NCERT Exemplar (Part 2): Atoms and Molecules | Extra Documents & Tests for Class 9 PDF Download

Q.16. Calcium chloride, when dissolved in water, dissociates into its ions according to the following equation.
CaCl₂ (aq) → Ca²⁺ (aq) + 2CI^- (aq)
Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water.
Ans. CaCl₂(aq) → Ca²⁺(aq) + 2Cl^- (aq)
1 mole of CaCl₂ = 111 g
111 g of CaCl₂ = 1 mole

1 mole of CaCl₂ gives 3 moles of ion
2 moles of CaCl₂ gives 6 moles of ion
= 6 x 6.022 x 10²³ ions
= 36.132 x 10²³ ions = 3.6132 x 10²⁴ ions
 
Q.17. The difference in the mass of 100 moles each of sodium atoms and sodium ions is 5.48002 g.
Compute the mass of an electron.
Ans.

100 moles of electrons weigh = 5.48002 g
100 x 6.022 x 10²³ electrons weigh = 5.48002 g

= 0.91 x 10^-25g = 9.1 x 10^-26g = 9.1 x 10^-29kg
It is not the real mass of electron.

Q.18. Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS? Molar mass of Hg and S are 200.6 g mol^-1 and 32 g mol^-1 respectively.
Ans. Molar mass of HgS = 200.6 + 32
= 232.6 g mol^-1 
232.6 g of HgS contains 200.6 g of pure Hg


Q.19. The mass of one steel screw is 4.11 g. Find the mass of one mole of these steel screws. Compare this value with the mass of the Earth (5.98 x 10²⁴ kg). Which one of the two is heavier and by how many times?
Ans. Mass of 1 steel screw = 4.11 g
Mass of 6.022 x 10²³ steel screws
= 4.11 x 6.022 x 10²³ g = 2.475 x 10²⁴ g
= 2.475 x 10²¹ kg
Therefore, mass of one mole of screws
= 2.475 x 10²¹ kg

= 2.4 x 10³ 
Mass of the earth is 2400 times the mass of 1 mole of screws.

Q.20. A sample of vitamin C is known to contain 2.58 x 10²⁴ oxygen atoms. How many moles of oxygen atoms are present in the sample?
Ans. Number of moles of oxygen atoms

= 4.28 moles.

Q.21. Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of the same weight.
(a) Whose container is heavier?
(b) Whose container has more number of atoms?
Ans. (a) 1 mole of carbon atoms = 12 g
5 moles of carbon atoms = 12 x 5 = 60 g
1 mole of Na = 23 g
5 moles of Na = 23 x 5 = 115 g
Thus, Krish’s container is heavier than that of Raunak.
(b) Both the containers have same number of atoms as both have the same number of moles and both elements are monoatomic.

Q.22. Fill in the missing data in the Table.

Ans. 1 mole of H₂O = 1 x 2 + 16 = 18 g
2 moles of H₂0 = 2 x 18 = 36 g
1 mole of H₂0 contains 6.022 x 10²³ molecule
2 moles of H₂0 contains 2 x 6.022 x 10²³ = 12.044 x 10²³ molecules
1 mole of CO₂ = 12 + 2 + 16 = 44 g
0.5 mole of CO₂ = 44 x 0.5 = 22 g
1 mole of CO₂ contains 6.022 x 10²³ molecules
0.5 mole of CO₂ will contain 0.5 x 6.022 x 10²³ = 3.011 x 10²³ molecules
1 mole of Na = 23 g
5 moles of Na = 23 x 5 = 115 g
1 mole of Na = 6.022 x 10²³ atoms
5 moles of Na = 5 x 6.022 x 10²³ = 30.110 x 10²³ atoms
1 mole of MgCl₂ = 24 + 2 x 35.5 = 24 + 71 = 95 g
0.5 mole of MgCl₂ = 0.5 x 95 = 47.5 g
1 mole of MgCl₂ = 6.022 X 10²³ 
0.5 mole of MgCl₂ = 0.5 x 6.022 x 10²³ = 3.011 x 10²³ formula units.


Q.23. The visible universe is estimated to contain 10²² stars. How many moles of stars are present in the visible universe?
Ans. Number of moles of stars =
= 0.167 x 10^-1 = 0.0167 mole.

Q.24. What is the SI prefix for each of the following multiples and sub-multiples of a unit?
(a) 10³ 
(b) 10^-1 
(c) 10^-2 
(d) 10^-6 
(e) 10^-9 
(f) 10^-12 
Ans. (a) kilo
(b) deci
(c) centi
(d) micro
(e) nano
(f) pico

25. Express each of the following in kilograms:
(a) 5.84 x 10^-3 mg
(b) 58.34 g
(c) 0.584 g
(d) 5.873 x 10^-21g
Ans. (a) 5.84 x 10^-3 mg x 10^-6 = 5.84 x 10^-9 kg [∵ 1 mg = 10^-3 g = 10^-6 kg]
(b) 58.34 g x 10^-3 kg = 5.834 x 10^-2 kg [∵ 1 g = 10^-3 kg]
(c) 0.584 g = 0.584 x 10^-3 kg = 5.84 x 10^-4 kg
(d) 5.873 x 10^-21 g x 10^-3 = 5.873 x 10^-24 kg

Q.26. Compute the difference in masses of 10³ moles each of magnesium atoms and magnesium ions. (Mass of an electron = 9.1 x 10^-31 kg)
Ans.

1 electron weighs = 9.1 x 10^-31 kg
∴ 2000 x 6.022 x 10²³ electrons weigh
= 9.1 x 10^-31 x 2000 x 6.022 x 10²³ kg
= 109.6004 x 10^-5 kg = 1.096004 x 10^-3 kg

Q.27. Which has more number of atoms? 100 g of N₂ or 100 g of NH₃ 
Ans. 1 mole of N₂ - 28 g
28 g of N₂ = 1 mole



= 43.01 x 10²³ atoms
= 4.30 x 10²⁴ atoms
1 mole of NH₃ = 17 g
17 g of NH₃ = 1 mole


Therefore, 100 g of NH₃ contains more number of atoms.

Q.28. Compute the number of ions present in 5.85 g of sodium chloride.
Ans. NaCl → Na⁺ + Cl^- 
1 mole of NaCl = 23 + 35.5 = 58.5 g
58.5 g of NaCl = 1 mole

1 mole of NaCl gives 2 moles of ions
0.1 mol of NaCl gives 2 x 0.1 = 0.2 mol
= 0.2 x 6.022 x 10²³ ions = 1.2044 x 10²³ ions

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 g1 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.

Across2. 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