## Thursday, 28 July 2011

### CBSE Class IX ( 9th) Science | Chapter 3. Atoms And Molecules | Lesson Exercises

Question 1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Mass of boron as given = 0.096 g
Mass of oxygen as given  = 0.144 g
Mass of sample as given   = 0.24 g
Percentage of boron by weight in the compound = (0.096 ÷ 0.24 )  x 100 = 40%
Percentage of oxygen by weight in the compound = (0.144 ÷ 0.24)  x 100 = 60%

Question 2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide willbe formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Answer : We know that in burning process carbon utilises oxygen to form Carbon Dioxide .
Carbon + Oxygen => Carbon Dioxide + Heat Energy

Here as given,  in burning, 3.0 g of carbon utilises 8.00 g of oxygen to form 11.00 g of carbon dioxide.When 3.00 g of carbon is burnt in 50.00 g of oxygen, it will use just 8.00 g of oxygen from  the total of 50 gm of Oxygen to produce 11 gm of Carbon Dioxide. 42 gm of Oxygen will be left behind without any change.
The above answer is governed by the following two laws of chemical combination:

1. Law of conservation of mass which states that mass can neither be created nor destroyed in a chemical reaction.
Here in terms of mass 3g of Carbon and 50 gm of Oxygen combines to give 11 gm of Carbon Dioxide and 42 gm of unused Oxygen keeping the total mass of combining

chemicals 53 gm (3+50 gm)  by weight constant before and after the reaction.

2. Law of constant proportions which states that In a chemical substance the elements are always present in definite proportions by mass.Here respective of the excess quantity of Oxygen available for burning of same weight of Carbon, the amount of carbon dioxide will remain same.

Question 3. What are polyatomic ions? Give examples.
Answer : Polyatomic ions are clusters or group of atoms that always have a  fixed  charge attach to them. This charge may be positive or negative .
Ammonium  NH4+
Hydroxide OH-
Nitrate NO3 -
Carbonate CO32-
Sulphate  SO42-

Question 4. Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.

(a) Magnesium chloride     => MgCl2
(b) Calcium oxide         => CaO
(c) Copper nitrate        => Cu(NO3)2.
(d) Aluminium chloride    => AlCl3
(e) Calcium carbonate.    => CaCO3

Question 5. Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.

 Compound Chemical Formula Elements (a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate. CaO HBr NaHCO3  K2SO4 Calcium, Oxygen Hydrogen, Bromine Sodium, Hydrogen, Carbon, Oxygen Potassium, Sulphur, Oxygen

Question 6. Calculate the molar mass of the following substances.
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO3

(a) Molar Mass of Ethyne, C2H2; =     2 × 12 + 2 × 1 =  26 g
(b) Molar Mass of Sulphur molecule, S8 =   8 × 32  =  256 g
(c) Molar Mass of Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)  =   4 × 31  =  124 g
(d) Molar Mass of Hydrochloric acid, HCl  =   1 + 35.5 =  36.5 g
(e) Molar Mass of Nitric acid, HNO3 =   1 + 14 + 3 × 16  =  63 g

Question 7. What is the mass of.
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium= 27)?
(c) 10 moles of sodium sulphite (Na2SO3)?

Answer :The mass of 1 mole of a substance is equal to its relative atomic or molecular mass in grams. The atomic mass of an element gives us the mass of one atom of that element in atomic mass units (u). To get the mass of 1 mole of atom of that element, that is, molar mass, we have to take the same numerical value but change the units from 'u' to 'g'.
Molar mass of atoms is also known as gram atomic mass. For example, atomic mass of hydrogen=1u. So, gram atomic mass of
hydrogen = 1 g.

(a) The mass of 1 mole of Nitrogen atoms = 14 g.

(b) The mass of 4 moles of Aluminium atoms is (4 × 27) g = 108 g

(c) The mass of 10 moles of Sodium Sulphite (Na2SO3) =10 × [2 × 23 + 32 + 3 × 16] g = 10 × 126 g = 1260 g

Question 8. Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide.

Answer: To get the mass of 1 mole of atom of that element, that is, molar mass, we have to take the same numerical value but change the units from 'u' to 'g'.
1 mole substance = Total Atomic mass of a substance (u)       Or
= Total gram atomic mass of substance (g)  Or
= Molar Mass
= 1 Mole of substance                                Or
=  6.022 × 1023 in number of particles of that substance
Here required mole of substance  =  Given mass of substance in 'g' ÷ Molar Mass

(a) 12 g of oxygen gas =  Given mass of oxygen in 'g' ÷ Molar Mass of Oxygen
= 12 ÷ 32
= 0.375 mole of oxygen

(b) 20 g of water = Given mass of water in 'g' ÷ Molar Mass of Water
= 20 ÷ 18
= 1.11 mole of water

(c) 22 g of carbon dioxide.= Given mass of Carbon Dioxide in 'g' ÷ Molar Mass of Carbon Dioxide
= 22 ÷ 44
= 0.5 mole of carbon dioxide

Question 9. What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?

Answer : Mass of atoms (m) in given moles (n) = Molar Mass of substance (M)× Given Moles of substance (n)

(a) ∴  Mass of 0.2 mole of oxygen atoms = Molar Mass of oxygen × Given Moles  =  16g × 0.2   = 3.2 g

(b)
Mass of 0.5 mole of water molecules =Molar Mass of water × Given Moles = 18 g × 0.5  = 9 g

Question 10. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

1 mole of Sulphur (S8 = 8 × 32 g = 256 g
also as
1 mole = 6.022 × 1023 in number of particles of that substance
So 256 g of  sulphur contains = 6.022 × 1023  molecules of sulphur
Therefor 16 g of sulphur will contains = 3.76 ×
1022  molecules (approx)

Question 11. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)
1 mole of Aluminium Oxide (Al2O3) = 2 × 27 + 3 × 16  = 102 g

We know, 102 g of
Aluminium Oxide (Al2O3) = 6.022 × 1023 molecules of Aluminium Oxide (Al2O3)
We know, The number of atoms present in given mass = (Given mass ÷ molar mass) × Avogadro number
Then, 0.051 g of Aluminium Oxide (Al2O3) contains
= (0.051 ÷ 102 ) × 6.022 × 1023 molecules of Aluminium Oxide (Al2O3)
= 3.011 × 1020 molecules of Aluminium Oxide (Al2O3)

The number of Aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Therefore, the number of Aluminium ions (
Al3+) present in 3.011 × 1020 molecules (0.051 g ) of Aluminium Oxide (Al2O3)
= 2 × 3.011 × 1020

= 6.022 ×
1020

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5. Question 6 part a. 12×2+2×1=26g

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