Intext Questions

Q.1      In a reaction 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide,0.9g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass. Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water
Sol.       In a reaction, sodium carbonate  reacts with ethanoic acid to produce sodium ethanoate, carbondioxide, and water.
Sodium      +      Ethanoic → Sodium    +     Carbon    +    Water
Carbonate            acid            ethanoate        dioxide
Mass of sodium carbonate = 5.3g (Given)
Mass of ethanoic acid = 6g (Given)
Mass of sodium ethanoate = 8.2g (Given)
Mass of carbon dioxide = 2.2 (Given)
Mass of water = 0.9g (Given)
Now, total mass before the reaction = (5.3 + 6)g
= 11. 3g
and total mass after the reaction = (8.2 + 2.2 + 0.9)g
= 11.3g
Therefore, Total mass before the reaction = Total mass after the reaction
Hence, the given observations are in agreement with the law of conservation of mass.

---

Page 33

Q.2      Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3g of hydrogen gas?
Sol.      It is given that the ratio of hydrogen and oxygen by mass to form water is 1:8. Then, the mass of oxygen gas required to react completely with 1g of hydrogen gas is 8g . Therefore, the mass of oxygen gas required to react completely with 3g of hydrogen gas is 8 × 3g = 24 g.

---

Q.3     Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Sol.      The postulate of Dalton’s atomic theory which is a result of the law of conservation of mass is : Atoms are indivisible particles, which can neither be created nor destroyed in a chemical reaction.

---

Q.4     Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Sol.      The postulate of Dalton’s atomic theory which can explain the law of definite proportion is: The relative number and kind of atoms in a given compound remains constant.

---

Page 35

Q.1     Define atomic mass unit.
Sol.      Mass unit equal to exactly one- twelfth the mass of one atom of carbon – 12 is called one atomic mass unit. It is written as ‘u’

---

Q.2     Why is it not possible to see an atom with naked eyes?
Sol.      The size of an atom is so small that it is not possible to see it with naked eyes. Also, atom of an element does not exist independently.

---

Page 39

Q.1     Write down the formula of
           (i) sodium oxide
           (ii) aluminium chloride
           (iii) sodium suphide
           (iv) magnesium hydroxide
Sol.     (i) Sodium oxide → Na2O
(ii) Aluminium chloride → AlCl3
(iii) Sodium suphide → Na2S
(iv) Magnesium hydroxide → Mg(OH)2

---

Q.2      Write down the names of compounds represented by the following formula:
            (i) Al2(SO4)3
            (ii) CaCl2
            (iii) K2SO4
            (iv) KNO3
            (v) CaCO3
Sol.      (i) Al(SO4)3→ Aluminium sulphate
(ii) CaCl2→ Calcium chloride
(iii) K2SO4→ Potassium sulphate
(iv) CaCO3→ Calcium carbonate

---

Q.3     What is meant by the term chemical formula?
Sol.      The chemical formula of a compound means the symbolic representation of the composition of
a compound. From the chemical formula of a compound, we can know the number and kinds of atoms of different elements that constitute the compound. For example, from the chemical formula Co2 of carbon dioxide, we come to know that one carbon atom and two oxygen atoms are chemically bonded together to form one molecule of the compound, carbon dioxide.

---

Q.4     How many atoms are present in a
(i) H2S molecule and
(ii) PO3−4 ion?
Sol.      (i) In an H2S molecule, three atoms are present; two of hydrogen and one of sulphur.
(ii) In a PO3−4– ion, five atoms are present; one of phosphorus and four of oxygen.

---

Page 40

Q.1     Calculate the molecular masses of H2,O2,Cl2,CO2,CH4,C2H6,C2H4,NH3,CH3OH.
Sol.      Molecular mass of H2 = 2 × Atomic mass of H
= 2 × 1
= 2u
Molecular mass of O2 = 2 × Atomic mass of O
= 2 × 16
= 32u
Molecular mass of Cl2 = 2 × Atomic mass of Cl
= 2 × 35.5
= 71 u
Molecular mass of CO2 = Atomic mass of C + 2 × Atomic mass of O
= 12 + 2 × 16
= 44 u
Molecular mass of CH4 = Atomic mass of C + 4 × Atomic mass of H
= 12 + 4 × 1
= 16 u
Molecular mass of C2H6 = 2 × Atomic mass of C + 6 × Atomic mass of H
= 2 × 12 + 6 × 1
= 30u
Molecular mass of C2H4 = 2 × Atomic mass of C + 4 × Atomic mass of H
= 2 × 12 + 4 × 1
= 28u
Molecular mass of NH3 = Atomic mass of N + 3 × Atomic mass of H
= 14 + 3 × 1
=17 u
Molecular mass of CH3OH Atomic mass of C + 4 × Atomic mass of H + Atomic mass of O
= 12 + 4 × 1 + 16
= 32 u

