The Language of Chemistry

Solutions for Chemistry, Class 9, ICSE

Exercise 1A

30 questions

Question 1(i)

The formula of a compound represents

  1. an atom
  2. a particle
  3. a molecule
  4. a combination
Exercise 1A

Answer:

a molecule

Reason — The formula of a compound employs symbols to denote the molecule of a compound.

Question 1(ii)

The correct formula of aluminium oxide is

  1. AlO3
  2. AlO2
  3. Al2O3
  4. Al3O2
Exercise 1A

Answer:

Al2O3

Reason — The formula of aluminium oxide can be determined as below:

Al3+ O2Al33  O2Al22 O33Al2O3\text{Al}^{3+} \space \text{O}^{2-} \\[1em] \overset{\phantom{3}{3}}{\text{Al}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{O}} \Rightarrow \underset{\phantom{2}{2}}{\text{Al}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{O}} \\[1em] \text{Al}_2\text{O}_3

Question 1(iii)

The valency of lead in lead oxide (PbO) is

  1. one
  2. two
  3. three
  4. four
Exercise 1A

Answer:

two

Reason — The valency of lead in lead oxide can be determined from the formula PbO as below:

PbO=Pb11  O1Pb11  O1\text{PbO} = \underset{\phantom{1}{1}}{\text{Pb}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{1}{\text{O}} \Rightarrow \overset{\phantom{1}{1}}{\text{Pb}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]

But valency of O is 2. Multiplying by 2, we get:

Pb1×2  O2×1Pb22  O2\overset{{1 \times 2}}{\text{Pb}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{2}{2}}{\text{Pb}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]

Therefore, valency of Lead is 2.

Question 1(iv)

Which of the following elements has a molecular mass that is four times its atomic mass?

  1. Oxygen
  2. Neon
  3. Phosphorus
  4. Sulphur
Exercise 1A

Answer:

Phosphorus

Reason

Atomic mass of P = 31 μ

Molecular form = P4 ⟶ Molecular mass = 31 × 4 = 124 μ

124 μ is 4 times the atomic mass

Hence, phosphorus has molecular mass four times its atomic mass.

Question 2

Match the following : (Refer common names in the beginning of the book)

CompoundFormula
(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
Exercise 1A

Answer:

CompoundFormula
(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) 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 3

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

Exercise 1A

Answer:

The basic and acidic radicals in the compounds are given in the table below:

S.
No.
CompoundBasic
radicals
Acidic
radicals
aMgSO4Mg2+SO42-
b(NH4)2SO4NH4+SO42-
cAl2(SO4)3Al3+SO42-
dZnCO3Zn2+CO32-
eMg(OH)2Mg2+OH-

Question 4

Give the formula and valency of :

(a) aluminate

(b) chromate

(c) aluminium

(d) cupric

Exercise 1A

Answer:

The formula and valency are given in the table below:

S.
No.
NameFormulaValency
aaluminateAlO33-3
bchromateCrO42-2
caluminiumAl3
dcupricCu2+2

Question 5

What do the following symbols stand for ?

(a) H

(b) H2

(c) 2H

(d) 2H2

Exercise 1A

Answer:

(a) H → One atom of Hydrogen.

(b) H2 → One molecule of Hydrogen.

(c) 2H → Two atoms of Hydrogen.

(d) 2H2 → Two molecules of Hydrogen.

Question 6

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

Exercise 1A

Answer:

(a) Ca3(PO4)2 → Calcium Phosphate

(b) K2CO3 → Potassium Carbonate

(c) K2MnO4 → Potassium Manganate

(d) Mn3(BO3)2 → Manganese (II) borate

(e) Mg(HCO3)2 → Magnesium Hydrogen Carbonate

(f) Na4Fe(CN)6 → Sodium Ferrocyanide

(g) Ba(ClO3)2 → Barium Chlorate

(h) Ag2SO3 → Silver Sulphite

(i) (CH3COO)2Pb → Lead Acetate

(j) Na2SiO3 → Sodium Silicate

Question 7

Write chemical formulae of the sulphates of Aluminium, Ammonium and Zinc.

Exercise 1A

Answer:

(a) Chemical formula of Aluminium Sulphate:

Al3+ SO42Al33  SO42Al22 SO433Al2(SO4)3\text{Al}^{3+} \space \text{SO}_4^{2-} \\[1em] \overset{\phantom{3}{3}}{\text{Al}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{SO}_4} \Rightarrow \underset{\phantom{2}{2}}{\text{Al}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{SO}_4} \\[1em] \text{Al}_2\text{(SO}_4)_3

∴ Chemical formula of Aluminium Sulphate is Al2(SO4)3

(b) Chemical formula of Ammonium Sulphate:

NH4+ SO42NH431  SO42NH422 SO431(NH4)2SO4\text{NH}_4^{+} \space \text{SO}_4^{2-} \\[1em] \overset{\phantom{3}{1}}{\text{NH}_4} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{SO}_4} \Rightarrow \underset{\phantom{2}{2}}{\text{NH}_4} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{1}}{\text{SO}_4} \\[1em] \text{(NH}_4)_2\text{SO}_4

∴ Chemical formula of Ammonium Sulphate is (NH4)2SO4

(c) Chemical formula of Zinc Sulphate:

Zn2+ SO42Dividing by H.C.F. it becomesZn1+ SO41Zn221  SO41Zn21 SO421ZnSO4\text{Zn}^{2+} \space \text{SO}_4^{2-} \\[1em] \text{Dividing by H.C.F. it becomes} \\[1em] \text{Zn}^{1+} \space \text{SO}_4^{1-}\\[1em] \overset{\phantom{22}{1}}{\text{Zn}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{SO}_4} \Rightarrow \underset{\phantom{2}{1}}{\text{Zn}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{1}}{\text{SO}_4} \\[1em] \text{ZnSO}_4

∴ Chemical formula of Zinc Sulphate is ZnSO4

Question 8

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.

Exercise 1A

Answer:

Formula of the compound formed by the combination of A and B:

A33  B2A22 B33A2B3\overset{\phantom{3}{3}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{B}} \Rightarrow \underset{\phantom{2}{2}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{B}} \\[1em] \text{A}_2\text{B}_3

∴ Formula of the compound is A2B3

Question 9

Give the names of the following compounds.

(a) KClO

(b) KClO2

(c) KClO3

(d) KClO4

Exercise 1A

Answer:

(a) KClO → Potassium hypochlorite

(b) KClO2 → Potassium Chlorite

(c) KClO3 → Potassium Chlorate

(d) KClO4 → Potassium Perchlorate

Question 10

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

(a) Chloride

(b) Oxide

(c) Phosphate

(d) Acetate

Exercise 1A

Answer:

Given, formula of sulphate is M2(SO4)3.

M2(SO4)3=M22  SO43M33  SO42\text{M}_2(\text{SO}_4)_3 = \underset{\phantom{2}{2}}{\text{M}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{3}{\text{SO}_4} \Rightarrow \overset{\phantom{3}{3}}{\text{M}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{SO}_4} \\[0.5em]

∴ Valency of M is 3.

(a) Formula of Chloride:

M3+ Cl1M33  Cl1M11 Cl33MCl3\text{M}^{3+} \space \text{Cl}^{1-} \\[1em] \overset{\phantom{3}{3}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{Cl}} \\[1em] \text{M}\text{Cl}_3

∴ Formula of chloride of element M is MCl3

(b) Formula of Oxide:

M3+ O2M33  O2M22 O33M2O3\text{M}^{3+} \space \text{O}^{2-} \\[1em] \overset{\phantom{3}{3}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{O}} \Rightarrow \underset{\phantom{2}{2}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{O}} \\[1em] \text{M}_2\text{O}_3

∴ Formula of oxide of element M is M2O3

(c) Formula of Phosphate:

M3+ PO43Dividing by H.C.F. it becomesM1+ PO41M31  PO41M21 PO431MPO4\text{M}^{3+} \space \text{PO}_4^{3-} \\[1em] \text{Dividing by H.C.F. it becomes} \\[1em] \text{M}^{1+} \space \text{PO}_4^{1-} \\[1em] \overset{\phantom{3}{1}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{PO}_4} \Rightarrow \underset{\phantom{2}{1}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{1}}{\text{PO}_4} \\[1em] \text{MPO}_4

∴ Formula of phosphate of element M is MPO4

(d) Formula of Acetate:

M3+ CH3COO1M33  CH3COOCHCOOO1M11 CH3COO33M(CH3COO)3\text{M}^{3+} \space \text{CH}_3\text{COO}^{1-} \\[1em] \overset{\phantom{3}{3}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \space \overset{\phantom{CHCOOO}1}{\text{CH}_3\text{COO}} \Rightarrow \underset{\phantom{1}{1}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{CH}_3\text{COO}} \\[1em] \text{M}\text{(CH}_3\text{COO})_3

∴ Formula of acetate of element M is M(CH3COO)3

Question 11

What is a symbol of an element ? What information does it convey ?

