XI Chemistry Chapter 01 Solved Numerical, Question Answer & Exercise

XI Chemistry Chapter 01 Solved Numerical, Question Answer & Exercise



Chemistry is the branch of science which deals with the properties, composition and structure of matter. It also deals with the changes in matter and with the principles according to which these changes occur.


The simplest form of matter which cannot be further simplified by any physical or chemical process is called element.

There are about 115 elements. Out of these, 92 are natural and others are artificial. Last natural element is URANIUM (At. No.92). Other examples are H, O, N, C, Na, etc.


Atom is the smallest particle of an element, having all the properties of that element and capable of taking part in a chemical reaction. It may or may not exist independently. Atoms of same elements may differ from each other in atomic mass and physical properties. They are called Isotopes.

Thus “H” stands for one atom of Hydrogen.

“N” stands for one atom of Nitrogen.


An ion is a charged atom, i.e. an ion is formed when an atom loses or gains electrons. They are of two types:

Cations are the positive ion formed by the loss of electron, e.g. Ex, Na + 1 , Mg. Anions are the negative ion formed by the gain of electron, e.g. Ex., Cl- 1 , O- 2 .


1) It is the smallest particle of an element. 1) It is the atom that has gained or lost electron. 2) It may or may not exist independently. 2) It exists independently. 3) It is neutral having equal number of electron 3) It is not neutral. Cation (+ve) has less and proton. electrons than protons. Anions (-ve)

Have more electrons than protons. 4) It does not move under influence 4) They move to anode and cathode of electric field. under the influence of electric field.


It is the smallest particle of a substanc e or compound. It can exist independently and can take part in a chemical reaction.


The smallest particle of a substance or compound is called molecule. It is formed by the combination of two or more, same or different atoms. When a molecule is made up of atoms of same elements, it is called HOMOATOMIC MOLECULE.

The number of atoms in one molecule is called ATOMICITY, e.g. Ex, H2 , O2 , N2 , Cl 2 , O3 (Ozone), etc.

When a molecule is made up of atoms of different elements it is called HETEROATOMIC MOLECULE, e.g. Ex, CO, CO2 , NH3 , H2SO4 , etc.

Formation of Molecule

H+H⮴H2 (One molecule of hydrogen)

2H+O⮴ H2O (one molecule of water)


The representation of a molecule OR a compound in symbols and figures is called FORMULA, e.g. formula of water is H2O. It represents one molecule of water.

Formula Unit

Ionic compounds; such as, Na + 1 , Cl- 1 do not exist as molecules. They exist as aggregate of ions. Na + 1 ,Cl- 1 represents the ratio of two ions and it is called a formula unit.

Atomic Number

Atomic number of an element is the number of protons present in its nucleus. The name was suggested by MOSELEY in 1911. It is denoted by Z.

Atomic Number = No of protons

Atomic Mass

Atomic mass is defined as the mass of one atom of the element as compared to the mass of one atom of C1 2 (The stable isotope of carbon.).

a. m. u (Atomic Mass Unit)

It is the unit of atomic mass or molecular mass.

1 a.m.u = 1.66 x10-24 gm, also 1 amu = 1/ Avogadro’s No.

It also represents mass of one atom or one molecule of a substance.


Molecular Mass of NaOH = 40a.m.u

Thus mass of 1 molecule of NaOH = 40 a.m.u

Molecular Mass

Molecular mass is defined as the mass of one molecule of a substance compared to the mass of one atom of C1 2 (the stable isotope of carbon).

It is equal to the sum of atomic masses of the atoms of all elements present in a molecule as shown by its molecular formula.


Molecular Mass of H2O = 18 a.m.u

Molecular Mass of H2SO4 = 98 a.m.u

Molar Mass

It is defined as the mass of 1 mole of a substance expressed in gram.


Molecular Mass of H2O = 18 a.m.u

Hence, Molar Mass of H2O = 18 gm

It is also defined as mass of 6.02 x 102 3 molecules of a substance expressed in gram.

