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NCERT Exemplar · Class 9 Science Atoms and Molecules

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Short Answer Questions 11–23 (part 2 of 5)

  1. Exercise 11

    Which of the following represents a correct chemical formula? Name it.
    (a)
    CaCl (b) BiPO4\displaystyle BiPO_{4}
    (c)
    NaSO4\displaystyle NaSO_{4}
    (d)
    NaS

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    NCERT’s answer
    (b)
    \(\displaystyle BiPO_{4}\)— Both ions are trivalent Bismuth phosphate
    Cross each pair's charges to check the subscripts: \[\text{Ca}^{2+}\text{Cl}^{-} \rightarrow \text{CaCl}_2 \] \[\text{Na}^{+}\text{SO}_4^{2-} \rightarrow \text{Na}_2\text{SO}_4 \] \[\text{Na}^{+}\text{S}^{2-} \rightarrow \text{Na}_2\text{S} \] \[\text{Bi}^{3+}\text{PO}_4^{3-} \rightarrow \text{BiPO}_4 \] Answer: \(\displaystyle \text{BiPO}_4\), bismuth phosphate — the only option with correctly balanced subscripts; (a), (c), (d) are missing theirs.

  2. Exercise 12

    Write the molecular formulae for the following compounds
    (a)
    Copper (II) bromide
    (b)
    Aluminium (III) nitrate
    (c)
    Calcium (II) phosphate
    (d)
    Iron (III) sulphide
    (e)
    Mercury (II) chloride
    (f)
    Magnesium (II) acetate

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    (a)
    \(\displaystyle CuBr_{2}\) (b) Al(\(\displaystyle NO_{3}\))$\displaystyle 3$ (c) \(\displaystyle Ca_{3}\)(\(\displaystyle PO_{4}\))$\displaystyle 2$ (d) \(\displaystyle Fe_{2}\)\(\displaystyle S_{3}\) (e) \(\displaystyle HgCl_{2}\) (f) Mg(\(\displaystyle CH_{3}\)COO) $\displaystyle 2$
    Cross each ion's charge to the other's subscript: \[\text{Cu}^{2+}\text{Br}^{-} \rightarrow \text{CuBr}_2 \] \[\text{Al}^{3+}\text{NO}_3^{-} \rightarrow \text{Al}(\text{NO}_3)_3 \] \[\text{Ca}^{2+}\text{PO}_4^{3-} \rightarrow \text{Ca}_3(\text{PO}_4)_2 \] \[\text{Fe}^{3+}\text{S}^{2-} \rightarrow \text{Fe}_2\text{S}_3 \] \[\text{Hg}^{2+}\text{Cl}^{-} \rightarrow \text{HgCl}_2 \] \[\text{Mg}^{2+}\text{CH}_3\text{COO}^{-} \rightarrow \text{Mg}(\text{CH}_3\text{COO})_2 \] Answer: (a) \(\displaystyle \text{CuBr}_2\) (b) \(\displaystyle \text{Al}(\text{NO}_3)_3\) (c) \(\displaystyle \text{Ca}_3(\text{PO}_4)_2\) (d) \(\displaystyle \text{Fe}_2\text{S}_3\) (e) \(\displaystyle \text{HgCl}_2\) (f) \(\displaystyle \text{Mg}(\text{CH}_3\text{COO})_2\).

  3. Exercise 13

    Write the molecular formulae of all the compounds that can be formed by the combination of following ions Cu2+\displaystyle Cu^{2+}, Na+\displaystyle Na^{+}, Fe3+\displaystyle Fe^{3+}, C1−\displaystyle C1^{-}, 2\displaystyle 2- S O , 3\displaystyle 3- PO

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    NCERT’s answer
    \(\displaystyle CuCl_{2}\)/ \(\displaystyle CuSO_{4}\)/ \(\displaystyle Cu_{3}\) (\(\displaystyle PO_{4}\))$\displaystyle 2$ NaCl/ \(\displaystyle Na_{2}\)\(\displaystyle SO_{4}\)/ \(\displaystyle Na_{3}\) \(\displaystyle PO_{4}\) \(\displaystyle FeCl_{3}\)/ \(\displaystyle Fe_{2}\)(SO $\displaystyle 4$)$\displaystyle 3$ / \(\displaystyle FePO_{4}\)
    Cross-multiply charges for all nine cation–anion pairs: \[\text{Cu}^{2+}:\ \text{CuCl}_2,\ \text{CuSO}_4,\ \text{Cu}_3(\text{PO}_4)_2 \] \[\text{Na}^{+}:\ \text{NaCl},\ \text{Na}_2\text{SO}_4,\ \text{Na}_3\text{PO}_4 \] \[\text{Fe}^{3+}:\ \text{FeCl}_3,\ \text{Fe}_2(\text{SO}_4)_3,\ \text{FePO}_4 \] Answer: nine compounds, as listed above.

