The Language of Chemistry Class 9 Chemistry

Selina ICSE Solutions for Class 9 Chemistry Chapter 4 The Language of Chemistry

Question 1.
What is a symbol? What information does it convey?
Solution:
A symbol is the short form which stands for the atom of a specific element or the abbreviations used for the names of elements.

1. It represents a specific element.
2. It represents one atom of an element.
3. A symbol represents how many atoms are present in its one gram (gm) atom.
4. It represents the number of times an atom is heavier than one atomic mass unit (amu) taken as a standard.

Question 2
Why is the symbol S for sulphur, but Na for sodium and Si for silicon?
Solution:
In most cases, the first letter of the name of the element is taken as the symbol for that element and written in capitals (e.g. for sulphur, we use the symbol S). In cases where the first letter has already been adopted, we use a symbol derived from the Latin name (e.g. for sodium/Natrium, we use the symbol Na). In some cases, we use the initial letter in capital together with a small letter from its name (e.g. for silicon, we use the symbol Si).

Question 3.
Write the full form of IUPAC. Name the elements represented by the following symbols:
Au, Pb, Sn, Hg
Solution:
The full form of IUPAC is International Union of Pure and Applied Chemistry.
Names of the elements:
Au – Gold
Pb – Lead
Sn – Tin
Hg – Mercury

Question 4.
If the symbol for Cobalt, Co, were written as CO, what would be wrong with it?
Solution:
Co stands for Cobalt. If we write CO, then it would mean that it is a compound containing two non-metal ions, i.e. carbon and oxygen, which forms carbon monoxide gas.

Question 5(d).
2H₂
Solution:
(a) H stands for one atom of hydrogen.
(b) H₂ stands for one molecule of hydrogen.
(c) 2H stands for two atoms of hydrogen.

Question 6.
What is meant by atomicity? Name the diatomic element.
Solution:
The number of atoms of an element that join together to form a molecule of that element is known as its atomicity. Diatomic molecules: H₂, O₂, N₂, Cl₂

Question 7(a).
Explain the terms ‘valency‘ and ‘variable valency‘.
Solution:

1. Valency of Na is +1 because it can lose one electron.
2. Valency of O is -2 because it can accept two electrons.

Variable valency: It is the combining capacity of an element in which the metal loses more electrons from a shell next to a valence shell in addition to electrons of the valence shell.

Question 7(b).
How are the elements with variable valency named? Explain with an example.
Solution:
If an element exhibits two different positive valencies, then

1. for the lower valency, use the suffix -OUS at the end of the name of the metal
2. for the higher valency, use the suffix -IC at the end of the name of the metal.

Example:

Element	Lower valency	Higher valency
Ferrum (Iron)	Ferrous (Fe2+)	Ferric (Fe3+)

Question 8.
Give the formula and valency of:

1. aluminate ………………… .
2. chromate ………….…….. .
3. aluminium ………………. .
4. cupric  ………………… .

Solution:

	Name	Formula	Valency
a.	Aluminate	AlO2	-2
b.	Chromate	CrO4	-2
c.	Aluminium	Al	+3
d.	Cupric	Cu	+2

Question 9.b
What is the significance of formula?
Solution:
Chemical formula: The chemical formula of a substance (element or compound) is a symbolic representation of the actual number of atoms present in one molecule of that substance.
It also indicates the fixed proportion by weight in which atoms combine.
Rules:
(i) The positive and negative radicals are written side by side (+ve first) with their charge as a superscript on the right side.
(ii) Charges are then interchanged and written as a subscript.
(iii) The final formula is written without the sign of charge, e.g. Hg₂O

1. Hg¹⁺O^2-
   
2. 
3. Hg₂O

Question 10(a).
What do you understand by the following terms?
Acid radical
Solution:
Acid radical: The electronegative or negatively charged radical is called an acid radical.
Examples: Cl^–, O^2-

Question 10(b).
What do you understand by the following terms? Basic radical
Solution:
Basic radical: The electropositive or positively charged radical is called a basic radical.
Examples: K⁺, Na⁺

Question 11.

