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NCERT Exemplar · Class 9 Science Atoms and Molecules

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Multiple Choice Questions 1–10 (part 1 of 5)

  1. Exercise 1

    Which of the following correctly represents 360\displaystyle 360 g of water?
    (i)
    2\displaystyle 2 moles of H2\displaystyle H_{2}0\displaystyle 0
    (ii)
    20\displaystyle 20 moles of water
    (iii)
    6.022\displaystyle 022 × 1023\displaystyle 10^{23} molecules of water
    (iv)
    1.2044\displaystyle 2044×1025\displaystyle 10^{25} molecules of water
    (i)
    (b)(i)
    and (iv)
    (ii)
    and (iii)
    (ii)
    and (iv)

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    NCERT’s answer
    (ii)
    $\displaystyle 20$ moles of water = $\displaystyle 20$ ×$\displaystyle 18$ g = $\displaystyle 360$ g of water, because mass of $\displaystyle 1$ mole of water is the same as its molar mass, i.e., $\displaystyle 18$ g. (iv) $\displaystyle 1.2044$ × \(\displaystyle 10^{25}\) molecules of water contains × A $\displaystyle 1.2044$ $\displaystyle 10$ N number of moles, \(\displaystyle N_{A}\) = $\displaystyle 6.023$×$\displaystyle 10$ $\displaystyle 23$ ∴ × = × $\displaystyle 1.2044$ $\displaystyle 20$ moles $\displaystyle 6.022$ $\displaystyle 20$ moles of water = $\displaystyle 20$×$\displaystyle 18$ g = $\displaystyle 360$ g of water.
    (d) (ii) and (iv). \[n = \dfrac{m}{M} = \dfrac{360\ \text{g}}{18\ \text{g mol}^{-1}} = 20\ \text{mol} \] \[N = nN_A = 20\ \text{mol} \times 6.022\times10^{23}\ \text{mol}^{-1} = 1.2044\times10^{25}\ \text{molecules} \] $\displaystyle 20$ mol of \(\displaystyle \text{H}_2\text{O}\) and \(\displaystyle 1.2044\times10^{25}\) molecules are the same amount; $\displaystyle 2$ mol ($\displaystyle 36$ g) and \(\displaystyle 6.022\times10^{23}\) molecules ($\displaystyle 18$ g, (iii)) are not.

  2. Exercise 2

    Which of the following statements is not true about an atom?
    (a)
    Atoms are not able to exist independently
    (b)
    Atoms are the basic units from which molecules and ions are formed
    (c)
    Atoms are always neutral in nature
    (d)
    Atoms aggregate in large numbers to form the matter that we can see,
    feel or touch

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    NCERT’s answer
    (a)
    Inert gases exist in monoatomic form.
    (a) Atoms are not able to exist independently. Noble-gas atoms (\(\displaystyle \text{He}\), \(\displaystyle \text{Ne}\), \(\displaystyle \text{Ar}\)) exist as free, independent atoms, so the blanket claim in (a) is false; (b), (c), (d) hold for atoms in general.

  3. Exercise 3

    The chemical symbol for nitrogen gas is
    (a)
    Ni
    (b)
    N2\displaystyle N_{2}
    (c)
    N+\displaystyle N^{+} (d) N

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    NCERT’s answer
    (b)
    (b) \(\displaystyle \text{N}_2\). Free nitrogen occurs only as the diatomic molecule \(\displaystyle \text{N}_2\); N is the symbol for a single nitrogen atom, not the gas.

  4. Exercise 4

    The chemical symbol for sodium is
    (a)
    So
    (b)
    Sd
    (c)
    NA
    (d)
    Na

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    NCERT’s answer
    (d)
    (d) \(\displaystyle \text{Na}\). Sodium's symbol comes from the Latin natrium; (a), (b), (c) are not valid element symbols.

  5. Exercise 5

    Which of the following would weigh the highest?
    (a)
    0.2\displaystyle 2 mole of sucrose (C12\displaystyle C_{12} H22\displaystyle H_{22} O11\displaystyle O_{11})
    (b)
    2\displaystyle 2 moles of CO2\displaystyle CO_{2}
    (c)
    2\displaystyle 2 moles of CaCO3\displaystyle CaCO_{3}
    (d)
    10\displaystyle 10 moles of H2\displaystyle H_{2}O