Exercise 1A

Answer:

A symbol is a short form that stands for the atom of a specific element or the abbreviation used for the name of an element.

A symbol conveys the following information:

  1. It represents the name of the element.
  2. It represents one atom of the element.
  3. It represents a definite mass of the element (equal to atomic mass expressed in grams).

Question 12

Why is the symbol S for sulphur, but Na for sodium and Si for silicon ?

Exercise 1A

Answer:

When the first letter of more than one element is the same the elements are denoted by two letters. Sulphur, Sodium and Silicon all have the first letter as S. Therefore, only sulphur is denoted by S, Silicon is denoted by two letter Si, Sodium is denoted by two letter Na taken from its latin name Natrium.

Question 13(a)

Write the full form of IUPAC.

Exercise 1A

Answer:

Full form of IUPAC is International Union of Pure and Applied Chemistry.

Question 13(b)

Name the elements represented by the following symbols :

Au, Pb, Sn, Hg.

Exercise 1A

Answer:

The elements represented by the following symbols are:

Au → Gold

Pb → Lead

Sn → Tin

Hg → Mercury

Question 14

If the symbol for Cobalt, Co, were written as CO, what would be wrong with it ?

Exercise 1A

Answer:

When writing symbols, we need to be careful about the case of the letters in the symbol. CO means the compound Carbon Monoxide not Cobalt.

Question 15

Explain the terms 'valency' and 'variable valency'.

Exercise 1A

Answer:

Valency is the combining capacity of an atom or of a radical. The valency of an element or of a radical is the number of hydrogen atoms that will combine with or displace one atom of that element or radical.
Variable valency is the ability of certain elements to have more than one valency or different combining capacities. An atom of an element can sometimes lose more electrons than are present in its valence shell, i.e., there is a loss of electrons from the penultimate shell too. Such element is said to exhibit variable valency.

Question 16

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

Exercise 1A

Answer:

If an element exhibits two different positive valencies, then the suffix "ous" is used for the lower valency and the suffix "ic" is used for the higher valency. Modern chemists use Roman numerals in place of these trivial names.
For example, Iron can exhibit a valency of +2 or +3. The ion with valency +2 is named as Ferrous ion (Fe2+) and with valency +3 is named as Ferric ion (Fe3+). Its oxides will be named as Ferrous oxide or Iron (II) oxide [FeO] and Ferric oxide or Iron (III) oxide [Fe2O3].

Question 17

(a) What is a chemical formula ?

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

Exercise 1A

Answer:

(a) A chemical formula also known as molecular formula employs symbols to denote the molecule of an element or of a compound.

(b) The chemical/molecular formula of a compound has quantitative significance. It represents:

  1. both the molecule and the molecular mass of the compound.
  2. the respective numbers of different atoms present in one molecule of the compound.
  3. the ratios of the respective masses of the elements present in the compound.

For example, the formula CO2 means that:

  1. the molecular formula of carbon dioxide is CO2.
  2. each molecule contains one carbon atom joined by chemical bonds with two oxygen atoms.
  3. the molecular mass of carbon dioxide is 44, given that the atomic mass of carbon is 12 and that of oxygen is 16.

Question 18

What do you understand by the following terms ?

(a) Acid radical

(b) Basic radical

Exercise 1A

Answer:

(a) An Acid radical is the radical that remains after an acidic molecule loses a hydrogen ion (H+). Acid radicals typically have a negative charge. They are also called electronegative radicals or anions.

(b) A Basic radical is the radical that remains after a base molecule loses a hydroxyl ion (OH-). Basic radicals typically have a positive charge. They are also called electropositive radicals or cations.

Question 19

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 manganate

(n) Potassium ferrocyanide

Exercise 1A

Answer:

S.
No.
CompoundBasic
radical
Acidic
radical
Formula
aBarium
sulphate
Ba2+SO42-BaSO4
bBismuth
nitrate
Bi3+NO3-Bi(NO3)3
cCalcium
bromide
Ca2+Br-CaBr2
dFerrous
sulphide
Fe2+S2-FeS
eChromium
sulphate
Cr3+SO42-Cr2(SO4)3
fCalcium
silicate
Ca2+SiO32-CaSiO3
gStannic
oxide
Sn4+O2-SnO2
hSodium
zincate
Na1+ZnO22-Na2ZnO2
iMagnesium
phosphate
Mg2+PO43-Mg3(PO4)2
jSodium
thiosulphate
Na1+S2O32-Na2S2O3
kStannic
phosphate
Sn4+PO43-Sn3(PO4)4
lNickel
bisulphate
Ni2+HSO4-Ni(HSO4)2
mPotassium
manganate
K1+MnO42-K2MnO4
nPotassium
ferrocyanide
K1+Fe(CN)64-K4[Fe(CN)6]

Question 20

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

(d) Ammonia

(e) Ammonium dichromate

Exercise 1A

Answer:

(a) Sodium sulphate
     Formula: Na2SO4
   ∴ It contains 2 atoms of Sodium (Na), 1 atom of Sulphur (S) and 4 atoms of Oxygen (O).

(b) Quick lime
     Formula: CaO
   ∴ It contains 1 atom of Calcium (Ca) and 1 atom of Oxygen (O).

(c) Baking soda
     Formula: NaHCO3
   ∴ It contains 1 atom of Sodium (Na), 1 atom of Hydrogen (H), 1 atom of Carbon (C) and 3 atoms of Oxygen (O).

(d) Ammonia
     Formula: NH3
   ∴ It contains 1 atom of Nitrogen (N) and 3 atoms of Hydrogen (H).

(e) Ammonium dichromate
     Formula: (NH4)2Cr2O7
   ∴ It contains 2 atoms of Nitrogen (N), 8 atoms of Hydrogen (H), 2 atoms of Chromium (Cr) and 7 atoms of Oxygen (O).

Question 21

The valency of an element A is 4. Write the formula of its:

(a) oxide

(b) nitrate

(c) phosphate

Exercise 1A

Answer:

(a) Chemical formula of Oxide of A:

A4+ O2Dividing by H.C.F. it becomesA2+ O1A2  O1A21 O22AO2\text{A}^{4+} \space \text{O}^{2-} \\[1em] \text{Dividing by H.C.F. it becomes} \\[1em] \text{A}^{2+} \space \text{O}^{1-}\\[1em] \overset{{2}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{O}} \Rightarrow \underset{\phantom{2}{1}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{2}}{\text{O}} \\[1em] \text{AO}_2

∴ Chemical formula of Oxide of A is AO2

(b) Chemical formula of Nitrate of A:

A4+ NO31A4  NO31A21 NO324A(NO3)4\text{A}^{4+} \space \text{NO}_3^{1-} \\[1em] \overset{{4}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{NO}_3} \Rightarrow \underset{\phantom{2}{1}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{4}}{\text{NO}_3} \\[1em] \text{A(NO}_3)_4

∴ Chemical formula of Nitrate of A is A(NO3)4

(c) Chemical formula of Phosphate of A:

A4+ PO43A4  PO43A23 PO424A3(PO4)4\text{A}^{4+} \space \text{PO}_4^{3-} \\[1em] \overset{{4}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \space \overset{3}{\text{PO}_4} \Rightarrow \underset{\phantom{2}{3}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{4}}{\text{PO}_4} \\[1em] \text{A}_3(\text{PO}_4)_4

∴ Chemical formula of Phosphate of A is A3(PO4)4

Question 22

Identify the cations (electropositive ions) and anions (electronegative ions) in the following compounds:

(a) CH3COONa

(b) NH4Cl

(c) PbCl2

(d) MgO

Exercise 1A

Answer:

(a) CH3COONa (Sodium acetate)
     Cation: Na+ (Sodium ion)
     Anion: CH3COO- (Acetate ion)

(b) NH4Cl (Ammonium chloride)
     Cation: NH4+ (Ammonium ion)
     Anion: Cl- (Chloride ion)

(c) PbCl2 (Lead(II) chloride)
     Cation: Pb2+ (Lead(II) ion)
     Anion: Cl- (Chloride ion)

(d) MgO (Magnesium oxide)
     Cation: Mg2+ (Magnesium ion)
     Anion: O2- (Oxide ion)

Question 23

Give an example of a divalent anion and a trivalent cation. Write the formula of the compound formed by their combination.

