ICSE Solutions for Chapter 1 The Language of Chemistry Class 9 Selina Chemistry

Exercise 1A

1. What is a symbol? What information does it convey?

Solution

A symbol is the short form for the atom of a specific element or the abbreviations used to refer to the names of elements

(i) It represents a specific element.

(ii) It represents one atom of an element.

(iii) A symbol represents how many atoms are present in its one gram (gm) atom.

(iv) It represents the number of times an atom is heavier than one atomic mass unit (amu) taken as a standard.

2. Why is the symbol S for Sulphur, but Na for Sodium and Si for Silicon?

Solution

For most of the element, 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 used for naming another element, we use a symbol derived from the Latin name of that element (e.g. for sodium/Atrium, 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).

3. Write the full form of IUPAC. Name the elements represented by the following Symbols: Au, Pb, Sn, Hg

Solution

IUPAC stands for The International Union of Pure and Applied Chemistry (IUPAC).

Au- Gold

Pb- Lead

Sn- Tin

Hg- Mercury

4. If the symbol for cobalt, Co were written as CO, What would be wrong with it?

Solution

Co represents Cobalt. If we write CO then it means it consists of two different non-metals namely Carbon and Oxygen which will represent Carbon-monoxide but not Cobalt.

C represents Carbon and O represents Oxygen.

5. What do the following symbols stand for?

(a) H

(b) H₂

(c) 2H

(d) 2H₂

Solution

(a) One Hydrogen atom

(b) Hydrogen molecules

(d) Two molecules of hydrogen.

6. What is meant by atomicity? Name the diatomic element.

Solution

The number of atoms of an element which join together to form a molecule is known as atomicity of that element.

Diatomic molecules: H₂, O₂, N₂, Cl₂

7. (a) Explain the terms ‘valence’ and ‘variable valency.

(b) How are the elements with variable valence named? Explain with an example.

Solution

(a) Valence of Na is +1 because it can lose one electron.

Valence of O is -2 because it can accept two electrons.

Variable valence: 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.

(b) If an element exhibits two different positive valences, then

for the lower valence, use the suffix -OUS at the end of the name of the metal

for the higher valency, use the suffix -IC at the end of the name of the metal.

Example:

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

8. Give the formula and valency of

(a) aluminate
(b) chromate
(c) aluminium
(d) cupric

Solution

Name	Formula	Valency
Aluminate	AlO₂	-2
Chromate	CrO₄	-2
Aluminium	Al	+3
Cupric	Cu	+2

9. (a) What is a chemical formula?
(b) What is the significance of a formula? Give an example to illustrate.

Solution

(a) A chemical formula is a symbolic representation of the number of atoms present in a molecule of that substance.

(b) Significance of the molecular formula:

It represents both molecule and molecular mass of the compound.
It represents the respective number of different atoms present in one molecule of the compound.
It represents the ratios of the respective masses of the elements present in the compound.

Example: Salt: NaCl, Ethanol C₂H₆O because the molecule of ethanol contains two Carbon, 6 Hydrogen and 1 Oxygen atom.

10. What do you understand by the following terms?

(a) Acid radical

(b) Basic radical

Solution

(a) Acid radical: The electronegative or negatively charged radical is called an acid radical.

Examples: Cl^–, O^2-

(b) Basic radical:The electropositive or positively charged radical is called a basic radical.

Examples: K⁺, Na⁺

11. Select the basic and acidic radicals in the following compounds.

(a) MgSO₄

(b) (NH₄)₂SO₄

(c) Al₂ (SO₄)₃

(d) ZnCO₃

(e) Mg (OH)₂

Solution

	Acidic radical	Basic radical
a. MgSO₄	SO₄^–	Mg⁺
b. (NH₄)₂SO₄	SO₄^–	NH₄⁺
c. Al₂(SO₄)₃	SO₄^–	Al³⁺
d. ZnCO₃	CO₃^–	Zn²⁺
e. Mg(OH)₂	OH^–	Mg²⁺

12. Write chemical formula of the sulphate of Aluminum, Ammonium and Zinc.

Solution

Valences of aluminum, ammonium and zinc are 3, 1 and 2, respectively. The valences of sulphate is 2.

