CLASS 11 CHAPTER 1- Some Basic concepts of Chemistry

Introduction

Chemistry deals with the composition, structure, properties and interaction of matter and is of much use to human beings in daily life.

Importance of Chemistry

Chemistry has a great impact on human life since it has a wide range of applications in different fields. These are mentioned below:

(A) In Agriculture and Food:

It has provided chemical fertilizers such as urea, calcium phosphate, carbamide, ammonium ion phosphate etc.
It has helped to guard the crops against insects and harmful bacteria, by the use of an improved variety of insecticides, fungicides and pesticides.
A wide variety of preservatives has helped to preserve food items like jam, butter, squashes etc. for longer periods.

(B) In Health and Sanitation:

Chemistry has provided a large number of life-saving drugs. Now, dysentery and pneumonia are curable because of the discovery of sulpha drugs and penicillin life-saving drugs. Cisplatin and taxol have been found to be very effective for cancer therapy and AZT (Azidothymidine) is used for AIDS patients.
Disinfectants such as phenol are used to kill the micro-organisms present in drains, toilets, floors etc.
A wide variety of disinfectants are available for the disinfection of drinking water.

Chemistry has also contributed to the development and growth of the nation. For example, the production of superconducting ceramics, conducting polymers, optical fibres etc.

Chemistry has led to the establishment of many industries which manufacture utility goods, like acids, alkali, dyes, polymers, drugs, metals, alloy etc.

Matter

Anything that has mass and occupies space is called matter.

Classification of matter

Matter can be classified in two ways:

Physical classification
Chemical classification

Physical Classification:

Matter can exist in three physical states:
1. Solids 2. Liquids 3. Gases

Solids:

In solids, the particles are held very close to each other in an orderly fashion and not much movement is possible.
Characteristics of solids: Solids have definite volume and definite shape.

Liquids:

In liquids, the particles are close to each other but the little movement is possible.
Characteristics of liquids: Liquids have definite volume but no definite shape. They take the shape of their container.

Gases:

In gases, the particles are far apart from each other as compared to those present in solid or liquid states. They are free to move due to the space between them.
Characteristics of Gases: Gases have neither definite volume nor definite shape. They completely occupy the container in which they are placed.

Chemical Classification:

On the basis of composition, matter can be divided into two main types:
1. Pure Substances 2. Mixtures.

Pure substances:

A pure substance may be defined as a single substance whose constituents are chemically the same in nature and cannot be separated by simple physical methods.
Pure substances can be further classified as (i) Elements (ii) CompoundS

Elements:

An element consists of only one type of particle. These particles may be atoms or molecules.
For example, sodium, cobalt, silver, hydrogen, nitrogen etc. They all contain one type of atom. However, atoms of different elements are different in nature.
Some elements such as sodium or copper occur in atomic form whereas some others like hydrogen, nitrogen and chlorine occur in molecular form(diatomic).

Compounds:

It may be defined as a pure substance obtained by a combination of two or more elements in a fixed proportion by weight and can be decomposed into the constituent elements by suitable chemical methods.
The properties of a compound are altogether different from the constituting elements.
The compounds can be classified into two types. These are:
Inorganic Compounds: Earlier it was believed that these are compounds that are obtained from non-living sources like rocks and minerals. But, chemically these are compounds of all the elements except hydrides of carbon and their derivatives.
examples: Common salt, marble, gypsum, washing soda etc.
Organic Compounds Similarly these compounds were believed to be derived from plants and animals or their dead remains. Later it was found that these organic compounds are hydrides of carbon and their derivatives. For example, carbohydrates, proteins, oils, fats etc.

Mixtures

These are composed of two or more pure substances which may be present in any ratio. They can possess variable composition and can be separated into their constituents by some physical methods. For example, milk, seawater, petrol, paint glass, cement etc

Mixtures are classified as follows:

Homogenous mixtures: A homogenous mixture has a uniform composition throughout and there are no visible boundaries of separation between the components. For example, sugar solution and air.
Heterogenous mixture: A heterogeneous mixture has a non-uniform composition throughout and sometimes different constituents are visible. For example, a mixture of sugar and salt, grain and pulses etc.