Molar Volume

Molar volume is defined as the volume occupied by 1mole of a gas at S.T.P or N.T.P.

It is equal to 22.4 dm3.


1 mole of CO2 = 44gms occupies or volume of 22.4dm3 at 00C and 1 atmospheric pressure.

Equivalent Weight (Eq. Wt.)

It is that weight of an element which combines with or gives out one power by weight of H or 8 parts by weight of ‘O’ or 35.5 parts by weight of ‘Cl’.


It is the combining power of an element. It is also defined as equal to number of H atoms with which one atom of the element combines.


In NH3 Valency of ‘N’ is 3.


It is the atom or group of atoms either neutral or charged and capable of taking part in chemical reactions as a single entity, i.e. Li + 1 , Ba + 2 , CO+ 2 , F- 1 , SO- 24, CO- 23 , etc. Following is the list of charged radicals:


Hydrogen H+ Hydride H- Thiocy onide SCN- 1 Lithium Li + 1 Fluoride F- 1 Bisulphide HS- 1 Sodium Na + 1 Chloride Cl - 1 Bisulphite HSO3- 1 Potassium K+ 1 Bro mide Br- 1 Bisulphate HSO4- 1 Silver Ag + 1 Iodide I- 1 Bicarbonate HCO3- Cuprous Cu + 1 Hydroxide OH- 1 Acetate CH3COO- 1

Mercurous Hg + 1 Nitrite NO2-1 Chlorate ClO3- 1 Ammonium NH4+ 1 Nitrate NO3- 1 Permanganate MnO4-1 Cy onide CN- 1


Cupric Cu + 2 Barium Ba + 2 Oxide O– 2 Chro mate CrO- 24 Mercuric Hg + 2 Zinc Zn +2 Sulphide S- 2 Dichro mate Cr 2O- 27 Ferrous Fe +2 Nickle Ni +2 Sulphite SO3- 2

Lead Pb + 2 Manganese Mn + 2 Sulphate SO4- 2

Stannous Sn + 2 Cad mium Cd +2 Carbonate CO3- 2

Calcium Ca + 2 Cobalt Co + 2 Thiosulphate S2O3- 2

Magnesium Mg + 2

Strontium Sr + 2

T R I V A L E N T + v e T R I V A L E N T – v e T ET R A V A L E N T + ve T E T R A V E L N T - v e Ferric Fe + 3 Nitride N- 3 Stannic Sn + 4 Carbide C- 4 Aluminum Al + 3 Plu mbic Pb + 4 Ferrocy anide [Fe(CN) 6 ]-4

Chromium Cr + 3 Phosphate PO4- 3

Bismuth Bi + 3 Ferricy anide [Fe(CN) 6 ]_ 3

Gold (Aurum)Au +3

Formula Mass

Formula mass is defined as the sum of the atomic mass as given in the simplest or empirical formula of the substance.


Simplest formula of Sodium Chloride is Na +Cl - 1

Thus, formula mass of Na +Cl- = 23 + 35.5 = 58.5 a.m.u


A mole is defined as gram atomic mass or gram molecular mass or gram formula mass of a substance which contains as many particles as there are in 12 gms of C12 which contains AVOGADRO’S number of particles 6.02 x 102 3 .


Mole is the atomic weight or molecular weight or formula weight of a substance expressed in grams.

1 mole of H atom = 1gm = 6.02 x 10 23 atoms

1 mole of H2 molecule = 2gms = 6.02 x 1023 molecules

1 mole of CH4 = 16gm = 6.02 x102 3 molecules

1 mole of Na + 1Cl - 1= 58.5 gm = 6.02 x 102 3 formula units

= 6.02 x 102 3 Na + 1 ions and 6.02 x 102 3 Cl - 1 ions.


Mole = Mass in grams

Atomic Mass or Molecular Mass or Formula Mass


Atomic mass of an element expressed in gram is called Gram Atomic Mass or Gram Atom. Now a days, it is called Mole.

A gram atom contains Avogadro’s No. (6.02 x 1023 ) of atoms of the element.