Hence, chemical formulae of the sulphates of aluminum, ammonium and zinc are Al₂(SO₄)₃, (NH₄)₂SO₄and ZnSO₄.

13. The valence 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

14. Match the following:

Compound	Formula
Boric acid	NaOH
Phosphoric acid	SiO₂
Nitrous acid	Na₂CO₃
Nitric acid	KOH
Sulphurous acid	CaCO₃
Sulphuric acid	NaHCO₃
Hydrochloric acid	H₂S
Silica (Sand)	H₂O
Caustic soda ( Sodium Hydroxide)	PH₃
Caustic potash( potassium hydroxide)	CH₄
Washing soda( Sodium carbonate)	NH₃
Baking Soda ( Sodium bicarbonate)	HCl
Limestone ( calcium carbonate)	H₂SO₃
Water	HNO₃
Hydrogen Sulphide	HNO₂
Ammonia	H₃BO₃
Phosphine	H₃PO₄
Methane	H₂SO₄

Solution

Compound	Formula
Boric acid	H₃BO₃
Phosphoric acid	H₃PO₄
Nitrous acid	HNO₂
Nitric acid	HNO₃
Sulphurous acid	H₂SO₃
Sulphuric acid	H₂SO₄
Hydrochloric acid	HCl
Silica (Sand)	SiO₂
Caustic soda ( Sodium Hydroxide)	NaOH
Caustic potash( potassium hydroxide)	KOH
Washing soda( Sodium carbonate)	Na₂CO₃
Baking Soda ( Sodium bicarbonate)	NaHCO₃
Limestone ( calcium carbonate)	CaCO₃
Water	H₂O
Hydrogen Sulphide	H₂S
Ammonia	NH₃
Phosphine	PH₃
Methane	CH₄

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

(a) Barium sulphate

(b) Bismuth nitrate

(c) Calcium bromide

(d) Ferrous sulphide

(e) Chromium sulphate

(f) Calcium silicate

(g) Stannic oxide

(h) Sodium zincate

(i) Magnesium phosphate

(j) Sodium thiosulphate

(k) Stannic phosphate

(l) Nickel bisulphate

(m) Potassium manganite

(n) Potassium Ferro cyanide

Solution

Compounds	Acidic radical	Basic radical	Chemical formulae
Barium sulphate	SO₄^2-	Ba²⁺	BaSO₄
Bismuth nitrate	NO₃^–	Bi³⁺	Bi(NO₃)₃
Calcium bromide	Br^–	Ca²⁺	CaBr₂
Ferrous sulphide	S^2-	Fe²⁺	FeS
Chromium sulphate	SO₄^2-	Cr³⁺	Cr₂(SO₄)₃
Calcium silicate	SiO₄^2-	Ca²⁺	Ca₂SiO₄
Potassium ferrocyanide	[Fe(CN)₆]^4-	K¹⁺	K₄[Fe(CN)₆]
Stannic oxide	O^2-	Sn²⁺	SnO₂
Magnesium phosphate	(PO₄)^3-	Mg²⁺	Mg₃(PO₄)₂
Sodium zincate	ZnO^2-	Na¹⁺	Na₂ZnO₂
Stannic phosphate	(PO₄)^3-	Sn⁴⁺	Sn₃(PO₄)₄
Sodium thiosulphate	(S₂O₃)^2-	Na¹⁺	Na₂S₂O₃
Potassium manganate	MnO₄^2-	K¹⁺	K₂MnO₄
Nickel bisulphate	HSO₄^1-	Ni³⁺	Ni(HSO₄)₃

16. Write the chemical names of the following compounds:

(a) Ca₃ (PO₄)₂

(b) K₂CO₃

(c)K₂MnO₄

(d) Mn₃ (BO₃)₂

(e) Mg (HCO₃)₂

(f) Na₄Fe (CN)₆

(g) Ba (ClO₃)₂

(h) Ag₂SO₃

(i) (CH₃COO)₂Pb

(j) Na₂SiO₃

Solution

Chemical names of compounds:

Ca₃(PO₄)₂– Calcium phosphate

K₂CO₃– Potassium carbonate

K₂MnO₄– Potassium manganite

Mn₃(BO₃)₂– Manganese (II) borate

Mg(HCO₃)₂– Magnesium hydrogen carbonate

Na₄Fe(CN)₆– Sodium Ferro cyanide

Ba(ClO₃)₂– Barium chlorate

Ag₂SO₃– Silver sulphite

(CH₃COO)₂Pb – Lead acetate

Na₂SiO₃– Sodium silicate

17. Give the names of the following compounds.

(a) KClO

(b) KClO₂

(c) KClO₃

(d) KClO₄

Solution

(a) Potassium hypochlorite

(b) Potassium chlorate

(d) Sodium perchlorate

18. Complete the following statements by selecting the correct option.

(a) The formula of a compound represents

an atom
a particle
a molecule
a combination

(b) The correct formula of aluminium oxide is

AlO₃
AlO₂
Al₂O₃
Al₃O₂

(c) The valency of Nitrogen in Nitrogen dioxide( NO₂) is

One
Two
Three
Four

Solution

(a) iii. The formula of a compound represents a molecule.

(b) iii. The correct formula of aluminum oxide is Al₂O₃.

19. Give the names of the elements and number of atoms of those elements present in the following compounds.

Sodium sulphate
Quick lime
Baking soda (NaHCO₃)
Ammonia
Ammonium dichromate

Solution

Sodium sulphate – Na₂SO₄

There are two sodium atoms, one Sulphur atom and four oxygen atoms.

Quick lime – CaO

There is one calcium atom and one oxygen atom.

Baking soda – NaHCO₃

There is one sodium, carbon and hydrogen atom and three oxygen atoms.

Ammonia – NH₃

There is one nitrogen atom and three hydrogen atoms.

Ammonium dichromate – (NH₄) Cr2O₇

There two ammonium atoms, two chromium atoms and seven oxygen atoms.

20. The formula of the sulphate of an element M is M₂(SO₄)₃. Write the formula of its:

Chloride
Oxide
Phosphate
Acetate

Solution

The valence of metal M is 3. So, the formulae are as follows:

Chloride – MCl₃
Oxide – M₂O₃
Phosphate – M(PO₄)
Acetate – M(CH₃COO)₃

Exercise 1B

1. What is a chemical equation? Why it is necessary to balance it?

Solution

A chemical equation is the symbolic representation of a chemical reaction using the symbols and formulae of the substances which are involved in the reaction.

Necessity for balancing a chemical equation:

A chemical equation needs to be balanced because a chemical reaction is just a rearrangement of atoms.

Atoms themselves are neither created nor destroyed during the course of a chemical reaction.

The chemical equation needs to be balanced to follow the law of conservation of mass.

2. State the information conveyed by the following equation:

Zn_(s) + 2HCl_(aq) → ZnCl_{2 (aq)}+ H₂↑

Solution

A solid metal zinc reacts with hydrochloric acid in the aqueous state to produce zinc chloride in the aqueous state and hydrogen gas.

3. What is the limitation of the reaction given in question 2?

Solution

The chemical equation given in question 2 does not give the time taken for the completion of the reaction.

Also, it does not give information about whether heat is absorbed or evolved during the reaction.