Properties of matter and their measurement

Physical properties – Properties that can be measured or observed without changing the identity or the composition of the substance.
Examples of physical properties are colour, odour, melting point, boiling point etc.
Chemical Properties: For measurement or observation of chemical properties a chemical change is required to occur. The examples of chemical properties are the typical characteristic reactions of different substances. These include acidity, basicity, combustibility etc.

The International System of Units

The International System of Units was established by the 11th General Conference on Weights and Measures.
The SI system has seven base units that pertain to the seven fundamental scientific quantities.
The other physical quantities, such as speed, volume, density, etc., can be derived from these base quantities.

Base physical quantities and their units

Base physical quantity	Symbol for quantity	Name of SI unit	symbol for SI unit
Length	l	Metre	M
Mass	m	Kilogram	Kg
Time	t	Second	s
Electric current	I	Ampere	A
Thermodynamic temperature	T	Kelvin	K
Amount of substance	n	Mole	Mol
Luminous intensity	I₀	candela	cd

Definitions of SI base units

Quantity	Unit	Definition
Length	Metre	It is defined as the length of the path travelled by light in a vacuum during a time interval of 1/299792458 of a second.
Mass	Kilogram	It is equivalent to the mass of the international prototype of the kilogram.
Time	Second	It is the duration of 9192631, 770 periods of radiation which correspond to the transition between the two hyperfine levels of the ground state of caesium- 133 atoms.
Electric current	Ampere	The ampere is that constant current which if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section and placed, 1 metre apart in vacuum, would produce between these conductors a force equal to 2 x 10-7 N per metre of length.
Thermodynamic temperature	Kelvin	It is the unit of thermodynamic temperature and is equal to 1/273.16 of the thermodynamic temperature of the triple point of water.
Amount of substance	Mole	It is the amount of substance which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon -12. Its symbol is ‘mol’.
Luminous intensity	candela	It may be defined as the luminous intensity in a given direction, from a source which emits monochromatic radiation of frequency 540 x 1012 Hz and that has a radiant intensity in that direction of 1/ 683 watt per steradian.

Prefixes used in the SI system

multiple	prefix	symbol
10^-24	Yocto	y
10^-21	zepto	z
10^-18	Atto	a
10^-15	Famto	f
10^-12	pico	p
10^-9	nano	n
10^-6	micro	µ
10^-3	milli	m
10^-2	centi	c
10^-1	deci	d
10	deca	da
10²	hecto	h
10³	kilo	k
10⁶	mega	M
10⁹	giga	G
10¹²	tera	T
10¹⁵	peta	P
10¹⁸	exa	E
10²¹	zeta	Z
10²⁴	yotta	Y

Mass & Weight

The mass of a substance is defined as the amount of matter present in it.
Weight is the force exerted b gravity on an object.
The mass of a substance is constant, while, its weight may vary from one place to another due to changes in gravity.
Mass of a substance can be accurately determined by using an analytical balance in a laboratory.
The SI unit of mass is Kilogram.

Volume

Volume is the space occupied by a substance. It has units of (length)³. In SI units, volume is expressed in metre³ (m³). However, a common unit of measuring volume, particularly in liquids is the litre (L) but it is not an SI unit.
Mathematically,
1L = 1000 mL = 1000 cm³ = 1dm³.
The volume of liquids can be measured with the help of different calibrated devices like burette, pipette, cylinder, measuring flask etc.

Temperature

There are three scales for measuring temperature – ^oC (degree Celsius), ^oF (degree Fahrenheit) and K (Kelvin). K is the SI unit.
Thermometers with Celsius scale are calibrated from 0° to 100°.
Thermometers with a Celsius scale are calibrated from 32^oto 212^o.

