The Language of Chemistry
Solutions for Chemistry, Class 9, ICSE
Exercise 1A
30 questionsAnswer:
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.
Match the following : (Refer common names in the beginning of the book)
Compound | Formula |
---|---|
(a) Boric acid | (i) NaOH |
(b) Phosphoric acid | (ii) SiO2 |
(c) Nitrous acid | (iii) Na2CO3 |
(d) Nitric acid | (iv) KOH |
(e) Sulphurous acid | (v) CaCO3 |
(f) Sulphuric acid | (vi) NaHCO3 |
(g) Hydrochloric acid | (vii) H2S |
(h) Silica (sand) | (viii) H2O |
(i) Caustic soda (sodium hydroxide) | (ix) PH3 |
(j) Caustic potash (potassium hydroxide) | (x) CH4 |
(k) Washing soda (sodium carbonate) | (xi) NH3 |
(l) Baking soda (sodium bicarbonate) | (xii) HCl |
(m) Lime stone (calcium carbonate) | (xiii) H2SO3 |
(n) Water | (xiv) HNO3 |
(o) Hydrogen sulphide | (xv) HNO2 |
(p) Ammonia | (xvi) H3BO3 |
(q) Phosphine | (xvii) H3PO4 |
(r) Methane | (xviii) H2SO4 |
Answer:
Compound | Formula |
---|---|
(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 |
Answer:
The basic and acidic radicals in the compounds are given in the table below:
S. No. | Compound | Basic radicals | Acidic radicals |
---|---|---|---|
a | MgSO4 | Mg2+ | SO42- |
b | (NH4)2SO4 | NH4+ | SO42- |
c | Al2(SO4)3 | Al3+ | SO42- |
d | ZnCO3 | Zn2+ | CO32- |
e | Mg(OH)2 | Mg2+ | OH- |
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
Answer:
(a) Chemical formula of Aluminium Sulphate:
∴ Chemical formula of Aluminium Sulphate is Al2(SO4)3
(b) Chemical formula of Ammonium Sulphate:
∴ Chemical formula of Ammonium Sulphate is (NH4)2SO4
(c) Chemical formula of Zinc Sulphate:
∴ Chemical formula of Zinc Sulphate is ZnSO4
Answer:
Given, formula of sulphate is M2(SO4)3.
∴ Valency of M is 3.
(a) Formula of Chloride:
∴ Formula of chloride of element M is MCl3
(b) Formula of Oxide:
∴ Formula of oxide of element M is M2O3
(c) Formula of Phosphate:
∴ Formula of phosphate of element M is MPO4
(d) Formula of Acetate:
∴ Formula of acetate of element M is M(CH3COO)3
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:
- It represents the name of the element.
- It represents one atom of the element.
- It represents a definite mass of the element (equal to atomic mass expressed in grams).
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.
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.
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].
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:
- both the molecule and the molecular mass of the compound.
- the respective numbers of different atoms present in one molecule of the compound.
- the ratios of the respective masses of the elements present in the compound.
For example, the formula CO2 means that:
- the molecular formula of carbon dioxide is CO2.
- each molecule contains one carbon atom joined by chemical bonds with two oxygen atoms.
- the molecular mass of carbon dioxide is 44, given that the atomic mass of carbon is 12 and that of oxygen is 16.
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.
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
Answer:
S. No. | Compound | Basic radical | Acidic radical | Formula |
---|---|---|---|---|
a | Barium sulphate | Ba2+ | SO42- | BaSO4 |
b | Bismuth nitrate | Bi3+ | NO3- | Bi(NO3)3 |
c | Calcium bromide | Ca2+ | Br- | CaBr2 |
d | Ferrous sulphide | Fe2+ | S2- | FeS |
e | Chromium sulphate | Cr3+ | SO42- | Cr2(SO4)3 |
f | Calcium silicate | Ca2+ | SiO32- | CaSiO3 |
g | Stannic oxide | Sn4+ | O2- | SnO2 |
h | Sodium zincate | Na1+ | ZnO22- | Na2ZnO2 |
i | Magnesium phosphate | Mg2+ | PO43- | Mg3(PO4)2 |
j | Sodium thiosulphate | Na1+ | S2O32- | Na2S2O3 |
k | Stannic phosphate | Sn4+ | PO43- | Sn3(PO4)4 |
l | Nickel bisulphate | Ni2+ | HSO4- | Ni(HSO4)2 |
m | Potassium manganate | K1+ | MnO42- | K2MnO4 |
n | Potassium ferrocyanide | K1+ | Fe(CN)64- | K4[Fe(CN)6] |
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).