  4. Exercise 14

    Write the cations and anions present (if any) in the following compounds
    (a)
    CH3\displaystyle CH_{3}COONa
    (b)
    NaCl
    (c)
    H2\displaystyle H_{2}
    (d)
    NH4\displaystyle NH_{4}NO3\displaystyle NO_{3}
    ATOMS AND MOLECULES

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    NCERT’s answer
    Anions Cations (a) \(\displaystyle CH_{3}\) \(\displaystyle COO^{-}\) \(\displaystyle Na^{+}\) (b) \(\displaystyle Cl^{-}\) \(\displaystyle Na^{+}\) (c) It is a covalent compound (d) NO − NH +
    Each ionic compound dissociates into its ions: \[\text{CH}_3\text{COONa} \rightarrow \text{Na}^{+} + \text{CH}_3\text{COO}^{-} \] \[\text{NaCl} \rightarrow \text{Na}^{+} + \text{Cl}^{-} \] \[\text{NH}_4\text{NO}_3 \rightarrow \text{NH}_4^{+} + \text{NO}_3^{-} \] \(\displaystyle \text{H}_2\) is a covalent molecule — no cation or anion. Answer: (a) \(\displaystyle \text{Na}^{+}\), \(\displaystyle \text{CH}_3\text{COO}^{-}\) (b) \(\displaystyle \text{Na}^{+}\), \(\displaystyle \text{Cl}^{-}\) (c) none (d) \(\displaystyle \text{NH}_4^{+}\), \(\displaystyle \text{NO}_3^{-}\).

  5. Exercise 15

    Give the formulae of the compounds formed from the following sets of
    elements
    (a)
    Calcium and fluorine
    (b)
    Hydrogen and sulphur
    (c)
    Nitrogen and hydrogen
    (d)
    Carbon and chlorine
    (e)
    Sodium and oxygen
    (f)
    Carbon and oxygen

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    NCERT’s answer
    (a)
    \(\displaystyle CaF_{2}\) (e) \(\displaystyle Na_{2}\)O (b) \(\displaystyle H_{2}\)S (f) CO, \(\displaystyle CO_{2}\) (c) \(\displaystyle NH_{3}\) (d) \(\displaystyle CCl_{4}\)
    Cross the standard valency of each element (a number, not a charge) to fix the subscripts: \[\text{Ca}^{(2)}\ \text{F}^{(1)} \Rightarrow \text{CaF}_2 \] \[\text{H}^{(1)}\ \text{S}^{(2)} \Rightarrow \text{H}_2\text{S} \] \[\text{N}^{(3)}\ \text{H}^{(1)} \Rightarrow \text{NH}_3 \] \[\text{C}^{(4)}\ \text{Cl}^{(1)} \Rightarrow \text{CCl}_4 \] \[\text{Na}^{(1)}\ \text{O}^{(2)} \Rightarrow \text{Na}_2\text{O} \] \[\text{C}^{(4)}\ \text{O}^{(2)} \Rightarrow \text{CO}_2,\qquad \text{C}^{(2)}\ \text{O}^{(2)} \Rightarrow \text{CO} \] Answer: (a) \(\displaystyle \text{CaF}_2\) (b) \(\displaystyle \text{H}_2\text{S}\) (c) \(\displaystyle \text{NH}_3\) (d) \(\displaystyle \text{CCl}_4\) (e) \(\displaystyle \text{Na}_2\text{O}\) (f) \(\displaystyle \text{CO}\), \(\displaystyle \text{CO}_2\).

  6. Exercise 16

    Which of the following symbols of elements are incorrect? Give their correct
    symbols
    (a)
    Cobalt
    CO (b) Carbon
    c (c) Aluminium
    AL (d) Helium
    He (e) Sodium

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    NCERT’s answer
    (a)
    Incorrect, the correct symbol of cobalt is Co (b) Incorrect, the correct symbol of carbon is C (c) Incorrect, the correct symbol of aluminium is Al (d) Correct (He) (e) Incorrect, the correct symbol of sodium is Na
    Symbol rule: first letter capital, second (if any) lowercase; several are Latin-derived. Answer: (a) \(\displaystyle \text{Co}\), not \(\displaystyle \text{CO}\) — that is carbon monoxide. (b) \(\displaystyle \text{C}\), not \(\displaystyle \text{c}\). (c) \(\displaystyle \text{Al}\), not \(\displaystyle \text{AL}\). (d) \(\displaystyle \text{He}\) is already correct. (e) \(\displaystyle \text{Na}\) (Latin natrium), not \(\displaystyle \text{So}\).