The temperatures on two scales are related to each other by the following relationship:

°F = 9/5 (^oC) + 32

The Kelvin scale is related to the Celsius scale as follows:

K = °C + 273.15

Note: temperature below 0^o i.e., negative values are possible on the Celsius scale but not possible on the Kelvin scale.

Density

The two properties Mass and volume are related as given below:
The density of a substance is defined as its mass per unit volume. SI unit of density can be derived as follows:

Density= Mass / Volume

SI unit of density = SI unit of mass / SI unit of volume

= kg / m³or kgm^-3

Reference Standard

Measuring devices used in chemistry are always calibrated or standardised against some reference.
The mass standard is the kilogram since 1889. It has been defined as the mass of platinum-iridium (Pt-Ir) cylinder that is stored in an airtight jar at

International Bureau of Weights and Measures in Sevres, France.

Pt-Ir was chosen for this standard because it is highly resistant to chemical attack and its mass will not change for an extremely long time.
The metre was originally defined as the length between two marks on a Pt-Ir bar kept at a temperature of 0°C (273.15 K).
In1960 the length of the metre was defined as 1.65076373 × 106 times the wavelength of light emitted by a krypton laser.
Although this was a cumbersome number, it preserved the length of the metre at its agreed value.
The metre was redefined in1983 by CGPM as the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second.

Scientific Notation

Since chemistry deals with the study of atoms and molecules, which have extremely low masses and are present in extremely large numbers, a chemist has to deal with numbers as big as 602, 200,000,000,000,000,000,000 for the molecules of 2 g of hydrogen gas or as small as 0.00000000000000000000000166 g mass of an H atom.
Doing even simple mathematical calculations with such numbers is a real challenge. This problem is solved by using scientific notation for such numbers, i.e., exponential notation is used in which any number can be represented in form N × 10ⁿ, where n is an exponent having positive or negative values and N can vary between 1 to 10.
For example, 0.00016 can be written as 1.6 × 10^–4 using scientific notation.

Significant Figures

Significant figures are meaningful digits which are known with certainty. There are some prescribed rules for determining the number of significant figures. These are given as below:
All non-zero digits are significant. For example, in 425 cm, there are three significant figures and in 0.42 mL, there are two significant figures 4 and 2.
2. Zeros preceding to first non-zero digit are not significant. Such zeros indicate the position of the decimal points.
For example, 0.05 has one significant figure and 0.0036 has two significant figures.
3. Zeros between two non-zero digits are significant. Thus, 4.002 has four significant figures.
4. Zeros at the end or right of a number are significantly provided they are on the right side of the decimal point. For example, 0.600 g has three significant figures.
5. Counting numbers of objects. For example, 2 balls or 20 eggs have infinite significant figures as these are exact numbers and can be represented by writing the infinite number of zeros after placing a decimal.
i.e., 5 = 5.000000
or 60 = 60.000000

Precision & Accuracy

Every experimental measurement has some amount of uncertainty associated with it.
Everyone wants the results to be precise and accurate.
Precision and Accuracy are often referred to whenever we talk about measurement.
Precision refers to the closeness of various measurements for the same quantity. It is independent of the true value.
Accuracy is the agreement of a particular value to the true value of the result. It basically depends on the true value.

Addition and subtraction of significant figures

The result cannot have more digits to the right of the decimal point than either of the original numbers.
For example: – (14.21 + 19.0+ 1.023) = 34.233 As 19.0 have only one digit after the decimal point, therefore, the result will be 34.2, one digit after the decimal point.

Multiplication and Division of Significant Figures

In the multiplication or division, the final result should be reported up to the same number of significant figures as present in the lowest precise number.

Multiplication of Numbers: 2.2120 x 0.011 = 0.024332

As per the rule the final result = 0.024

Division of Numbers: 4.2211÷3.76 = 1.12263

The correct answer = 1.12

Dimensional analysis

During calculations, it is often required to convert units from one system to the other.
The method used for the conversion is called the factor label method or unit factor method or dimensional analysis.