Exercise 1A

Answer:

Aluminium is trivalent cation with a valency of 3 → Al3+

Sulphate is divalent anion with a valency of 2 → SO42-

Formula:

Al3+ SO42Al3  SO42Al2  SO43Al2(SO4)3\text{Al}^{3+} \space \text{SO}_4^{2-} \\[1em] \overset{3}{\text{Al}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{SO}_4} \Rightarrow \underset{2}{\text{Al}} \space {\swarrow}\mathllap{\searrow} \space \underset{3}{\text{SO}_4} \\[1em] \text{Al}_2(\text{SO}_4)_3

Hence, the formula is : Al2(SO4)3

Question 24

Write the Latin names of sodium, mercury and iron.

Exercise 1A

Answer:

NameLatin Name
SodiumNatrium
MercuryHydrargyrum
IronFerrum

Question 26

State the number of atoms present in each of the following:

(a) PO43-

(b) P2O5

(c) H2SO4

(d) SO4-2

Exercise 1A

Answer:

(a) PO43-

P = 1 atom

O = 4 atoms

Total atoms = 1 + 4 = 5 atoms

(b) P2O5

P = 2 atoms

O = 5 atoms

Total atoms = 2 + 5 = 7 atoms

(c) H2SO4

H = 2 atoms

S = 1 atom

O = 4 atoms

Total atoms = 2 + 1 + 4 = 7 atoms

(d) SO4-2

S = 1 atom

O = 4 atoms

Total atoms = 1 + 4 = 5 atoms

Question 27

Depending on the number of oxygen atoms in the anions, the names of some acids and their formulae are given. Write the formula of their corresponding salts.

AcidFormulaSaltFormula
Perchloric acidHClO4Sodium perchlorateNaClO4
Chloric acidHClO3Sodium chlorate...............
Chlorous acidHClO2Potassium chlorite...............
Hypochlorous acidHClOSodium hypochlorite...............
Exercise 1A

Answer:

AcidFormulaSaltFormula
Perchloric acidHClO4Sodium perchlorateNaClO4
Chloric acidHClO3Sodium chlorateNaClO3
Chlorous acidHClO2Potassium chloriteKClO2
Hypochlorous acidHClOSodium hypochloriteNaClO

Exercise 1B

5 questions

Question 1

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

Exercise 1B

Answer:

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

An equation must be balanced in order to comply with the "Law of Conservation of Matter", which states that matter is neither created nor destroyed in the course of a chemical reaction. An unbalanced equation would imply that atoms have been created or destroyed.

Question 2

State the information conveyed by the following equation.

Zn(s) + 2HCl (aq) ⟶ ZnCl2 (aq) + H2

Exercise 1B

Answer:

The given equation tells us:

  1. About the reactants involved and the products formed as a result of the reaction and their state. Zinc (Zn) and Hydrochloric acid (HCl) are the reactants. Zinc is in solid state and HCl is in aqueous solution state. The product Zinc Chloride (ZnCl2) is also in aqueous solution state and Hydrogen is in gaseous state.
  2. About the number of molecules of each substance taking part and formed in the reaction. Here one molecule of Zinc and two molecules of Hydrochloric acid react to give one molecule of Zinc chloride and one molecule of Hydrogen gas.
  3. About chemical composition of respective molecules. For example, one molecule of Zinc chloride contains one atom of Zinc and two atoms of Chlorine.
  4. About molecular mass; that 65 parts by weight of Zinc reacts with 73 parts by weight of Hydrochloric acid to produce 136 parts by weight of Zinc chloride and 2 parts by weight of Hydrogen.
    Zn65(s)+2HCl73(aq)ZnCl2136(aq)+H22\underset{65}{\text{Zn}}\text{(s)} + \underset{73}{\text{2HCl}}\text{(aq)} \longrightarrow \underset{136}{\text{ZnCl}_2}\text{(aq)} + \underset{2}{\text{H}_2}\uparrow
  5. 65 g of Zinc on treatment with 73 g of HCl, will produce 22.4 litres of hydrogen gas at S.T.P.
  6. It also proves the law of conservation of mass. According to the above equation, 138 gram of reactants are producing 138 gram of products.

Question 3

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

Exercise 1B

Answer:

The equation of the given reaction does not tell us:

  1. the time taken for the completion of the reaction.
  2. whether heat is given out or absorbed during the reaction.
  3. the respective concentrations of the reactants and the products.
  4. the rate at which the reaction proceeds.
  5. whether the reaction is completed or it is not completed.

Question 4

Write 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.

Exercise 1B

Answer:

The balanced chemical equations for the word equations are given below:

(a) C + O2 ⟶ CO2

(b) N2 + O2 ⟶ 2NO

(c) 3Ca + N2 ⟶ Ca3N2

(d) CaO + CO2 ⟶ CaCO3

(e) Mg + H2SO4 ⟶ MgSO4 + H2

(f) 2Na + 2H2O ⟶ 2NaOH + H2

Question 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

Exercise 1B

Answer:

(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 + 2H2O + 2SO2

(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 Descriptive Type

7 questions

Question 1

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.

Exercise 1C Descriptive Type

Answer:

(a) NaCl + AgNO3 ⟶ AgCl + NaNO3

(b) The equation is balanced.

(c) Calculating the weights of reactants and products from the equation:

NaCl+AgNO3AgCl+NaNO323+35.5+108+14+3(16)108+35.5+23+14+3(16)58.5+170143.5+85228.5228.5\begin{matrix} \text{NaCl} & + & \text{AgNO}_3 & \longrightarrow & \text{AgCl} & + & \text{NaNO}_3 \\ 23 + 35.5 & + & 108 + 14 + 3(16) & & 108 + 35.5 & + & 23 + 14 + 3(16) \\ \Rightarrow 58.5 & + & 170 & & 143.5 & + & 85 \\ \Rightarrow & 228.5 & & & & 228.5 & \end{matrix}

Thus, weights of reactants = weights of products = 228.5 g

(d) The equation satisfies the law of conservation of mass. It states that the total mass of the substances on either side of the equation is the same.

Question 2

What information does the following chemical equations convey ?

(a) Zn + H2SO4 ⟶ ZnSO4 + H2

(b) Mg + 2HCl ⟶ MgCl2 + H2

Exercise 1C Descriptive Type

Answer:

(a) Zn65+H2SO498ZnSO4161+H22\text{(a) }\underset{65}{\text{Zn}} + \underset{98}{\text{H}_2\text{SO}_4} \longrightarrow \underset{161}{\text{Zn}\text{SO}_4} + \underset{2}{\text{H}_2} \\[1em]

The above balanced equation conveys the following information:

  1. One molecule of zinc reacts with one molecule of sulphuric acid to produce one molecule of zinc Sulphate and one molecule of hydrogen.
  2. 65 g of zinc reacts with 98 g of sulphuric acid to produce 161 g of zinc Sulphate and 2 g of hydrogen.
  3. 65 g of zinc reacts with 98 g of sulphuric acid to produce 22.4 litres of hydrogen gas at S.T.P.
  4. It also proves the law of conservation of mass. According to the above equation, 163 gram of reactants are producing 163 gram of products.