Match the following:

Compound	Formula
(a) Boric acid	i. NaOH
(b) Phosphoric acid	ii. SiO2
(c) Nitrous acid	iii. Na2CO3
(d) Nitric acid	iv. KOH
(e) Sulphurous acid	v. CaCO3
(f) Sulphuric acid	vi. NaHCO3
(g) Hydrochloric acid	vii. H2S
(h) Silica (sand)	viii. H2O
(i) Caustic soda (sodium hydroxide)	ix. PH3
(j) Caustic potash (potassium hydroxide)	x. CH4
(k) Washing soda(sodium carbonate)	xi. NH3
(l) Baking soda(sodium bicarbonate)	xii. HCl
(m) Lime stone.(calcium carbonate)	xiii. H2SO3
(n) Water	xiv. HNO3
(o) Hydrogen sulphide	xv. HNO2
(p) Ammonia	xvi. H3BO3
(q) Phosphine	xvii. H3PO4
(r) Methane	xviii. H2SO4

Solution:

Compound	Formula (Ans)
(a) Boric acid	xvi. H3BO3
(b) Phosphoric acid	xvii. H3PO4
(c) Nitrous acid	xv. HNO2
(d) Nitric acid	xiv. HNO3
(e) Sulphurous acid	xiii. H2SO3
(f) Sulphuric acid	xviii. H2SO4
(g) (a) Hydrochloric acid	xii. HCl
(h) Silica (sand)	ii. SiO2
(i) Caustic soda (sodium hydroxide)	i. NaOH
(j) Caustic potash (potassium hydroxide)	iv. KOH
(k) Washing soda (sodium carbonate)	iii. Na2CO3
(l) Baking soda (sodium bicarbonate)	vi. NaHCO3
(m) Lime stone (calcium carbonate)	v. CaCO3
(n) Water	viii. H2O
(o) Hydrogen sulphide	vii. H2S
(p) Ammonia	xi. NH3
(q) Phosphine	ix. PH3
(r) Methane	x. CH4

Question 12.
Select the basic and acidic radicals in the following compounds.

1. MgSO₄
2. (NH₄)₂SO₄
3. Al₂(SO₄)₃
4. ZnCO₃
5. Mg(OH)₂

Solution:

	Acidic radical	Basic radical
MgSO4	SO4–	Mg+
(NH4)2SO4	SO4–	NH4+
Al2(SO4)3	SO4–	Al3+
ZnCO3	CO3–	Zn2+
Mg(OH)2	OH–	Mg2+

Question 13.
Write chemical formula of the sulphate of Aluminium, Ammonium and Zinc.
Solution:
Valencies of aluminium, ammonium and zinc are 3, 1 and 2, respectively.
The valency of sulphate is 2.
Hence, chemical formulae of the sulphates of aluminium, ammonium and zinc are Al₂(SO₄)₃, (NH₄)₂SO₄ and ZnSO₄.

Question 14.
The valency of an element A is 3 and that of element B is 2. Write the formula of the compound formed by the combination of A and B
Solution:
Formula of the compound = A₂B₃

Question 15.
Write the chemical names of the following compounds:

1. Ca₃(PO₄)₂
2. K₂CO₃
3. K₂MnO₄
4. Mn₃(BO₃)₂
5. Mg(HCO₃)₂
6. Na₄Fe(CN)₆
7. Ba(ClO₃)₂
8. Ag₂SO₃
9. (CH₃COO)₂Pb
10. Na₂SiO₃

Solution:
Chemical names of compounds:

1. Ca₃(PO₄)₂ – Calcium phosphate
2. K₂CO₃ – Potassium carbonate
3. K₂MnO₄ – Potassium manganate
4. Mn₃(BO₃)₂ – Manganese (II) borate
5. Mg(HCO₃)₂ – Magnesium hydrogen carbonate
6. Na₄Fe(CN)₆ – Sodium ferrocyanide
7. Ba(ClO₃)₂ – Barium chlorate
8. Ag₂SO₃ – Silver sulphite
9. (CH₃COO)₂Pb – Lead acetate
10. Na₂SiO₃ – Sodium silicate

Question 16
Write the basic radicals and acidic radicals of the following and then write the chemical formulae of these compounds.