---

Q.2     Calculate the formula unit masses of ZnO, Na2O,K2CO3, given masses of Zn = 65u, Na= 23u, K =39u, C = 12u, and O = 16u.
Sol.      Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O
= 65 + 16
= 81 u
Formula unit mass of Na2O = 2 × Atomic mass of Na + Atomic mass of O
= 2 × 23 + 16
= 62u
Formula unit mass of K2CO3 = 2 × Atomic mass of K + Atomic mass of C + 3 ×
Atomic mass of O
= 2 × 39 + 12 + 3 × 16
= 138u

---

Page 42

Q.1     If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?
Sol.      One mole of carbon atoms weighs 12g (Given)
i.e., mass of 1 mole of carbon atoms = 12g
Then, mass of 6.022× 1023 number of carbon atoms = 12g
Therefore, mass of 1 atom of carbon = 126.022×1023g
= 1.9926 × 10−23g

---

Q.2     Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23u, Fe =56 u)?
Sol.      Atomic mass of Na = 23u (Given)
Then, gram atomic mass of Na = 23g
Now, 23g of Na contains = 6.022×1023 number of atoms
Thus, 100g of Na contains = 6.022×102323×100 number of atoms
= 2.6182 × 1024 number of atoms
Again, atomic mass of Fe = 56u (Given)
Then, gram atomic mass of Fe = 56g
Now, 56 g of Fe contains = 6.022×1023 number of atoms
Thus, 100 g of Fe 6.022×102356×100 number of atoms
= 1.0753 × 1024 number of atoms
Therefore, 100 grams of sodium contain more number of atoms than 100 grams of iron.

Exercise

Q.1     A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g if boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Sol.     Mass of boron = 0.096g(Given)
Mass of oxygen = 0.144g (Given)
Mass of sample = 0.24g (Given)
Thus, percentage of boron by weight in the compound = 0.0960.24×100%
= 40%
Thus, percentage of oxygen by weight in the compound = 0.1440.24×100%
= 60 %

---

Q.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 will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen ? Which law of chemical combinations will govern your answer ?
Sol.      Carbon + Oxygen → Carbon dioxide 3g of carbon reacts with 8 g of oxygen to produce 11g of carbon dioxide. If 3g of carbon is burnt in 50g of oxygen, then 3g of carbon will react with 8 g of oxygen. The remaining 42 g of oxygen will be left un-reactive. In this case also, only 11g of carbon dioxide will be formed.  The above answer is governed by the law of constant proportions.

---

Q.3     What are polyatomic ions? Give examples?
Sol.     A polyatomic ion is a group of atoms carrying a charge (positive or negative).For example, ammonium ion (NH+4), hydroxide ion (OH−), carbonate ion (CO2−3),sulphateion (SO2−4).

---

Q.4      Write the chemical formula of the following:
            (a) Magnesium chloride
            (b) Calcium oxide
            (c) Copper nitrate
            (d) Aluminium chloride
            (e) Calcium carbonate
Sol.      (a) Magnesium chloride →MgCl2
(b) Calcium oxide →CaO
(c) Copper nitrate →Cu(NO3)2
(d) Aluminium chloride →AlCl3
(e) Calcium carbonate →CaCO3

---

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

Compound	Chemical formula	Elements present
Quick lime	Cao	Calcium,oxygen
Hydrogen bromide	HBr	Hydrogen,bromine
Baking powder	NaHCO3	Sodium,hydrogen, carbon, oxygen
Potassium sulphate	k2SO4	Potassium,sulphur,oxygen

 

---

Q.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
Sol.     (a) Molar mass of ethyne, C2H2 = 2 × 12 + 2 × 1 = 28g
(b) Molar mass of sulphur molecule, S8 = 8 × 32 = 256g
(c) Molar mass of phosphorus molecule,P4 = 4 × 31 = 124g
(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5g
(e) Molar mass of nitric acid, HNO3 = 1 + 14 + 3 × 16 = 63g

---

Q.7     What is the mass of —
           (a) 1 mole of nitrogen atoms?
           (b) 4 mole of aluminium atoms (Atomic mass of aluminium = 27)?
           (c) 10 moles of sodium sulphite (Na2SO3) ?
Sol.     (a) The mass of 1 mole of nitrogen atoms is 14g.
(b) The mass of 4 moles of aluminium atoms is (4 × 27)g = 108g
(c) The mass of 10 moles of sodium sulphite (Na2SO3) is
10 × [2 × 23 + 32 + 3 × 16]g = 10 × 126g = 1260g

---

Q.8     Convert into mole.
           (a) 12g of oxygen gas 
           (b) 12g of water
           (c) 22g of carbon dioxide
Sol.     (a) 32 g of oxygen gas = 1 mole
Then, 12g of oxygen gas = 1232 mole = 0.375 mole
(b) 18g of water = 1 mole
Then, 20 g of water = 2018 mole = 1.11 moles (approx)
(c) 44g of carbon dioxide = 1 mole
Then, 22g of carbon dioxide = 2244 mole = 0.5 mole

---

Q.9     What is the mass of :
           (a) 0.2 mole of oxygen atoms?
           (b) 0.5 mole of water molecules?
Sol.     (a) Mass of one mole of oxygen atoms = 16g
Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2g
(b) Mass of one mole of water molecule = 18g
Then, mass of 0.5 mole of water molecules = 0.5 × 18g = 9g