(b) Mg24+2HCl73MgCl295+H22\text{(b) }\underset{24}{\text{Mg}} + \underset{73}{\text{2HCl}} \longrightarrow \underset{95}{\text{MgCl}_2} + \underset{2}{\text{H}_2} \\[1em]

The above balanced equation conveys the following information:

  1. One molecule of magnesium reacts with two molecules of hydrochloric acid to produce one molecule of magnesium chloride and one molecule of hydrogen.
  2. 24 g of magnesium reacts with 73 g of hydrochloric acid to produce 95 g of magnesium chloride and 2 g of hydrogen.
  3. 24 g of magnesium reacts with 73 g of hydrochloric acid to produce 22.4 litres of hydrogen gas at S.T.P.
  4. It also proves the law of conservation of mass. According to the above equation, 97 gram of reactants are producing 97 gram of products.

Question 3

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

(b) Name and state the fundamental law that every equation must fulfill.

Exercise 1C Descriptive Type

Answer:

(a) Polyatomic ions are ions composed of two or more atoms that are covalently bonded together and carry a net electrical charge. Examples — Nitrate ion (NO3-) and Sulphate ion (SO42-).

(b) Every equation must fulfill the "Law of Conservation of Matter". It states that matter is neither created nor destroyed in the course of a chemical reaction. Thus, the total mass of the substances on either side of the equation must be the same.

Question 4

Give the information conveyed by the chemical formula of a compound.

Exercise 1C Descriptive Type

Answer:

By looking at the chemical formula, we understand the ratio in which different atoms are united to form the molecule.

Question 5

Write the significance of a molecular formula.

Exercise 1C Descriptive Type

Answer:

The molecular formula of a compound has quantitative significance. It represents:

  1. both the molecule and the molecular mass of the compound.
  2. the respective numbers of different atoms present in one molecule of the compound.
  3. the ratios of the respective masses of the elements present in the compound.

For example, the formula CO2 means that:

  1. the molecular formula of carbon dioxide is CO2
  2. each molecule contains one carbon atom joined by chemical bonds with two oxygen atoms;
  3. the molecular mass of carbon dioxide is 44, given that atomic mass of carbon is 12 and that of oxygen is 16.

Question 6

What do you understand by radicals : What are basic and acidic radicals ? Explain with examples.

Exercise 1C Descriptive Type

Answer:

A radical is an atom or a group of atoms of the same or of different elements that behaves as a single unit with a positive or a negative charge.

An Acid radical is the radical that remains after an acidic molecule loses a hydrogen ion (H+). Acid radicals typically have a negative charge. They are also called electronegative radicals or anions.

A Basic radical is the radical that remains after a base molecule loses a hydroxyl ion (OH-). Basic radicals typically have a positive charge. They are also called electropositive radicals or cations.

For example, in the compound ammonium carbonate (NH4)2CO3, ammonium (NH4+) is a basic radical with combining power 1 and carbonate (CO32-) is an acidic radical with combining power 2.

Question 7

Give four examples of compounds with variable valency.

Exercise 1C Descriptive Type

Answer:

MetalValencyName of compound formedFormula
Iron2
3
Ferrous - [Iron (II)] oxide
Ferric - [Iron (III)] oxide
FeO
Fe2O3
Copper1
2
Cuprous - [Copper (I)] oxide
Cupric- [Copper (II)] oxide
Cu2O
CuO
Mercury1
2
Mercurous - [Mercury (I)] oxide
Mercuric - [Mercury(II)] oxide
Hg2O
HgO
Lead2
4
Plumbous - [Lead (II)] oxide
Plumbic - [Lead (IV)] oxide
PbO
PbO2

Exercise 1C Multiple Choice Type

24 questions

Question 1(i)

Modern atomic symbols are based on the method proposed by

  1. Bohr
  2. Dalton
  3. Berzelius
  4. Alchemist
Exercise 1C Multiple Choice Type

Answer:

Berzelius

Reason — Berzelius suggested that the initial letter of an element written in capitals should represent that particular element. This method suggested by him laid the basis of the IUPAC system of chemical symbols and formulae.

Question 1(ii)

The number of carbon atoms in a hydrogen carbonate radical is

  1. one
  2. two
  3. three
  4. four
Exercise 1C Multiple Choice Type

Answer:

one

Reason — Hydrogen carbonate radical is represented as HCO3-. Thus, it shows that one carbon atom is present in a hydrogen carbonate radical.

Question 1(iii)

The formula of iron (III) sulphate is

  1. Fe3SO4
  2. Fe(SO4)3
  3. Fe2(SO4)3
  4. FeSO4
Exercise 1C Multiple Choice Type

Answer:

Fe2(SO4)3

Reason — Chemical formula of iron (III) sulphate:

Fe3+ SO42Fe223  SO42Fe22 SO423Fe2(SO4)3\text{Fe}^{3+} \space \text{SO}_4^{2-} \\[1em] \overset{\phantom{22}{3}}{\text{Fe}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{SO}_4} \Rightarrow \underset{\phantom{2}{2}}{\text{Fe}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{3}}{\text{SO}_4} \\[1em] \text{Fe}_2\text{(SO}_4)_3

∴ Chemical formula of iron (III) sulphate is Fe2(SO4)3

Question 1(iv)

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

  1. 1 : 8
  2. 1 : 16
  3. 1 : 32
  4. 1 : 64
Exercise 1C Multiple Choice Type

Answer:

1 : 8

Reason — Hydrogen-to-oxygen mass ratio in water can be found as below:

Molecualr formula of water = H2O

∴ Hydrogen-to-oxygen mass ratio in water

   = 216\dfrac{2}{16} = 18\dfrac{1}{8}

   = 1 : 8

Question 1(v)

The formula of sodium carbonate is Na2CO3 and that of calcium hydrogen carbonate is

  1. CaHCO3
  2. Ca(HCO3)2
  3. Ca2HCO3
  4. Ca(HCO3)3
Exercise 1C Multiple Choice Type

Answer:

Ca(HCO3)2

Reason — Chemical formula of calcium hydrogen carbonate:

Ca2+ HCO31Ca222  HCO31Ca21 HCO322Ca(HCO3)2\text{Ca}^{2+} \space \text{HCO}_3^{1-} \\[1em] \overset{\phantom{22}{2}}{\text{Ca}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{HCO}_3} \Rightarrow \underset{\phantom{2}{1}}{\text{Ca}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{2}}{\text{HCO}_3} \\[1em] \text{Ca}\text{(HCO}_3)_2

∴ Chemical formula of calcium hydrogen carbonate is Ca(HCO3)2

Question 1(vi)

The correct atomic symbols for carbon, calcium, copper and cadmium respectively are:

  1. Ca, C, Cu, Cd
  2. Ca, C, CO, Cd
  3. C, Ca, Cu, Cd
  4. Ca, Cl, Co, Cd
Exercise 1C Multiple Choice Type

Answer:

C, Ca, Cu, Cd

Reason — Carbon is denoted by the first letter of its name C, Calcium by the first two letters of its name Ca, Cadmium by first and third letters Cd and Copper is denoted by the first two letters of its latin name Cuprum so Cu.

Question 1(vii)

Hydrargyrum is the Latin name of:

  1. Gold
  2. Silver
  3. Mercury
  4. Platinum
Exercise 1C Multiple Choice Type

Answer:

Mercury

Reason — The term "Hydrargyrum" is derived from the Greek words "hydr-" meaning water and "argyros" meaning silver. In Latin, "Hydrargyrum" is used to refer to the element mercury, which is a silvery, liquid metal.

