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NCERT Exemplar · Class 9 Science Atoms and Molecules

48 questions · 48 still being checked

Long Answer Questions 24–33 (part 3 of 5)

  1. Exercise 24

    Verify by calculating that
    (a)
    5\displaystyle 5 moles of CO2\displaystyle CO_{2} and 5\displaystyle 5 moles of H2\displaystyle H_{2}O do not have the same mass.
    (b)
    240\displaystyle 240 g of calcium and 240\displaystyle 240 g magnesium elements have a mole ratio of

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    NCERT’s answer
    (a)
    \(\displaystyle CO_{2}\) has molar mass = 44g \(\displaystyle mol^{-1}\) $\displaystyle 5$ moles of \(\displaystyle CO_{2}\) have molar mass = $\displaystyle 44$ × $\displaystyle 5$ = $\displaystyle 220$ g \(\displaystyle H_{2}\)O has molar mass = $\displaystyle 18$ g \(\displaystyle mol^{-1}\) $\displaystyle 5$ moles of H 2O have mass = $\displaystyle 18$ × $\displaystyle 5$ g = $\displaystyle 90$ g (b) Number of moles in 240g Ca metal = = Number of moles in 240g of Mg metal = = Ratio $\displaystyle 6$:$\displaystyle 10$ $\displaystyle 3$: $\displaystyle 5$
    (a) \[M(\text{CO}_2) = 12 + 2(16) = 44\ \text{g mol}^{-1} \] \[M(\text{H}_2\text{O}) = 2(1) + 16 = 18\ \text{g mol}^{-1} \] \[m = nM \Rightarrow m(\text{CO}_2) = 5\times44 = 220\ \text{g}, \quad m(\text{H}_2\text{O}) = 5\times18 = 90\ \text{g} \] (b) \[n = \dfrac{m}{M} \Rightarrow n(\text{Ca}) = \dfrac{240}{40} = 6\ \text{mol}, \quad n(\text{Mg}) = \dfrac{240}{24} = 10\ \text{mol} \] \[n(\text{Ca}):n(\text{Mg}) = 6:10 = 3:5 \] Answer: (a) $\displaystyle 220$ g \(\displaystyle \ne\) $\displaystyle 90$ g — unequal despite equal moles. (b) Mole ratio confirmed, $\displaystyle 3$:5.

  2. Exercise 25

    Find the ratio by mass of the combining elements in the following
    compounds. (You may use Appendix-III)
    (a)
    CaCO3\displaystyle CaCO_{3}
    (d)
    C2\displaystyle C_{2}H5\displaystyle H_{5}OH
    (b)
    MgCl2\displaystyle MgCl_{2}
    (e)
    NH3\displaystyle NH_{3}
    (c)
    H2\displaystyle H_{2}SO4\displaystyle SO_{4}
    (f)
    Ca(OH)2\displaystyle 2

