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NCERT Exemplar · Class 9 Science Atoms and Molecules

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Long Answer Questions 34–43 (part 4 of 5)

  1. Exercise 34

    What is the SI prefix for each of the following multiples and submultiples
    of a unit?
    (a)
    103\displaystyle 10^{3}
    (b)
    10−1\displaystyle 10^{-1}
    (c)
    10−2\displaystyle 10^{-2}(d) 10−6\displaystyle 10^{-6}
    (e)
    10−9\displaystyle 10^{-9}(f) 10−12\displaystyle 10^{-12}

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    NCERT’s answer
    (a)
    kilo (b) deci (c) centi (d) micro (e) nano (f) pico
    \[(a)\ 10^{3} = \text{kilo (k)}\] \[(b)\ 10^{-1} = \text{deci (d)}\] \[(c)\ 10^{-2} = \text{centi (c)}\] \[(d)\ 10^{-6} = \text{micro (}\mu\text{)}\] \[(e)\ 10^{-9} = \text{nano (n)}\] \[(f)\ 10^{-12} = \text{pico (p)}\] Answer: (a) kilo (b) deci (c) centi (d) micro (e) nano (f) pico.

  2. Exercise 35

    Express each of the following in kilograms
    (a)
    5.84\displaystyle 84×10−3\displaystyle 10^{-3} mg
    (b)
    58.34\displaystyle 34 g
    (c)
    0.584g
    (d)
    5.873\displaystyle 873×10−21\displaystyle 10^{-21}g

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    NCERT’s answer
    (a)
    5.$\displaystyle 84$ ×\(\displaystyle 10^{-9}\) kg (b) $\displaystyle 5.834$ ×\(\displaystyle 10^{-2}\) kg (c) $\displaystyle 5.84$ ×\(\displaystyle 10^{-4}\) kg (d) $\displaystyle 5.873$ ×\(\displaystyle 10^{-24}\) kg
    Convert each mass to kilograms via \(\displaystyle 1\ \text{mg}=10^{-6}\ \text{kg}\) and \(\displaystyle 1\ \text{g}=10^{-3}\ \text{kg}\). \[\text{(a) } 5.84\times10^{-3}\ \text{mg} \times 10^{-6}\ \dfrac{\text{kg}}{\text{mg}} = 5.84\times10^{-9}\ \text{kg}\] \[\text{(b) } 58.34\ \text{g} \times 10^{-3}\ \dfrac{\text{kg}}{\text{g}} = 5.834\times10^{-2}\ \text{kg}\] \[\text{(c) } 0.584\ \text{g} \times 10^{-3}\ \dfrac{\text{kg}}{\text{g}} = 5.84\times10^{-4}\ \text{kg}\] \[\text{(d) } 5.873\times10^{-21}\ \text{g} \times 10^{-3}\ \dfrac{\text{kg}}{\text{g}} = 5.873\times10^{-24}\ \text{kg}\] Answer: (a) \(\displaystyle 5.84\times10^{-9}\ \text{kg}\) (b) \(\displaystyle 5.834\times10^{-2}\ \text{kg}\) (c) \(\displaystyle 5.84\times10^{-4}\ \text{kg}\) (d) \(\displaystyle 5.873\times10^{-24}\ \text{kg}\).

  3. Exercise 36

    Compute the difference in masses of 103\displaystyle 10^{3} moles each of magnesium atoms and magnesium ions. (Mass of an electron = 9.1\displaystyle 9.1×10−31\displaystyle 10^{-31} kg)

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    NCERT’s answer
    A \(\displaystyle Mg^{2+}\) ion and Mg atom differ by two electrons. \(\displaystyle 10^{3}\) moles of Mg $\displaystyle 2$+ and Mg atoms would differ by \(\displaystyle 10^{3}\) × $\displaystyle 2$ moles of electrons Mass of $\displaystyle 2$ ×\(\displaystyle 10^{3}\) moles of electrons = $\displaystyle 2$×\(\displaystyle 10^{3}\) × $\displaystyle 6.023$ ×\(\displaystyle 10^{23}\) × $\displaystyle 9.1$ ×\(\displaystyle 10^{-31}\) kg ⇒ $\displaystyle 2$×$\displaystyle 6.022$ × $\displaystyle 9.1$×$\displaystyle 10$ -5kg ⇒ $\displaystyle 109.6004$ ×\(\displaystyle 10^{-5}\) kg ⇒ $\displaystyle 1.096$ × \(\displaystyle 10^{-3}\)kg
    Mg atoms form \(\displaystyle \text{Mg}^{2+}\) by losing two electrons each, so the mass gap is the mass of those electrons over all \(\displaystyle 10^{3}\) moles. \[N = 10^{3}\ \text{mol} \times 6.022\times10^{23}\ \text{mol}^{-1} = 6.022\times10^{26}\ \text{atoms}\] \[\Delta m = N \times 2 \times m_e = 6.022\times10^{26} \times 2 \times 9.1\times10^{-31}\ \text{kg}\] \[\Delta m = 1.096\times10^{-3}\ \text{kg}\] Answer: \(\displaystyle \Delta m \approx 1.096\times10^{-3}\ \text{kg}\); the atoms are heavier than the ions by this mass.