Question 1(viii)

In Zn3(PO4)2, the valency of Zn and Phosphate respectively is:

  1. 3 and 2
  2. 2 and 6
  3. 4 and 6
  4. 2 and 3
Exercise 1C Multiple Choice Type

Answer:

2 and 3

Reason — Given the formula Zn3(PO4)2 :

Zn3(PO4)2=Zn23  PO42Zn32  PO43\text{Zn}_3(\text{PO}_4)_2 = \underset{\phantom{2}{3}}{\text{Zn}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{PO}_4} \Rightarrow \overset{\phantom{3}{2}}{\text{Zn}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{3}{\text{PO}_4} \\[0.5em]

∴ Valency of Zinc is 2 and Phosphate is 3.

Question 1(ix)

In Ca3(PO4)x, the value of x is:

  1. 1
  2. 2
  3. 3
  4. 4
Exercise 1C Multiple Choice Type

Answer:

2

Reason — Chemical formula of calcium phosphate:

Ca2+ PO43Ca222  PO43Ca23 PO422Ca3(PO4)2\text{Ca}^{2+} \space \text{PO}_4^{3-} \\[1em] \overset{\phantom{22}{2}}{\text{Ca}} \space {\swarrow}\mathllap{\searrow} \space \overset{3}{\text{PO}_4} \Rightarrow \underset{\phantom{2}{3}}{\text{Ca}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{2}}{\text{PO}_4} \\[1em] \text{Ca}_3\text{(PO}_4)_2

∴ Chemical formula of calcium phosphate is Ca3(PO4)2 hence x = 2.

Question 1(x)

In C2H4, the valency of carbon is:

  1. 2
  2. 1
  3. 4
  4. 3
Exercise 1C Multiple Choice Type

Answer:

4

Reason — Given the formula C2H4 :

C2H4=C22  H4C34  H2\text{C}_2\text{H}_4 = \underset{\phantom{2}{2}}{\text{C}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{4}{\text{H}} \Rightarrow \overset{\phantom{3}{4}}{\text{C}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{H}} \\[0.5em]

∴ Valency of Carbon is 4.

Question 1(xi)

Which of the following represent the correct chemical formula of a compound :

  1. NaSO4
  2. AlSO4
  3. NH4PO4
  4. Ca3N2
Exercise 1C Multiple Choice Type

Answer:

Ca3N2

Reason — Calcium (Ca) forms a +2 ion (Ca+2) and nitride (N-3) has a –3 charge.

Ca2+ N3Ca222  N3Ca23 N22Ca3N2\text{Ca}^{2+} \space \text{N}^{3-} \\[1em] \overset{\phantom{22}{2}}{\text{Ca}} \space {\swarrow}\mathllap{\searrow} \space \overset{3}{\text{N}} \Rightarrow \underset{\phantom{2}{3}}{\text{Ca}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{2}}{\text{N}} \\[1em] \text{Ca}_3\text{N}_2

It is perfectly balanced. Therefore, Ca3N2 is correct.

Question 1(xii)

A chemical equation is always balanced to satisfy the conditions of :

  1. Law of constant proportions.

  2. Gay Lussac's law

  3. Law of conservation of mass

  4. Law of mass action.

Exercise 1C Multiple Choice Type

Answer:

Law of conservation of mass

Reason — Law of conservation of mass states that the total mass of the substances on either side of the equation is the same.

Question 1(xiii)

One "μ" stands for :

  1. One atom of any element
  2. 1/12th of hydrogen atom
  3. An atom of carbon (C-12)
  4. 1/12th the mass of carbon atom (C-12).
Exercise 1C Multiple Choice Type

Answer:

1/12th the mass of carbon atom (C-12)

Reason — Atomic mass is expressed in atomic mass units [a.m.u.] or "μ". Atomic mass unit is defined as 1/12 the mass of carbon atom C-12. (The mass of an atom of carbon-12 isotope was given the atomic mass of 12 units, i.e., 12 amu or simply 12 μ ).

Question 1(xiv)

In the compound magnesium nitride (Mg3N2):

P — Mg is trivalent, N is divalent.

Q — Mg is divalent. N is trivalent.

R — Mg is divalent, N is divalent.

Which of the statements given above is/are true?

  1. Only P

  2. Only Q

  3. Only R

  4. Both P and Q

Exercise 1C Multiple Choice Type

Answer:

Only Q

Reason — The valency of elements in compound magnesium nitride (Mg3N2) is,

Mg3N2=Mg23  N2Mg32  N3\text{Mg}_3\text{N}_2 = \underset{\phantom{2}{3}}{\text{Mg}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{N}} \Rightarrow \overset{\phantom{3}{2}}{\text{Mg}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{3}{\text{N}} \\[0.5em]

Hence, Mg is divalent and N is trivalent.

Question 1(xv)

Copper shows variable valency and forms two different compounds with oxygen — Cu2O and CuO.

P — A is cuprous oxide, B is cupric oxide.

Q — A is cupric oxide, B is cuprous oxide.

R — A is copper (I) oxide, B is copper (II) oxide.

  1. Only P

  2. Only Q

  3. Only R

  4. Both P and R

Exercise 1C Multiple Choice Type

Answer:

Both P and R

Reason — Certain elements exhibit more than one valency and they show variable valency.

For example, copper shows valency of 1 and 2 and forms compounds like, cuprous [copper(I)] oxide and cupric [copper(II)] oxide, respectively.

Hence, both P and R are true.

Question 1(xvi)

The figure given below shows the molecule of an element, where 'a' denotes the atom with atomic mass 32.

The figure given below shows the molecule of an element, where *'a'* denotes the atom with atomic mass 32. Language of Chemistry, Concise Chemistry Solutions ICSE Class 9.

P — Element is tetratomic with molecular mass 128.

Q — Element is octatomic with molecular mass 256.

R — Element is crown-shaped with molecular mass 256.

  1. Only P

  2. Only Q

  3. Only R

  4. Both P and R

Exercise 1C Multiple Choice Type

Answer:

Only Q

Reason — Given element 'a' is having atomic mass 32. The figure shows there are eight 'a' atoms forming a octatomic molecule. So when we calculate the molecular mass it will be

32 x 8 = 256

Hence, element 'a' is octatomic molecule with molecular mass 256.

Question 2(i)

Assertion (A): The atomic mass of sodium is 23 amu.

Reason (R): An atom of sodium is 23 times heavier than an atom of carbon with mass 12 amu.

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Exercise 1C Multiple Choice Type

Answer:

A is true but R is false.

Explanation — The atomic mass of sodium is 23 amu. Hence, the assertion (A) is true.

An atom of sodium is 23 times heavier than one-twelfth the mass of a carbon-12 atom, not 23 times heavier than an atom of carbon with mass 12 amu. Hence, reason (R) is false.

Question 2(ii)

Assertion (A): An atom is the smallest part of matter which can take part in a chemical reaction.

Reason (R): Atoms of every element can exist independently.

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Exercise 1C Multiple Choice Type

Answer:

A is true but R is false.

Explanation — An atom is the smallest particle of an element and can take part in a chemical reaction. Hence, the assertion (A) is true.

Atoms may or may not exist independently.
For example:

Noble gas atoms like He, Ne can exist freely.

But atoms like H, O, and N do not exist alone under normal conditions; they exist as diatomic molecules (H2, O2, N2). Hence, reason (R) is false.

Question 2(iii)

Assertion (A): All equations need to be balanced.

Reason (R): An unbalanced equation would imply that atoms have been created or destroyed.

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Exercise 1C Multiple Choice Type

Answer:

Both A and R are true and R is the correct explanation of A.

Explanation — An equation must be balanced in order to comply with the "Law of Conservation of Matter", which states that matter is neither created nor destroyed in the course of a chemical reaction. Hence, the assertion (A) is true.
An unbalanced equation would imply that atoms have been created or destroyed. Hence, both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Question 2(iv)

Assertion (A): PCl3, is known as phosphorus trichloride while AlCl3, is aluminium chloride and not aluminium trichloride.

Reason (R) : Phosphorus shows variable valency. Aluminium does not show variable valency

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Exercise 1C Multiple Choice Type

Answer:

Both A and R are true and R is the correct explanation of A.