1. Barium sulphate
2. Bismuth nitrate
3. Calcium bromide
4. Ferrous sulphide
5. Chromium sulphate
6. Calcium silicate
7. Potassium ferrocyanide
8. Stannic oxide
9. Magnesium phosphate
10. Sodium zincate
11. Stannic phosphate
12. Sodium thiosulphate
13. Potassium manganate
14. Nickel bisulphate

Solution:

Compounds	Acidic radical	Basic radical	Chemical formulae
Barium sulphate	SO42-	Ba2+	BaSO4
Bismuth nitrate	NO3–	Bi3+	Bi(NO3)3
Calcium bromide	Br–	Ca2+	CaBr2
Ferrous sulphide	S2-	Fe2+	FeS
Chromium sulphate	SO42-	Cr3+	Cr2(SO4)3
Calcium silicate	SiO42-	Ca2+	Ca2SiO4
Potassium ferrocyanide	[Fe(CN)6]4-	K1+	K4[Fe(CN)6]
Stannic oxide	O2-	Sn2+	SnO2
Magnesium phosphate	(PO4)3-	Mg2+	Mg3(PO4)2
Sodium zincate	ZnO2-	Na1+	Na2ZnO2
Stannic phosphate	(PO4)3-	Sn4+	Sn3(PO4)4
Sodium thiosulphate	(S2O3)2-	Na1+	Na2S2O3
Potassium manganate	MnO42-	K1+	K2MnO4
Nickel bisulphate	HSO41-	Ni3+	Ni(HSO4)3

Question 16.
Give the names of the following compounds.

1. NaClO
2. NaClO₂
3. NaClO₃
4. NaClO₄

Solution:

1. NaClO – Sodium hypochlorite
2. NaClO₂ – Sodium chlorite
3. NaClO₃ – Sodium chlorate
4. NaClO₄ – Sodium perchlorate

Question 18(a).
Complete the following statements by selecting the correct option :
The formula of a compound represents
i. an atom
ii. a particle
iii. a molecule
iv. a combination
Solution:
iii. The formula of a compound represents a molecule.

Question 18(b).
Complete the following statements by selecting the correct option :
The correct formula of aluminium oxide is
i. AlO₃
ii. AlO₂
 iii. Al₂O₃
Solution:
iii. The correct formula of aluminium oxide is Al₂O₃.

Question 18(c).
Complete the following statements by selecting the correct option :
The valency of nitrogen in nitrogen dioxide (NO₂) is
i. one
ii. two
iii. three
iv. four
Solution:
iv. The valency of nitrogen in nitrogen dioxide (NO₂) is four.

Page No: 13

Question 1.
Balance the following equations:

1. Fe + H₂O → Fe₃O₄ + H₂
2. Ca + N₂ → Ca₃N₂
3. Zn + KOH → K₂ZnO₂ + H₂
4. Fe₂O₃ + CO → Fe + CO₂
5. PbO + NH₃ → Pb + H₂O + N₂
6. Pb₃O₄ → PbO + O₂
7. PbS + O₂ → PbO + SO₂
8. S + H₂SO₄ → SO₂ + H₂O
9. S + HNO₃ → H₂SO₄ + NO₂ + H₂O
10. MnO₂ + HCl → MnCl₂ + H₂O + Cl₂
11. C + H₂SO₄ → CO₂ + H₂O + SO₂
12. KOH + Cl₂ → KCl + KClO + H₂O
13. NO₂ +H₂O → HNO₂ + HNO₃
14.  Pb₃O₄ + HCl → PbCl₂ + H₂O + Cl₂
15.  H₂O + Cl₂ → HCl + O₂
16.  NaHCO₃ → Na₂CO₃ + H₂O + CO₂
17. HNO₃ + H₂S → NO₂ + H₂O + S
18. P + HNO₃ → NO₂ + H₂O + H₃PO₄
19. Zn + HNO₃  → Zn(NO₃)₂ + H₂O + NO₂

Solution:
Balanced chemical equations:

1. 3Fe + 4H₂O → Fe₃O₄ + 4H₂
2. 3Ca + N₂ → Ca₃N₂
3. Zn + 2KOH → K₂ZnO₂ + H₂
4. Fe₂O₃ + 3CO → 2Fe + 3CO₂
5. 3PbO + 2NH₃ → 3Pb + 3H₂O + N₂
6. 2Pb₃O₄ → 6PbO + O₂
7. 2PbS + 3O₂ → 2PbO + 2SO₂
8. S + 2H₂SO₄ → 3SO₂ + 2H₂O
9. S + 6HNO₃ → H₂SO₄ + 6NO₂ + 2H₂O
10. MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
11. C + 2H₂SO₄ → CO₂ + H₂O + SO₂
12. 2KOH + Cl₂ → KCl + KClO + H₂O
13. 2NO₂ + H₂O → HNO₂ + HNO₃
14. Pb₃O₄ + 8HCl → 3PbCl₂ + 4H₂O + Cl₂
15. 2H₂O + 2Cl₂ → 4HCl + O₂
16. 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂
17. 2HNO₃ + H₂S → 2NO₂ + 2H₂O + S
18. P + 5HNO₃ → 5NO₂ + H₂O + H₃PO₄
19. Zn + 4HNO₃  → Zn(NO₃)₂ + 2H₂O + 2NO₂

Page No: 17

Question 1.
Fill in the blanks:

1. Dalton used symbol _____ for oxygen _____ for hydrogen.
2. Symbol represents _____ atom(s) of an element.
3. Symbolic expression for a molecule is called _____.  .
4. Sodium chloride has two radicals. Sodium is a _____ radical while chloride is _____ radical.
5. Valency of carbon in CH₄ is _____ , in C₂H₆ _____, in C₂H₄ ___ and in C₂H₂ is ____.
6. Valency of Iron in FeCl₂ is _____  and in FeCl₃ it is ____ .
7. Formula of iron (ill) carbonate is _____ .

Solution:

1. Dalton used symbol [O] for oxygen,[H] for hydrogen.
2. Symbol represents gram atom(s) of an element.
3. Symbolic expression for a molecule is called molecular formula.
4. Sodium chloride has two radicals. Sodium is a basic radical, while chloride is an acid radical.
5. Valency of carbon in CH₄ is 4, in C₂H₆4, in C₂H₄4 and in C₂H₂ is 4.
6. Valency of iron in FeCl₂ is 2 and in FeCl₃ it is 3.
7. Formula of iron (III) carbonate is Fe₂[CO₃]₃.

Question 2.
Complete the following table.

Acid Radicals   Basic Radicals	Chloride	Nitrate	Sulphate	Carbonate	Hydroxide	Phosphate
Magnesium	MgCl2	Mg(NO3)2	MgSO4	MgCO3	Mg(OH)2	Mg3(PO4)2
Sodium
Zinc
Silver
Ammonium
Calcium
Iron (II)
Potassium

Solution:

Question 3.
Sodium chloride reacts with silver nitrate to produce silver chloride and sodium nitrate

1. Write the equation.
2. Check whether it is balanced, if not balance it.
3. Find the weights of reactants and products.
4. State the law which this equation satisfies.

Solution:
(a) NaCl+ AgNO₃ → NaNO₃ + AgCl↓
(b) It is a balanced equation.
(c) Weights of reactants:NaCl – 58.44, AgNO₃ – 169.87
Weights of products: NaNO₃ – 84.99, AgCl – 143.32
NaCl + AgNO₃ → NaNO₃  + AgCl
(23+35.5) + (108+14+48) → (23+14+48) + (108+35.5)
58.5 + 170 → 85 + 143.5
228.5 g → 228.5 g
(d) Law of conservation of mass: Matter is neither created nor destroyed in the course of a chemical reaction.

Question 4(a).
What information does the following chemical equation convey? Zn + H₂SO₄ → ZnSO₄+ H₂
Solution:
(a) This equation conveys the following information:

1. The actual result of a chemical change.
2. Substances take part in a reaction, and substances are formed as a result of the reaction.
3. Here, one molecule of zinc and one molecule of sulphuric acid react to give one molecule of zinc sulphate and one molecule of hydrogen.
4. Composition of respective molecules, i.e. one molecule of sulphuric acid contains two atoms of hydrogen, one atom of sulphur and four atoms of oxygen.
5. Relative molecular masses of different substances, i.e. molecular mass of
   Zn = 65
   H₂SO₄ = (2+32+64) = 98
   ZnSO₄ = (65+32+64) = 161
   H₂ = 2
6. 22.4 litres of hydrogen are formed at STP.

