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Mole Concept and Stoichiometry — ICSE Class 10 Chemistry Important Questions

13 hand-picked ICSE Class 10 Chemistry important questions for Mole Concept and Stoichiometry, each with a full model answer — the formats and topics most likely to appear in your board exam.

13
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6
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32
Total marks
₹0
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Quick answer

High-yield ICSE Mole Concept questions use 111 mole =6.022×1023=6.022\times10^{23}=6.022×10^23 particles === molar mass in grams =22.4 dm3=22.4\ \text{dm}^3=22.4 dm^3 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): CISCE (Council for the Indian School Certificate Examinations)

About Mole Concept and Stoichiometry

In the ICSE Class 10 Chemistry chapter Mole Concept and Stoichiometry you use Avogadro's number 6.022×10236.022\times10^{23}6.022×10^23 and the molar volume 22.4 dm322.4\ \text{dm}^322.4 dm^3 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}6.022×10^23 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 dm^3. 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 dm^3.

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 particles6.022×10^23=volume at STP/22.4 dm^3.

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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Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

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}3.011×10^23

  2. (b)

    6.022×10236.022\times10^{23}6.022×10^23

  3. (c)

    1.204×10241.204\times10^{24}1.204×10^24

  4. (d)

    1.5×10231.5\times10^{23}1.5×10^23

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Answer: (a) 3.011×10233.011\times10^{23}3.011×10^23.

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}×6.022×10^23=0.5×6.022×10^23=3.011×10^23.

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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 dm^3

  2. (b)

    5.6 dm35.6\ \text{dm}^35.6 dm^3

  3. (c)

    11.2 dm311.2\ \text{dm}^311.2 dm^3

  4. (d)

    22.4 dm322.4\ \text{dm}^322.4 dm^3

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Answer: (b) 5.6 dm35.6\ \text{dm}^35.6 dm^3.

Moles of O2=832=0.25O_2=\dfrac{8}{32}=0.25O_2=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 dm^3.

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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_2OCH_2O

  2. (b)

    CH3OCH_3OCH_3O

  3. (c)

    C2H4OC_2H_4OC_2H_4O

  4. (d)

    CHOCHOCHO

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Answer: (a) CH2OCH_2OCH_2O.

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_2OCH_2O.

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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_2H_2

  2. (b)

    1 g1\ \text{g}1 g of O2O_2O_2

  3. (c)

    1 g1\ \text{g}1 g of CO2CO_2CO_2

  4. (d)

    1 g1\ \text{g}1 g of N2N_2N_2

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Answer: (a) 1 g1\ \text{g}1 g of H2H_2H_2.

For a fixed mass, moles =1M=\dfrac{1}{M}=1/M is largest for the smallest molar mass. H2H_2H_2 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 dm^3 of any gas at STP contains 6.022×10236.022\times10^{23}6.022×10^23 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 dm^3 at STP, it contains 6.022×10236.022\times10^{23}6.022×10^23 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 dm^3 (i.e. 22400 cm322400\ \text{cm}^322400 cm^3) 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=44CO_2=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}=0.25×6.022×10^23=1.51×10^23 molecules.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Calculate the percentage of nitrogen in ammonium nitrate, NH4NO3NH_4NO_3NH_4NO_3. (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=80NH_4NO_3=(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_3CH_3 (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=(CH_3)_2=C_2H_6 (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\uparrowCaCO_3 → (Δ) CaO + CO_2

100 g100\ \text{g}100 g of CaCO3CaCO_3CaCO_3 (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_3CaCO_3 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 + H_2SO_4 → ZnSO_4 + H_2

Calculate (i) the mass of zinc required to produce 2.24 dm32.24\ \text{dm}^32.24 dm^3 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_2H_2: 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:H_2:ZnSO_4=1:1:1, so moles of Zn=0.1Zn=0.1Zn=0.1 and moles of ZnSO4=0.1ZnSO_4=0.1ZnSO_4=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)=161ZnSO_4=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\uparrowNa_2CO_3 + 2HCl → 2NaCl + H_2O + CO_2

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)=106Na_2CO_3=(2×23)+12+(3×16)=106.

(i) Moles of Na2CO3Na_2CO_3Na_2CO_3 =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_3Na_2CO_3 gives 111 mol CO2CO_2CO_2, so moles of CO2=0.05CO_2=0.05CO_2=0.05.

Volume of CO2=0.05×22.4=1.12 dm3CO_2=0.05\times22.4=1.12\ \text{dm}^3CO_2=0.05×22.4=1.12 dm^3 at STP.

(iii) 111 mol Na2CO3Na_2CO_3Na_2CO_3 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

A student is given a sample of 4.4 g4.4\ \text{g}4.4 g of carbon dioxide gas. Using the relationships that 111 mole of CO2CO_2CO_2 has a mass of 44 g44\ \text{g}44 g, occupies 22.4 dm322.4\ \text{dm}^322.4 dm^3 at STP and contains 6.022×10236.022\times10^{23}6.022×10^23 molecules, answer the following. (C=12, O=16C=12,\ O=16C=12, O=16)

(i) Calculate the number of moles of CO2CO_2CO_2.

(ii) Calculate the volume of the gas at STP.

(iii) Calculate the number of molecules present.

(iv) Calculate the number of oxygen atoms present.

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(i) Moles =4.444=0.1 mol=\dfrac{4.4}{44}=0.1\ \text{mol}=4.4/44=0.1 mol.

(ii) Volume at STP =0.1×22.4=2.24 dm3=0.1\times22.4=2.24\ \text{dm}^3=0.1×22.4=2.24 dm^3.

(iii) Number of molecules =0.1×6.022×1023=6.022×1022=0.1\times6.022\times10^{23}=6.022\times10^{22}=0.1×6.022×10^23=6.022×10^22.

(iv) Each CO2CO_2CO_2 molecule has 222 oxygen atoms, so oxygen atoms =2×6.022×1022=1.204×1023=2\times6.022\times10^{22}=1.204\times10^{23}=2×6.022×10^22=1.204×10^23.

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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

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    Yes — they are aligned to the CISCE latest syllabus syllabus for ICSE Class 10 Chemistry, so nothing here is outside the current course.
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  • What types of questions are covered for Mole Concept and Stoichiometry?
    A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the ICSE paper is covered.

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