Explanation — Phosphorus can show variable valency (like 3 and 5), so we name its compounds with prefixes like phosphorus trichloride (PCl3).
Aluminium, however, shows a fixed valency of 3, so AlCl3 is simply called aluminium chloride without needing a prefix. Hence, both assertion (A) and reason (R) are true, and reason (R) correctly explains assertion (A).

Question 2(v)

Assertion (A): In a chemical reaction, the total mass of the products remains the same as that of the reactants.

Reason (R): A chemical reaction involves a simple exchange of partners and no new species are formed.

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Exercise 1C Multiple Choice Type

Answer:

Both A and R are true but R is not the correct explanation of A.

Explanation — In a chemical reaction, the total mass of the substances on either side of the equation is the same, this is referred to be as law of conservation of mass. Hence, the assertion (A) is true.
In a chemical reaction the products are formed by the rearrangement of atoms or elements in the reactants. Hence, the reason (R) is true but reason (R) is not the correct explanation for assertion (A).

Question 2(vi)

Assertion (A): Magnesium phosphate is written as Mg3(PO4)2.

Reason (R): The valencies of magnesium and phosphate are three and two respectively.

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Exercise 1C Multiple Choice Type

Answer:

A is true but R is false.

Explanation — The molecular formula of Magnesium phosphate is Mg3(PO4)2. Hence, the given assertion (A) is true.

Given the formula Mg3(PO4)2 :

Mg3(PO4)2=Mg23  PO42Mg32  PO43\text{Mg}_3(\text{PO}_4)_2 = \underset{\phantom{2}{3}}{\text{Mg}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{PO}_4} \Rightarrow \overset{\phantom{3}{2}}{\text{Mg}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{3}{\text{PO}_4} \\[0.5em]

∴ Valency of magnesium (Mg) is 2 and Phosphate is 3. Hence, the reason (R) is false.

Question 2(vii)

Assertion (A): A chemical equation is written to represent molecules.

Reason (R): This is necessary for the law of conservation of mass

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Exercise 1C Multiple Choice Type

Answer:

Both A and R are true but R is not the correct explanation of A.

Explanation — A chemical equation represents molecules and it is necessary for the chemical equation to follow the law of conservation of mass. Hence, both assertion (A) and reason (R) are true.
However, balancing the equation is what ensures the law of conservation of mass. Therefore, reason (R) is not the correct explanation of assertion (A).

Question 2(viii)

Assertion (A): Chemical combination always takes place between two elements.

Reason (R): A single substance is formed in a chemical combination.

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Exercise 1C Multiple Choice Type

Answer:

A is false but R is true.

Explanation — Chemical combination may take place between two or more elements. Hence, the assertion (A) is false.
A compound is a substance formed by the chemical combination of two or more elements in a fixed proportion. So, a single substance is formed in a chemical combination. Hence, the reason (R) is true.

Exercise 1C Numericals

10 questions

Question 1

Calculate the molecular mass of the following :

(i) Na2SO4.10H2O

(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

Exercise 1C Numericals

Answer:

(i) Molecular mass of Na2SO4.10H2O
     = 2 x 23 + 32 + 4 x 16 + 10(2 x 1 + 16)
     = 46 + 32 + 64 + 180
     = 322 amu

(ii) Molecular mass of (NH4)2CO3
     = 2(14 + 4 x 1) + 12 + 3 x 16
     = 36 + 12 + 48
     = 96 amu

(iii) Molecular mass of (NH2)2CO
     = 2(14 + 2 x 1) + 12 + 16
     = 32 + 12 + 16
     = 60 amu

(iv) Molecular mass of Mg3N2
     = 3 x 24 + 2 x 14
     = 72 + 28
     = 100 amu

Question 2

Calculate the relative molecular masses of

(a) CHCl3

(b) (NH4)2Cr2O7

(c) CuSO4.5H2O

(d) (NH4)2SO4

(e) CH3COONa

(f) Potassium chlorate, KClO3

(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, Cr = 52]

Exercise 1C Numericals

Answer:

(a) The relative molecular mass of CHCl3
     = 12 + 1 + 3 x 35.5
     = 12 + 1 + 106.5
     = 119.5 amu

(b) The relative molecular mass of (NH4)2Cr2O7
     = 2(14 + 4 x 1) + 2 x 52 + 7 x 16
     = 36 + 104 + 112
     = 252 amu

(c) The relative molecular mass of CuSO4.5H2O
     = 63.5 + 32 + 4 x 16 + 5(2 x 1 + 16)
     = 63.5 + 32 + 64 + 90
     = 249.5 amu

(d) The relative molecular mass of (NH4)2SO4
     = 2(14 + 4 x 1) + 32 + 4 x 16
     = 36 + 32 + 64
     = 132 amu

(e) The relative molecular mass of CH3COONa
     = 12 + 3 + 12 + 16 + 16 + 23
     = 82 amu

(f) The relative molecular mass of KClO3
     = 39 + 35.5 + 3 x 16
     = 39 + 35.5 + 48
     = 122.5 amu

(g) The relative molecular mass of (NH4)2PtCl6
     = 2(14 + 4 x 1) + 195 + 6 x 35.5
     = 36 + 195 + 213
     = 444 amu

Question 3

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

Exercise 1C Numericals

Answer:

Relative molecular mass of MgSO4.7H2O
   = 24 + 32 + 4 x 16 + 7(2 x 1 + 16)
   = 24 + 32 + 64 + 126
   = 246 amu

246 g of Epsom salt contains 126 g of water of crystallisation

∴ 100 g of Epsom salt contains 126×100246\dfrac{126 \times 100}{246}

= 12600246\dfrac{12600}{246} = 51.22 g of water of crystallisation.

∴ % mass of water in Epsom salt, MgSO4.7H2O = 51.22%

Question 4

Calculate the percentage of phosphorus in:

(a) Calcium hydrogen phosphate, Ca(H2PO4)2

(b) Calcium phosphate, Ca3(PO4)2

Exercise 1C Numericals

Answer:

(a) Relative molecular mass of Ca(H2PO4)2
     = 40 + 2(2 x 1 + 31 + 4 x 16)
     = 40 + 2(2 + 31 + 64)
     = 40 + 194
     = 234 amu

Wt. of P in Ca(H2PO4)2 = 2 x 31 = 62 g

% of P = Wt. of PTotal Wt. of Ca(H2PO4)2×100\dfrac{\text{Wt. of P}}{\text{Total Wt. of Ca(H}_2\text{PO}_4)_2}\times 100

= 62234\dfrac{62}{234} x 100 = 26.5%

Phosphorus in Calcium hydrogen phosphate is 26.5%

(b) Relative molecular mass of Calcium phosphate, Ca3(PO4)2
     = 3 x 40 + 2(31 + 4 x 16)
     = 120 + 2(31 + 64)
     = 120 + 190
     = 310 amu

Wt. of P in Ca3(PO4)2 = 2 x 31 = 62 g

% of P = Wt. of PTotal Wt. of Ca3(PO4)2×100\dfrac{\text{Wt. of P}}{\text{Total Wt. of Ca}_3\text{(PO}_4)_2}\times 100

= 62310\dfrac{62}{310} x 100 = 20%

Phosphorus in Calcium phosphate is 20%

Question 5

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

Exercise 1C Numericals

Answer:

Relative molecular mass of KClO3
   = 39 + 35.5 + 3 x 16
   = 39 + 35.5 + 48
   = 122.5 amu

122.5 g of KClO3 contains 39 g of Potassium

∴ 100 g of KClO3 contains 39×100122.5\dfrac{39 \times 100}{122.5} g of Potassium

= 390001225\dfrac{39000}{1225} = 31.83 g of Potassium

122.5 g of KClO3 contains 35.5 g of Chlorine

∴ 100 g of KClO3 contains 35.5×100122.5\dfrac{35.5 \times 100}{122.5} g of Chlorine

= 355001225\dfrac{35500}{1225} = 28.98 g of Chlorine

122.5 g of KClO3 contains 48 g of Oxygen

∴ 100 g of KClO3 contains 48×100122.5\dfrac{48 \times 100}{122.5} g of Oxygen

= 480001225\dfrac{48000}{1225} = 39.18 g of Oxygen

In KClO3 : K = 31.83%, Cl = 28.98% and O = 39.18%

Question 6

Urea is a very important nitrogenous fertilizer. Its formula is CON2H4. Calculate the percentage of carbon in urea.