---

Q.10     Calculate the number of molecules of sulphur (S8) present in 16g of solid sulphur.
Sol.        1 mole of solid sulphur (S8) = 8 × 32g = 256g
i.e., 256g of solid sulphur contains = 6.022 × 1023 molecules
Then, 16g of solid sulpur contains 6.022×1023256×16 molecules
= 3.76 × 1022 molecules (approx)

---

Q.11     Calculate the number of aluminium ions present in 0.051g 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 = 27u)
Sol.        1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16 = 102g
i.e., 102g of Al2O3 = 6.022 × 1023 molecules of Al2O3
Then, 0.051 g of  Al2O3 contains = 6.022×1023102×0.051 molecules
= 3.011 ×  1020  molecules of  Al2O3
The number of aluminium ions (Al3+) present in one  molecules of aluminium oxide is 2.
Therefore, The number of aluminium ions  (Al3+) present in
3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020
= 6.022 × 1020

 

Exemplar

 

Multiple Choice Questions 

Q.1 Which of the following correctly represents 360 g of water?

(i) 2 moles of H₂0

(ii) 20 moles of water

(iii) 6.022 × 10²³ molecules of water

(iv) 1.2044 ×10²⁵ molecules of water

(a) (i)

(b) (i) and (iv)

(c) (ii) and (iii)

(d) (ii) and (iv)

Sol. (d)

(ii) 20 moles of water = 20 ×18 g = 360 g of water, because mass of 1 mole of water is the same as its molar mass, i.e., 18 g. Molarmassis the mass of one mole of a substance.

The molecular formula of water is H₂O, which means that water is made up of two hydrogen and one oxygen atoms.

The standard atomic weight of hydrogen is 1.00794 and that of oxygen is 15.9994. So the molecular weight will be:

Molecular Weight of H₂O = (2 × 1.00794) + (1×15.994) = 18.00988

So the molar mass of water is 18.00988 gm or 0.01800988 Kg.

(iv) 1.2044 × 10²⁵ molecules of water contains

1.2044 × 10²⁵/N_A number of moles,

Where N_A = 6.023 × 10²³

Therefore,

1.2044 × 10²⁵ /6.023×10²³

= 0.199 x 10² moles  = 20 moles

---

Q.2 Which of the following statements is not true about an atom?

(a) Atoms are not able to exist independently

(b) Atoms are the basic units from which molecules and ions are formed

(c) Atoms are always neutral in nature

(d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch

Sol. (a)

---

Q.3 The chemical symbol for nitrogen gas is

(a) Ni

(b) N₂

(c) N⁺

(d) N

Sol. (b)

---

Q.4 The chemical symbol for sodium is

(a) So

(b) Sd

(c) NA

(d) Na

Sol. (d)

---

Q.5 Which of the following would weigh the highest?

(a) 0.2 mole of sucrose (C₁₂ H₂₂ O₁₁)

(b) 2 moles of CO₂

(c) 2 moles of CaCO₃

(d) 10 moles of H₂O

Sol. (c)

Weight of a sample in gram = number of moles × molar mass

(a) 0.2 moles of C₁₂H₂₂O₁₁ = 0.2 × 342 = 68.4 g

(b) 2 moles of CO₂ = 2 × 44 = 88 g

12.011 + 2(16.00) = 12.011 + 32.00 = 44.011g/mol.

(c) 2 moles of CaCO₃ = 2 × 100 = 200 g

2(40 + 12 + 3(16) = 2(40 +  12 + 48)  = 2 × 100 = 200g

(d) 10 moles of H₂O = 10  × 18 = 180 g

10(1 + 1 + 16) = 10(18) = 10(18)  = 180 g

---

Q.6 Which of the following has maximum number of atoms?

(a) 18g of H₂O

(b) 18g of O₂

(c) 18g of CO₂

(d) 18g of CH₄

Sol. (d)

Number of atoms = Mass of substance × Number of atoms in the molecule/ Molar mass    × N_A

(a) 18 g of water =18 x³/18 ×N_A = 3 N_A

(b) 18 g of oxygen = 18 x² /32 × N_A = 1.12 N_A

(c) 18 g of CO₂ = 18 x³/44 × N_A = 1.23 N_A

(d) 18 g of CH₄ =18 x⁵ /16 × N_A = 5.63 N_A

---

Q.7 Which of the following contains maximum number of molecules?

