ICSE Selina Solution for Class 9 Chemistry Chapter 1 : The Language Of Chemistry

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ICSE Selina Solution for Class 9 Chemistry Chapter 1 – The Language Of Chemistry provides you with detailed answers to all the exercise questions provided in Selina publication ICSE Class 9 Chemistry textbook. This solution helps you to attain perfection in the topics involved in the chapter, The Language Of Chemistry. To score good marks in the Chemistry examination, students need to focus on this study material thoroughly.

This Class 9 Selina Solutions helps you to attain a basic knowledge of Chemistry such as symbols, valency, radicals, compounds, chemical equations, atomic weights, molecular weight and calculation of relative molecular mass and percentage composition of a compound.

ICSE Selina solutions act as a perfect study material that is required for your ICSE examination. Studying this ICSE Selina Solution for Class 9 Chemistry will also help you to get a basic knowledge of introduction to Chemistry.

Important topics covered in ICSE Selina Solution for Class 9 Chemistry Chapter 1 The Language Of Chemistry

• Introduction
• Symbols
• Valency
• Formula
• Radicals
• Writing chemical formulae
• Naming certain compounds
• Calculating the valency from the formula
• Chemical equation
• Relative atomic mass (Atomic weight)
• Relative molecular mass (Molecular weight)
• Percentage composition
• Empirical formula of a compound

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Exercise 1A Page No: 9-10

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

Solution:

Short form of atom of specific element or the abbreviations used to refer names of the element is known as symbol.

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.

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

Solution:

While naming an element first letter of the element is taken and written in capital (e.g. for sulphur, we use the symbol S). In case if the letter is already adopted. We use a symbol derived from Latin word of the element 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).

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 was written as CO. What would be wrong with it?

Solution:

If we write CO it means it consist of two non-metals namely Carbon and Oxygen and it would represent Carbon- monoxide but not Cobalt.

5. What do the following symbols stand for?

a) H b) H₂ c) 2H d) 2H₂

Solution:

a) H stands for one atom of Hydrogen

b) H₂ stands for one molecule of Hydrogen

c) 2H stands for 2 atoms of Hydrogen

d) 2H₂ stands for 2 molecules of Hydrogen.

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

Solution:

A set of atoms of the same type together forms a molecule of the element. The number of atoms in a molecule of an element is called its atomicity.

Examples of diatomic elements are H₂ – Hydrogen, O₂ – Oxygen, N₂ – Nitrogen

7. a) Explain the terms valency and variable valency.

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

Solution:

a) Valency is the capacity of an atom to lose, gain or share atoms during a chemical reaction is called its valency. Sometimes atom of an element can lose more electron than they are present which means they lose electron from its penultimate shell. Such an element is said to exhibit variable valency.

b) 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.

8. Give the formula and valency of

a) aluminate

b) chromate

c) aluminium

d) cupric

Solution:

9. a) What is a chemical formula?

b) What is the significance of a formula? Give an illustrate.

Solution:

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

b) Significance of Chemical Formula

Chemical formula is very important in finding information about chemical compounds as it tells us about the elements and the number of atoms in a substance

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 following terms?

a) Acid radical b) Basic radical

Solution:

a) Negatively charged radical is called as acidic radical.

b) Positively charged radical is called as basic radical.

11. Select the basic radical in the following compounds

a) MgSO₄

b) (NH₄)₂

c) Al2(SO₄)₃

d) ZnCO₃

e) Mg(OH)₂

Solution:

12. Write the chemical formulae of sulphates 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, ZnSO₄

13. The valency of 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 compound having valency of elements are 3 and 2 is A₂B₃.