(C= 12, O = 16, N = 14 and H = 1)

Exercise 1C Numericals

Answer:

Relative molecular mass of CON2H4
   = 12 + 16 + 2 x 14 + 4 x 1
   = 12 + 16 + 28 + 4
   = 60 amu

Wt. of C in CON2H4 = 12 g

% of C = Wt. of CTotal Wt. of CON2H4×100\dfrac{\text{Wt. of C}}{\text{Total Wt. of CON}_2\text{H}_4}\times 100

= 1260\dfrac{12}{60} x 100 = 20%

Carbon in Urea is 20%

Question 7

Calculate the percentage of iron in Ferric oxide Fe2O3.

Exercise 1C Numericals

Answer:

Relative molecular mass of Ferric oxide (Fe2O3)
  = 56 x 2 + 16 x 3
  = 112 + 48
  = 160 amu

Since 160 g of Ferric oxide (Fe2O3 ) contains 112 g of iron

∴ 100 g of contains Ferric oxide (Fe2O3 ) contains 112160\dfrac{112}{160} x 100 = 70%

∴ Percentage of Iron in Ferric oxide (Fe2O3 ) is 70%

Question 8

Calculate amount of Nitrogen in one bag (50 kg) of Urea.

Exercise 1C Numericals

Answer:

Molecular formula of Urea is CON2H4

Relative molecular mass of CON2H4
   = 12 + 16 + 2 x 14 + 4 x 1
   = 12 + 16 + 28 + 4
   = 60 amu

Wt. of N in CON2H4 = 28 g

% of N = Wt. of NTotal Wt. of CON2H4×100\dfrac{\text{Wt. of N}}{\text{Total Wt. of CON}_2\text{H}_4}\times 100

= 2860\dfrac{28}{60} x 100 = 46.67%

Wt. of N in 50 kg CON2H4

= 46.67100\dfrac{46.67}{100} x 50 = 23.33 kg

∴ Amount of Nitrogen in one bag (50 kg) of Urea is 23.33 kg.

Question 9

Calculate percentage of platinum in [(NH4)2 PtCl6]. Ammonium chloride platinate in nearest whole number.

Exercise 1C Numericals

Answer:

The relative molecular mass of (NH4)2 PtCl6
     = 2(14 + 4 x 1) + 195 + 6 x 35.5
     = 36 + 195 + 213
     = 444 amu

Wt. of Pt in [(NH4)2 PtCl6] = 195 g

∴ Percentage of platinum in (NH4)2PtCl6 is

    = 195444\dfrac{195}{444} x 100

    = 43.91 %

∴ Rounding to the nearest whole number, the percentage of platinum in ammonium chloride platinate (NH4)2 PtCl6 is 44%.

Question 10

Calculate percentage composition of the elements in calcium phosphate Ca3(PO4)2.

Exercise 1C Numericals

Answer:

Relative molecular mass of Calcium phosphate, Ca3(PO4)2
     = 3 x 40 + 2(31 + 4 x 16)
     = 120 + 2(31 + 64)
     = 120 + 190
     = 310 amu

310 g of Ca3(PO4)2 contains 120 g of Calcium

∴ 100 g of Ca3(PO4)2 contains 120×100310\dfrac{120 \times 100}{310} g of Calcium

= 12000310\dfrac{12000}{310} = 38.70 % of Calcium.

310 g of Ca3(PO4)2 contains 62 g of Phosphorus

∴ 100 g of Ca3(PO4)2 contains 62×100310\dfrac{62 \times 100}{310} g of Phosphorus

= 6200310\dfrac{6200}{310} = 20 % of Phosphorus.

310 g of Ca3(PO4)2 contains 128 g of Oxygen.

∴ 100 g of Ca3(PO4)2 contains 128×100310\dfrac{128 \times 100}{310} g of Oxygen.

= 12800310\dfrac{12800}{310} = 41.3 % of Oxygen.

In Ca3(PO4)2 : Ca = 38.70%, P = 20% and O = 41.3%

Exercise 1C Short Answer Type

7 questions

Question 1

What is the valency of :

(a) fluorine in CaF2

(b) sulphur in SF6

(c) phosphorus in PH3

(d) carbon in CH4

(e) Manganese in MnO2

(f) Copper in Cu2O

(g) Magnesium in Mg3N2

(h) nitrogen in the following compounds :

(i) N2O3 (ii) N2O5 (iii) NO2 (iv) NO

Exercise 1C Short Answer Type

Answer:

(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 manganese in MnO2 is +4.

(f) Valency of copper in Cu2O is +1.

(g) Valency of magnesium in Mg3N2 is +2.

(h) Valency of nitrogen in the given compounds :

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

Question 2

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

Exercise 1C Short Answer Type

Answer:

An equation must be balanced in order to comply with the "Law of Conservation of Matter", which states that matter is neither created nor destroyed in the course of a chemical reaction. An unbalanced equation would imply that atoms have been created or destroyed.

For example, consider the following unbalanced equation:

KNO3 ⟶ KNO2 + O2

In this equation, there are 3 oxygen atoms on the left side, but 4 oxygen atoms on the right side. This means that the equation is not balanced and does not satisfy the law of conservation of mass.

The balanced form of the equation is:

2KNO3 ⟶ 2KNO2 + O2

Now there are 6 oxygen atoms on both the sides. This means that the law of conservation of mass is satisfied and the equation correctly represents the reaction.

Question 3(a)

Define Atomic mass unit.

Exercise 1C Short Answer Type

Answer:

Atomic mass unit is defined as 1⁄12 the mass of an carbon atom C-12.

Question 3(b)

Define Basic radical.

Exercise 1C Short Answer Type

Answer:

A Basic radical is the radical that remains after a base molecule loses a hydroxyl ion (OH-). Basic radicals typically have a positive charge. They are also called electropositive radicals or cations.

Question 4

Balance the following equations.

(a) Al2O3 + H2SO4 ⟶ Al2(SO4)3 + H2O

(b) NH3 + Cl2 ⟶ NH4Cl + N2

(c) C5H12 (pentane) + O2 ⟶ CO2 + H2O

(d) C4H10 (butane) + O2 ⟶ CO2 + H2O

(e) FeSO4 + H2SO4 + Cl2 ⟶ Fe2(SO4)3 + HCl

(f) NaCl + MnO2 + H2SO4 ⟶ NaHSO4+ MnSO4 + H2O + Cl2

(g) Na2Cr2O7 + HCl ⟶ NaCl + CrCl3 + H2O + Cl2

(h) KMnO4 + H2SO4 + H2S ⟶ K2SO4 + MnSO4 + H2O + S

Exercise 1C Short Answer Type

Answer:

(a) Al2O3 + 3H2SO4 ⟶ Al2(SO4)3 + 3H2O

(b) 8NH3 + 3Cl2 ⟶ 6NH4Cl + N2

(c) C5H12 (pentane) + 8O2 ⟶ 5CO2 + 6H2O

(d) 2C4H10 (butane) + 13O2 ⟶ 8CO2 + 10H2O

(e) 2FeSO4 + H2SO4 + Cl2 ⟶ Fe2(SO4)3 + 2HCl

(f) 2NaCl + MnO2 + 3H2SO4 ⟶ 2NaHSO4 + MnSO4 + 2H2O + Cl2

(g) Na2Cr2O7 + 14HCl ⟶ 2NaCl + 2CrCl3 + 7H2O + 3Cl2

(h) 2KMnO4 + 3H2SO4 + 5H2S ⟶ K2SO4 + 2MnSO4 + 8H2O + 5S

Question 5

Atomic mass of calcium is 40 μ. What does it signify ?

Exercise 1C Short Answer Type

Answer:

The atomic mass of calcium being 40 amu signifies that, on average, a calcium atom weighs 40 times the mass of one twelfth of a carbon-12 atom.