Question 4(b).
What information do the following chemical equations convey? Mg + 2HCl → MgCl₂+ H₂
Solution:
(b) This equation conveys the following information:

1. Magnesium reacts with hydrochloric acid to form magnesium chloride and hydrogen gas.
2. 24 g of magnesium reacts with 2(1 + 35.5) = 73 g of hydrochloric acid to produce (24 + 71), i.e. 95 g of magnesium chloride.
3. Hydrogen produced at STP is 22.4 litres.

Question 5(a).
What are polyatomic ions? Give two examples.
Solution:
(a) A poly-atomic ion is a charged ion composed of two or more atoms covalently bounded that can be carbonate (CO₃^2-) and sulphate (SO₄^2-)

Question 5(b).
Name the fundamental law that is involved in every equation.
Solution:
(b) The fundamental laws which are involved in every equation are:

1. A chemical equation consists of formulae of reactants connected by plus sign (+) and arrow (→) followed by the formulae of products connected by plus sign (+).
2. The sign of an arrow (→) is to read ‘to form’. It also shows the direction in which reaction is predominant.

Question 6(a).
What is the valency of : fluorine in CaF₂
Solution:
(a) Valency of fluorine in CaF₂ is -1.

Question 6(b).
What is the valency of :
sulphur in SF₆
Solution:

(b) Valency of sulphur in SF₆ is -6.

Question 6(c).

What is the valency of :

phosphorus in PH₃
Solution:
(c) Valency of phosphorus in PH₃ is +3.

Question 6(d).
What is the valency of : carbon in CH₄

Solution:

(d) Valency of carbon in CH₄ is +4.

Question 6(e).
What is the valency of :
nitrogen in the following compounds:
(i) N₂O₃ (ii) N₂O₅ (iii) NO₂ (iv) NO
Solution:
(e) Valency of nitrogen in the given compounds:

1. N₂O₃  = N is +3
2. N₂O₅ = N is +5
3. NO₂ = N is +4
4. NO = N is +2

Question 7.
Why should an equation be balanced? Explain with the help of a simple equation.
Solution:
According to law of conservation of mass, “matter can neither be created nor be destroyed in a chemical reaction”. This is possible only, if total number of atoms on the reactants side is equals to total number of atoms on products side. Thus, a chemical reaction should be always balanced.
Let us consider an example,
Fe + H₂O → Fe₃O₄ + H₂
In this equation number of atoms on both sides is not the same, the equation is not balanced.
The balanced form of this equation is given by,
3Fe + 4H₂O  → Fe₃O₄ + 4H₂

Question 8(a).
Write the balanced chemical equations of the following reactions. sodium hydroxide + sulphuric acid → sodium sulphate + water
Solution:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Question 8(b).
Write the balanced chemical equations of the following reactions. potassium bicarbonate + sulphuric acid → potassium sulphate + carbon dioxide + water
Solution:
2KHCO₃ + H₂SO₄ → K₂SO₄ + 2CO₂ + 2H₂O

Question 8(c).
Write the balanced chemical equations of the following reactions. iron + sulphuric acid → ferrous sulphate + hydrogen.
Solution:
Fe + H₂SO₄ → FeSO₄ + H₂

Question 8(d).
Write the balanced chemical equations of the following reactions. chlorine + sulphur dioxide + water →  sulphuric acid + hydrogen chloride
Solution:
Cl₂ + SO₂ + 2H₂O → H₂SO₄ + 2HCl

Question 8(e).
Write the balanced chemical equations of the following reactions. silver nitrate → silver + nitrogen dioxide + oxygen”
Solution:
2AgNO₃ → 2Ag + 2NO₂ + O₂

Question 8(f).
Write the balanced chemical equations of the following reactions.
copper + nitric acid → copper nitrate + nitric oxide + water
Solution:
3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O

Question 8(g).
Write the balanced chemical equations of the following reactions.
ammonia + oxygen → nitric oxide + water
Solution:
  

Question 8(h).
Write the balanced chemical equations of the following reactions.
barium chloride + sulphuric acid → barium sulphate + hydrochloric acid
Solution:
BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

Question 8(i).
Write the balanced chemical equations of the following reactions.
zinc sulphide + oxygen → zinc oxide + sulphur dioxide
Solution:
2ZnS + 3O₂ → 2ZnO + 2SO₂