  7. Exercise 17

    Give the chemical formulae for the following compounds and compute the
    ratio by mass of the combining elements in each one of them. (You may
    use appendix-III).
    (a)
    Ammonia
    (b)
    Carbon monoxide
    (c)
    Hydrogen chloride
    (d)
    Aluminium fluoride
    (e)
    Magnesium sulphide

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    NCERT’s answer
    (a)
    \(\displaystyle NH_{3}\) (b) CO (c) HCI (d) \(\displaystyle AlF_{3}\) (e) Mg S N : H × $\displaystyle 3$ C : O H : Cl Al : F × $\displaystyle 3$ Mg : S $\displaystyle 14$ : $\displaystyle 1$ × $\displaystyle 3$ $\displaystyle 12$ : $\displaystyle 16$ $\displaystyle 1$ : $\displaystyle 35.5$ $\displaystyle 27$ : $\displaystyle 19$ × $\displaystyle 3$ $\displaystyle 24$ : $\displaystyle 32$ $\displaystyle 14$ : $\displaystyle 3$ $\displaystyle 3$ : $\displaystyle 4$ $\displaystyle 2$ : $\displaystyle 71$ $\displaystyle 9$ : $\displaystyle 19$ $\displaystyle 3$ : $\displaystyle 4$
    Formula, then mass ratio from atomic masses: \[\text{u: H=1, C=12, N=14, O=16, F=19, Mg=24, Al=27, S=32, Cl=35.5} \] \[\text{NH}_3:\ \text{N}:\text{H} = 14:(3\times1) = 14:3 \] \[\text{CO}:\ \text{C}:\text{O} = 12:16 = 3:4 \] \[\text{HCl}:\ \text{H}:\text{Cl} = 1:35.5 = 2:71 \] \[\text{AlF}_3:\ \text{Al}:\text{F} = 27:(3\times19) = 27:57 = 9:19 \] \[\text{MgS}:\ \text{Mg}:\text{S} = 24:32 = 3:4 \] Answer: \(\displaystyle \text{NH}_3\) \(\displaystyle 14:3\); \(\displaystyle \text{CO}\) \(\displaystyle 3:4\); \(\displaystyle \text{HCl}\) \(\displaystyle 2:71\); \(\displaystyle \text{AlF}_3\) \(\displaystyle 9:19\); \(\displaystyle \text{MgS}\) \(\displaystyle 3:4\).

  8. Exercise 18

    State the number of atoms present in each of the following chemical species
    (a)
    CO3\displaystyle CO_{3}
    2\displaystyle 2- (b) PO4\displaystyle PO_{4}
    3\displaystyle 3- (c) P2\displaystyle P_{2}O5\displaystyle O_{5}
    (d)
    CO

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    NCERT’s answer
    (a)
    $\displaystyle 4$ (b) (c) $\displaystyle 7$ (d)
    Count atoms in each formula unit, ions included: \[\text{CO}_3^{2-}:\ 1+3=4\ \text{atoms} \] \[\text{PO}_4^{3-}:\ 1+4=5\ \text{atoms} \] \[\text{P}_2\text{O}_5:\ 2+5=7\ \text{atoms} \] \[\text{CO}:\ 1+1=2\ \text{atoms} \] Answer: (a) $\displaystyle 4$ (b) $\displaystyle 5$ (c) $\displaystyle 7$ (d) 2.

  9. Exercise 19

    What is the fraction of the mass of water due to neutrons?

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    NCERT’s answer
    $\displaystyle 8$/$\displaystyle 18$ Mass of one mole (Avogadro Number) of neutrons ~ $\displaystyle 1$ g Mass of one neutron = A g Avogadro Number (N ) Mass of one molecule of water = A A Molar mass $\displaystyle 18$ g N N = There are $\displaystyle 8$ neutrons in one atom of oxygen Mass of $\displaystyle 8$ neutrons = A N Fraction of mass of water due to neutrons ~
    In \(\displaystyle \text{H}_2\text{O}\), H (mass number $\displaystyle 1$, $\displaystyle 1$ proton) has \(\displaystyle 1-1=0\) neutrons; O (mass number $\displaystyle 16$, $\displaystyle 8$ protons) has \(\displaystyle 16-8=8\) neutrons. \[\text{mass from neutrons} = 2(0) + 8 = 8\ u \] \[\text{total molecular mass} = 2(1) + 16 = 18\ u \] \[\text{fraction} = \dfrac{8}{18} = \dfrac{4}{9} \] Answer: \(\displaystyle \dfrac{4}{9}\) of the mass of water comes from neutrons.

  10. Exercise 20

    Does the solubility of a substance change with temperature? Explain with the help of an example.