Question 6

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 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

Exercise 1C Short Answer Type

Answer:

(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 ⟶ 4NO + 6H2O

(h) BaCl2 + H2SO4 ⟶ BaSO4 + 2HCl

(i) 2ZnS + 3O2 ⟶ 2ZnO + 2SO2

(j) Al4C3 + 12H2O ⟶ 4Al(OH)3 + 3CH4

(k) 4FeS2 + 11O2 ⟶ 2Fe2O3 + 8SO2

(l) 2KMnO4 + 16HCl ⟶ 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 + 6HNO3 ⟶ H2SO4 + 6NO2 + 2H2O

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

Exercise 1C Structuredapplicationskill Type

2 questions

Question 1

Elements X, Y and Z have 3, 7 and 6 electrons in their valence shells respectively. Write the formula of the compound formed between :

(a) X and Y

(b) X and Z

Exercise 1C Structuredapplicationskill Type

Answer:

(a) Valency of X is +3 and Y is -1. Formula of the compound formed between X and Y is :

X3+ Y1X223  Y1X21 Y3XY3\text{X}^{3+} \space \text{Y}^{1-} \\[1em] \overset{\phantom{22}{3}}{\text{X}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{Y}} \Rightarrow \underset{\phantom{2}{1}}{\text{X}} \space {\swarrow}\mathllap{\searrow} \underset{{3}}{\text{Y}} \\[1em] \text{X}\text{Y}_3

∴ Chemical formula of the compound formed between X and Y is XY3

(b) Valency of X is +3 and Z is -2. Formula of the compound formed between X and Z is :

X3+ Z2X223  Z2X22 Z3X2Z3\text{X}^{3+} \space \text{Z}^{2-} \\[1em] \overset{\phantom{22}{3}}{\text{X}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{Z}} \Rightarrow \underset{\phantom{2}{2}}{\text{X}} \space {\swarrow}\mathllap{\searrow} \underset{{3}}{\text{Z}} \\[1em] \text{X}_2\text{Z}_3

∴ Chemical formula of the compound formed between X and Z is X2Z3

Question 2

The following figure represents the structural formula of a chemical compound.

The following figure represents the structural formula of a chemical compound. How many Carbon and Hydrogen atoms are present in the compound. Language of Chemistry, Concise Chemistry Solutions ICSE Class 9.

Answer the questions given below:

(a) How many Carbon and Hydrogen atoms are present in the compound.

(b) Write the molecular and empirical formulae of the compound.

(c) Calculate the percentage composition of all the elements present in the compound. [At. wt. of C = 12, H = 1]

Exercise 1C Structuredapplicationskill Type

Answer:

(a) 6 Carbon and 12 Hydrogen atoms are present.

(b) Its molecular formula is C6H12.
As the simplest ration between C and H is 1:2, hence the empirical formula is CH2

(c) Relative molecular mass of C6H12
   = (6 x 12) + (1 x 12)
   = 72 + 12
   = 84 g

84 g of C6H12 contains 72 g of Carbon

∴ 100 g of C6H12 contains 72×10084\dfrac{72 \times 100}{84} g of Carbon

= 720084\dfrac{7200}{84} = 85.7 g of Carbon.

84 g of C6H12 contains 12 g of Hydrogen

∴ 100 g of C6H12 contains 12×10084\dfrac{12 \times 100}{84} g of Hydrogen

= 120084\dfrac{1200}{84} = 14.3 of Hydrogen.

∴ In C6H12, C = 85.7% and H = 14.3%

Exercise 1C Very Short Type

5 questions

Question 1

Fill in the blanks:

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

(b) Symbol represents ............... atom 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 phosphorus in PCl3 is ............... and in PCl5 is ............... .

(f) Valency of Iron in FeCl2 is ............... and in FeCl3 it is ............... .

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

Exercise 1C Very Short Type

Answer:

(a) Dalton used symbol \bigcirc for oxygen and  {\bigcirc}\mathllap{\bullet\phantom{\space}} for hydrogen.

(b) Symbol represents one atom 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 a acidic radical.

(e) Valency of phosphorus in PCl3 is 3 and 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 .

Question 2

Complete the following table.

Acid Radicals →
Basic Radicals ↓
ChlorideNitrateSulphateCarbonateHydroxidePhosphate
MagnesiumMgCl2Mg(NO3)2MgSO4MgCO3Mg(OH)2Mg3(PO4)2
Sodium      
Zinc      
Silver      
Ammonium      
Calcium      
Iron (II)      
Potassium      
Exercise 1C Very Short Type

Answer:

The completed table is given below:

Acid Radicals →
Basic Radicals ↓
ChlorideNitrateSulphateCarbonateHydroxidePhosphate
MagnesiumMgCl2Mg(NO3)2MgSO4MgCO3Mg(OH)2Mg3(PO4)2
SodiumNaClNaNO3Na2SO4Na2CO3NaOHNa3PO4
ZincZnCl2Zn(NO3)2ZnSO4ZnCO3Zn(OH)2Zn3(PO4)2
SilverAgClAgNO3Ag2SO4Ag2CO3AgOHAg3PO4
AmmoniumNH4ClNH4NO3(NH4)2SO4(NH4)2CO3NH4OH(NH4)3PO4
CalciumCaCl2CaNO3CaSO4CaCO3Ca(OH)2Ca3(PO4)2
Iron (II)FeCl2Fe(NO3)2FeSO4FeCO3Fe(OH)2Fe3(PO4)2
PotassiumKClKNO3K2SO4K2CO3KOHK3PO4

Question 3

Name the following compounds and state how many atoms of each kind are present in one molecule of these compounds.

(a) H2SO4

(b) HClO4

(c) K2Cr2O7

(d) KMnO4

(e) K4[Fe(CN)6]

(f) Na2CrO4

(g) Mn3(BO4)2

(h) HNO2

Exercise 1C Very Short Type

Answer:

(a) Atoms in H2SO4 are

H ⟶ 2

S ⟶ 1

O ⟶ 4

(b) Atoms in HClO4

H ⟶ 1

Cl ⟶ 1

O ⟶ 4

(c) Atoms in K2Cr2O7

K ⟶ 2

Cr ⟶ 2

O ⟶ 7

(d) Atoms in KMnO4

K ⟶ 1

Mn ⟶ 1

O ⟶ 4

(e) Atoms in K4[Fe(CN)6]

K ⟶ 4

Fe ⟶ 1

C ⟶ 6

N ⟶ 6

(f) Atoms in Na2CrO4

Na ⟶ 2

Cr ⟶ 1

O ⟶ 4

(g) Atoms in Mn3(BO4)2

Mn ⟶ 3

B ⟶ 2

O ⟶ 8

(h) Atoms in HNO2

H ⟶ 1

N ⟶ 1

O ⟶ 2

Question 4

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.

Exercise 1C Very Short Type

Answer:

(a) A molecular formula represents the molecule of an element or of a compound.

(b) Molecular formula of water is H2O.

(c) A molecule of sulphur is octatomic.

(d) CO represents Carbon Monoxide and Co represents cobalt.

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

Question 5

Give the empirical formula of :

  1. Benzene (C6H6)
  2. Glucose (C6H12O6)
  3. Acetylene (C2H2)
  4. Acetic acid (CH3COOH)
Exercise 1C Very Short Type

Answer:

  1. Ratio of C and H atoms in Benzene (C6H6) = 6 : 6
    Simplest Ratio = 1 : 1
    ∴ Empirical formula of Benzene (C6H6) = CH
  2. Ratio of C, H and O atoms in Glucose (C6H12O6) = 6 : 12 : 6
    Simplest Ratio = 1 : 2 : 1
    ∴ Empirical formula of Glucose (C6H12O6) = CH2O
  3. Ratio of C and H atoms in Acetylene (C2H2) = 2 : 2
    Simplest Ratio = 1 : 1
    ∴ Empirical formula of Acetylene (C2H2) = CH
  4. Ratio of C, H and O atoms in Acetic acid (CH3COOH) = 2 : 4 : 2
    Simplest Ratio = 1 : 2 : 1
    ∴ Empirical formula of Acetic acid (CH3COOH) = CH2O