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    NCERT’s answer
    (a)
    Ca \(\displaystyle CO_{3}\) (b) \(\displaystyle MgCl_{2}\) (c) \(\displaystyle H_{2}\) \(\displaystyle SO_{4}\) Ca : C:O × $\displaystyle 3$ Mg : Cl × $\displaystyle 2$ H × $\displaystyle 2$ : S : O × $\displaystyle 4$ $\displaystyle 40$ : $\displaystyle 12$ : $\displaystyle 16$×$\displaystyle 3$ $\displaystyle 24$ : $\displaystyle 35.5$ × $\displaystyle 2$ $\displaystyle 1$ × $\displaystyle 2$ : $\displaystyle 32$ : $\displaystyle 16$ × $\displaystyle 4$ $\displaystyle 40$ : $\displaystyle 12$ : $\displaystyle 48$ $\displaystyle 24$ : $\displaystyle 71$ $\displaystyle 2$ : $\displaystyle 32$ : $\displaystyle 64$ $\displaystyle 10$ : $\displaystyle 3$: $\displaystyle 12$ $\displaystyle 1$ : $\displaystyle 16$ : $\displaystyle 32$ (d) \(\displaystyle C_{2}\) \(\displaystyle H_{5}\) OH (e) \(\displaystyle NH_{3}\) (f) Ca (OH)$\displaystyle 2$ C × $\displaystyle 2$ : H × $\displaystyle 6$: O N : H × $\displaystyle 3$ Ca : O × $\displaystyle 2$ : H × $\displaystyle 2$ $\displaystyle 12$ × $\displaystyle 2$ : $\displaystyle 1$ × $\displaystyle 6$: $\displaystyle 16$ $\displaystyle 14$: $\displaystyle 1$ × $\displaystyle 3$ $\displaystyle 40$ : $\displaystyle 16$ ×$\displaystyle 2$ : $\displaystyle 1$ × $\displaystyle 2$ $\displaystyle 24$ : $\displaystyle 6$ : $\displaystyle 16$ $\displaystyle 14$:$\displaystyle 3$ $\displaystyle 40$ : $\displaystyle 32$ : $\displaystyle 2$ $\displaystyle 12$ : $\displaystyle 3$ : $\displaystyle 8$ $\displaystyle 20$ : $\displaystyle 16$ : $\displaystyle 1$
    (a) \(\displaystyle \text{CaCO}_3\) \[Ca:C:O = 40:12:3(16) = 40:12:48 = 10:3:12 \] (b) \(\displaystyle \text{MgCl}_2\) \[Mg:Cl = 24:2(35.5) = 24:71 \] (c) \(\displaystyle \text{H}_2\text{SO}_4\) \[H:S:O = 2(1):32:4(16) = 2:32:64 = 1:16:32 \] (d) \(\displaystyle \text{C}_2\text{H}_5\text{OH}\) \[C:H:O = 2(12):6(1):16 = 24:6:16 = 12:3:8 \] (e) \(\displaystyle \text{NH}_3\) \[N:H = 14:3(1) = 14:3 \] (f) \(\displaystyle \text{Ca(OH)}_2\) \[Ca:O:H = 40:2(16):2(1) = 40:32:2 = 20:16:1 \] Answer: ratios $\displaystyle 10$:$\displaystyle 3$:$\displaystyle 12$, $\displaystyle 24$:$\displaystyle 71$, $\displaystyle 1$:$\displaystyle 16$:$\displaystyle 32$, $\displaystyle 12$:$\displaystyle 3$:$\displaystyle 8$, $\displaystyle 14$:$\displaystyle 3$, $\displaystyle 20$:$\displaystyle 16$:1.

  3. Exercise 26

    Calcium chloride when dissolved in water dissociates into its ions according to the following equation. CaCl2\displaystyle CaCl_{2} (aq) → Ca2+\displaystyle Ca^{2+} (aq) + 2Cl−\displaystyle 2Cl^{-} (aq) Calculate the number of ions obtained from CaCl2\displaystyle CaCl_{2} when 222\displaystyle 222 g of it is dissolved in water.

