Request a call back

Join NOW to get access to exclusive study material for best results

Class 9 SELINA Solutions Chemistry Chapter 1 - The Language of Chemistry

The Language of Chemistry Exercise Ex. 1(A)

Solution 1

A symbol is the short form which stands for the atom of a specific element or the abbreviations used for the names of elements.

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

 

Solution 2

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

Solution 3

The full form of IUPAC is International Union of Pure and Applied Chemistry.

Names of the elements:

Au - Gold 

Pb - Lead

Sn - Tin

Hg - Mercury

Solution 4

Co stands for Cobalt. If we write CO, then it would mean that it is a compound containing two non-metal ions, i.e. carbon and oxygen, which forms carbon monoxide gas. 

Solution 5

a. H - Hydrogen atom

 

b. H2 - Dihydrogen or hydrogen gas

 

c. 2H - Two hydrogen atoms

 

d. 2H2 - Two dihydrogen molecule

 

Solution 6

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

Diatomic molecules: H2, O2, N2, Cl2

Solution 7(a)

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

 

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

Solution 7(b)

If an element exhibits two different positive valencies, then  

 

 i. for the lower valency, use the suffix -OUS at the end of the name of the metal

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

 

Example:

 

Element

Lower valency

Higher valency

Ferrum (Iron)

Ferrous (Fe2+)

Ferric (Fe3+)

 

Solution 8

 

 

 

Name

Formula

Valency 

a.

Aluminate 

AlO2

-2

b.

Chromate

CrO4

-2

c.

Aluminium 

Al

+3

d.

Cupric

Cu

+2

 

 

Solution 9

  1. Chemical formula: The chemical formula of a substance (element or compound) is a symbolic representation of the actual number of atoms present in one molecule of that substance.
  2. Significance of the molecular formula:

 i. It represents both molecule and molecular mass of the compound.

 ii. It represents the respective number of different atoms present in one molecule of the compound.

 iii. It represents the ratios of the respective masses of the elements present in the compound.e.g. Hg2O 1. Hg1+O2-

2.

  

3. Hg2O

Solution 10(a)

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

Examples: Cl-, O2- 

Solution 10(b)

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

Examples: K+, Na+ 

Solution 11

 

 

 

Acidic radical

Basic radical

  1. MgSO4 

SO4-

Mg+

  1. (NH4)2SO4 

SO4-

NH4+

  1. Al2(SO4)3 

SO4-

Al3+

  1. ZnCO3 

CO3-

Zn2+

  1. Mg(OH)2 

OH-

Mg2+

 

 

Solution 12

Valencies of aluminium, ammonium and zinc are 3, 1 and 2, respectively.

The valency of sulphate is 2.

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

Solution 13

 

Formula of the compound = A2B3

 

Solution 14

 

 

Compound

Formula (Ans)

  1. Boric acid

xvi. H3BO3

  1. Phosphoric acid

xvii. H3PO4

  1. Nitrous acid

xv. HNO2

  1. Nitric acid

xiv. HNO3

  1. Sulphurous acid

xiii. H2SO3

  1. Sulphuric acid

xviii. H2SO4

  1. Hydrochloric acid

xii. HCl

  1. Silica (sand)

ii. SiO2

  Caustic soda

     (sodium hydroxide)

i. NaOH

Caustic potash (potassium hydroxide)

iv. KOH

 Washing soda

(sodium carbonate)

iii. Na2CO3

 Baking soda

(sodium bicarbonate)

vi. NaHCO3

 Lime stone

(calcium carbonate)

v. CaCO3

  1. Water

viii. H2O

  1. Hydrogen sulphide

vii. H2S

  1. Ammonia

xi. NH3

  1. Phosphine 

ix. PH3

  1. Methane

x. CH4

 

 

Solution 15

 

Compounds

Acidic

radical

Basic

radical

Chemical formulae

Barium sulphate

SO42-

Ba2+

BaSO4

Bismuth nitrate

NO3-

Bi3+

Bi(NO3)3

Calcium bromide

Br-

Ca2+

CaBr2

Ferrous sulphide

S2-

Fe2+

FeS 

Chromium sulphate

SO42-

Cr3+

Cr2(SO4)3

Calcium silicate

SiO42-

Ca2+

Ca2SiO4

Potassium ferrocyanide

[Fe(CN)6]4-

K1+

K4[Fe(CN)6]