Practice problem:

A jug contains 4 litre of milk. Calculate the volume of the milk in m³.

Solution: since 1 L = 1000 cm³

and 1 m = 100 cm, from this we get

(1 m / 100 cm) = 1 = 100 cm / 1m

To get m3 from the above unit factors, the first unit factor is taken and it is cubed.

(1 m / 100cm)³= (1m³/ 10⁶cm³) = 1³ = 1

Now 4 L = 4×1000 cm³

The above will be multiplied by the unit factor

2 ×1000 cm³ 1 m³/ 10⁶cm³= 2 m³/ 10³= 2 ×10 ^-3

How many seconds are there in 2 days?

Here we know that 1 day = 24 hour or

1 day/24 h = 1 = 24 h /1 day

Than 1 h = 60 min

Or 1 h / 60 min = 1 = 60 min/ 1 h

So, for converting 2 days to seconds, the unit factor can be multiplied as follows

2 day × 24 h/ 1 day × 60 min/ 1 h × 60 s/ 1 min

= 2 × 24 × 60 × 60 s

= 172800 s

Laws of Chemical Combination

Law of Conservation of mass

This law states that matter can neither be created nor be destroyed.
This law was put forward by Antoine Lavoisier in 1789.
He performed carefully planned experiments for reaching this conclusion.

Law of Definite proportion

This law was proposed by a French chemist, Joseph Proust.
This law states that a given compound always contains exactly the same proportion of elements by weight.
He worked on two samples of Cupric carbonate, one was of natural origin and the other was synthetic.
He found that both the samples contained same composition of constituent elements irrespective of the sources.
This law is also known as the Law of Definite Composition.

Law of multiple proportions

This law was given by Dalton in 1803.
According to this law, if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers.
For example, hydrogen combines with oxygen to form two compounds, water and hydrogen peroxide.

Hydrogen + Oxygen → Water

2g 16g 18g

Hydrogen + Oxygen → Hydrogen Peroxide

2g 32g 34g

Here, the masses of oxygen (i.e., 16 g and 32 g), which combine with a fixed mass of hydrogen (2g) bears a simple ratio, i.e., 16:32 or 1: 2.

Gay Lussac’s Law of Gaseous Volumes

Gay Lussac’s law was given by Gay Lussac in 1808.
He observed that when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same temperature and pressure.

For example: – H (Hydrogen) + O (Oxygen) → Water

(100ml) (50ml) (100 ml)

The volumes of hydrogen (H) and oxygen (O) which combine together (i.e.,100mL and 50mL) bear a simple ratio of 2:1.

Avogadro Law

In 1811, Avogadro proposed that equal volumes of gases at the same temperature and pressure should contain an equal number of molecules.
He made a distinction between atoms and molecules.

Dalton’s Atomic Theory

In 1808, Dalton published ‘A New System of Chemical Philosophy’ in which he proposed the following:
1. Matter consists of indivisible atoms.
2. All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.
3. Compounds are formed when atoms of different elements combine in a fixed ratio.
4. Chemical reactions involve the reorganisation of atoms. These are neither created nor destroyed in a chemical reaction.

Atomic mass

Atomic mass refers to the mass of an atom.
Today, sophisticated techniques like mass spectrometry are used to determine the atomic mass of an atom.
But in the 19th-century atomic mass was determined relative to Hydrogen by using stoichiometry. But by doing so the atomic masses of most of the elements came out to be in fractions.
Consequently, Carbon-12 was taken as the reference for the determination of atomic masses.
The isotope Carbon-12 is assigned a mass of exactly 12 atomic mass unit (amu) and masses of all other atoms are given relative to this standard.
One atomic mass unit is the mass exactly equal to one-twelfth of the mass of one atom of carbon- 12.
The atomic mass of an element is the number of times an atom of that element is heavier than an atom of carbon taken as 12.