14. Match the following

Solution:

15. Write the basic 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-bi-silphate

m) Potassium mangnate

n) Potassium ferrocynide

Solution:

16. Write 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(Cl₃)₂

h) Ag₂SO₃

i) (CH₃COO)₂Pb

j) Na₂SiO₃

Solution:

a) Calcium phosphate

b) Potassium carbonate

c) Potassium manganate

d) Manganese(II) Borate

e) Magnesium bicarbonate.

f) Sodium ferrocyanide

g) Barium Chlorate

h) Silver sulfite

i) Lead(II) acetate

j) Sodium silicate

17. Give the names of the following compounds

a) KClO

b) KClO₂

c) KClO₃

d) KClO₄

Solution:

a) Potassium hypochlorite

b) Potassium chlorite

c) Potassium chlorate

d) Potassium per chlorate

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

a) The formula of a compound represents

i) an atom

ii) a particle

iii) a molecule

iv) a combination

b) The correct formula of aluminium oxide is

i) AlO₃

ii) AlO₂

iii) Al₂O₃

iv) Al₃O₂

c) The valency of Nitrogen in Nitrogen di oxide (NO₂) is

i) One

ii) Two

iii) Three

iv) Four

Solution:

a) The formula of a compound represents a molecule

b) The correct formula of aluminium oxide is Al₂O₃

c) The valency of Nitrogen in Nitrogen di oxide (NO₂) is four

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

a) Sodium sulphate

b) Quick lime

c) Baking soda (NaHCO3)

d) Ammonia

e) Ammonium dichromate

Solution:

a) Sodium sulphate

Chemical formula is Na₂SO₄

Atoms – 2 sodium, one Sulphur and 4 oxygen atoms.

b) Quick lime

Chemical formula is CaO

Atoms – one Calcium atom and 1 oxygen atom

c) Baking soda (NaHCO₃)

Chemical formula of is NaHCO₃

Atoms – 1 Sodium, 1 hydrogen, 1 carbon and 3 oxygen atoms.

d) Ammonia

Chemical formula is NH₃

Atoms – 3 hydrogen and 1 nitrogen atom.

e) Ammonium dichromate

Chemical formula is (NH₄)₂Cr₂O₇.

Atoms – 2 ammonium, 2 chromium and 7 oxygen atoms.

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

a) Chloride

b) Oxide

c) Phosphate

d) Acetate

Solution:

Answer is a) Chloride

Exercise 1B Page No: 15

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

Solution:

A chemical equation is a symbolic representation of a chemical reaction. Here we use symbols and formulas of the substance involved in the reaction.

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.

2. State the information conveyed by the following equation.

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂

Solution:

This chemical equation shows ‘single displacement reaction’, in which a non-metal is displaced by a metal. Here, non-metal is hydrogen which is evolved as gas. It is displaced by the metal zinc. In the given equation – Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g), Zinc (Zn) is a reductant metal that displaces hydrogen (H₂) from aqueous solution of Hydrochloric acid (HCl).

3. Write the limitation of reaction given in question 2.

Solution:

HCl will be the limiting reagent in the reaction and Zn will be excess reagent.

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

a) Carbon + Oxygen → Carbon-di-oxide

b) Nitrogen + Oxygen → Nitrogen monoxide

c) Calcium + Nitrogen → Calcium nitride

d) Calcium oxide + carbon dioxide → Calcium carbonate

e) Magnesium + Sulphuric acid → Magnesium sulphate + Hydrogen

Solution:

a) 2C + O₂ → CO₂

b) N₂ + O₂ → 2NO

c) 3Ca_(s) + N_2(g) → Ca₃N₂

d) CaO + CO₂ → CaCO₃

e) Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)

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:

a) 3Fe + 4H₂O → Fe₃O₄ + 4H₂

b) 3Ca + N₂ → Ca₃N₂

c) Zn + 2KOH → K₂ZnO₂ + H₂

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 Page No: 18-20

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 a ___ radical.

e) Valency of Phosphorous in PCl₃ is _____, in PCl₅ is _____.

f) Valency of iron in FeCl₂ is ____ and in FeCl₃ it is _____.

g) Formula of iron (III) carbonate is ______.

Solution:

a) Dalton used symbol [O] for oxygen, [H] for hydrogen.

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

c) Symbolic expression for a molecule is called molecular formula.

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

e) Valency of Phosphorous in PCl3 is 3, in PCl5 is 5,

f) Valency 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 that 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

Thus, 228.5 g of reactants → 228.5 g of products

d) This equation states law of conservation of mass where mass is neither created nor destroyed.

4. What information does the following chemical equations convey?

a) Zn + H₂SO₄ → ZnSO₄ + H₂

b) Mg + 2HCl → MgCl₂ + H₂

Solution:

a) This equation shows the result of a chemical change. When one molecule of zinc and one molecule of sulphuric acid reacts, it results in the production of one molecule of zinc sulphate and one molecule of hydrogen.

b) This equation shows reaction of Magnesium with HCl which gives magnesium chloride and liberated Hydrogen gas.