Question 8(j).
Write the balanced chemical equations of the following reactions.
aluminium carbide + water → aluminium hydroxide + methane
Solution:
Al₄C₃ + 12H₂O → 4Al(OH)₃ + 3CH₄

Question 8(k).
Write the balanced chemical equations of the following reactions.
iron pyrites(FeS₂) + oxygen → ferric oxide + sulphur dioxide
Solution:
 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

Question 8(l).
Write the balanced chemical equations of the following reactions.
potassium permanganate + hydrochloric acid → potassium chloride + manganese chloride + chlorine + water
Solution:
2KMnO₄ + HCl → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O

Question 8(m).
Write the balanced chemical equations of the following reactions.
aluminium sulphate + sodium hydroxide → sodium sulphate + sodium meta aluminate + water.
Solution:
Al₂(SO₄)₃ + 8NaOH → 3Na₂SO₄ + 2NaAlO₂ + 4H₂O

Question 8(n).
Write the balanced chemical equations of the following reactions.
aluminium + sodium hydroxide + water → sodium meta aluminate + hydrogen
Solution:
2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂

Question 8(o).
Write the balanced chemical equations of the following reactions.
potassium dichromate + sulphuric acid → potassium sulphate + chromium sulphate + water + oxygen.
Solution:
2K₂Cr₂O₇ + 8H₂SO₄ → 2K₂SO₄ + 2Cr₂(SO₄)₃ + 8H₂O + 3O₂

Question 8(p).
Write the balanced chemical equations of the following reactions.
potassium dichromate + hydrochloric acid → Potassium chloride + chromium chloride + water + chlorine
Solution:
K₂Cr₂O₇ + 14HCl → 2KCl + 2CrCl₃ + 7H₂O + 3Cl₂

Question 8(q).
Write the balanced chemical equations of the following reactions.
sulphur + nitric acid → sulphuric acid + nitrogen dioxide + water.
Solution:
S + HNO₃ → H₂SO₄ + NO₂ + H₂O

Question 8(r).
Write the balanced chemical equations of the following reactions.
sodium chloride + manganese dioxide + sulphuric acid → sodium hydrogen sulphate  + manganese sulphate + water + chlorine.
Solution:
2NaCl + MnO₂ + 3H₂SO₄ → 2NaHSO₄ + MnSO₄ + 2H₂O + Cl₂

Question 9(a).
Define atomic mass unit.
Solution:
Atomic mass unit (amu) is equal to one-twelfth the mass of an atom of carbon-12 (atomic mass of carbon taken as 12).

Question 9(b)(ii)
Calculate the molecular mass of the following:
(NH₄)₂CO₃
Given atomic mass of Cu = 63·5, H = 1, O= 16, C = 12, N = 14, Mg = 24, S = 32
Solution:
Molecular mass of (NH₄)₂CO₃
= (2 × 14) + (8 × 1) + 12 + (3 × 16)
= 28 + 8 + 12 + 48
= 96

Question 9(b)(iii)
Calculate the molecular mass of the following:
(NH₂)₂CO
Given atomic mass of Cu = 63·5, H = 1, O= 16, C = 12, N = 14, Mg = 24, S = 32
Solution:
Molecular mass of (NH₂)₂CO
= (14 × 2) + (4 × 1) + 12 + 16
= 28 + 4 + 12 + 16
= 60

Question 9(b)(iv)
Calculate the molecular mass of the following:
Mg₃N₂
Given atomic mass of Cu = 63·5, H = 1, O = 16, C = 12, N = 14, Mg = 24, S = 32
Solution:
Molecular mass of Mg₃N₂
= (3 × 24) + (2 × 14)
= 72 + 28
= 100

Question 10(a).
Choose the correct answer from the options given below.
Modern atomic symbols are based on the method proposed by
i. Bohr
ii. Dalton
iii. Berzelius
iv. Alchemist
Solution:
iii. Berzelius

Question 10(b).
Choose the correct answer from the options given below.
The number of carbon atoms in a hydrogen carbonate radical is
i. One
ii. Two
iii. Three
iv. Four
Solution:
One

Question 10(c).
Choose the correct answer from the options given below.
The formula of iron (III) sulphate is
i. Fe₃SO₄
ii. Fe(SO₄)₃
iii. Fe₂(SO₄)₃
iv. FeSO₄
Solution:
iii. Fe₂(SO₄)₃