Average Atomic Mass

Most of the elements exist as more than one isotope which are different atoms of the same element with different mass numbers and the same atomic number.
Therefore, the average atomic mass can be calculated by taking into account the existence of these isotopes and their relative abundance.
For example, carbon has three isotopes and their relative abundance and masses are as follows:

Isotopes	Relative abundance	Atomic mass (amu)
¹²C	98.892	12
¹³C	1.108	13.00335
¹⁴C	2× 10^-10	14.00317

Average atomic mass can be calculated as:

(0.98892) (12 u) + (0.01108) (13.00335 u) + (2 × 10^–12) (14.00317 u) =

= 12.011 u

Note: Nowadays, ‘amu’ has been replaced by ‘u’, which is known as unified mass.

Molecular mass

Molecular mass is the sum of the atomic masses of the elements present in a molecule.
For example, the Molecular mass of methane can be calculated as

CH₄= (12.001 u) + 4(1.008u)

= 16.043 u

Similarly,

Molecular mass of NH₃= atomic mass of N + 3 × atomic mass of H

= 14.0067 u + 3× (1.008u)

= 17.0307 u

Formula mass

Some substances like sodium chloride do not contain discrete molecules as constituent units, instead the positive (sodium ions) and the negative (chloride ions) entities are arranged in a three-dimensional structure as given below:

In sodium chloride, one Na⁺ion is surrounded by six Cl^–ions and vice – versa.
In such cases, the formula mass is calculated in place of the molecular mass.
Thus, the formula mass of sodium chloride is the atomic mass of sodium + atomic mass of chlorine

= 23.0 u + 35.5 u = 58.5 u

Mole concept and molar masses

Atoms and molecules are extremely tiny in size and their numbers even in a very small quantity of a substance is very large. To work with such a large number, a convenient unit is required.
Just as we use one dozen for 12 items, a score for 20 items, similarly we use mole to count entities at the microscopic level.
In SI system, the seventh base quantity was introduced for the amount of substance as
One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope.
One mole contains exactly 6.02214076× 1023 elementary entities.

The mass of a carbon–12 atom was found to be equal to 1.992648 × 10–23 g. Knowing that one mole of carbon weighs 12 g, the number of atoms in it is equal to:

= (12 g / mol ¹²C) / (1.992684 × 10 ^-23 gm / ¹²C atom)

= 6.0221367 × 10²³atoms/mol

This number of entities in 1 mol is given a separate name and symbol. It is known as ‘Avogadro constant’, or Avogadro number denoted by N_A in honour of Amedeo Avogadro.
One mole of any substance contains 602213670000000000000000 entities.
One mole of water molecules = 6.022 × 10²³

One mole of ammonia molecules = 6.022 × 10²³

Molar mass

The mass of one mole of a substance in grams is called its molar mass.
The molar mass in grams is numerically equal to atomic/molecular/

formula mass in u.

Molar mass of water (H₂O) = 18.02g.
Molar mass of sodium chloride (NaCl) = 58.5g.

Percentage composition

The percentage composition of a given compound is defined as the ratio of the amount of each element to the total amount of the individual elements present in the compound multiplied by 100.
For example, let’s take a methane molecule.

Molar mass of Carbon = 12 g

Molar mass of 4 Hydrogen = 4 g

Molar mass of CH₄ = 16 g

% Composition of H = (4 × 1.008 × 100) / 16.04 = 25.13 %

% Composition of C = (1 × 12.0107 × 100) / 16.04 = 74.87%

Empirical Formula for Molecular formula

The Empirical formula of the compound gives the simplest whole-number ratio of the atoms of various elements present in one molecule of the compound.
The molecular formula is the exact number of different types of atoms present in one molecule of a compound.
If the mass per cent of various elements present in a compound is known, its empirical formula can be calculated. The molecular formula can also be determined if the molar mass is known.

Practice problem

A molecule with a molecular weight of 180.18 g/mol is analysed and found to contain 40.00% carbon, 6.72% hydrogen and 53.28% oxygen.