4. Write the chemical equations for the following word equations and balance them.

(a)Carbon + Oxygen → Carbon dioxide

(b)Nitrogen + Oxygen → Nitrogen monoxide

(c) Calcium + Nitrogen → Calcium nitride

(d)Calcium oxide + Carbon dioxide → Calcium carbonate

(e)Magnesium+ Sulphuric acid → Magnesium sulphate + Hydrogen

(f) Sodium reacts with water to form sodium hydroxide and hydrogen

Solution

(a) C + O₂→ CO₂

(b) N₂ + O₂→ 2NO

(d) CaO + CO₂→ CaCO₃

(e) Mg + H₂SO₄→ MgSO₄ + H₂↑

(f) Na + H₂O → NaOH + H₂↑

5. Balance the following equations:

(a) Fe + H₂O → Fe₃O₄ + H₂

(b) Ca + N₂ → Ca₃N₂

(c) Zn + KOH → K₂ZnO₂ + H₂

(d) Fe₂O₃ + CO → Fe + CO₂

(e) PbO + NH₃ → Pb + H₂O + N₂

(f) Pb₃O₄ → PbO + O₂

(g) PbS + O₂ → PbO + SO₂

(h) S + H₂SO₄→ SO₂ + H₂O

(i) S + HNO₃ → H₂SO₄ + NO₂ + H₂O

(j) MnO₂ + HCl → MnCl₂ + H₂O + Cl₂

(k) C + H₂SO₄ → CO₂ + H₂O + SO₂

(l) KOH + Cl₂ → KCl + KClO + H₂O

(m) NO₂ +H₂O → HNO₂ + HNO₃

(n) Pb₃O₄ + HCl → PbCl₂ + H₂O + Cl₂

(o) H₂O + Cl₂ → HCl + O₂

(p) NaHCO₃ → Na₂CO₃ + H₂O + CO₂

(q) HNO₃ + H₂S → NO₂ + H₂O + S

(r) P + HNO₃ → NO₂ + H₂O + H₃PO₄

(s) Zn + HNO₃ → Zn (NO₃)₂ + H₂O + NO₂

Solution

Balanced chemical equations:

(a) 3Fe + 4H₂O → Fe₃O₄ + 4H₂

(b) 3Ca + N₂ → Ca₃N₂

(d)Fe₂O₃ + 3CO → 2Fe + 3CO₂

(e) 3PbO + 2NH₃ → 3Pb + 3H₂O + N₂

(f) 2Pb₃O₄ → 6PbO + O₂

(g) 2PbS + 3O₂ → 2PbO + 2SO₂

(h)S + 2H₂SO₄→ 3SO₂ + 2H₂O

(i)S + 6HNO₃ → H₂SO₄ + 6NO₂ + 2H₂O

(j) MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂

(k) C + 2H₂SO₄ → CO₂ + H₂O + SO₂

(l) 2KOH + Cl₂ → KCl + KClO + H₂O

(m) 2NO2 + H₂O → HNO₂ + HNO₃

(n) Pb₃O₄ + 8HCl → 3PbCl₂ + 4H₂O + Cl₂

(O) 2H2O + 2Cl₂ → 4HCl + O₂

(p) 2NaHCO3 → Na₂CO₃ + H₂O + CO₂

(q) 2HNO3 + H₂S → 2NO₂ + 2H₂O + S

(r) P + 5HNO₃ → 5NO₂ + H₂O + H₃PO₄

(s) Zn + 4HNO₃ → Zn (NO₃)₂ + 2H₂O + 2NO₂

Exercise 1C

1 Fill in the blanks:

(a) Dalton used symbol _____ for oxygen _____ for hydrogen.

(b) Symbol represents _____ atom(s) of an element.

(c) Symbolic expression for a molecule is called _____. .

(d) Sodium chloride has two radicals. Sodium is a _____ radical while chloride is _____ radical.

(e) Valence of carbon in CH₄ is _____, in C₂H₆_____, in C₂H₄ ___ and in C₂H₂ is ____.

(f) Valence of Iron in FeCl₂ is _____ and in FeCl₃ it is ____.

(g) Formula of iron (ill) carbonate is _____.

Solution

(a) Dalton used symbol ‘O’ for oxygen, ‘H’ for hydrogen.

(b) Symbol represents gram atom(s) of an element.

(d) Sodium chloride has two radicals. Sodium is a basic radical, while chloride is an acid radical.

(e) Valence of carbon in CH₄ is 4, in C₂H₆4, in C₂H₄4 and in C₂H₂ is 4.