  4. Exercise 37

    Which has more number of atoms? 100g of N2\displaystyle N_{2} or 100\displaystyle 100 g of NH3\displaystyle NH_{3}

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    NCERT’s answer
    (i)
    $\displaystyle 100$ g of \(\displaystyle N_{2}\) = $\displaystyle 100$ $\displaystyle 28$ moles Number of molecules = $\displaystyle 100$ $\displaystyle 28$ × $\displaystyle 6.022$ ×\(\displaystyle 10^{23}\) Number of atoms = $\displaystyle 2$ ×$\displaystyle 100$ × $\displaystyle 6.022$ ×$\displaystyle 10$ = $\displaystyle 43.01$×\(\displaystyle 10^{23}\) (ii) $\displaystyle 100$ g of \(\displaystyle NH_{3}\) = $\displaystyle 100$ $\displaystyle 17$ moles = $\displaystyle 100$ $\displaystyle 17$ × $\displaystyle 6.022$×\(\displaystyle 10^{23}\) molecules $\displaystyle 100$ × $\displaystyle 6.022$ ×$\displaystyle 10$ × $\displaystyle 4$ atoms = = $\displaystyle 141.69$ ×\(\displaystyle 10^{23}\) \(\displaystyle NH_{3}\) would have more atoms
    \[n = \dfrac{m}{M} \] For \(\displaystyle \text{N}_2\) (\(\displaystyle M=28\ \text{g mol}^{-1}\)): \[n(\text{N}_2) = \dfrac{100\ \text{g}}{28\ \text{g mol}^{-1}} = 3.571\ \text{mol} \] \[N(\text{atoms}) = n\times2\times N_A = 3.571\ \text{mol}\times2\times6.022\times10^{23}\ \text{mol}^{-1} = 4.30\times10^{24} \] For \(\displaystyle \text{NH}_3\) (\(\displaystyle M=17\ \text{g mol}^{-1}\)): \[n(\text{NH}_3) = \dfrac{100\ \text{g}}{17\ \text{g mol}^{-1}} = 5.882\ \text{mol} \] \[N(\text{atoms}) = n\times4\times N_A = 5.882\ \text{mol}\times4\times6.022\times10^{23}\ \text{mol}^{-1} = 1.417\times10^{25} \] Answer: $\displaystyle 100$ g of \(\displaystyle \text{NH}_3\) has more atoms (\(\displaystyle 1.417\times10^{25}\) against \(\displaystyle 4.30\times10^{24}\) for \(\displaystyle \text{N}_2\)).

  5. Exercise 38

    Compute the number of ions present in 5.85\displaystyle 5.85 g of sodium chloride.

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    NCERT’s answer
    5.$\displaystyle 85$ g of NaCl = $\displaystyle 5.85$ = 0.1moles $\displaystyle 58.5$ or $\displaystyle 0.1$ moles of NaCl particle Each NaCl particle is equivalent to one \(\displaystyle Na^{+}\) one \(\displaystyle Cl^{-}\) ⇒ $\displaystyle 2$ ions Total moles of ions = $\displaystyle 0.1$ × $\displaystyle 2$ $\displaystyle 0.2$ moles No. of ions= $\displaystyle 0.2$ × $\displaystyle 6.022$ ×\(\displaystyle 10^{23}\) ⇒ × $\displaystyle 1.2042$ ions
    \(\displaystyle \text{NaCl}\) (molar mass \(\displaystyle 58.5\ \text{g mol}^{-1}\)) gives \(\displaystyle \text{Na}^{+}\) and \(\displaystyle \text{Cl}^{-}\), two ions per formula unit. \[n=\dfrac{5.85\ \text{g}}{58.5\ \text{g mol}^{-1}}=0.1\ \text{mol}\] \[\text{formula units}=n\times N_A=0.1\times6.022\times10^{23}=6.022\times10^{22}\] \[\text{ions}=2\times6.022\times10^{22}=1.2044\times10^{23}\] Answer: \(\displaystyle 1.2044\times10^{23}\) ions in all (\(\displaystyle 6.022\times10^{22}\) each of \(\displaystyle \text{Na}^{+}\) and \(\displaystyle \text{Cl}^{-}\)).