(a) 1g CO₂

(b) 1g N₂

(c) 1g H₂

(d) 1g CH₄

Sol. (c)

1 g of H₂ = ½ x N_A = 0.5 N_A = 0.5 × 6.022 × 10²³  = 3.011  × 10²³

---

Q.8 Mass of one atom of oxygen is

Sol. (a)

Mass of one atom of oxygen = Atomic mass/N_A  = 16/6.023  × 10²³ g

---

Q.9 3.42 g of sucrose are dissolved in 18g of water in a beaker. The numbers of oxygen atoms in the solution are

(a) 6.68 × 10²³

(b) 6.09 × 10²²

(c) 6.022 × 10²³

(d) 6.022 × 10²¹

Sol. (a)

= 3.42 g/342 g mol^-1

= 0.01mol

1 mol of sucrose( C₁₂H₂₂O₁₁) contains = 11× N_A atoms of oxygen,  where N_A = 6.023×10²³

0.01 mol of sucrose (C₁₂ H₂₂ O₁₁) contains = 0.01 × 11 × N_A atoms of oxygen

= 0.11× N_A atoms of oxygen

= 18 g/(1x²+ 16)gmol^-1

=18 g /18 gmol^-1

= 1mol

1mol of water (H₂O) contains 1×NA atom of oxygen

Total number of oxygen atoms =

Number of oxygen atoms from sucrose + Number of oxygen   atoms from water

= 0.11 N_A + 1.0 N_A = 1.11N_A

Number of oxygen atoms in solution = 1.11 × Avogadro’s number      

= 1.11 × 6.022 ×1023 = 6.68 × 10²³

---

Q.10 A change in the physical state can be brought about

(a) Only when energy is given to the system

(b) Only when energy is taken out from the system

(c) When energy is either given to, or taken out from the system

(d) Without any energy change

Sol. (c)

---

Q.11 Which of the following represents a correct chemical formula? Name it.

(a) CaCl

(b) BiPO₄

(c) NaSO4

(d) NaS

Sol. (b) BiPO₄— Both ions are trivalent Bismuth phosphate(Bi³⁺– Trivalent anion. anion is an ion that is negatively charged)

(PO₄^3- -Trivalent cation. A cation has a net positive charge)

---

Q.12 Write the molecular formulae for the following compounds

(a) Copper (II) bromide

(b) Aluminium (III) nitrate

(c) Calcium (II) phosphate

(d) Iron (III) sulphide

(e) Mercury (II) chloride

(f) Magnesium (II) acetate

(a) Copper (II) bromide-  CuBr₂

(b) Aluminium (III) nitrate =  Al(NO₃) ₃           

(c) Calcium (II) phosphate – Ca₃(PO₄) ₂

(d) Iron (III) sulphide – Fe₂S₃

(e) Mercury (II) chloride – HgCl₂

(f) Magnesium (II) acetate-  Mg(CH₃COO)₂

---

Q.13 Write the molecular formulae of all the compounds that can be formed by the combination of following ions Cu²⁺, Na⁺, Fe³⁺, C1^– , SO₄ ^2-, PO₄^3-

CuCl₂/ CuSO₄/ Cu₃ (PO₄) ₂

NaCl/ Na₂SO₄/ Na₃ PO₄

FeCl₃/ Fe₂(SO₄) ₃ / FePO₄

---

Q.14 Write the cations and anions present (if any) in the following compounds

(a) CH₃COONa

(b) NaCl

(c) H₂

(d) NH₄NO₃

Anions                                    Cations

(a) CH₃ COO-                                 Na

(b) Cl^–                                               Na⁺

(c) It is a covalent compound

(d) NO ₃ ⁻                                          NH₄ ⁺ 

---

Q.15 Give the formulae of the compounds formed from the following sets of elements

(a) Calcium and fluorine

(b) Hydrogen and sulphur

(c) Nitrogen and hydrogen

(d) Carbon and chlorine

(e) Sodium and oxygen

(f) Carbon and oxygen

(a) Calcium and fluorine –   CaF₂

(b) Hydrogen and sulphur-   H₂S

(c) Nitrogen and hydrogen-   NH₃

(d) Carbon and chlorine –  CCl₄

(e) Sodium and oxygen – Na₂O

(f) Carbon and oxygen- CO₂ ; CO

---

Q.16 Which of the following symbols of elements are incorrect? Give their correct symbols

(a) Cobalt CO

(b) Carbon c

(c) Aluminium AL

(d) Helium He

(e) Sodium So

(a) Cobalt CO -Incorrect, the correct symbol of Cobalt is Co

(b) Carbon c – Incorrect, the correct symbol of Carbon is C

(c) Aluminium AL-Incorrect, the correct symbol of Aluminium is Al

(d) Helium He – Correct

(e) Sodium So- Incorrect, the correct symbol of Sodium is Na

---

Q.17 Give the chemical formulae for the following compounds and compute the ratio by mass of the combining elements in each one of them. (You may use appendix-III).

(a) Ammonia

(b) Carbon monoxide

(c) Hydrogen chloride

(d) Aluminium fluoride

(e) Magnesium sulphide

(a) Ammonia

NH₃

N  :  H × 3

14  : 1 × 3

14 :  3

(b) Carbon monoxide

CO

C : O

12: 16

3 : 4

(c) Hydrogen chloride

HCl

H : Cl

1: 35.5

(e)Aluminium fluoride

AlF₃

Al  : F × 3

27 : 19 × 3

27 : 57

9 : 19

(f) Magnesium sulphide

MgS

Mg : S

24 : 32

3: 4

---

Q.18 State the number of atoms present in each of the following chemical species

(a) CO₃ ^2–

(b) PO₄ ^3–

(c) P₂O₅

(d) CO

(a) CO₃ ^2–  = 1+3 =4

(b) PO₄ ^3–  = 1 + 4= 5

(c) P₂O₅  = 2 + 5 = 7

(d) CO = 1 + 1 =2

---

Q.19 What is the fraction of the mass of water due to neutrons?