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    NCERT’s answer
    $\displaystyle 1$ mole of calcium chloride = 111g ∴ 222g of \(\displaystyle CaCl_{2}\) is equivalent to $\displaystyle 2$ moles of \(\displaystyle CaCl_{2}\) Since $\displaystyle 1$ formula unit \(\displaystyle CaCl_{2}\) gives $\displaystyle 3$ ions, therefore, $\displaystyle 1$ mol of \(\displaystyle CaCl_{2}\) will give $\displaystyle 3$ moles of ions $\displaystyle 2$ moles of \(\displaystyle CaCl_{2}\) would give $\displaystyle 3$×$\displaystyle 2$=$\displaystyle 6$ moles of ions. No. of ions = No. of moles of ions × Avogadro number = $\displaystyle 6$ × $\displaystyle 6.022$ ×$\displaystyle 10$ $\displaystyle 23$ = $\displaystyle 36.132$ ×$\displaystyle 10$ $\displaystyle 23$ = $\displaystyle 3.6$ $\displaystyle 132$ ×\(\displaystyle 10^{24}\) ions
    \[M(\text{CaCl}_2) = 40 + 2(35.5) = 111\ \text{g mol}^{-1} \] \[n(\text{CaCl}_2) = \dfrac{222}{111} = 2\ \text{mol} \] Each mole gives $\displaystyle 1$ \(\displaystyle Ca^{2+}\) and $\displaystyle 2$ \(\displaystyle Cl^{-}\): \[n(Ca^{2+}) = 2\times1 = 2\ \text{mol}, \quad n(Cl^{-}) = 2\times2 = 4\ \text{mol} \] \[n(\text{ions}) = 2+4 = 6\ \text{mol} \] \[N(\text{ions}) = nN_A = 6\times6.022\times10^{23} = 3.6132\times10^{24} \] Answer: \(\displaystyle 3.6132\times10^{24}\) ions.

  4. Exercise 27

    The difference in the mass of 100\displaystyle 100 moles each of sodium atoms and sodium ions is 5.48002\displaystyle 5.48002 g. Compute the mass of an electron.

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    \[N(e^{-}) = n(\text{Na})\,N_A = 100\ \text{mol} \times 6.022\times10^{23}\ \text{mol}^{-1} = 6.022\times10^{25} \] One electron is lost per \(\displaystyle \text{Na}\to\text{Na}^{+}\), so the mass lost equals the mass of these electrons: \[m(e^{-}) = \dfrac{\Delta m}{N(e^{-})} \] On the printed data, \(\displaystyle \Delta m = 5.48002\ \text{g}\): \[m(e^{-}) = \dfrac{5.48002\ \text{g}}{6.022\times10^{25}} = 9.1\times10^{-26}\ \text{g} \] This is \(\displaystyle 100\times\) the accepted electron mass, so \(\displaystyle \Delta m\) must be \(\displaystyle 5.48\times10^{-2}\ \text{g}\) (\(\displaystyle 0.0548\ \text{g}\)), not \(\displaystyle 5.48002\ \text{g}\): \[m(e^{-}) = \dfrac{5.48\times10^{-2}\ \text{g}}{6.022\times10^{25}} = 9.1\times10^{-28}\ \text{g} = 9.1\times10^{-31}\ \text{kg} \] Answer: mass of one electron \(\displaystyle =9.1\times10^{-28}\ \text{g} = 9.1\times10^{-31}\ \text{kg}\).NCERT prints: Mass of one electron \(\displaystyle = \dfrac{5.48002}{100\times6.022\times10^{23}} = 9.1\times10^{-28}\ \text{g}\) — the division itself is wrong: \(\displaystyle 5.48002 \div 6.022\times10^{25} = 9.1\times10^{-26}\ \text{g}\), not \(\displaystyle 10^{-28}\).

  5. Exercise 28

    Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225\displaystyle 225 g of pure HgS? Molar mass of Hg and S are 200.6\displaystyle 200.6 g mol−1\displaystyle mol^{-1} and 32\displaystyle 32 g mol−1\displaystyle mol^{-1} respectively.

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    NCERT’s answer
    Molar mass of HgS = $\displaystyle 200.6$ + $\displaystyle 32$ = $\displaystyle 232.6$ g \(\displaystyle mol^{-1}\) Mass of Hg in $\displaystyle 232.6$ g of HgS = $\displaystyle 200.6$ g Mass of Hg in $\displaystyle 225$ g of HgS $\displaystyle 200.6$ $\displaystyle 232.6$ = × $\displaystyle 225$ = 194.04g
    \[M(\text{HgS}) = 200.6+32 = 232.6\ \text{g mol}^{-1} \] \[m(\text{Hg}) = 225\times\dfrac{200.6}{232.6} = 194.05\ \text{g} \] Answer: \(\displaystyle \approx194.05\ \text{g}\) of mercury.