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    NCERT’s answer
    (c)
    Weight of a sample in gram = number of moles × molar mass (a) $\displaystyle 0.2$ moles of \(\displaystyle C_{12}\)\(\displaystyle H_{22}\)\(\displaystyle O_{11}\) = $\displaystyle 0.2$ × $\displaystyle 342$ = $\displaystyle 68.4$ g (b) $\displaystyle 2$ moles of \(\displaystyle CO_{2}\) = $\displaystyle 2$×$\displaystyle 44$ = $\displaystyle 88$ g (c) $\displaystyle 2$ moles of \(\displaystyle CaCO_{3}\) = $\displaystyle 2$×$\displaystyle 100$ = $\displaystyle 200$ g (d) $\displaystyle 10$ moles of \(\displaystyle H_{2}\)O = $\displaystyle 10$×$\displaystyle 18$ = $\displaystyle 180$ g
    (c) $\displaystyle 2$ moles of \(\displaystyle \text{CaCO}_3\). \[m = nM \] \[\text{(a)}\ 0.2\ \text{mol} \times 342\ \text{g mol}^{-1} = 68.4\ \text{g} \] \[\text{(b)}\ 2\ \text{mol} \times 44\ \text{g mol}^{-1} = 88\ \text{g} \] \[\text{(c)}\ 2\ \text{mol} \times 100\ \text{g mol}^{-1} = 200\ \text{g} \] \[\text{(d)}\ 10\ \text{mol} \times 18\ \text{g mol}^{-1} = 180\ \text{g} \] $\displaystyle 200$ g is the largest of the four.

  6. Exercise 6

    Which of the following has maximum number of atoms?
    (a)
    18g of H2\displaystyle H_{2}O
    (b)
    18g of O2\displaystyle O_{2}
    (c)
    18g of CO2\displaystyle CO_{2}
    (d)
    18g of CH4\displaystyle CH_{4}

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    NCERT’s answer
    (d)
    Massof substance × Number of atoms in the molecule Number of atoms = × NA Molar mass ∴ (a) $\displaystyle 18$ g of water = A A N $\displaystyle 3$ N × × = (b) $\displaystyle 18$ g of oxygen = A A N $\displaystyle 1.12$ N × × = (c) $\displaystyle 18$ g of \(\displaystyle CO_{2}\) = A A N 1.23N × × = (d) $\displaystyle 18$ g of \(\displaystyle CH_{4}\) = A A N 5.63N × × =
    (d) $\displaystyle 18$ g of \(\displaystyle \text{CH}_4\). \[n = \dfrac{18\ \text{g}}{M},\qquad \text{atoms} = n \times (\text{atoms per molecule}) \times N_A \] \[\text{H}_2\text{O}:\ \dfrac{18}{18}\times3 = 3\,N_A \qquad \text{O}_2:\ \dfrac{18}{32}\times2 = 1.125\,N_A \] \[\text{CO}_2:\ \dfrac{18}{44}\times3 = 1.227\,N_A \qquad \text{CH}_4:\ \dfrac{18}{16}\times5 = 5.625\,N_A \] \(\displaystyle \text{CH}_4\) has the lightest molar mass and five atoms per molecule, so it wins on both counts.

  7. Exercise 7

    Which of the following contains maximum number of molecules?
    (a)
    1g CO2\displaystyle CO_{2}
    (b)
    1g N2\displaystyle N_{2}
    (c)
    1g H2\displaystyle H_{2}
    (d)
    1g CH4\displaystyle CH_{4}

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    NCERT’s answer
    (c)
    $\displaystyle 1$ g of \(\displaystyle H_{2}\) = A A N 0.5N $\displaystyle 2$ × = = $\displaystyle 0.5$×$\displaystyle 6.022$×\(\displaystyle 10^{23}\) = $\displaystyle 3.011$×\(\displaystyle 10^{23}\)
    (c) $\displaystyle 1$ g of \(\displaystyle \text{H}_2\). \[n = \dfrac{1\ \text{g}}{M} \] \[\text{CO}_2:\ \dfrac{1}{44} = 0.0227\ \text{mol} \qquad \text{N}_2:\ \dfrac{1}{28} = 0.0357\ \text{mol} \] \[\text{H}_2:\ \dfrac{1}{2} = 0.5\ \text{mol} \qquad \text{CH}_4:\ \dfrac{1}{16} = 0.0625\ \text{mol} \] Molecules \(\displaystyle = nN_A\); the lightest molar mass gives the most moles per gram, hence the most molecules.