  6. Exercise 39

    A gold sample contains 90\displaystyle 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold?

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    NCERT’s answer
    One gram of gold sample will contain $\displaystyle 90$ 0.9g of gold $\displaystyle 100$ = Number of moles of gold = Massof gold Atomicmassof gold = $\displaystyle 0.9$ $\displaystyle 0.0046$ $\displaystyle 197$ = One mole of gold contains \(\displaystyle N_{A}\) atoms = $\displaystyle 6.022$ ×\(\displaystyle 10^{23}\) ∴ $\displaystyle 0.0046$ mole of gold will contain = $\displaystyle 0.0046$ × $\displaystyle 6.022$ ×\(\displaystyle 10^{23}\) = $\displaystyle 2.77$×$\displaystyle 1021$
    Copper is set aside; only the gold fraction contributes atoms. \[m_{\text{Au}} = 90\% \times 1\ \text{g} = 0.9\ \text{g} \] \[n_{\text{Au}} = \frac{m_{\text{Au}}}{M_{\text{Au}}} = \frac{0.9\ \text{g}}{197\ \text{g mol}^{-1}} = 4.57\times10^{-3}\ \text{mol} \] \[N_{\text{Au}} = n_{\text{Au}} N_A = 4.57\times10^{-3}\ \text{mol} \times 6.022\times10^{23}\ \text{mol}^{-1} = 2.75\times10^{21}\ \text{atoms} \] Answer: \(\displaystyle 2.75\times10^{21}\) atoms of gold.

  7. Exercise 40

    What are ionic and molecular compounds? Give examples.

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    NCERT’s answer
    Atoms of different elements join together in definite proportions to form molecules of compounds. Examples— water, ammonia, carbondioxide. Compounds composed of metals and non-metals contain charged species. The charged species are known as ions. An ion is a charged particle and can be negatively or positively charged. A negatively charged ion is called an anion and the positively charged ion is called cation. Examples— sodium chloride, calcium oxide.
    Ionic compounds form between a metal and a non-metal, held together as oppositely charged ions, e.g. \(\displaystyle \text{NaCl}\), \(\displaystyle \text{CaO}\). Molecular compounds form between atoms of non-metals joined in fixed proportion, e.g. \(\displaystyle \text{H}_2\text{O}\), \(\displaystyle \text{NH}_3\), \(\displaystyle \text{CO}_2\). Answer: Ionic -- metal + non-metal as charged ions, e.g. \(\displaystyle \text{NaCl}\); Molecular -- non-metal atoms in fixed proportion, e.g. \(\displaystyle \text{H}_2\text{O}\).

  8. Exercise 41

    Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions. (Mass of an electron is 9.1\displaystyle 9.1×10−28\displaystyle 10^{-28} g). Which one is heavier?

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    NCERT’s answer
    Mass of $\displaystyle 1$ mole of aluminium atom = the molar mass of aluminium = $\displaystyle 27$ g \(\displaystyle mol^{-1}\) An aluminium atom needs to lose three electrons to become an ion, \(\displaystyle Al^{3+}\) For one mole of \(\displaystyle A1^{3+}\) ion, three moles of electrons are to be lost. The mass of three moles of electrons = $\displaystyle 3$ × ($\displaystyle 9.1$×\(\displaystyle 10^{-28}\)) × $\displaystyle 6.022$×\(\displaystyle 10^{23}\) g = $\displaystyle 27.3$ × $\displaystyle 6.022$ ×\(\displaystyle 10^{-5}\) g = $\displaystyle 164.400$ ×$\displaystyle 10$ -$\displaystyle 5$ g = $\displaystyle 0.00164$ g Molar mass of \(\displaystyle Al^{3+}\) = ($\displaystyle 27$-$\displaystyle 0.00164$) g \(\displaystyle mol^{-1}\) = $\displaystyle 26.9984$ g \(\displaystyle mol^{-1}\) Difference = $\displaystyle 27$ - $\displaystyle 26.9984$ = $\displaystyle 0.0016$ g
    Forming \(\displaystyle \text{Al}^{3+}\) removes three electrons per atom; the mass difference is only that of the lost electrons. \[\Delta m = 3 \times N_A \times m_e \] \[\Delta m = 3 \times 6.022\times10^{23}\ \text{mol}^{-1} \times 9.1\times10^{-28}\ \text{g} \] \[\Delta m = 1.644\times10^{-3}\ \text{g} \] Answer: one mole of \(\displaystyle \text{Al}\) atoms is heavier than one mole of \(\displaystyle \text{Al}^{3+}\) ions by \(\displaystyle 1.644\times10^{-3}\ \text{g}\).