  6. Exercise 29

    The mass of one steel screw is 4.11g. Find the mass of one mole of these steel screws. Compare this value with the mass of the Earth (5.98\displaystyle 5.98 × 1024\displaystyle 10^{24}kg). Which one of the two is heavier and by how many times?

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    NCERT’s answer
    One mole of screws weigh = $\displaystyle 2.475$ ×\(\displaystyle 10^{24}\)g = $\displaystyle 2.475$×\(\displaystyle 10^{21}\) kg $\displaystyle 5.98$ ×$\displaystyle 10$ kg Mass of the Earth = $\displaystyle 2.4$ ×$\displaystyle 10$ Mass of $\displaystyle 1$ mole of screws $\displaystyle 2.475$× $\displaystyle 10$ kg = Mass of earth is $\displaystyle 2.4$×\(\displaystyle 10^{3}\) times the mass of screws The earth is $\displaystyle 2400$ times heavier than one mole of screws.
    \[m_{1\,\text{mol}} = 4.11\ \text{g} \times 6.022\times10^{23}\ \text{mol}^{-1} = 2.475\times10^{24}\ \text{g} \] \[m_{1\,\text{mol}} = 2.475\times10^{24}\ \text{g}\times\frac{1\ \text{kg}}{1000\ \text{g}} = 2.475\times10^{21}\ \text{kg} \] \[\frac{M_{\text{Earth}}}{m_{1\,\text{mol}}} = \frac{5.98\times10^{24}\ \text{kg}}{2.475\times10^{21}\ \text{kg}} = 2.42\times10^{3} \] The Earth is heavier.Answer: \(\displaystyle m_{1\,\text{mol}} = 2.475\times10^{21}\ \text{kg}\); the Earth is heavier, by about \(\displaystyle 2.42\times10^{3}\) times.

  7. Exercise 30

    A sample of vitamin C is known to contain 2.58\displaystyle 2.58 ×1024\displaystyle 10^{24} oxygen atoms. How many moles of oxygen atoms are present in the sample?

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    NCERT’s answer
    $\displaystyle 1$ mole of oxygen atoms = $\displaystyle 6.023$×\(\displaystyle 10^{23}\) atoms ∴Number of moles of oxygen atoms = $\displaystyle 2.58$ ×$\displaystyle 10$ $\displaystyle 6.022$×$\displaystyle 10$ = $\displaystyle 4.28$ mol $\displaystyle 4.28$ moles of oxygen atoms.
    \[n = \frac{N}{N_A} = \frac{2.58\times10^{24}}{6.022\times10^{23}\ \text{mol}^{-1}} = 4.28\ \text{mol} \]Answer: \(\displaystyle 4.28\) mol.

  8. Exercise 31

    Raunak took 5\displaystyle 5 moles of carbon atoms in a container and Krish also took 5\displaystyle 5 moles of sodium atoms in another container of same weight. (a) Whose container is heavier? (b) Whose container has more number of atoms?

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    NCERT’s answer
    (a)
    Mass of sodium atoms carried by Krish = ($\displaystyle 5$ × $\displaystyle 23$) g = $\displaystyle 115$ g While mass of carbon atom carried by Raunak = ($\displaystyle 5$ ×$\displaystyle 12$) g = 60g Thus, Krish’s container is heavy (b) Both the bags have same number of atoms as they have same number of moles of atoms
    \[m_{\text{C}} = n M = 5\ \text{mol}\times12\ \text{g mol}^{-1} = 60\ \text{g} \] \[m_{\text{Na}} = n M = 5\ \text{mol}\times23\ \text{g mol}^{-1} = 115\ \text{g} \] Empty containers weigh the same, so Krish's (sodium) container is heavier. \[N = nN_A = 5\ \text{mol}\times 6.022\times10^{23}\ \text{mol}^{-1} = 3.011\times10^{24} \] Both took $\displaystyle 5$ mol, so both containers hold the same number of atoms.Answer: (a) Krish's (sodium) container is heavier (b) both hold the same number of atoms, \(\displaystyle 3.011\times10^{24}\).