  8. Exercise 8

    Mass of one atom of oxygen is
    (a)
    166.023×1023 g\displaystyle \dfrac{16}{6.023\times10^{23}}\ \text{g}
    (b)
    326.023×1023 g\displaystyle \dfrac{32}{6.023\times10^{23}}\ \text{g}
    (c)
    16.023×1023 g\displaystyle \dfrac{1}{6.023\times10^{23}}\ \text{g}
    (d)
    8u\displaystyle 8u

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    NCERT’s answer
    (a)
    Mass of one atom of oxygen = Atomic mass/\(\displaystyle N_{A}\) = × g $\displaystyle 6.022$ $\displaystyle 10$
    (a) \[m(\text{O}) = \dfrac{M}{N_A} \]
    \[m(\text{O}) = \dfrac{16\ \text{g mol}^{-1}}{6.023\times10^{23}\ \text{mol}^{-1}} = \dfrac{16}{6.023\times10^{23}}\ \text{g} \]
    (d)
    is wrong: one \(\displaystyle \text{O}\) atom is \(\displaystyle 16u\), not \(\displaystyle 8u\).
    Answer: (a) \(\displaystyle \dfrac{16}{6.023\times10^{23}}\ \text{g}\).

  9. Exercise 9

    3.42\displaystyle 42 g of sucrose are dissolved in 18g of water in a beaker. The number of
    oxygen atoms in the solution are
    (a)
    6.68\displaystyle 68 × 1023\displaystyle 10^{23}
    (b)
    6.09\displaystyle 09 × 1022\displaystyle 10^{22}
    (c)
    6.022\displaystyle 022 × 1023\displaystyle 10^{23}
    (d)
    6.022\displaystyle 022 × 1021\displaystyle 10^{21}

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    NCERT’s answer
    (a)
    Number of moles of sucrose = Massof substance Molar mass $\displaystyle 3.42$ g 0.01mol $\displaystyle 342$ g \(\displaystyle mol^{−}\) = = $\displaystyle 1$ mol of sucrose (\(\displaystyle C_{12}\) \(\displaystyle H_{22}\) \(\displaystyle O_{11}\)) contains = $\displaystyle 11$× \(\displaystyle N_{A}\) atoms of oxygen $\displaystyle 0.01$ mol of sucrose (\(\displaystyle C_{12}\) \(\displaystyle H_{22}\) \(\displaystyle O_{11}\)) contains = $\displaystyle 0.01$ × $\displaystyle 11$ × \(\displaystyle N_{A}\) atoms of oxygen = $\displaystyle 0.11$× \(\displaystyle N_{A}\) atoms of oxygen Number of moles of water 18g 1mol 18g \(\displaystyle mol^{−}\) = = $\displaystyle 1$ mol of water (\(\displaystyle H_{2}\)O) contains $\displaystyle 1$×N A atom of oxygen Total number of oxygen atoms = Number of oxygen atoms from sucrose + Number of oxygen atoms from water = $\displaystyle 0.11$ \(\displaystyle N_{A}\) + $\displaystyle 1.0$ \(\displaystyle N_{A}\) = 1.\(\displaystyle 11N_{A}\) Number of oxygen atoms in solution = $\displaystyle 1.11$ × Avogadro’s number = $\displaystyle 1.11$ × $\displaystyle 6.022$ ×\(\displaystyle 10^{23}\) = $\displaystyle 6.68$ × \(\displaystyle 10^{23}\)
    (a) \(\displaystyle 6.68\times10^{23}\). \[n_{\text{sucrose}} = \dfrac{3.42\ \text{g}}{342\ \text{g mol}^{-1}} = 0.01\ \text{mol}, \qquad n_{\text{O, sucrose}} = 0.01 \times 11 = 0.11\ \text{mol} \] \[n_{\text{water}} = \dfrac{18\ \text{g}}{18\ \text{g mol}^{-1}} = 1\ \text{mol}, \qquad n_{\text{O, water}} = 1 \times 1 = 1\ \text{mol} \] \[n_{\text{O, total}} = 0.11 + 1 = 1.11\ \text{mol} \] \[N_{\text{O}} = 1.11\ \text{mol} \times 6.022\times10^{23}\ \text{mol}^{-1} = 6.68\times10^{23}\ \text{atoms} \]

  10. Exercise 10

    A change in the physical state can be brought about
    (a)
    only when energy is given to the system
    (b)
    only when energy is taken out from the system
    (c)
    when energy is either given to, or taken out from the system
    (d)
    without any energy change

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    NCERT’s answer
    (c)
    (c) when energy is either given to, or taken out from the system. \[\text{solid} \xrightarrow{+\text{heat}} \text{liquid} \xrightarrow{+\text{heat}} \text{gas} \] Melting and boiling take energy in; freezing and condensation give it out, so a state change always exchanges energy.