Question 10(d).
Choose the correct answer from the options given below.
In water, the hydrogen-to-oxygen mass ratio is
i. 1: 8
ii. 1: 16
iii. 1: 32
iv. 1: 64
Solution:
i. 1:8

Question 10(e).
Choose the correct answer from the options given below.
The formula of sodium carbonate is Na₂CO₃ and that of calcium hydrogen carbonate is
i. CaHCO₃
ii. Ca(HCO₃)₂
iii. Ca₂HCO₃
iv. Ca(HCO₃)₃
Solution:
i. Ca(HCO₃)₂

Solution 11.
(a) A molecular formula represent The Molecule of an element or of a Compound.
(b) The molecular formula of water (H₂O) represents 18 parts by mass of water.
(c) A balanced equation obeys the law of conservation of mass wherever unbalanced equation does not obey this law.
(d) CO and Co represent carbon-monoxide and cobalt respectively.

Solution 12.

1. Relative molecular mass of CHCl3
   = 12 + 1 + (3 × 35.5)
   = 12 + 1 + 106.5
   = 119.5
2. Relative molecular mass of (NH4)2 Cr2O7
   = (14 × 2) + (1× 8) + (52 × 2) + (16 × 7)
   = 28 + 8 + 104 + 112
   = 252
3. Relative molecular mass of CuSO4· 5H2O
   = 63.5 + 32 + (16 × 4) + 5(2 + 16)
   = 159.5 + 90
   = 249.5
4. Relative molecular mass of (NH4)2SO4
   = (2 × 14) + (8 × 1) + 32 + (4 × 16)
   = 28 + 8 + 32 + 64
   = 132
5. Relative molecular mass of CH3COONa
   = (12 × 2) + (1× 3) + (16 × 2) + 23
   = 24 + 3 + 32 + 23
   = 82
6. Potassium chlorate (KClO3)
   = 39.1+ 35.5 + (16 × 3)
   = 39.1+ 35.5 + 48
   = 122.6
7. Ammonium chloroplatinate (NH4)2PtCl6
   = (14 × 2) + (1 × 8) + 195.08 + (35.5 × 6)
   = 28 + 8 + 195.08 + 213
   = 444.08

Solution 13.

Compound	Empirical formula
(a) Benzene (C6H6)	CH
CompoundEmpirical formula (b) Glucose (C6H12O6)	CH2O
CompoundEmpirical formula (c) Acetylene (C2H2)	CH
CompoundEmpirical formula (d) Acetic acid (CH3COOH)	CH2O

Solution 14.
Relative molecular mass of MgSO₄·7H₂O
=24 + 32 + (16 × 4) + 7(2 + 16)
=24 + 32 + 64 + 126
=246
26 g of Epsom salt contains 126 g of water of crystallisation.
Hence, 100 g of Epsom salt contains

The % of H₂O in MgSO₄·7H₂O = 51.2

Solution 15.
(a)  Relative molecular mass of Ca(H₂PO₄)₂
= 40.07 + (1 × 4) + (30.9 × 2) + (16 × 8)
= 40.07 + 4 + 61.8 + 128
= 233.87
233.87 g Ca(H₂PO₄)₂ contains 61.8 g P
So, 100 g Ca(H₂PO₄)₂ contains

The % of P in Ca(H₂PO₄)₂ is 26.42%.
(b) Relative molecular mass of Ca₃(PO₄)₂
= (40.07 × 3) + (30.9 × 2) + (16 × 8)
= 120.21 + 61.8 + 128
= 310.01
310.01 g Ca₃(PO₄)₂ contains 61.8 g P
So, 100 g Ca(H₂PO₄)₂ contains
(IMAGE)
The % of P in Ca(H₂PO₄)₂ is 19.93%.

Solution 16.
Relative molecular mass of KClO₃
= 39.09 + 35.5 + (3 × 16)
= 122.59 g

The percentages of K, Cl and O in KClO₃ are 31.9%, 28.9% and 39.1%, respectively.

Solution 17.
Relative molecular mass of urea is

Element	No. of atoms	Atomic mass	Total
N	2	14	28
C	1	12	12
H	4	1	4
O	1	16	16

[12 + 16 + 28 + 4] = 60
Hence, relative molecular mass of urea = 60