The mass of 1 mole of a substance is equal to its relative atomic or molecular mass in grams.

Mass of one mole (Avogadro Number) of neutrons =1g

Mass of one neutron = 1/ Avogadro  number(N_A) g

Mass of one molecule of water = Molar mass / N_A = 18/ N_A  g

The molar mass of water is 18.015 g/mol. This was calculated by multiplying the atomic weight of hydrogen (1.008) by two and adding the result to the weight for one oxygen (15.999)

Mass of one molecule of water = Molar mass / N_A = 18/ N_A  g

Avogadro  number(N_A)  =6.022 x 10²³ mol¯1

There are 8 neutrons in one atom of oxygen

Number of neutrons in oxygen= number of oxygen – Atomic number of oxygen

Oxygen’s atomic weight= 15.9994
increases with an increase in temperature.
Therefore the mass is 16

Therefore number of neutrons= 16 – 8 = 8

Mass of one neutron = 1/ Avogadro  number(N_A) g

Mass of 8 neutrons = 8/ Avogadro  number(N_A) g

Fraction of mass of water due to neutrons = 8/18 g

---

Q.20 Does the solubility of a substance change with temperature? Explain with the help of an example

Solubility refers to the ability of a given volume of solvent to dissolve a substance.

The solubility of a given solute in a given solvent typically depends on temperature. Solubility of solids and liquids usually increase as the temperature increases, but the solubility of gases decrease with increasing temperature.

For example,

(i)Take some salt  in a bowl.  Bring the temperature of the water to 20°C by adding some ice cubes to the water. Once the temperature of the water has stabilized, 100ml of the water is measured using the measurement cylinder and poured a beaker.

(ii)Sugar is added, 1 mg at a time, to the beaker and it is mixed using the spatula. The sugar is continuously added until it is not able to dissolve in the water anymore. The amount of sugar that has been added is calculated and recorded.

(iii)The beaker is now placed on the hot plate and the temperature of the solution is brought to 40°C. The salt is again added, 10mg a time, until it does not dissolve in the water anymore. The total salt added (including the amount of salt added previously) is calculated and recorded

(iv)Step iii is again repeated at 60°C, 80°C and 100°C and the results are recorded.

The results show that as the temperature of the water increases, the amount of salt dissolving in the solution is also higher.

---

Q.21 Classify each of the following on the basis of their atomicity.

(a) F₂

(b) NO₂

(c) N₂O

(d) C₂H₆

(e) P₄

(f) H₂O₂

(g) P₄O₁₀

(H) O₃

(i) HCl

(j) CH₄ (k) He (l) Ag

Atomicity is the number of atoms present in a molecule

(a) 2- diatomic

(b) NO₂  = 1+ 2 = 3. Triatomic

(c)N₂O = 2 + 1 = 3. Triatomic

(d) C₂H₆  = 2 + 6 =  8 Octa atomic

(e) P₄  =  4 Tetra

(f) H₂O₂  = 2 + 2  = 4. Tetra

(g) P₄O₁₀ =  4 +  10 =  14

(h) O₃ = 3

(i) HCl = 1+ 1 = 2

(j) CH₄ = 1+ 4= 5

(k) 1 (monatomic (inert gases)do not combine and exist as monoatomic gases)

(l) Polyatomic.

---

Q.22 You are provided with a fine white coloured powder which is either sugar or salt. How would you identify it without tasting?

(i)On heating the powder, if the white powder is sugar, it will melt down to a liquid form (sucrose has a decomposition point and melting point at temperatures between 190 to 192 degrees Celsius) and turned a light brown colour. Once the sugar had been heating for awhile, it should have turned black and started producing wisps of smoke.

Salt (sodium chloride) has a melting point of 841 degrees Celsius and 1545.8 degrees Fahrenheit. If we don’t heat it to that point nothing much happens.

(ii)Alternatively, the powder may be dissolved in water and checked for its conduction of electricity. If it conducts electricity, it is a salt. When table salt is dissolved in water, the solution conducts very well, because the solution contains ions. The ions come from the table salt, whose chemical name is sodium chloride. Sodium chloride contains sodium ions, which have a positive charge, and chloride ions, which have a negative charge. Because sodium chloride is made up of ions, it is called an ionic substance.
Sugar is made of uncharged particles called molecules. When sugar is dissolved in water, the solution does not conduct electricity, because there are no ions in the solution.

---

Q.23 Calculate the number of moles of magnesium present in a magnesium ribbon weighing 12 g. Molar atomic mass of magnesium is 24g mol^–1.

Number of moles = w/ atomic weight  =  12/24 =  0.5 mol

---

Q.24 Verify by calculating that

(a) 5 moles of CO₂ and 5 moles of H₂O do not have the same mass.

(b) 240 g of calcium and 240 g magnesium elements have a mole ratio of 3:5.

(a) CO₂ has molar mass = 12.011 + 2(16.00) = 12.011 + 32.00 = 44.011g/mol.