Atmoic Mass of C = 12 amu

12 gm C = 1gm-atom C = 6.02 x1023 atoms of C


Avogadro’s number 6.02 x1023 is defined as number of particles, i.e. atoms or molecules or formula units present in one mole of any substance.

This number 6.02 x 1023 was determined by Avogadro’s and is denoted by NA.


1 mole C (= 12gm) = 6.02 x 1023 atoms of C

1 mole H2O (=18gm) = 6.02 x1023 molecules of H2O

Expression for Calculations

No. of particles = moles x 6.023 x 1023


No. of particles = Mass in gra m x6.02x102 3 Atomic Mass or Molecular Mass or Formula Mass


The departure of a measurement from is accurate value is called ERROR.

Systemic errors is also known as DETERMINATE ERROR. These are those which are caused by

i) Defects in analytical methods

ii) Improper functioning of the instruments

Systemic errors for one experiment may not stands valid for the other.


These are the errors which are caused by some uncertainty in physical measurement due to human judgment. They are UNAVOIDABLE. A random error may be positive or negative. Hence, to minimize it, we take several measurements.


The numbers expressed as power of some base (as 10) are called exponential items or notation.


103, 10- 9 , 105 5 , etc.

The power to the base 10 is called EXPONENT. Thus, in exponential notation

10-6, -6 is called exponent. Any number can also be used as base.

Advantages of Exponential Notation

i) It helps in simplifying many arithmetical calculations.

ii) It minimizes errors in calculations.


Empirical formula represents the simplest ratio of the atoms of each element present in one molecule of a compound.


Formula of Glucose (C6 H1 2O6 ) is CH2O.


Formula CH O represents that one molecule of glucose has C H and O in a ratio of 1: 2 : 1

Empirical formula of two components can be same.


Empirical formula of Benzene is CH

Empirical formula of Acetylene is CH.

It represents covalent as well as ionic compounds. There are many compounds (especially ionic) which have same empirical formula.


NaCl, CaCO3 , MgO (Ionic) CH4 , H2O, NH3 (Covalent)










Molecular formula represents the actual number of atoms of each element present in one molecule of a compound.


Molecular formula of sucrose is C1 2H2 2O1 1 . Molecular formula C6 H1 2 O6 represents that each molecule of glucose is made up of 6 atoms of C, 12atoms of H and 6 atoms of O.


Molecular formula of two different compounds cannot be the same. It represents only covalent compounds.



Significant figures in a number are the digits which are known with certainty and are reliable.

189 has 3 Significant figures.

10000 has 1 Significant figure.

1. All non-zero digits are significant.


3754 has 4 significant figures.

7 has 1 significant figure.

2. Zero between two numbers is significant.


4007 has 4 significant figures.

3.06 has 3 significant figures.

3. Zero before a decimal is non significant provided it is not between two digits.


0.32 has 2 significant figures.

4. Zero after decimal is non significant provided it is not between two digits.


0.007 has 1 significant figure.

5. Zero at the last end of a decimal figure is significant.


3.2400 has 5 sign2.00 has 3 significant figures.
6 Zero at the last end of a non decimal figure is non significant.


35400 has 5 significant figures.


The study of quantitative relationship between reaction and product as conveyed by a chemical equation is called STOICHIOMETRY.


The reaction Zn + 2HCl ⮴ ZnCl2 +H2 conveys that.

1 mole of Zn reacts with 2 moles of HCl to give 1 mole ZnCl2


65 p/m of Zn reacts with 73 p/m HCl to give 136p/m ZnCl2 (p/m = parts of mass)

From this relationship, we can calculate mass of any species from mass of a given species. (Note: p/m = parts by mass)

Types of Relationships In Stoichiometric Calculations

There are three types of relationship involved in the stoichiometric substances: 

1) Mass – Mass relationship

2) Mass – Volume relationship

3) Volume – Volume relationship

Mass –Mass Relationship

In these relationships, we find an unknown mass of reactant and product from a given mass of a substance in a chemical reaction with the help of balanced chemical equation.


