SolveItNCERT · CBSE Boards

NCERT Exemplar · Class 9 Science Atoms and Molecules

48 questions · 48 still being checked

Long Answer Questions 44–48 (part 5 of 5)

  1. Exercise 44

    Fill in the blanks
    (a)
    In a chemical reaction, the sum of the masses of the reactants and
    products remains unchanged. This is called ————.
    (b)
    A group of atoms carrying a fixed charge on them is called ————.
    (c)
    The formula unit mass of Ca3\displaystyle Ca_{3} (PO4\displaystyle PO_{4})2\displaystyle 2 is ————.
    (d)
    Formula of sodium carbonate is ———— and that of ammonium
    sulphate is ————.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (a)
    Law of conservation of mass (b) Polyatomic ion (c) ($\displaystyle 3$ × atomic mass of Ca) + ($\displaystyle 2$ × atomic mass of phosphorus) + ($\displaystyle 8$ × atomic mass of oxygen) = $\displaystyle 310$ (d) \(\displaystyle Na_{2}\) \(\displaystyle CO_{3}\); (\(\displaystyle NH_{4}\))$\displaystyle 2$ \(\displaystyle SO_{4}\)
    (a) Law of conservation of mass.(b) Polyatomic ion.(c) \[M = 3M_{\text{Ca}} + 2M_{\text{P}} + 8M_{\text{O}} \] \[M = 3(40) + 2(31) + 8(16) \] \[M = 120 + 62 + 128 = 310\ \text{u} \](d) \(\displaystyle \text{Na}_2\text{CO}_3\); \(\displaystyle (\text{NH}_4)_2\text{SO}_4\)Answer: (a) Law of conservation of mass (b) Polyatomic ion (c) \(\displaystyle 310\ \text{u}\) (d) \(\displaystyle \text{Na}_2\text{CO}_3\); \(\displaystyle (\text{NH}_4)_2\text{SO}_4\).

  2. Exercise 45

    Complete the following crossword puzzle (Fig. 3.1\displaystyle 3.1) using the names of the chemical elements. Use the data given below.Across 2. The element used by Rutherford during his α-scattering experiment 3. An element which forms rust on exposure to moist air 5. A very reactive non-metal stored under water 7. Zinc metal when treated with dilute hydrochloric acid produces a gas of this element which when tested with a burning splinter produces a pop sound.Down 1. A white lustrous metal used for making ornaments and which tends to get tarnished black in the presence of moist air 4. Both brass and bronze are alloys of the element 6. The metal which exists in the liquid state at room temperature 8. An element with symbol Pb NCERT_Question_Class9_Science_Exemplar_Ch3_Q45

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    2. Gold — the foil used in Rutherford's α-scattering experiment. 3. Iron — rusts (forms hydrated iron(III) oxide) on exposure to moist air. 5. Phosphorus — a highly reactive non-metal kept under water so it cannot catch fire in air. 7. Hydrogen — \[\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2\uparrow \] the gas gives the pop test with a burning splinter.1. Silver — a lustrous metal for ornaments; tarnishes black (\(\displaystyle \text{Ag}_2\text{S}\)) in moist air. 4. Copper — the common metal of both brass (Cu–Zn) and bronze (Cu–Sn). 6. Mercury — the only metal that is liquid at room temperature. 8. Lead (Pb).Answer: Across: $\displaystyle 2$ Gold, $\displaystyle 3$ Iron, $\displaystyle 5$ Phosphorus, $\displaystyle 7$ Hydrogen. Down: $\displaystyle 1$ Silver, $\displaystyle 4$ Copper, $\displaystyle 6$ Mercury, $\displaystyle 8$ Lead.

  3. Exercise 46

    (a)
    In this crossword puzzle (Fig. 3.2\displaystyle 3.2), names of 11\displaystyle 11 elements are hidden. Symbols of these are given below. Complete the puzzle.
    1. Cl 7. He
    2. H 8. F
    3. Ar 9. Kr
    4. O 10. Rn
    5. Xe 11. Ne
    6. N(b) Identify the total number of inert gases, their names and symbols from this crossword puzzle.
    NCERT_Question_Class9_Science_Exemplar_Ch3_Q46

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (a)
    (b)
    Six : Helium (He); Neon ( Ne); Argon (Ar); Krypton (Kr); Xenon (Xe); Radon (Rn).
    (a) $\displaystyle 1$ Cl — Chlorine $\displaystyle 2$ H — Hydrogen $\displaystyle 3$ Ar — Argon $\displaystyle 4$ O — Oxygen $\displaystyle 5$ Xe — Xenon $\displaystyle 6$ N — Nitrogen $\displaystyle 7$ He — Helium $\displaystyle 8$ F — Fluorine $\displaystyle 9$ Kr — Krypton $\displaystyle 10$ Rn — Radon $\displaystyle 11$ Ne — Neon.(b) Six of the eleven are inert (noble) gases — Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe) and Radon (Rn).Answer: (a) as listed above; (b) six inert gases — He, Ne, Ar, Kr, Xe, Rn.