(f) Valence of iron in FeCl₂ is 2 and in FeCl₃ it is 3.

(g) Formula of iron (III) carbonate is Fe₂ (CO₃)₃.

2. Complete the following table.

Solution

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

(a) Write the equation.

(b)Check whether it is balanced, if not balance it.

(c)Find the weights of reactants and products.

(d) State the law which this equation satisfies.

Solution

(a) NaCl + AgNO₃→ NaNO₃ + AgCl↓

(b) It is a balanced equation.

Weights of products: NaNO₃ – 84.99, AgCl – 143.32

(d) Law of conservation of mass: Matter is neither created nor destroyed in the course of a chemical reaction.

4. What information does the following chemical equation convey?

(a) Zn + H₂SO₄ → ZnSO₄+ H₂

(b) Mg + 2HCl → MgCl₂+ H₂

Solution

(a) The actual result of a chemical change.

Substances take part in a reaction, and substances are formed as a result of the reaction.

Here, one molecule of zinc and one molecule of sulphuric acid react to give one molecule of zinc sulphate and one molecule of hydrogen.
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.
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

22.4 liters of hydrogen are formed at STP.

(b) This equation conveys the following information:

Magnesium reacts with hydrochloric acid to form magnesium chloride and hydrogen gas.
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.
Hydrogen produced at STP is 22.4 liters.

5. (a)What are polyatomic ions? Give two examples.

(b) Name the fundamental law that is involved in. every equation

Solution

(a) A polyatomic ion is a charged ion composed of two or more covalently bounded atoms.

(b) Fundamental laws which are involved in every equation:

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

6. What is the valence of?

(a)Fluorine in CaF₂

(b) Sulphur in SF₆

(c) Phosphorus in PH₃

(d) Carbon in C₄

(e) Nitrogen in the following compounds:

(i) N₂O₅(ii) N₂O₃(iii) NO₂(iv) NO

(f) Manganese in MnO₂

(g) Copper is Cu₂O

(h) Magnesium in Mg₃

Solution

(a) Fluorine in CaF₂is -1.

(b) Sulphur in SF₆ is -6.

(d) Carbon in CH₄ is +4.

(e) Valence of nitrogen in the given compounds:

(i) N₂O₃= N is +3

(ii) N₂O₅= N is +5

(iii) NO₂ = N is +4

(iv) NO = N is +2

7. Why should an equation be balanced? Explain with the help of a simple equation.

Solution

According to the law of conservation of mass, ‘matter can neither be created nor can it be destroyed’. This is possible only if the total number of atoms on the reactants side is equal to the total number of atoms on the products side. Thus, a chemical reaction should always be balanced.

Example: KNO₃ → KNO₂ + O₂

In this equation, the number of atoms on both sides is not the same, and the equation is not balanced. The balanced form of this equation is:

2KNO₃ → 2KNO₂ + O₂

8. Write the balanced chemical equations of the following word equation.

(a) Sodium hydroxide + sulphuric acid → sodium sulphate + water

(b) potassium bicarbonate + sulphuric acid → potassium sulphate + carbon dioxide + water

(c) iron + sulphuric acid → ferrous sulphate + hydrogen.

(d) chlorine + sulphur dioxide + water → sulphuric acid + hydrogen chloride

(e) silver nitrate → silver + nitrogen dioxide + oxygen

(f) copper + nitric acid → copper nitrate + nitric oxide + water

(g) ammonia + oxygen → nitric oxide + water

(h) barium chloride + sulphuric acid → barium sulphate + hydrochloric acid

(i) zinc sulphide + oxygen → zinc oxide + sulphur dioxide

(j) aluminium carbide + water → aluminium hydroxide + methane

(k) iron pyrites(FeS₂) + oxygen → ferric oxide + sulphur dioxide

(l) potassium permanganate + hydrochloric acid → potassium chloride + manganese chloride + chlorine + water

(m) aluminium sulphate + sodium hydroxide → sodium sulphate + sodium meta aluminate + water.

