ICSE · Class 9 · Chemistry

Class 9 The Language of Chemistry Mock Test (ICSE)

6 practice questions from The Language of Chemistry, 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 The Language of Chemistry

  • Symbols and valency
  • Radicals and writing formulae
  • Chemical equations and balancing
  • Atomic mass and molecular mass
  • Percentage composition

The Language of Chemistry practice questions with answers

  1. Q1Multiple choice1 mark
    What is the correct chemical formula for calcium phosphate?
    1. (a)Ca2(PO4)3\text{Ca}_2(\text{PO}_4)_3
    2. (b)Ca3PO4\text{Ca}_3\text{PO}_4
    3. (c)CaPO4\text{CaPO}_4
    4. (d)Ca3(PO4)2\text{Ca}_3(\text{PO}_4)_2
    Show answer and step marks
    1. Correct option: (d) Ca3(PO4)2\text{Ca}_3(\text{PO}_4)_2+1
    Answer: (d)
  2. Q2Multiple choice1 mark
    What is the relative molecular mass of calcium carbonate (CaCO3\text{CaCO}_3)? [Given atomic masses: Ca=40\text{Ca} = 40, C=12\text{C} = 12, O=16\text{O} = 16]
    1. (a)6868
    2. (b)5252
    3. (c)100100
    4. (d)8484
    Show answer and step marks
    1. Correct option: (c) 100100+1
    Answer: (c)
  3. Q3Multiple choice1 mark
    The percentage of oxygen in carbon dioxide (CO2\text{CO}_2) is: [Atomic masses: C=12\text{C} = 12, O=16\text{O} = 16]
    1. (a)27.27%27.27\%
    2. (b)50.00%50.00\%
    3. (c)72.73%72.73\%
    4. (d)32.00%32.00\%
    Show answer and step marks
    1. Correct option: (c) 72.73%72.73\%+1
    Answer: (c)
  4. Q4Multiple choice1 mark
    Which of the following elements exhibits variable valency?
    1. (a)Fluorine
    2. (b)Iron
    3. (c)Sodium
    4. (d)Aluminium
    Show answer and step marks
    1. Correct option: (b) Iron+1
    Answer: (b)
  5. Q5Short answer2 marks
    Balance the following chemical equation: Fe+H2O→Fe3O4+H2\text{Fe} + \text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2
    Show answer and step marks
    1. Fe: 11 on the left, 33 in Fe3O4\text{Fe}_3\text{O}_4, so write 3Fe3\text{Fe}; O: 44 in Fe3O4\text{Fe}_3\text{O}_4, so write 4H2O4\text{H}_2\text{O}.+1
    2. H: 88 on the left, so write 4H24\text{H}_2: 3Fe+4H2O→Fe3O4+4H23\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2+1
    Answer: 3Fe+4H2O→Fe3O4+4H23\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2
  6. Q6Short answer2 marks
    Balance the following chemical equation: Pb(NO3)2→PbO+NO2+O2\text{Pb}(\text{NO}_3)_2 \rightarrow \text{PbO} + \text{NO}_2 + \text{O}_2
    Show answer and step marks
    1. Balance N with 2NO22\text{NO}_2: Pb(NO3)2→PbO+2NO2+O2\text{Pb}(\text{NO}_3)_2 \rightarrow \text{PbO} + 2\text{NO}_2 + \text{O}_2 leaves O at 66 on the left and 1+4+2=71 + 4 + 2 = 7 on the right.+1
    2. Double the lead nitrate, lead oxide and nitrogen dioxide: 2Pb(NO3)2→2PbO+4NO2+O22\text{Pb}(\text{NO}_3)_2 \rightarrow 2\text{PbO} + 4\text{NO}_2 + \text{O}_2 (O: 12=2+8+212 = 2 + 8 + 2).+1
    Answer: 2Pb(NO3)2→2PbO+4NO2+O22\text{Pb}(\text{NO}_3)_2 \rightarrow 2\text{PbO} + 4\text{NO}_2 + \text{O}_2

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