  9. Exercise 42

    A silver ornament of mass ‘m’ gram is polished with gold equivalent to 1\displaystyle 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.

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    NCERT’s answer
    Mass of silver = m g Mass of gold = m g Number of atoms of silver = A Mass N Atomic mass × = A m N $\displaystyle 108$ × Number of atoms of gold = A m N $\displaystyle 100$ $\displaystyle 197$× × Ratio of number of atoms of gold to silver = Au : Ag = A A m m N : N $\displaystyle 100$ $\displaystyle 197$ × × × = $\displaystyle 108$ : $\displaystyle 100$×$\displaystyle 197$ = $\displaystyle 108$ : $\displaystyle 19700$ = $\displaystyle 1$ : $\displaystyle 182.41$
    Equal moles give equal numbers of atoms through \(\displaystyle N_A\), so it cancels out of the ratio. \[m_{\text{Ag}} = m\ \text{g}, \qquad m_{\text{Au}} = 1\% \times m = 0.01m\ \text{g} \] \[n_{\text{Au}} = \frac{0.01m}{197}\ \text{mol}, \qquad n_{\text{Ag}} = \frac{m}{108}\ \text{mol} \] \[\frac{N_{\text{Au}}}{N_{\text{Ag}}} = \frac{n_{\text{Au}}}{n_{\text{Ag}}} = \frac{0.01m/197}{m/108} = \frac{0.01\times108}{197} = \frac{1.08}{197} \] \[\frac{N_{\text{Au}}}{N_{\text{Ag}}} = 5.48\times10^{-3} \approx \frac{27}{4925} \] Answer: \(\displaystyle N_{\text{Au}} : N_{\text{Ag}} \approx 27 : 4925\) (about $\displaystyle 1$ gold atom per $\displaystyle 182$ silver atoms).

  10. Exercise 43

    A sample of ethane (C2\displaystyle C_{2}H6\displaystyle H_{6}) gas has the same mass as 1.5\displaystyle 1.5 ×1020\displaystyle 10^{20} molecules of methane (CH4\displaystyle CH_{4}). How many C2\displaystyle C_{2}H6\displaystyle H_{6} molecules does the sample of gas contain?

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    NCERT’s answer
    Mass of $\displaystyle 1$ molecule of \(\displaystyle CH_{4}\) = A 16g N Mass of $\displaystyle 1.5$ ×\(\displaystyle 10^{20}\) molecules of methane = A $\displaystyle 1.5$ $\displaystyle 16$ g N × × Mass of $\displaystyle 1$ molecule of \(\displaystyle C_{2}\)\(\displaystyle H_{6}\) = A $\displaystyle 30$ g N Mass of molecules of \(\displaystyle C_{2}\)\(\displaystyle H_{6}\) is = A $\displaystyle 1.5$ $\displaystyle 16$ g N × × ∴ Number of molecules of ethane = A A $\displaystyle 1.5$ N $\displaystyle 0.8$ N × × × = ×
    Same mass, different molar mass — the heavier gas gives fewer moles. \[n_{\text{CH}_4} = \frac{N_{\text{CH}_4}}{N_A} = \frac{1.5\times10^{20}}{6.022\times10^{23}\ \text{mol}^{-1}} = 2.49\times10^{-4}\ \text{mol} \] \[m = n_{\text{CH}_4} \times M_{\text{CH}_4} = 2.49\times10^{-4}\ \text{mol} \times 16\ \text{g mol}^{-1} = 3.99\times10^{-3}\ \text{g} \] \[n_{\text{C}_2\text{H}_6} = \frac{m}{M_{\text{C}_2\text{H}_6}} = \frac{3.99\times10^{-3}\ \text{g}}{30\ \text{g mol}^{-1}} = 1.33\times10^{-4}\ \text{mol} \] \[N_{\text{C}_2\text{H}_6} = n_{\text{C}_2\text{H}_6} N_A = 1.33\times10^{-4}\ \text{mol} \times 6.022\times10^{23}\ \text{mol}^{-1} = 8\times10^{19} \] Answer: \(\displaystyle 8\times10^{19}\) molecules of \(\displaystyle \text{C}_2\text{H}_6\).