(n) aluminium + sodium hydroxide + water → sodium meta aluminate + hydrogen

(o) potassium dichromate + sulphuric acid → potassium sulphate + chromium sulphate + water + oxygen.

(p) potassium dichromate + hydrochloric acid → Potassium chloride + chromium chloride + water + chlorine

(q)sulphur + nitric acid → sulphuric acid + nitrogen dioxide + water.

(r) sodium chloride + manganese dioxide + sulphuric acid → sodium hydrogen sulphate + manganese sulphate + water + chlorine.

Solution

(a) 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

(b) 2KHCO₃ + H₂SO₄ → K₂SO₄ + 2CO₂ + 2H₂O

(c)Fe + H₂SO₄ → FeSO₄ + H₂

(d) Cl₂ + SO₂ + 2H₂O → H₂SO₄ + 2HCl

(e) 2AgNO3 → 2Ag + 2NO₂ + O₂

(f) 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O

(h) BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

(i) 2ZnS + 3O₂ → 2ZnO + 2SO₂

(j) Al₄C₃ + 12H₂O → 4Al(OH)₃ + 3CH₄

(k) 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

(l) 2KMnO₄ + HCl → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O

(m) Al₂(SO₄)₃+ 8NaOH → 3Na₂SO₄ + 2NaAlO₂ + 4H₂O

(n) 2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂

(o) 2K₂Cr₂O₇ + 8H₂SO₄ → 2K₂SO₄ + 2Cr₂(SO₄)₃ + 8H₂O + 3O₂

(p) K₂Cr₂O₇ + 14HCl → 2KCl + 2CrCl₃ + 7H₂O + 3Cl₂.

(q) S + HNO₃ → H₂SO₄ + NO₂ + H₂O

(r) 2NaCl + MnO₂ + 3H₂SO₄ → 2NaHSO₄ + MnSO₄ + 2H₂O + Cl₂

9. (a)Define atomic mass unit.

(b) Calculate the molecular mass of the following:

(I) Na₂SO_4.10H₂O

(ii) (NH₄)₂CO₃

(iii) (NH₂)₂CO

(iv) Mg₃N₂

Given atomic mass of Na = 23, H = 1, O = 16, C = 12, N = 14, Mg = 24, S = 32.

Solution

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

(b) (i) The relative molecular mass of CuSO₄.5H₂O

= 63.5 + 32 + (16×4) + 5(2+16)

= 159.5 + 90

= 249.5

(ii) The relative molecular mass of (NH₄)₂CO₃ = N₂H₈CO₃

= (14×2) + (1×8) + 12 + (3×16)

= 28 + 8 + 12 + 48

(iii) The relative molecular mass of (NH₂)₂CO = N₂H₄CO

= (2×14) + (1×4) + 12 + 16

= 28 + 4 + 12 + 16 = 60

(iv) The relative molecular mass of Mg₃N₂

= (3×24) + (2×14)

= 72 + 28

= 100

10. Choose the correct answer from the options given below.

(a) Modern atomic symbols are based on the method proposed by

(i) Bohr

(ii) Dalton

(iii) Berzelius

(iv) Alchemist

(b) The number of carbon atoms in a hydrogen carbonate radical is

(i) One

(ii) Two

(iii) Three

(iv) Four

(c) The formula of iron (III) sulphate is

(i) Fe₃SO₄

(ii) Fe (SO₄)₃

(iii) Fe₂ (SO₄)₃

(iv) FeSO₄

(d) In water, the hydrogen-to-oxygen mass ratio is

(ii) 1: 8

(ii) 1: 16

(iii) 1: 32

(iv) 1: 64

(e) 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

(a) iii. Berzelius

(b) i. One

(d) i. 1:8

(e) ii. Ca(HCO₃)₂

11. Correct the following Statement

(a)A molecular formula represent an elements.

(b) Molecular formula of water is H₂O₂.

(c) A molecule of Sulphur is monoatomic.

(d) CO and Co both represent cobalt.

(e) Formula of iron (III) oxide is FeO.

Solution

(a) Molecular formula represents molecule of an element or a compound.