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    NCERT’s answer
    Yes, it is a temperature dependent property. The solubility generally, increases with increase in temperature. For example, you can dissolve more sugar in hot water than in cold water.
    Yes — for most solids dissolved in water, solubility rises as temperature rises: \(\displaystyle T\uparrow \Rightarrow\) solubility \(\displaystyle \uparrow\). For example, a saturated solution of potassium chloride takes up more salt when warmed; on cooling back to room temperature, the extra salt separates out as crystals. Answer: Solubility changes with temperature — it increases on heating and decreases on cooling, as shown by potassium chloride.

  11. Exercise 21

    Classify each of the following on the basis of their atomicity.
    (a)
    F2\displaystyle F_{2}
    (b)
    NO2\displaystyle NO_{2}
    (c)
    N2\displaystyle N_{2}O
    (d)
    C2\displaystyle C_{2}H6\displaystyle H_{6}
    (e)
    P4\displaystyle P_{4}
    (f)
    H2\displaystyle H_{2}O2\displaystyle O_{2}
    (g)
    P4\displaystyle P_{4}O10\displaystyle O_{10}
    (H)
    O3\displaystyle O_{3}
    (i)
    HCl
    (j) CH4\displaystyle CH_{4}
    (k) He
    (l) Ag

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    NCERT’s answer
    (a)
    $\displaystyle 2$ (b) $\displaystyle 3$ (c) $\displaystyle 3$ (d) $\displaystyle 8$ (e) $\displaystyle 4$ (f) $\displaystyle 4$ (g) $\displaystyle 14$ (h) $\displaystyle 3$ (i) $\displaystyle 2$ (j) $\displaystyle 5$ (k) $\displaystyle 1$ (Noble gases do not combine and exist as monoatomic gases) (l) Polyatomic. It is difficult to talk about the atomicity of metals as any measurable quantity will contain millions of atoms bound by metallic bond (about which you would learn later).
    Atomicity is the number of atoms per molecule. Monatomic ($\displaystyle 1$ atom): \(\displaystyle \text{He}\). Diatomic ($\displaystyle 2$ atoms): \(\displaystyle \text{F}_2\), \(\displaystyle \text{HCl}\). Triatomic ($\displaystyle 3$ atoms): \(\displaystyle \text{NO}_2\), \(\displaystyle \text{N}_2\text{O}\), \(\displaystyle \text{O}_3\). Poly-atomic ($\displaystyle 4$ or more atoms): \(\displaystyle \text{C}_2\text{H}_6\) ($\displaystyle 8$), \(\displaystyle \text{P}_4\) ($\displaystyle 4$), \(\displaystyle \text{H}_2\text{O}_2\) ($\displaystyle 4$), \(\displaystyle \text{P}_4\text{O}_{10}\) ($\displaystyle 14$), \(\displaystyle \text{CH}_4\) ($\displaystyle 5$). \(\displaystyle \text{Ag}\) is a metal: its atoms sit in one giant lattice, not a fixed-size molecule, so it too is classed poly-atomic. Answer: Monatomic -- \(\displaystyle \text{He}\); Diatomic -- \(\displaystyle \text{F}_2\), \(\displaystyle \text{HCl}\); Triatomic -- \(\displaystyle \text{NO}_2\), \(\displaystyle \text{N}_2\text{O}\), \(\displaystyle \text{O}_3\); Poly-atomic -- \(\displaystyle \text{C}_2\text{H}_6\)($\displaystyle 8$), \(\displaystyle \text{P}_4\)($\displaystyle 4$), \(\displaystyle \text{H}_2\text{O}_2\)($\displaystyle 4$), \(\displaystyle \text{P}_4\text{O}_{10}\)($\displaystyle 14$), \(\displaystyle \text{CH}_4\)($\displaystyle 5$), \(\displaystyle \text{Ag}\).

  12. Exercise 22

    You are provided with a fine white coloured powder which is either sugar or salt. How would you identify it without tasting?

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    NCERT’s answer
    On heating the powder, it will char if it is a sugar. Alternatively, the powder may be dissolved in water and checked for its conduction of electricity. If it conducts, it is a salt.
    Heat a pinch of the powder on a metal spatula over a flame, gently and away from the face.\[\text{sugar} \xrightarrow{\Delta} \text{melts, chars black, burnt smell} \] \[\text{NaCl (salt)} \xrightarrow{\Delta} \text{no melting, no charring, stays white} \] Answer: Charring with a burnt smell identifies sugar; an unchanged white powder identifies salt.

  13. Exercise 23

    Calculate the number of moles of magnesium present in a magnesium ribbon weighing 12\displaystyle 12 g. Molar atomic mass of magnesium is 24g mol−1\displaystyle mol^{-1}.

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    NCERT’s answer
    Number of moles = = $\displaystyle 0.5$ mol Long Answer Questions
    \[n = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{12\ \text{g}}{24\ \text{g mol}^{-1}} = 0.5\ \text{mol} \] Answer: $\displaystyle 0.5$ mol of magnesium.