Stannic oxide

O2-

Sn2+

SnO2

Magnesium phosphate

(PO4)3-

Mg2+

Mg3(PO4)2

Sodium zincate

ZnO2-

Na1+

Na2ZnO2

 Stannic phosphate

(PO4)3-

Sn4+

Sn3(PO4)4

 Sodium thiosulphate

(S2O3)2-

Na1+

Na2S2O3

Potassium manganate

MnO42-

K1+

K2MnO4

Nickel bisulphate

HSO41-

Ni3+

Ni(HSO4)3

 

 

Solution 16

Chemical names of compounds:

 

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

 

Solution 17

  1. KClO- Potassium hypochlorite
  2. KClO2- Potassium chlorite
  3. KClO3- Potassium chlorate
  4. KClO4- Potassium perchlorate

Solution 18(a)

iii. The formula of a compound represents a molecule.

 

Solution 18(b)

iii. The correct formula of aluminium oxide is Al2O3.

Solution 18(c)

iv. The valency of nitrogen in nitrogen dioxide (NO2) is four.

 

Solution 19

  1. Sodium sulphate - Na2SO4
    There are two sodium atoms, one sulphur atom and four oxygen atoms.
  1. Quick lime - CaO
    There is one calcium atom and one oxygen atom.
  1. Baking soda - NaHCO3
    There is one sodium, carbon and hydrogen atom and three oxygen atoms.
  1. Ammonia - NH3
    There is one nitrogen atom and three hydrogen atoms.
  1. Ammonium dichromate - (NH4)Cr2O7
    There two ammonium atoms, two chromium atoms and seven oxygen atoms.

Solution 20

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

  1. Chloride - MCl3
  2. Oxide - M2O3
  3. Phosphate - M(PO4)
  4. Acetate - M(CH3COO)3 

The Language of Chemistry Exercise Ex. 1(C)

Solution 6(f)

Valency of Manganese in MnO2 is +4.

Solution 6(g)

Valency of Copper in Cu2O is +1

Solution 6(h)

Valency of Magnesium in Mg3N2 is +2

Solution 9(b)(i)

Molecular mass of Na2SO4.10H2O

= (2⨯23) + 32 +(4⨯16)+(10⨯18)

= 46 + 32 + 64 + 180

= 322

Solution 11

The correct statements are -

  1. The number of elements present in a molecule is represented by molecular formula.
  2. The molecular formula of water is H O. H O is the molecular formula of hydrogen peroxide.
  3. A molecule of sulfur is not monoatomic. It consists of 8 atoms.
  4. Co represents cobalt and CO represents carbon monoxide.
  5. The formula of iron (III) oxide is Fe2O3.

Solution 12

  1. CHCl3 = (C)12 + (H)1+ (3Cl)3 x 35.5 = 119.5
  2. (NH4)2Cr2O7 = (2N)28 + (8H)8 + (2Cr)2 x 51.9+ (7O)7 x 16 = 252
  3. CuSO4.5H2O = (Cu)63.5 + (S)32 + (4O)64 + (5H2O)5 x 18 = 249.5
  4. (NH4)2SO4 = (2N)28 + (8H)8 + (S)32 + (4O)64 = 132
  5. CH3COONa = (C)12 + (3H)3 + (C)12 + (2O)32 + (Na)23 = 82
  6. Potassium chlorate KClO3 = (K)39 + (Cl)35.5 + (3O)48 = 122.5
  7. Ammonium chloroplatinate, (NH4)2PtCl6 = (2N)28 + (8H)8 + (Pt)195 + (6Cl)35.5 x 6 = 444

Solution 13

The empirical formula is:

  1. Benzene (C6H6) = CH
  2. Glucose (C6H12O6) = CH6O
  3. Acetylene (C2H2) = CH
  4. Acetic acid (CH3COOH) = C2H2O

Solution 14

Given, 

Mass of epsom salt, MgSO4.7H2O.= 246.36 g/mole

Mass of water = 18 g/mole

Mass of 7H2O = 7⨯18 = 126 g/mole

Now we have to calculate the percentage mass of water in the epsom salt.

