=6.5 g=0.1\times65=6.5\ \text{g}=0.1×65=6.5 g.

(ii) Mass of zinc sulphate: molar mass of ZnSO4=65+32+(4×16)=161ZnSO_4=65+32+(4\times16)=161ZnSO₄=65+32+(4×16)=161.

Mass =0.1×161=16.1 g=0.1\times161=16.1\ \text{g}=0.1×161=16.1 g.

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Q12Long AnswerHOTS5 marks

5.3 g5.3\ \text{g}5.3 g of anhydrous sodium carbonate is completely reacted with dilute hydrochloric acid:

Na2CO3+2HCl→2NaCl+H2O+CO2↑Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2\uparrowNa₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂↑

Calculate (i) the number of moles of sodium carbonate, (ii) the volume of carbon dioxide liberated at STP, and (iii) the mass of sodium chloride formed. (Na=23, C=12, O=16, Cl=35.5Na=23,\ C=12,\ O=16,\ Cl=35.5Na=23, C=12, O=16, Cl=35.5)

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Molar mass of Na2CO3=(2×23)+12+(3×16)=106Na_2CO_3=(2\times23)+12+(3\times16)=106Na₂CO₃=(2×23)+12+(3×16)=106.

(i) Moles of Na2CO3Na_2CO_3Na₂CO₃ =5.3106=0.05 mol=\dfrac{5.3}{106}=0.05\ \text{mol}=5.3/106=0.05 mol.

(ii) From the equation, 111 mol Na2CO3Na_2CO_3Na₂CO₃ gives 111 mol CO2CO_2CO₂, so moles of CO2=0.05CO_2=0.05CO₂=0.05.

Volume of CO2=0.05×22.4=1.12 dm3CO_2=0.05\times22.4=1.12\ \text{dm}^3CO₂=0.05×22.4=1.12 dm³ at STP.

(iii) 111 mol Na2CO3Na_2CO_3Na₂CO₃ gives 222 mol NaClNaClNaCl, so moles of NaCl=0.10NaCl=0.10NaCl=0.10.

Molar mass of NaCl=23+35.5=58.5NaCl=23+35.5=58.5NaCl=23+35.5=58.5; mass =0.10×58.5=5.85 g=0.10\times58.5=5.85\ \text{g}=0.10×58.5=5.85 g.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

Read the passage and answer the questions.

100 cm3100\text{ cm}^3100 cm³ of carbon monoxide is burnt in 80 cm380\text{ cm}^380 cm³ of oxygen. All volumes are measured at the same temperature and pressure.
2CO+O2→2CO22CO+O_2\rightarrow2CO_22CO+O₂→2CO₂
(i) State Gay-Lussac's law of combining volumes.

(ii) What volume of oxygen is used up?

(iii) What volume of carbon dioxide is formed?

(iv) What is the total volume of gas left after the reaction?

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(i) When gases react, they do so in volumes that bear a simple whole-number ratio to one another and to the volumes of the gaseous products, all measured at the same temperature and pressure.

(ii) CO:O2=2:1CO:O_2=2:1CO:O₂=2:1, so 100 cm3100\text{ cm}^3100 cm³ of COCOCO uses 50 cm350\text{ cm}^350 cm³ of O2O_2O₂.

(iii) CO:CO2=2:2CO:CO_2=2:2CO:CO₂=2:2, so 100 cm3100\text{ cm}^3100 cm³ of CO2CO_2CO₂ forms.

(iv) Unused oxygen =80−50=30 cm3=80-50=30\text{ cm}^3=80-50=30 cm³. Total gas left =100+30=130 cm3=100+30=130\text{ cm}^3=100+30=130 cm³.

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All ICSE Class 10 Chemistry Chapters

Periodic Table, Periodic Properties and Variations of PropertiesChemical BondingAcids, Bases and SaltsAnalytical Chemistry: Uses of Ammonium Hydroxide and Sodium HydroxideMole Concept and StoichiometryElectrolysisMetallurgyStudy of Compounds: Hydrogen ChlorideStudy of Compounds: AmmoniaStudy of Compounds: Nitric AcidSulphuric AcidOrganic ChemistryPractical Work

Frequently asked questions

  • What are the important questions in Mole Concept and Stoichiometry ICSE Class 10?
    Converting between mass, moles, gas volume at STP and number of molecules, percentage composition, empirical and molecular formulae with vapour density, and mass and volume calculations from equations. This page has 13 such numericals with answers.
  • How do you convert grams of a gas to volume at STP?
    Find the moles, then multiply by 22.4 dm322.4\ \text{dm}^322.4 dm³. 888 g of oxygen is 832=0.25\tfrac{8}{32}=0.258/32=0.25 mol, so it occupies 0.25×22.4=5.6 dm30.25\times22.4=5.6\ \text{dm}^30.25×22.4=5.6 dm³ at STP. 111111 g of carbon dioxide is also 0.250.250.25 mol, about 1.5×10231.5\times10^{23}1.5×10²³ molecules.
  • How do you find a molecular formula from percentage composition and vapour density?
    Divide each percentage by the atomic mass and simplify the ratio to get the empirical formula, then use molecular mass =2×=2\times=2× vapour density. 80%80\%80% C and 20%20\%20% H gives CH3CH_3CH₃; with vapour density 151515 the molecular mass is 303030, so the formula is C2H6C_2H_6C₂H₆.
  • How much calcium oxide comes from heating 50 g of calcium carbonate?
    From CaCO3→CaO+CO2CaCO_3\rightarrow CaO+CO_2CaCO₃→ CaO+CO₂, 100100100 g of calcium carbonate gives 565656 g of calcium oxide. So 505050 g gives 282828 g. Work from the balanced equation's molar masses, then scale.
  • Which ICSE Class 9 chapter prepares you for the mole concept?
    Study of Gas Laws (Class 9 Chemistry, Chapter 7) for gas volumes and S.T.P., and The Language of Chemistry (Chapter 1) for formulae and molecular mass. In Class 10, Mole Concept is Chapter 5, followed by Electrolysis (Chapter 6).
  • How do I avoid mistakes in stoichiometry numericals?
    Always write the balanced equation, convert the given quantity to moles, use the mole ratio from the equation, and only then convert to the required mass or volume. Writing units at every step catches most slips, such as mixing grams and moles.
  • Can I download Mole Concept ICSE Class 10 numericals?
    Yes. Enter your email in the free PDF box on this page to open all 13 mole-concept questions with step-by-step solutions as a printable worksheet.

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