NCERT Exemplar (Part 3): Atoms and Molecules | Extra Documents & Tests for Class 9 PDF Download

Q.29. A gold sample contains 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold?
Ans. Amount of gold in 1 g of 90% pure gold

1 mole of Au = 197 g
1 mole of Au = 6.022 x 10²³ atoms
197 g of gold contains 6.022 x 10²³ atoms

= 2.75 x 10²¹ atoms

Q.30. What are ionic and molecular compounds? Give examples.
Ans. Ionic compounds are those compounds which are solid and form ions in aqueous solution, have high melting and boiling points, do not conduct electricity in solid state but conduct electricity in molten state or in aqueous solution, e.g. NaCl, KCl, MgO, CaO, etc Molecular compounds may be solids, liquids or gases, do not form ions in aqueous solution, have low melting and boiling points, do not conduct electricity e.g. CH₄, CCl₄, NH₃, PH₃, etc.

Q.31. 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 x 10^-28 g).
Which one is heavier?
Ans.

Mass of 1 electron = 9.1 x 10^-28 g
Mass of 3 x 6.022 x 10²³ electrons
= 9.1 x 10^-28 x 3 x 6.022 x 10²³ 
Mass of 3 moles of electrons = 164.400 x 10^-5 g
= 0.00164 g
Molar mass of Al³⁺ ions = 27 - 0.00164 g
= 26.9984 g mol^-1 
Difference in mass between Al and Al³⁺=0.00164 g
Al atom is heavier than Al³⁺.

Q.32. A silver ornament of mass ‘m’ gram is polished with gold equivalent to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.
Ans. Mass of silver in the ornament = m gram
Mass of gold in the ornament

108 g of Ag contains 6,022 x 10²³ atoms
m gram of Ag contain

197 g of Au contains 6.022 x 10²³ atoms

Ratio of number of atoms of gold and silver
= Au : Ag

= 108 : 19700 = 1 : 182.41

Q.33. A sample of ethane (C₂H₆) gas has the same mass as 1.5 x 10²⁰ molecules of methane (CH₄). How many C₂H₆ molecules does the sample of gas contain?
Ans. 1 mole of CH₄ = 16 g
1 mole of CH₄ contains 6.022 x 10²³ molecules 6.022 x 10²³ molecules of CH₄ has mass = 16 g
1.5 x 10²⁰ molecules of CH₄ has mass

Now, 1 mole of C₂H₆ = 30 g
1 mole of C₂H₆ = 6.022 x 10²³ molecules
30 g of C₂H₆ contains 6.022 x 10²³ moles


= 8 x 10¹⁹ molecules

Q.34. 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 m ass of Ca₃(PO₄)₂ is _______ . [Ca = 40 u, P = 31 u, O = 16 u]
(d) Formula of sodium carbonate is _______. and that of ammonium sulphate is______.
Ans. (a) Law of conservation of mass
(b) Polyatomic ions (radicals)
(c) 40 x 3 + 2 x 31 + 8 x 16
= 120 + 62 + 128 = 310 u
(d) Na₂CO₃; (NH₄)₂SO₄ 

Q.35. Complete the following crossword puzzle by using the name of the chemical elements. Use the data given in Table.

Across
2. The element used by Rutherford during his a-scattering experiment (4)
3. An element which forms rust on exposure to moist air (4)
5. A very reactive non-metal stored under water (10)
7. Zinc metal, when treated with dilute hydrochloric acid, produces a gas of this element which when tested with burning splinter produces a pop sound. (8)
Down
1. A white lustrous metal used for making ornaments and which tends to get tarnished black in the presence of moisture (6)
4. Both brass and bronze are alloys of the element (6)
6. The metal which exists in the liquid state at room temperature (7)
8. An element with symbol Pb (4)
Ans. > Across :
2. GOLD
3. IRON
5. PHOSPHORUS
7. HYDROGEN
> Down :
1. SILVER
4. COPPER
6. MERCURY
8. LEAD

Q.36. (a) In the given crossword puzzle, names of 11 elements are hidden. The symbols of these elements are given below. Complete the puzzle.
1. C    
2. H    
3. Ar
4. O    
5. Xe    
6. N
7. He    
8. F    
9. Kr
10. Rn    
11. Ne

(b) Identify the total number of inert gases, their names and symbols from this crossword puzzle.
Ans. (a)
1. Cl-CHLORINE
2. H-HYDROGEN
3. Ar-ARGON
4. O-OXYGEN
5. Xe-XENON
6. N-NITROGEN
7. He-HELIUM
8. F-FLUORINE
9. Kr-KRYPTON
10. Rn-RADON
11. Ne-NEON

(b) Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe) and Radon (Rn) are six inert gases.

Q.37. 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
Ans. (a) Caustic potash is KOH, Molecular mass
= 39 + 16 + 1 = 56 u
(b) Baking powder is NaHC0₃, Molecular mass
= 23 + 1 + 12 + 48 - 84 u
(c) Lime stone is CaC0₃, Molecular mass
= 40 + 12 + 48 - 100 u
(d) Caustic soda is NaOH, Molecular mass
= 23 + 16 + 1 = 40 u
(e) Ethanol is C₂H₅OH, Molecular mass
= 24 + 5 + 16 + 1 = 46 u
(f) Common salt is NaCl, Molecular mass
= 23 + 35.5 = 58.5 u

Q.38. In photosynthesis, 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 C₆H₁₂O₆. How many grams of water would be required to produce 18 g of glucose? Compute the volume of water so consumed, assuming the density of water to be 1 g cm^-3.
Ans. 

Molar mass of C₆H₁₂O₆ 
= 6 x 12 + 12 x 1 + 6 x 16
= 72 + 12 + 96 = 180 g mol^-1 
180 g of C₆H₁₂O₆ needs 108 g of H₂0

Volume of water = 10.8 cm³