  4. Exercise 47

    Write the formulae for the following and calculate the molecular mass for
    each one of them.
    (a)
    Caustic potash
    (b)
    Baking powder
    (c)
    Lime stone
    (d)
    Caustic soda
    (e)
    Ethanol
    (f)
    Common salt

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (a)
    KOH ($\displaystyle 39$ +$\displaystyle 16$+$\displaystyle 1$) = $\displaystyle 56$ g \(\displaystyle mol^{-1}\) (b) \(\displaystyle NaHCO_{3}\) $\displaystyle 23$ + $\displaystyle 1$ + $\displaystyle 12$ + ($\displaystyle 3$ × $\displaystyle 16$) = $\displaystyle 84$ g \(\displaystyle mol^{-1}\) (c) \(\displaystyle CaCO_{3}\) $\displaystyle 40$ + $\displaystyle 12$ + ($\displaystyle 3$ × $\displaystyle 16$) = $\displaystyle 100$ g \(\displaystyle mol^{-1}\) (d) NaOH $\displaystyle 23$+$\displaystyle 16$+$\displaystyle 1$ = $\displaystyle 40$ g \(\displaystyle mol^{-1}\) (e) \(\displaystyle C_{2}\)\(\displaystyle H_{5}\)OH = \(\displaystyle C_{2}\)\(\displaystyle H_{6}\)O $\displaystyle 2$ × $\displaystyle 12$ + ($\displaystyle 6$ × $\displaystyle 1$) + $\displaystyle 16$ = $\displaystyle 46$ g \(\displaystyle mol^{-1}\) (f) NaCl $\displaystyle 23$ + $\displaystyle 35.5$ = $\displaystyle 58.5$ g \(\displaystyle mol^{-1}\)
    (a) Caustic potash — potassium hydroxide, \(\displaystyle \text{KOH}\) \[M = 39 + 16 + 1 = 56\ \text{u} \](b) Baking powder (active ingredient baking soda) — sodium hydrogen carbonate, \(\displaystyle \text{NaHCO}_3\) \[M = 23 + 1 + 12 + 3(16) = 23 + 1 + 12 + 48 = 84\ \text{u} \](c) Lime stone — calcium carbonate, \(\displaystyle \text{CaCO}_3\) \[M = 40 + 12 + 3(16) = 40 + 12 + 48 = 100\ \text{u} \](d) Caustic soda — sodium hydroxide, \(\displaystyle \text{NaOH}\) \[M = 23 + 16 + 1 = 40\ \text{u} \](e) Ethanol, \(\displaystyle \text{C}_2\text{H}_5\text{OH}\) \[M = 2(12) + 6(1) + 16 = 24 + 6 + 16 = 46\ \text{u} \](f) Common salt — sodium chloride, \(\displaystyle \text{NaCl}\) \[M = 23 + 35.5 = 58.5\ \text{u} \]Answer: (a) \(\displaystyle \text{KOH}\), $\displaystyle 56$ u (b) \(\displaystyle \text{NaHCO}_3\), $\displaystyle 84$ u (c) \(\displaystyle \text{CaCO}_3\), $\displaystyle 100$ u (d) \(\displaystyle \text{NaOH}\), $\displaystyle 40$ u (e) \(\displaystyle \text{C}_2\text{H}_5\text{OH}\), $\displaystyle 46$ u (f) \(\displaystyle \text{NaCl}\), $\displaystyle 58.5$ u

  5. Exercise 48

    In photosynthesis, 6\displaystyle 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex series of reactions to give a molecule of glucose having a molecular formula C6\displaystyle C_{6} H12\displaystyle H_{12} O6\displaystyle O_{6}. How many grams of water would be required to produce 18\displaystyle 18 g of glucose? Compute the volume of water so consumed assuming the density of water to be 1\displaystyle 1 g cm−3\displaystyle cm^{-3}.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle 6CO_{2}\) + $\displaystyle 6$ \(\displaystyle H_{2}\)O Chlorophyll Sunlight  → \(\displaystyle C_{6}\) \(\displaystyle H_{12}\) \(\displaystyle O_{6}\) + \(\displaystyle 6O_{2}\) $\displaystyle 1$ mole of glucose needs $\displaystyle 6$ moles of water $\displaystyle 180$ g of glucose needs ($\displaystyle 6$×$\displaystyle 18$) g of water $\displaystyle 1$ g of glucose will need $\displaystyle 180$ g of water. $\displaystyle 18$ g of glucose would need $\displaystyle 180$ × $\displaystyle 18$ g of water = $\displaystyle 10.8$ g Volume of water used Mass Density = -$\displaystyle 3$ $\displaystyle 10.8$ g 1g cm = =$\displaystyle 10.8$ \(\displaystyle cm^{3}\) .
    \[6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \] Molar masses: \(\displaystyle \text{H}_2\text{O} = 18\text{ g mol}^{-1}\), \(\displaystyle \text{C}_6\text{H}_{12}\text{O}_6 = 180\text{ g mol}^{-1}\). \[180\text{ g glucose needs } 6 \times 18 = 108\text{ g H}_2\text{O} \] \[18\text{ g glucose needs } \frac{18}{180} \times 108 = 10.8\text{ g H}_2\text{O} \] \[V = \frac{m}{\rho} = \frac{10.8\text{ g}}{1\text{ g cm}^{-3}} = 10.8\text{ cm}^3 \]Answer: \(\displaystyle 10.8\) g of water; volume \(\displaystyle 10.8\ \text{cm}^3\).