% mass of water =    

Now put all the given values in this formula, we get the percentage mass of water in the epsom salt.

% mass of water =    

Therefore, the percentage mass of water in the epsom salt is, 51.14 % 

Solution 15

  1. Calcium hydrogen phosphate Ca(H2PO4)2

Molecular mass of Ca(H2PO4)2 = 234 g/mole

Atomic weight of P = 31

% of P in Ca(H2PO4)2  =   

= 26.49 % of P

  1. Calcium phosphate Ca3(PO4)2

Molecular mass of Ca3(PO4)2 = 310.17 g/mole

Atomic weight of P = 31

% of P in Ca(H2PO4)2 =  = 19.98 % of P 

Solution 16

Molecular mass of KClO3 = 122.5 g

% of K = 39 /122.5 = 31.8%

% of Cl = 35.5/122.5 = 28.98%

% of O = 3 × 16/122.5 = 39.18%

Solution 17

Molar mass of urea; CON2H4 = 60 g

So, % of Carbon = 12 × 100/60 = 20%

Solution 1

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

 

Solution 2

 

 

 

 Acid Radicals 

 

 

 

Basic Radicals

Chloride

Nitrate

Sulphate 

Carbonate

Hydroxide

Phosphate

Magnesium

MgCl2

Mg(NO3)2

MgSO4

MgCO3

Mg(OH)2

 Mg3(PO4)2

Sodium

NaCl 

NaNO3

Na2SO4

Na2CO3

NaOH 

Na3PO4

Zinc

ZnCl2

Zn(NO3)2

Zn(SO4)2

ZnCO3

Zn(OH)2

Zn3(PO4)2

Silver

AgCl 

AgNO3

Ag2SO4

AgCO3

AgOH 

Ag3PO4

Ammonium

NH4Cl

NH4NO3

(NH4)2SO4

(NH4)2CO3

NH4OH

(NH4)3PO4

Calcium

CaCl2

CaCO3

CaSO4

CaCO3

Ca(OH)2

Ca3(PO4)2

Iron (II)

FeCl2

Fe(NO3)2

FeSO4

FeCO3

Fe(OH)2

Fe3(PO4)2

Potassium

KCl 

KNO3

K2SO4

K2CO3

KOH

K3PO4

 

 

Solution 3

 

a. NaCl + AgNO3 → NaNO3 + AgCl

 

b. It is a balanced equation.

 

c. Weights of reactants: NaCl - 58.44, AgNO3 - 169.87 

    Weights of products: NaNO3 - 84.99, AgCl - 143.32

    NaCl + AgNO3  NaNO3  + AgCl

    (23+35.5) + (108+14+48) → (23+14+48) + (108+35.5)

    58.5 + 170  85 + 143.5

    228.5 g  228.5 g

 

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

Solution 4(a)

 

This equation conveys the following information:

 

  1. The actual result of a chemical change.
  2. Substances take part in a reaction, and substances are formed as a result of the reaction.
  3. Here, one molecule of zinc and one molecule of sulphuric acid react to give one molecule of zinc sulphate and one molecule of hydrogen.
  4. Composition of respective molecules, i.e. one molecule of sulphuric acid contains two atoms of hydrogen, one atom of sulphur and four atoms of oxygen.
  5. Relative molecular masses of different substances, i.e. molecular mass of

Zn = 65

H2SO4 = (2+32+64) = 98

ZnSO4 = (65+32+64) = 161  

H2 = 2

  1. 22.4 litres of hydrogen are formed at STP.

 

Solution 4(b)

This equation conveys the following information:

 

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

Solution 5(a)

A polyatomic ion is a charged ion composed of two or more covalently bounded atoms. Examples: Carbonate (CO32-) and sulphate (SO42-)

Solution 5(b)

Fundamental laws which are involved in every equation:

  1. 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 (+).
  2. 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'.

Solution 6(a)

Valency of fluorine in CaF2 is -1. 

Solution 6(b)

Valency of sulphur in SF6 is -6.

Solution 6(c)

Valency of phosphorus in PH3 is +3.