5 moles of CO₂ have molar mass = 44 × 5 = 220 g

Molar mass of H₂O  = H × 2 = 1.01588 x 2 = 2. 01588 g
O × 1 = 1 x 15.9994 g = 15.9994 g

Molar mass of H₂O = 2. 01588 + 15.9994 = 18.0153 g/mol.

5 moles of H₂O have mass = 18 × 5 g = 90 g

(b) Number of moles = w/ atomic weight

Atomic weight of Ca= 40 amu

Number of moles in 240g Ca metal 240/ 40 = 6

Number of moles in 240g of Mg metal 240/ 24 = 10

Atomic weight of Mg = 24amu

Ratio 6:10

3: 5

---

Q.25 Find the ratio by mass of the combining elements in the following compounds. (You may use Appendix-III)

(a) CaCO₃

(b) MgCl₂

(c) H₂SO₄

(d) C₂H₅OH

(e) NH₃

(f) Ca(OH)₂

(a) CaCO₃

Ca: C : O × 3

40 : 12 : 16 × 3

40: 12 : 48

10 : 3 : 12

(b) MgCl₂

Mg : Cl × 2

24: 35.5 × 2

24: 71

(c) H₂SO₄

H x 2 : S : O × 4

2: 32 : 16 × 4

2 : 32 : 64

1: 16: 32

(d) C₂H₅OH

C × 2 : H × 6 : O

12 × 2 : 1 × 6 : 16

24 : 6 : 16

12 : 3 : 8

(e) NH₃

N : H × 3

14 : 1 × 3

14: 3

(f) Ca(OH)₂

Ca : O × 2 : H × 2

40 : 16 × 2 : 1 × 2

40 : 32 : 2

20 : 16 : 1

---

Q.26 Calcium chloride when dissolved in water dissociates into its ions according to the following equation. CaCl₂ (aq) → Ca² + (aq) + 2Cl^– (aq) Calculate the number of ions obtained from CaCl² when 222 g of it is dissolved in water

Molar mass of CaCl² = 40.078 + 35.453 × 2  = 110.984 g/mol

1 mole of calcium chloride = 111g

∴  222g of CaCl² is equivalent to 2 moles of CaCl²

Since 1 formula unit CaCl² gives 3 ions(one Ca² + cation and two Cl- anions),

Therefore, 1 mol of CaCl² will give 3 moles of ions

2 moles of CaCl² would give 3 × 2 = 6 moles of ions.

No. of ions = No. of moles of ions × Avogadro number

= 6 × 6.022 ×10²³ = 36.132 ×10²³ = 3.6132 ×10²⁴ ions

---

Q.27 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.

A sodium atom and ion, differ by one electron. Sodium loses one electron to form sodium ion.

For 100 moles each of sodium atoms and ions there would be a difference of 100 moles of electrons. Mass of 100 moles of electrons= 5.48002 g

As 1 mole of electron have 6.022 × 10²³ electrons(1 mole = 6.022 ×10²³ electrons)

Therefore mass of 6.022 × 10²³ electrons = 5.48002/ 100 g

Mass of one electron = 5.48002/ 100 × 6.022 ×10 = 9.1 ×10^-28  g = 9.1×10^-31  kg

---

Q.28 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.

Molar mass of HgS = 200.6 + 32 = 232.6 g mol^–1

1molecule of HgS contains 1 atom of Hg

232.6 g of HgS contains   200.6 g of Hg

Therefore, Mass of Hg in 225 g of HgS =( 200.6 / 232.6)  × 225 = 194.04g

---

Q.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 × 1024kg). Which one of the two is heavier and by how many times?

One mole of screws weigh = 2.475 ×10²⁴g = 2.475×10²¹ k

Mass of the Earth/ Mass of 1 mole of screws  = 5.98 ×10²⁴ kg / 2.475 ×10²¹  kg = 2.4 x 10³

Therefore, we can say that the mass of earth is 2.4×10³ times the mass of screws.

The earth is 2400 times heavier than one mole of screws.

---

Q.30 A sample of vitamin C is known to contain 2.58 ×10²⁴ oxygen atoms. How many moles of oxygen atoms are present in the sample?

1 mole of oxygen atoms = 6.023×10²³ atoms

Therefore,

Number of moles of oxygen atoms = 2.58 × 10²⁴/6.023 ×10²³  = 4.28 mol

4.28 moles of oxygen atoms

---

Q.31 Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of same weight.

(a) Whose container is heavier?

(b) Whose container has more number of atoms?

(a) Mass of sodium atoms carried by Krish = (5 ×23) g = 115 g

Atomic weight of Na = 23

While mass of carbon atom carried by Raunak = (5 ×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

---

Q.32 Fill in the missing data in the Table .

---

Q.33 The visible universe is estimated to contain 10²² stars. How many moles of stars are present in the visible universe?

Number of moles of stars = 10²² / 6.023 ×10²³  = 0.0166 moles

---

Q.34 What is the SI prefix for each of the following multiples and submultiples of a unit?

