ICSE · Class 10 · Chemistry

Class 10 Mole Concept and Stoichiometry Mock Test (ICSE)

6 practice questions from Mole Concept and Stoichiometry, in the style of your board paper, each with a worked answer and the marks for every step. Then take a full chapter test on paper, snap the pages and get every step marked.

Topics in Mole Concept and Stoichiometry

  • Gay-Lussac's law of combining volumes
  • Avogadro's law
  • Mole and molar volume
  • Empirical and molecular formula
  • Calculations from chemical equations

Mole Concept and Stoichiometry practice questions with answers

  1. Q1Multiple choice1 mark
    What is the mass of 5.6 dm35.6\text{ dm}^3 of carbon dioxide (CO2\text{CO}_2) gas at STP?
    1. (a)44 g44\text{ g}
    2. (b)22 g22\text{ g}
    3. (c)11 g11\text{ g}
    4. (d)5.5 g5.5\text{ g}
    Show answer and step marks
    1. Correct option: (c) 11 g11\text{ g}. Moles = 5.6 dm322.4 dm3 mol−1=0.25 mol\frac{5.6\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.25\text{ mol}, Mass = 0.25 mol×44 g mol−1=11 g0.25\text{ mol} \times 44\text{ g mol}^{-1} = 11\text{ g}.+1
    Answer: (c)
  2. Q2Multiple choice1 mark
    An organic compound has an empirical formula CHO\text{CHO} and its vapour density is 2929. What is its molecular formula?
    1. (a)CHO\text{CHO}
    2. (b)C2H2O2\text{C}_2\text{H}_2\text{O}_2
    3. (c)C2H4O2\text{C}_2\text{H}_4\text{O}_2
    4. (d)C3H3O3\text{C}_3\text{H}_3\text{O}_3
    Show answer and step marks
    1. Correct option: (b) C2H2O2\text{C}_2\text{H}_2\text{O}_2. Molecular mass = 2×vapour density=582 \times \text{vapour density} = 58. Empirical formula mass = 2929. n=58/29=2n = 58 / 29 = 2. Molecular formula = (CHO)2=C2H2O2(\text{CHO})_2 = \text{C}_2\text{H}_2\text{O}_2.+1
    Answer: (b)
  3. Q3Multiple choice1 mark
    According to Gay-Lussac's law of combining volumes, gases react in volumes which bear a simple whole number ratio to one another and to the volume of the gaseous products, provided all volumes are measured at:
    1. (a)Standard temperature and 1 atm1\text{ atm} pressure
    2. (b)Same temperature and pressure
    3. (c)Same temperature and different pressures
    4. (d)Different temperatures and same pressure
    Show answer and step marks
    1. Correct option: (b) Same temperature and pressure.+1
    Answer: (b)
  4. Q4Multiple choice1 mark
    What volume of oxygen gas at STP is required to burn completely 4 g4\text{ g} of methane gas? (Reaction: CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O})
    1. (a)11.2 dm311.2\text{ dm}^3
    2. (b)22.4 dm322.4\text{ dm}^3
    3. (c)44.8 dm344.8\text{ dm}^3
    4. (d)5.6 dm35.6\text{ dm}^3
    Show answer and step marks
    1. Correct option: (a) 11.2 dm311.2\text{ dm}^3. Moles of CH4=4/16=0.25 mol\text{CH}_4 = 4 / 16 = 0.25\text{ mol}. Moles of O2\text{O}_2 required = 0.25×2=0.5 mol0.25 \times 2 = 0.5\text{ mol}. Volume at STP = 0.5×22.4=11.2 dm30.5 \times 22.4 = 11.2\text{ dm}^3.+1
    Answer: (a)
  5. Q5Short answer2 marks
    State Avogadro's Law. Using this law, deduce the atomicity of hydrogen, given that hydrogen reacts with chlorine to form hydrogen chloride gas in the volume ratio 1:1:21 : 1 : 2 under similar conditions of temperature and pressure.
    Show answer and step marks
    1. State Avogadro's law: Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.+1
    2. Apply Avogadro's law to the reaction: 1 molecule of hydrogen+1 molecule of chlorine→2 molecules of hydrogen chloride1\text{ molecule of hydrogen} + 1\text{ molecule of chlorine} \rightarrow 2\text{ molecules of hydrogen chloride}.+0.5
    3. Since 1 molecule of hydrogen1\text{ molecule of hydrogen} gives 2 molecules of hydrogen chloride2\text{ molecules of hydrogen chloride}, each containing 11 atom of hydrogen, 1 molecule of hydrogen1\text{ molecule of hydrogen} must contain 22 atoms. Hence, atomicity is 22.+0.5
    Answer: Avogadro's law: equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules. 1 molecule of hydrogen gives 2 molecules of HCl, each with 1 hydrogen atom, so the atomicity of hydrogen is 2.
  6. Q6Short answer2 marks
    State Avogadro's law. If 100 mL100\text{ mL} of nitrogen gas contains xx molecules, how many molecules will be present in 300 mL300\text{ mL} of carbon dioxide gas under the same conditions of temperature and pressure?
    Show answer and step marks
    1. State Avogadro's law: Equal volumes of all gases under similar conditions of temperature and pressure contain equal numbers of molecules.+1
    2. Apply the law: Since 300 mL300\text{ mL} is three times 100 mL100\text{ mL}, it contains 3×x=3x3 \times x = 3x molecules.+1
    Answer: Avogadro's law: equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules. 300 mL is three times 100 mL, so it contains 3x3x molecules.

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