C + O2 ⮴ CO2

12p/m 32p/m 44p/m

In the above reaction, 12 parts by mass (p/m) of C combines with 32 p/m of O2 to form 44 p/m of CO2 .

Hence, from this mass relationship, we can calculate mass of any reactant or product provided mass of any one substance is given.

Mass- Volume Relationship

In these relationships, we calculate the volume of a gas from a given mass or vice versa. These calculations are based on Avogadro’s Law. This law gives a quantitative relationship between mass and volume of a gas under standard conditions, i.e. S.T.P or N.T.P.
According to Avogadro’s Law,

“One gram mole of any gas at S.T.P or N.T.P (P =1atmosphere , T=O0C) occupies a volume of 22.4 dm3 (Molar Volume).

Also one ounce – mole (oz-mole) of any gas of STP occupies a volume 22.4 cubic-foot (Cu.ft).



Zn + 2HCl ⮴ ZnCl2 + H2

65p/m 1 mole = 22.4 dm3

The above reaction shows that 65 p/m or gm of Zn reacts with HCl to produce 22.4dm3 of H2 at S.T.P. Thus, we can calculate volume of H2 for any given mass of Zn or HCl.


In such calculations, the volume of a gas is calculated from a given volume of another gas in a chemical reaction. These calculations are based on GAY –LUSSAC’S LAW OF COMBINING VOLUME.

This law gives a quantitative relationship between volumes of gases in a chemical change.

According to GAY _LUSSAIS LAW,

“Gases react in the ratio of small whole numbers of volume under similar conditions of temperature and pressure.” Example

N2 (g) + 3H2 (g) ⮴ 2NH3 (g)

This reaction shows:

i) 1 volume of N2 reacts with 3 Volume of H2 of given 2 Volume of NH3 ii) 1 mole of N2 reacts with 3 mole of H2 of given 2 mole of NH3 iii) 22.4dm3 of N2 reacts with 3(22.4dm3) of H2 of given 2 (22.4dm3) of NH3 at STP.

Thus, we can calculate volume of any gas provided that volume of any gas is given in the reaction.


In “Stiochiometry”, when more than one reactant is involved in a chemical reaction, it is not so simple to get actual result of the stiochiometric problem by making relationship between any one of the reactant and product, which are involved in the chemical reaction. As we know that when any one of the reactant is completely used or consumed, the reaction is stopped, no matter the other reactants are present in very large quantity. This reactant which is totally consumed during the chemical reaction due to which the reaction is stopped is called limiting reactant.Limiting reactant helps us in calculating the actual amount of product formed during a chemical reaction. To understand the concept of limiting reactant, consider the following calculation:

We are provided 50gm of H2 and 50gm of N2 . Calculate how many gm of NH3 will be formed when the reaction is irreversible.

The equation for the reaction is as follows:

N2 + 3H2 2NH3


In this problem moles of N2 and H2 are as follows:

Moles of N2 = Mass of N2 = 50 = 1.79 mol

Mol. Mass of N2 28

Moles of H2 = Mass of H2 = 50 = 25 mol

Mol. Mass of H2 2

So, the provided moles for the reaction are nitrogen = 1.79 moles and Hydrogen = 25moles but in the equation of the process 1 mole of nitrogen requires 3 moles of hydrogen. Therefore, the provided moles of nitrogen, i.e. 1.79 require 1.79 x 3 moles of hydrogen, i.e. 5.37 moles. Although 25 moles of H2 are provided but when nitrogen is consumed, the reaction will be stopped and the remaining hydrogen is useless for the reaction. So in this problem, N2 is a limiting reactant by which we can calculate the actual amount of product formed during the reaction.

N2 + 3H2 ⮴ 2NH3

1 mo le 2 mo le

1 mole of N2 gives 2 mole of NH3

1.79 mole of N2 gives 2 x 1. 79 mole of NH3

3.58 mole of NH3

Mass of NH3 = Moles of NH3 x Mole Mass

= 3.58 x 17

= 60.86 gm of NH3