Solution 6(d)

Valency of carbon in CH4 is +4.

Solution 6(e)

Valency of nitrogen in the given compounds:

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

 

Solution 7

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.

e.g.  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

 

Solution 8(a)

2NaOH + H2SO4 → Na2SO4 + 2H2O

Solution 8(b)

2KHCO3 + H2SO4 → K2SO4 + 2CO2 + 2H2O

Solution 8(c)

Fe + H2SO4 → FeSO4 + H2

Solution 8(d)

Cl2 + SO2 + 2H2O → H2SO4 + 2HCl

Solution 8(e)

2AgNO3 → 2Ag + 2NO2 + O2

Solution 8(f)

3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

Solution 8(g)

   

Solution 8(h)

BaCl2 + H2SO4 → BaSO4 + 2HCl

Solution 8(i)

2ZnS + 3O2 → 2ZnO + 2SO2

Solution 8(j)

Al4C3 + 12H2O → 4Al(OH)3 + 3CH4

Solution 8(k)

 4FeS2 + 11O2 → 2Fe2O3 + 8SO2

Solution 8(l)

2KMnO4 + HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O

Solution 8(m)

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

Solution 8(n)

2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2

Solution 8(o)

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

Solution 8(p)

K2Cr2O7 + 14HCl → 2KCl + 2CrCl3 + 7H2O + 3Cl2

Solution 8(q)

S + HNO3 → H2SO4 + NO2 + H2O 

Solution 8(r)

2NaCl + MnO2 + 3H2SO4 → 2NaHSO4 + MnSO4 + 2H2O + Cl2

Solution 9(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).

Solution 9(b)(ii)

Molecular mass of (NH4)2CO3

= (2 × 14) + (8 × 1) + 12 + (3 × 16)

= 28 + 8 + 12 + 48

= 96

Solution 9(b)(iii)

Molecular mass of (NH2)2CO

= (14 × 2) + (4 × 1) + 12 + 16

= 28 + 4 + 12 + 16

= 60 

Solution 9(b)(iv)

Molecular mass of Mg3N2

= (3 × 24) + (2 × 14)

= 72 + 28

= 100

Solution 10(a)

iii. Berzelius

Solution 10(b)

i. One

Solution 10(c)

iii. Fe2(SO4)3

Solution 10(d)

i. 1:8

 

Solution 10(e)

ii. Ca(HCO3)2

The Language of Chemistry Exercise Ex. 1(B)

Solution 1

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

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.

Solution 2

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

Solution 3

  • 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.

Solution 4

  1. C + O2 CO2
  2. N2 + O2 2NO
  3. 3Ca + N2 Ca3N2
  4. CaO + CO2 CaCO3
  5. Mg + H2SO4 MgSO4 + H2 
  6. Na + H2O NaOH + H2

Solution 5

Balanced chemical equations:

 

  1. 3Fe + 4H2O → Fe3O4 + 4H2
  2. 3Ca + N2 → Ca3N2
  3. Zn + 2KOH → K2ZnO2 + H2
  4. Fe2O3 + 3CO → 2Fe + 3CO2
  5. 3PbO + 2NH3 → 3Pb + 3H2O + N2
  6. 2Pb3O4 → 6PbO + O2
  7. 2PbS + 3O2 → 2PbO + 2SO2
  8. S + 2H2SO4 → 3SO2 + 2H2O
  9. S + 6HNO3 → H2SO4 + 6NO2 + 2H2O
  10. MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
  11. C + 2H2SO4 → CO2 + H2O + SO2
  12. 2KOH + Cl2KCl + KClO + H2O
  13. 2NO2 + H2O → HNO2 + HNO3
  14. Pb3O4 + 8HCl → 3PbCl2 + 4H2O + Cl2
  15. 2H2O + 2Cl2 → 4HCl + O2
  16. 2NaHCO3 → Na2CO3 + H2O + CO2
  17. 2HNO3 + H2S → 2NO2 + 2H2O + S
  18. P + 5HNO3 → 5NO2 + H2O + H3PO4 
  19. Zn + 4HNO3  → Zn(NO3)2 + 2H2O + 2NO2
Season Break Offer!
Get 60% Flat off instantly.
Avail Now
×