5. a) What are poly-atomic ions? Give two examples

b) Name the fundamental law involved in every equation

Solution:

a) A charged ion that consists of two or more covalently bounded atoms are called as polyatomic ions. Eg: CaCO₃, MgSO₄

b) Fundamental law involved in every equation is “the law of conservation of mass”.

6. What is the valency of?

a) Fluorine in CaF₂

b) Sulphur in SF₆

c) Phosphorous in PH₃

d) Carbon in CH₄

e) Nitrogen in the following compound

i) N₂O₃ ii) N₂O₅ iii) NO₂ iv) NO

f) Manganese in MnO₂

g) Copper in Cu₂O

h) Magnesium in Mg₃N₂

Solution:

a) Valency of fluorine in CaF₂ is -1

b) Valency of sulphur in SF₆ is -6

c) Valency of phosphorus in PH₃ is +3

d) Valency of carbon in CH₄ is +4

e) Valency of nitrogen in the given compounds:

i) N₂O₃ = +3

ii) N₂O₅ = +5

iii) NO₂ = +4

iv) NO = +2

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

Solution:

An equation should n be balanced to make it comply with the law of conservation of matter which states that matter is neither created nor destroyed in the course if a chemical reaction. An unbalanced equation either deletes or adds extra atoms in the equation.

e.g. KNO₃ → KNO₂ + O₂

In this equation number of atoms in left and right side are not equal hence the balanced equation will be written as.

2KNO₃ → 2KNO₂ + O₂

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

a) Sodium hydroxide + Sulphuric acid → Sodium Sulphate + Water

b) Potassium bicarbonate + Sulphuric acid → Potassium Sulphate + Carbon di oxide + Water

c) Iron + Sulphuric acid → Ferrous sulphate + Hydrogen

d) Chlorine + Sulphur di oxide + Water → Sulphuric acid + Hydrogen Chloride

e) Silver Nitrate → Silver + Nitrogen di oxide + 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 + Oxygen → Ferric oxide + Sulphur di oxide

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

g) 4NH₃ + 5O₂ → 6H₂O + 4NO

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 → KCl + 2CrCl₃ + 7H₂O + 3Cl₂

q) S + 6HNO₃ → H₂SO₄ + 6NO₂ + 2H₂O

r) 2KI + 2MnO₂ + 4H₂SO₄ → I₂ + 2KHSO₄ + 2MnSO₄ + 4H₂O

9. a) Define atomic mass unit

b) Calculate the molecular mass of the following

i) Na₂SO₄.10H₂O ii) (NH₄)₂CO₃ iii) (NH₂)₂CO iv) Mg₃N₂

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

Solution:

a) The atomic mass unit (amu) is defined as 1/12^th of the mass of an atom of carbon

1 a.m.u. = 1.67 x 10^-24m =1.67 x 10²⁷ kg

1 gm mass = 6.02 x 10²³ a.m.u. and 1 kg mass = 6.02 x 10²⁶ a.m.u.

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

= 63.5 + 32 + (16 x 4) + 5 (2 + 16)

= 159.5 + 90

= 249.5 2

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

= 14 x 2 + 1 x 8 + 12 + 3 x 16

= 28 + 8 + 12 + 48

= 96

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

= 2 x 14 + 1 x 4 +12 + 16

= 28 + 4 + 12 + 16

= 60

iv) The relative molecular mass of Mg₃N₂

= 3 x 24 + 2 x 14

= 72 + 28

= 100

10. Choose the correct answer from the options given below

a) Modern atomic symbols are based on the methods 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 (ii) three (iv) four

c) The formula of Iron(III) Sulphate is

(i)Fe₃SO₄ (ii) Fe(SO)₃ (iii) Fe(SO₄)₃ (iv)FeSO₄

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

(i)1:8 (ii) 1:16 (iii) 1:32 (iv) 1:64

(e) The formula of sodium carbonate is Na, CO, and that of calcium hydrogen carbonate is (i)CaHCO₃ (ii) Ca(HCO₃)₂ (iii) CaHCO₃ (iv) Ca(HCO)₃