  9. Exercise 32

    Fill in the missing data in Table 3.1.Species: H2O\displaystyle \text{H}_2\text{O}, CO2\displaystyle \text{CO}_2, Na atom, MgCl2\displaystyle \text{MgCl}_2 No. of moles: 2\displaystyle 2, —, —, 0.5\displaystyle 0.5 No. of particles: —, 3.011×1023\displaystyle 3.011\times10^{23}, —, — Mass: 36\displaystyle 36 g, —, 115\displaystyle 115 g, —

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    NCERT’s answer
    Species \(\displaystyle H_{2}\)O \(\displaystyle CO_{2}\) Na atom \(\displaystyle MgCl_{2}\) Property No. of moles $\displaystyle 0.5$ $\displaystyle 0.5$ No of particles $\displaystyle 1.2044$ ×\(\displaystyle 10^{24}\) $\displaystyle 3.011$×\(\displaystyle 10^{23}\) $\displaystyle 3.011$×\(\displaystyle 10^{24}\) $\displaystyle 3.011$×\(\displaystyle 10^{23}\) Mass 36g 22g 115g 47.5g
    \[n_{\text{H}_2\text{O}} = \frac{m}{M} = \frac{36\ \text{g}}{18\ \text{g mol}^{-1}} = 2\ \text{mol} \]\[N_{\text{H}_2\text{O}} = nN_A = 2\ \text{mol}\times 6.022\times10^{23}\ \text{mol}^{-1} = 1.2044\times10^{24} \]\[n_{\text{CO}_2} = \frac{N}{N_A} = \frac{3.011\times10^{23}}{6.022\times10^{23}\ \text{mol}^{-1}} = 0.5\ \text{mol} \]\[m_{\text{CO}_2} = nM = 0.5\ \text{mol}\times44\ \text{g mol}^{-1} = 22\ \text{g} \]\[n_{\text{Na}} = \frac{m}{M} = \frac{115\ \text{g}}{23\ \text{g mol}^{-1}} = 5\ \text{mol} \]\[N_{\text{Na}} = nN_A = 5\ \text{mol}\times 6.022\times10^{23}\ \text{mol}^{-1} = 3.011\times10^{24} \]\[N_{\text{MgCl}_2} = nN_A = 0.5\ \text{mol}\times 6.022\times10^{23}\ \text{mol}^{-1} = 3.011\times10^{23} \]\[m_{\text{MgCl}_2} = nM = 0.5\ \text{mol}\times95\ \text{g mol}^{-1} = 47.5\ \text{g} \]Answer: \(\displaystyle \text{H}_2\text{O}\): \(\displaystyle n=2\) mol, \(\displaystyle N=1.2044\times10^{24}\); \(\displaystyle \text{CO}_2\): \(\displaystyle n=0.5\) mol, \(\displaystyle m=22\) g; Na: \(\displaystyle n=5\) mol, \(\displaystyle N=3.011\times10^{24}\); \(\displaystyle \text{MgCl}_2\): \(\displaystyle N=3.011\times10^{23}\), \(\displaystyle m=47.5\) g.

  10. Exercise 33

    The visible universe is estimated to contain 1022\displaystyle 10^{22} stars. How many moles of stars are present in the visible universe?

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    NCERT’s answer
    Number of moles of stars = $\displaystyle 6.023$×$\displaystyle 10$ = $\displaystyle 0.0166$ mols
    \[n = \frac{N}{N_A} = \frac{10^{22}}{6.022\times10^{23}\ \text{mol}^{-1}} = 1.66\times10^{-2}\ \text{mol} \]Answer: \(\displaystyle 1.66\times10^{-2}\ \text{mol}\).