(a) 10³

(b) 10^–1

(c) 10^–2

(d) 10^–6

(e) 10^–9

(f) 10^–12

(a) 10³ = 1000= kilo

(b) 10^–1 =1/10= 0.1= deci

(c) 10^–2 =1/100 = 0.01= centi

(d) 10^–6 = 0.000 001= micro

(e) 10^–9 =0.000 000 001 = nano

(f) 10^–12=0.000 000 000 001 = pico

---

Q.35 Express each of the following in kilograms

(a) 5.84 × 10^–3 mg

(b) 58.34 g

(c) 0.584g

(d) 5.873 × 10^-21g

(a) 5.84 × 10^–3 mg  = 5.84 ×10^–9 kg

(b) 58.34 g =5.834 ×10^–2 kg

(c) 0.584g =5.84 ×10^–4 kg

(d) 5.873×10^-21g=5.873 ×10^–24 kg

---

Q.36 Compute the difference in masses of 103 moles each of magnesium atoms and magnesium ions. (Mass of an electron = 9.1 × 10^–31 kg)

A Mg²⁺ ion and Mg atom differ by two electrons.

10³ moles of Mg²⁺ and Mg atoms would differ by 10³ × 2 moles of electrons

Mass of 2 ×10³ moles of electrons = 2×10³ × 6.023 ×10²³ × 9.1 ×10^–31 kg

= 2 × 6.022 × 9.1 × 10^–5kg  = 109.6004 ×10^–5 kg  = 1.096 × 10^–3kg

---

Q.37 Which has more number of atoms? 100g of N₂ or 100 g of NH₃

No of moles of atoms = w/ atomic weight

100 g of N₂ = 100/ 2 x 14 moles

100 g of N₂ =100/28 moles

Number of molecules = 100/ 28 × 6.022 ×1023

Molar mass of N₂= 2 × molar mass of monatomic N = 2 × 14.0067

Molar mass of N₂ = 28.01340

Molar mass of N₂ = 28

To find number of molecules, number of moles X multiply by Avogadro’s number

Number of molecules = 100/ 28 × 6.022 ×10²³

Number of atoms in N₂ =  1 + 1 = 2 = 2 ×100/ 28 × 6.022 ×10²³ = 43.01×10²³

(ii) 100 g of NH₃ = 100/ 17 moles

To find number of molecules, number of moles × Avogadro’s number

Number of molecules = 100/ 17 × 6.022 × 10²³ molecules

Number of atoms in NH₃  =  1 +  3 =  4 =  100/ 17× 6.022 ×10²³ × 4 atoms =  141.69 ×10²³

Therefore,NH₃ would have more atoms

---

Q.38 Compute the number of ions present in 5.85 g of sodium chloride.

Atomic weight; Na = 22.99 g, Cl = 35.45 g

Atomic weight of NaCl= 23 + 35.5 = 58.5g.

5.85 g of NaCl = 5.85/ 58.5 = 0.1moles

Each NaCl particle is equivalent to one Na⁺ and one Cl^– = 2 ions

Total moles of ions = 0.1 × 2= 0.2 moles

Number of ions= total number of moles of ions X Avogadro’s number

Therefore,

No. of ions= 0.2 × 6.022 ×10²³

1.2042 ×10²³ ions

---

Q.39 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?

Given, the sample contains  90% of  Gold (Au).

One gram of gold sample will contain 90% of Gold i.e. 90/100=0.9g of gold

Number of moles of gold = Mass of gold/ Atomic weight of gold

Atomic weight of gold = 196.967 grams/mol= 197 grams/mol

Therefore,

Number of moles of gold = 0.9/197 = 0.0046

One mole of gold contains = 6.022 ×10²³

∴ 0.0046 mole of gold will contain =0.0046  × 6.022 ×10²³  = 2.77 ×10²¹

---

Q.40 What are ionic and molecular compounds? Give examples.

Chemical elements can join with each other to form chemical compounds.

Ionic compounds: Ionic compounds are formed by the attraction between positive and negative ions. Positively charged ions are known as cations and negatively charged ions are known as anions. Ionic compounds are formed. The compounds formed by the transfer of electrons are called as ionic compound. The bond found in these compounds are ionic in nature.

These compounds are pure substances which are formed be a metal and a non-metal. Cations are usually formed by metal atoms and the anions are formed by nonmetal atoms.  They are good conductors of electricity when in a molten or aqueous state, have high melting point,  soluble in water. Ionic compounds exist as crystals. Examples of ionic compounds: sodium chloride,  sodium bromide, potassium chloride etc.

Ex : – 2Na + Cl₂ → 2Na⁺ Cl^– → 2NaCl (sodium chloride- common salt.) Sodium is a group 1 metal, thus forms a +1 charged cation. Chlorine is a nonmetal, and has the ability to form a -1 charged anion.

Molecular Compounds: Molecular compounds are formed by uncharged atoms. They are also called as covalent compounds, as they are formed by sharing of electrons between the two atoms and the elements are held together by covalent bonds.

These, are the pure substances, formed by non-metals. It has a low melting point and cannot conduct electricity regardless of state. Molecular compounds can exist as either solid, liquid or gaseous state. Eg: carbon dioxide, carbon monoxide, hydrogen chloride, water, methane etc.