Answer:
(a) Chemical formula of Oxide of A:
∴ Chemical formula of Oxide of A is AO2
(b) Chemical formula of Nitrate of A:
∴ Chemical formula of Nitrate of A is A(NO3)4
(c) Chemical formula of Phosphate of A:
∴ Chemical formula of Phosphate of A is A3(PO4)4
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)
Answer:
Aluminium is trivalent cation with a valency of 3 → Al3+
Sulphate is divalent anion with a valency of 2 → SO42-
Formula:
Hence, the formula is : Al2(SO4)3
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
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.
Acid | Formula | Salt | Formula |
---|---|---|---|
Perchloric acid | HClO4 | Sodium perchlorate | NaClO4 |
Chloric acid | HClO3 | Sodium chlorate | ............... |
Chlorous acid | HClO2 | Potassium chlorite | ............... |
Hypochlorous acid | HClO | Sodium hypochlorite | ............... |
Answer:
Acid | Formula | Salt | Formula |
---|---|---|---|
Perchloric acid | HClO4 | Sodium perchlorate | NaClO4 |
Chloric acid | HClO3 | Sodium chlorate | NaClO3 |
Chlorous acid | HClO2 | Potassium chlorite | KClO2 |
Hypochlorous acid | HClO | Sodium hypochlorite | NaClO |
Exercise 1B
5 questionsAnswer:
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.
Answer:
The given equation tells us:
- 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.
- 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.
- About chemical composition of respective molecules. For example, one molecule of Zinc chloride contains one atom of Zinc and two atoms of Chlorine.
- 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.
- 65 g of Zinc on treatment with 73 g of HCl, will produce 22.4 litres of hydrogen gas at S.T.P.
- It also proves the law of conservation of mass. According to the above equation, 138 gram of reactants are producing 138 gram of products.
Answer:
The equation of the given reaction does not tell us:
- the time taken for the completion of the reaction.
- whether heat is given out or absorbed during the reaction.
- the respective concentrations of the reactants and the products.
- the rate at which the reaction proceeds.
- whether the reaction is completed or it is not completed.
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.
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↑
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
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 questionsAnswer:
(a) NaCl + AgNO3 ⟶ AgCl + NaNO3
(b) The equation is balanced.
(c) Calculating the weights of reactants and products from the equation:
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.
Answer:
The above balanced equation conveys the following information:
- One molecule of zinc reacts with one molecule of sulphuric acid to produce one molecule of zinc Sulphate and one molecule of hydrogen.
- 65 g of zinc reacts with 98 g of sulphuric acid to produce 161 g of zinc Sulphate and 2 g of hydrogen.
- 65 g of zinc reacts with 98 g of sulphuric acid to produce 22.4 litres of hydrogen gas at S.T.P.
- It also proves the law of conservation of mass. According to the above equation, 163 gram of reactants are producing 163 gram of products.
The above balanced equation conveys the following information:
- One molecule of magnesium reacts with two molecules of hydrochloric acid to produce one molecule of magnesium chloride and one molecule of hydrogen.
- 24 g of magnesium reacts with 73 g of hydrochloric acid to produce 95 g of magnesium chloride and 2 g of hydrogen.
- 24 g of magnesium reacts with 73 g of hydrochloric acid to produce 22.4 litres of hydrogen gas at S.T.P.
- It also proves the law of conservation of mass. According to the above equation, 97 gram of reactants are producing 97 gram of products.
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.
Answer:
The molecular formula of a compound has quantitative significance. It represents:
- both the molecule and the molecular mass of the compound.
- the respective numbers of different atoms present in one molecule of the compound.
- the ratios of the respective masses of the elements present in the compound.
For example, the formula CO2 means that:
- the molecular formula of carbon dioxide is CO2
- each molecule contains one carbon atom joined by chemical bonds with two oxygen atoms;
- the molecular mass of carbon dioxide is 44, given that atomic mass of carbon is 12 and that of oxygen is 16.
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.
Answer:
Metal | Valency | Name of compound formed | Formula |
---|---|---|---|
Iron | 2 3 | Ferrous - [Iron (II)] oxide Ferric - [Iron (III)] oxide | FeO Fe2O3 |
Copper | 1 2 | Cuprous - [Copper (I)] oxide Cupric- [Copper (II)] oxide | Cu2O CuO |
Mercury | 1 2 | Mercurous - [Mercury (I)] oxide Mercuric - [Mercury(II)] oxide | Hg2O HgO |
Lead | 2 4 | Plumbous - [Lead (II)] oxide Plumbic - [Lead (IV)] oxide | PbO PbO2 |
Exercise 1C Multiple Choice Type
24 questionsAnswer:
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.
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.
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.
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.
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 ).
Answer:
Only Q
Reason — The valency of elements in compound magnesium nitride (Mg3N2) is,
Hence, Mg is divalent and N is trivalent.
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.