Solution:

a) Answer is (iii) Berzelius

b) Answer is (i) One

c) Answer is (iii) Fe₂(SO₄)₃

d) Answer is (i) 1: 8

e) Answer is (ii) Ca(HCO₃)₂

11. Correct the following statements

(a) A molecular formula represents an element

(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

c) A molecule of Sulphur is diatomic

d) CO represents carbon monoxide and Co represent cobalt

e) Formula of Iron(III) oxide is Fe₂O₃

12. Calculate the relative molecular masses of:

(a) CHCI₃

(b) (NH₄)₂Cr₂O₇

(c) CuSO₄.5H₂O

(d) (NH₄)₂SO₄

(e) CH₃COONa

(f) Potassium chlorate

(g) Ammonium chloroplatinate (NH₄)₂PtCl₆

[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 CHCI₃

= 12 + 1 + (3 x 35.5)

= 12 + 1 + 106.5

= 119.5 2

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

= (14 x 2) + (1 x 8) + (52 x 2) + (16 x 7)

= 28 + 8 + 104 + 112

= 252

(c) Relative molecular mass of CuSO₄.5H₂0

= 63.5 + 32 + (16 x 4) + 5(2 + 16)

= 159.5 + 90

= 249.5

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

= (2 x 14) + (8 x 1) + 32 + (4 x 16)

= 28 + 8 + 32 + 64

= 132

(e) Relative molecular mass of CH₃COONa

= (12 x 2) + (1 x 3) + (16 x 2) + 23

= 24 + 3 + 32 + 23

= 82

(f) Potassium chlorate (KClO₃)

= 39.1 + 35.5 + 16 x 3

= 39.1 + 35.5 + 48

= 122.6

(g) Ammonium chloroplatinate (NH₄)₂PtCl₆

= (14 x 2) + (1 x 8) + 195.08 + (35.5 x 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

(c) Acetylene – CH

(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 x 4) + 7(2 + 16)

= 24 + 32 + 64 + 126

= 246

26 g of Epsom salt contains 126 g of water of crystallisation.

So, 100 g of Epsom salt contains (100 x 126/246) g of water

Thus, percentage mass 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)

= 40.07 + (1 x 4) + (30.9 x 2) + (16 x 8)

= 40.07 + 4 + 61.8 + 128

= 233.87

Now, 233.87 g of Ca(H₂PO₄) contains 61.8 g of P

So, 100 g Ca(H₂PO₄) contains

(100 x 61.8)/233.87 = 26.42 g

Thus, the percentage of phosphorous in Ca(H₂PO₄) is 26.42%

(b) Relative molecular mass of Ca₃(PO₄)₂

= (40.07 x 3) + (30.9 x 2) + (16 x 8)

= 120.21 + 61.8 + 128

= 310.01

Now, 310.01 g of Ca₃(PO₄)₂ contains 61.8 g of P

So, 100 g Ca₃(PO₄)₂ contains

(100 x 61.8)/310.01 = 19.93 g

Thus, the percentage of phosphorous in Ca₃(PO₄)₂ is 19.93%

16. Calculate the percentage composition of each element in Potassium chlorate, KClO₃.

Solution:

Relative molecular mass of KClO₃

= 39.09 + 35.5 + (3 x 16)

= 122.59 g

So, 122.59 g of KClO₃ contains 39.09 g of K

Hence, 100 g of KClO₃ contains

= (100 x 39.09)/122.59

= 31.9 g

Also, 122.59 g of KClO₃ contains 35.5 g of Cl

Hence, 100 g of KClO₃ contains

= (100 x 35.5)/122.59

= 28.9 g

And, 122.59 g of KClO₃ contains 48 g of O

Hence, 100 g of KClO₃ contains

= (100 x 48)/122.59

= 39.1 g

Therefore, the percentage composition 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 CH₄N₂O. Calculate the percentage of carbon in urea. (C = 12, O = 16, N = 14 and H = 1)

Solution:

⇒ 12 + 16 + 28 + 4 = 60

Hence, relative molecular mass of urea = 60

Thus,

Percentage of carbon = Weight of carbon/Total weight of urea x 100

= 12/60 x 100

= 20%

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