(b) Molecular formula of water is H₂O.

(d) Formula of iron(III)oxide is Fe₂O₃.

12. Calculate the relative molecular masses of:

(a) CHCl₃

(b) (NH₄)₂Cr₂O₇

(c) CuSO₄.5H₂O

(d) (NH₄)₂SO₄

(e) CH₃COONa

(f) potassium chlorate

(g) Ammonium chloroplatinate = (NH₄)₂SO₄

[At. Mass: C = 12, H =1, O = 16, Cl = 35.5, N = 14, Cu = 63.5, S =32, Na = 23, K = 39, Pt = 195, Ca = 40, P =31 Mg = 24,]

Solution

(a) Relative molecular mass of CHCl₃

= 12 + 1 + (3×35.5)

= 12 + 1 + 106.5

= 119.5

(b) Relative molecular mass of (NH₄)₂ Cr₂O₇

= (14×2) + (1×8) + (52×2) + (16×7)

=28 + 8 + 104 + 112

=252

(c) Relative molecular mass of CuSO₄.5H2O

=63.5 + 32 + (16×4) + 5(2 + 16)

= 159.5 + 90

= 249.5

(d) Relative molecular mass of (NH₄)2SO₄

= (2×14) + (8×1) + 32 + (4×16)

=28 + 8 + 32 + 64

= 132

(e) Relative molecular mass of CH₃COONa

= (12×2) + (1×3) + (16×2) + 23

= 24 + 3 + 32 + 23

= 82

(f) Potassium chlorate (KClO₃)

= 39.1 + 35.5 + (16×3)

= 39.1 + 35.5 + 4

= 122.6

(g) Ammonium chloroplatinate (NH₄)₂PtCl₆

= (14×2) + (1×8) + 195.08 + (35.5×6)

= 28+ 8 +195.08 + 213

= 444.08

13. Give the empirical formula of:

(a) Benzene(C₆H₆)

(b) Glucose(C₆H₁₂O₆)

(c) Acetylene(C₂H₂)

(d) Acetic acid(CH₃COOH)

Solution

(a) Benzene-CH

(b) Glucose- CH₂O

(d) Acetic acid-CH₂O

14. Find the percentage mass of water in Epsom salt MgSO₄.7H₂O.

Solution

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 100×126/246.

The % of H₂O in MgSO4.7H₂O = 51.2

15. Calculate the percentage of phosphorus in:

(a) Calcium hydrogen phosphate Ca(H₂PO₄)₂

(b) Calcium phosphate Ca₃(PO₄)₂

Solution

(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), contains 61.8 g P.

So, 100 g Ca(H₂PO₄)₂ contains

The % of Pin 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

The % of P in Ca₃(PO₄)₂ is 19.93%.

16. Calculate the percentage composition of each clement in Potassium chlorate, KCIO.

Solution

Relative molecular mass of KClO₃

= 39.09 + 35.5 + (3×16)

= 122.59 g

122.59 g KClO₃ contains 39.09g K

Hence, 100 g KClO₃ contains

122.59 g KClO₃ contains 35.5 g Cl

Hence, 100 g KClO₃ contains

122.59 g KClO₃ contains 48 g O.

Hence, 100 g KClO₃ contains

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

17. Urea is a very important nitrogenous fertilizer. Its formula is CON₂H₄.Caleulate the percentage of carbon in urea. (C=12, O=16, N=14 and H=1)

Solution

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

Percentage of carbon = weight of carbon/Total weight of urea ×100

= 12/60 ×100

= 20%

ICSE Solutions for Chapter 1 The Language of Chemistry Class 9 Selina Chemistry

Exercise 1A

Exercise 1B

Exercise 1C

About Us

Contact Form

Selina Concise Chapter 1 The Language of Chemistry ICSE Solutions Class 9 Chemistry

ICSE Solutions for Chapter 1 The Language of Chemistry Class 9 Selina Chemistry

Exercise 1A

Exercise 1B

Exercise 1C

You might like

Contact Form