Only P
Only Q
Only R
Both P and R
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.
The figure given below shows the molecule of an element, where 'a' denotes the atom with atomic mass 32.

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.
Only P
Only Q
Only R
Both P and R
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.
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.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
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.
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.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
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.
Assertion (A): All equations need to be balanced.
Reason (R): An unbalanced equation would imply that atoms have been created or destroyed.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
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).
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
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
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).
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.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
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).
Assertion (A): Magnesium phosphate is written as Mg3(PO4)2.
Reason (R): The valencies of magnesium and phosphate are three and two respectively.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
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 :
∴ Valency of magnesium (Mg) is 2 and Phosphate is 3. Hence, the reason (R) is false.
Assertion (A): A chemical equation is written to represent molecules.
Reason (R): This is necessary for the law of conservation of mass
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
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).
Assertion (A): Chemical combination always takes place between two elements.
Reason (R): A single substance is formed in a chemical combination.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
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 questionsAnswer:
(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
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]
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
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
= = 51.22 g of water of crystallisation.
∴ % mass of water in Epsom salt, MgSO4.7H2O = 51.22%
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 =
= 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 =
= x 100 = 20%
∴ Phosphorus in Calcium phosphate is 20%
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 g of Potassium
= = 31.83 g of Potassium
122.5 g of KClO3 contains 35.5 g of Chlorine
∴ 100 g of KClO3 contains g of Chlorine
= = 28.98 g of Chlorine
122.5 g of KClO3 contains 48 g of Oxygen
∴ 100 g of KClO3 contains g of Oxygen
= = 39.18 g of Oxygen
∴ In KClO3 : K = 31.83%, Cl = 28.98% and O = 39.18%
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 =
= x 100 = 20%
∴ Carbon in Urea is 20%
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 x 100 = 70%
∴ Percentage of Iron in Ferric oxide (Fe2O3 ) is 70%
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 =
= x 100 = 46.67%
Wt. of N in 50 kg CON2H4
= x 50 = 23.33 kg
∴ Amount of Nitrogen in one bag (50 kg) of Urea is 23.33 kg.
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
= x 100
= 43.91 %
∴ Rounding to the nearest whole number, the percentage of platinum in ammonium chloride platinate (NH4)2 PtCl6 is 44%.
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 g of Calcium
= = 38.70 % of Calcium.
310 g of Ca3(PO4)2 contains 62 g of Phosphorus
∴ 100 g of Ca3(PO4)2 contains g of Phosphorus
= = 20 % of Phosphorus.
310 g of Ca3(PO4)2 contains 128 g of Oxygen.
∴ 100 g of Ca3(PO4)2 contains g of Oxygen.
= = 41.3 % of Oxygen.
∴ In Ca3(PO4)2 : Ca = 38.70%, P = 20% and O = 41.3%
Exercise 1C Short Answer Type
7 questionsAnswer:
(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 :
- N2O3 = +3
- N2O5 = +5
- NO2 = +4
- NO = +2
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.
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
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
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
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 questionsAnswer:
(a) Valency of X is +3 and Y is -1. Formula of the compound formed between X and Y is :
∴ 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 :
∴ Chemical formula of the compound formed between X and Z is X2Z3
The following figure represents the structural formula of a chemical compound.

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]
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 g of Carbon
= = 85.7 g of Carbon.
84 g of C6H12 contains 12 g of Hydrogen
∴ 100 g of C6H12 contains g of Hydrogen
= = 14.3 of Hydrogen.
∴ In C6H12, C = 85.7% and H = 14.3%
Exercise 1C Very Short Type
5 questionsFill 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 ............... .
Answer:
(a) Dalton used symbol for oxygen and 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 .
Answer:
The completed table is given below:
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 | ZnSO4 | ZnCO3 | Zn(OH)2 | Zn3(PO4)2 |
Silver | AgCl | AgNO3 | Ag2SO4 | Ag2CO3 | AgOH | Ag3PO4 |
Ammonium | NH4Cl | NH4NO3 | (NH4)2SO4 | (NH4)2CO3 | NH4OH | (NH4)3PO4 |
Calcium | CaCl2 | CaNO3 | 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 |
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
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.
Answer:
- Ratio of C and H atoms in Benzene (C6H6) = 6 : 6
Simplest Ratio = 1 : 1
∴ Empirical formula of Benzene (C6H6) = CH - Ratio of C, H and O atoms in Glucose (C6H12O6) = 6 : 12 : 6
Simplest Ratio = 1 : 2 : 1
∴ Empirical formula of Glucose (C6H12O6) = CH2O - Ratio of C and H atoms in Acetylene (C2H2) = 2 : 2
Simplest Ratio = 1 : 1
∴ Empirical formula of Acetylene (C2H2) = CH - 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