Ex: 2C + O₂ → 2CO ( Carbon monoxide)

---

Q.41 Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions. (Mass of an electron is 9.1×10^–28 g). Which one is heavier?

Mass of 1 mole of aluminium atom = the molar mass of aluminium = 27 g mol^–1

An aluminium atom loses three electrons to become an ion, Al³⁺

For one mole of A1³⁺ ion, three moles of electrons are to be lost

Therefore, total number of electrons lost =3

The mass of three moles of electrons = 3 × (9.1×10^–28) × 6.022×10²³ g

= 27.3 × 6.022 ×10^–5 g

= 164.400 ×10^–5 g

= 0.00164 g

Molar mass of Al³⁺ = molar mass of aluminium – mass of electrons

(As it loses electrons to become Al³⁺) =(27–0.00164) g mol^–1 = 26.9984 g mol^–1

Therefore, difference in masses of one mole each of aluminium atoms and one mole of its ions

= 27 – 26.9984 = 0.0016 g

---

Q.42 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.

Mass of silver = ‘m’ g                         … (i)

Mass of gold= m/100 g                     … (ii)

Number of atoms of silver(Ag) = Mass / Atomic mass × Avogadro’s number(N_A)

= m/108 × N_A                                (from (i)

Atomic weight of silver = 108 g/mol

Number of atoms of gold(Au) = Mass / Atomic mass × Avogadro’s number(N_A)

= m/100 ×197 × N_A                  (from (ii)

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

= m/100 × 197 × N_A   :  m/108 × N_A

= 108 : 100 × 197

= 108 : 19700

= 1 : 182.41

---

Q.43 A sample of ethane (C₂H₆) gas has the same mass as 1.5 ×10²⁰ molecules of methane (CH₄). How many C₂H₆ molecules does the sample of gas contain?

Mass of 1 molecule of CH₄ = 16 g/ N_A

The atomic weight of carbon is 12, the atomic weight of hydrogen is 1

weight of CH₄ = 12 +  4 × 1 = 16

Mass of 1.5 ×10²⁰ molecules of methane  = 1.5 ×10²⁰  × 16/ N_A

Mass of 1 molecule of C₂H₆ =30 g/ N_AC

C- 12 ;  H-1 ;  Therefore C₂H₆= 2 × 12 + 6 × 1 =  30

Mass of molecules of C₂H₆ is  = Mass of CH₄(given)

Therefore,

Mass of molecules of C₂H₆ =1.5 ×10²⁰ x 16/ N_A

Number of molecules of C₂H₆ = {1.5 ×10²⁰ x 16/ N_A}/ 30 g/ N_A

= {1.5 ×10²⁰  ×16/ N_A} × N_A/30

= 0.8 × 10²⁰

---

Q.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 (PO₄)² is _________.

(d) Formula of sodium carbonate is _______ and that of ammonium sulphate is ________.

(a) Law of conservation of mass.

(b) ions.

(c) 310

3 × atomic mass of Ca+  2 × atomic mass of phosphorus + 8 × atomic mass of oxygen) = 310

3 × 40 + 2 × 31 + 8 × 16 = 120 + 62 + 128 = 310

(d)  Na₂ CO₃      (NH₄) ₂ SO₄

---

Q.45 Complete the following crossword puzzle Figure by using the name of the chemical elements. Use the data given in Table.

Down

Silver

Copper

Mercury

Lead

Across

Gold

Iron

Phosphorous

Hydrogen

---

Q.46 (a) In this crossword puzzle Figure, names of 11 elements are hidden. Symbols of these are given below. Complete the puzzle.

Cl 7. He

H 8. F

Ar 9. Kr

O 10. Rn

Xe 11. Ne

N

(a)

(b) Six : Helium (He); Neon ( Ne); Argon (Ar); Krypton (Kr); Xenon (Xe); Radon (Rn).

---

Q.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

(a) KOH = (39 +16+1) = 56 g mol^–1

(b) NaHCO₃ = 23 + 1 + 12 + (3 × 16) = 84 g mol^–1

(c) CaCO₃ = 40 + 12 + (3 × 16) = 100 g mol^–1

(d) NaOH = 23+16+1 = 40 g mol^–1

(e) C₂H₅OH = C₂H₆O = 2 × 12 + (6 × 1) + 16 = 46 g mol^–1

(f) NaCl = 23 + 35.5 = 58.5 g mol^–1

---

Q.48 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 C6 H₁₂ O₆. 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

6CO₂(g) + 6 H₂O(l)  → C₆ H₁₂ O₆ + 6O₂(g)

1 mole of glucose needs 6 moles of water

Therefore, 180 g of glucose needs (6×18) g of water

Molar mass of water= 2+16= 18

Number of moles= given mass /molar mass

1 g of glucose will need 108/ 180 g of water.

C₆ H₁₂ O₆ = C – 6 × 12 + 12 × 1 + 6 × 16 = 180g

18 g of glucose would need=   (108/180) × 18g of water = 10.8g

Volume = Mass /density

Therefore, Volume = 10.8 g /1 g cm^–3  = 10.8 cm³