Solution

Step I Conversion of mass percent to grams

The molecule contains 40.00% carbon, 6.72% hydrogen and 53.28% oxygen. This means a 100-gram sample contains:

40 gm of Carbon

6.72 gm of Hydrogen

53.28 gm of Oxygen

Step 2. Convert into number moles of each element

Now, We need to find the number of moles of each element in the 100 gm sample. For this, divide the masses obtained above by respective atomic masses of various elements.

Moles of C = 40/ 12.01 = 3.33

Moles of H = 6.72 / 1.008 = 6.66

Moles of O = 53.28 / 16 = 3.33

Step 3: Find the ratios between the number of moles of each element.

Divide each of the mole values obtained above by the smallest number amongst them.

Since 3.33 is the smallest value, division by it gives a ratio of 1:2:1

In case the ratios are not whole numbers, then they can be converted into a whole number by multiplying by a suitable coefficient.

Step 4: write the empirical formula by writing the numbers alongwith the symbols of the elements.

The empirical formula obtained is CH₂O

Step 5: Find the molecular weight of the empirical formula.

Determine empirical formula mass by adding the atomic masses of various

atoms present in the empirical formula.

For CH₂O the empirical formula mass is

12.01 + (2 × 1.008) + 16 = 30.03 g/mol

Step 5: Find the number of empirical formula units in the molecular formula.

Divide molar mass by the empirical formula mass.

Molar mass / empirical formula mass = 180.08 / 30.03

number of empirical formula units = 6

Step 6: Write the molecular formula

It takes six empirical formula units to make the compound, so multiply each number in the empirical formula by 6.

Molecular formula = 6 × CH₂O = C₆H₁₂O₆

The empirical formula is CH₂O

The molecular formula is C₆H₁₂O₆

Stoichiometry and Stoichiometric Calculations

The word ‘stoichiometry’ is derived from two Greek words — stoicheion (meaning, element) and metron (meaning, measure).
Stoichiometry basically deals with the calculation of masses (sometimes volumes also) of the reactants and the products involved in a chemical reaction.
For this, information available from the balanced chemical equation of a reaction is used.
A chemical reaction occurs whenever a chemical change takes place.
Reactants are substances undergoing a chemical change in the reaction.
Products are the new substances formed as a result of the reaction.
Let’s take the example of the combustion of methane, for which a balanced chemical equation can be written as:

CH₄(g)+ 2O₂(g)→ CO₂(g)+ 2H₂O (g)

Here methane and oxygen are the reactants while carbon dioxide and water are the products.
All the reactants and products are in gaseous form.
The coefficients 2 for O2 and H2O are called stoichiometric coefficients.

Thus, according to the above chemical reaction,

One mole of CH4 (g) reacts with two moles of O2 (g) to give one mole of CO2 (g) and two moles of H2O(g)
One molecule of CH4 (g) reacts with 2 molecules of O2 (g) to give one molecule of CO2 (g) and 2 molecules of H2O(g)
22.7 L of CH4 (g) reacts with 45.4 L of O2 (g) to give 22.7 L of CO2 (g) and 45.4 L of H2O(g)
16 g of CH4 (g) reacts with 2×32 g of O2 (g) to give 44 g of CO2 (g) and 2×18 g of H2O (g).

From these relationships, the given data can be interconverted as follows: Mass Volume = Density

Limiting Reactant

In a chemical reaction, reactant which is present in less quantity gets consumed after some time and no further reaction takes place after that whatever be the amount of the other reactant present. Hence, the reactant which gets consumed limits the amount of product formed and is, therefore, called the limiting reagent. The other reactant that is left is called the excess reactant.

Practice problem: In a reaction: – A + B₂ → AB₂

Identify the limiting reagent, if any, in the following reaction mixtures.

(i) 250 atoms of A + 150 molecules of B

(ii) 3 mol A + 4 mol B

(iii) 300 atoms of A + 300 molecules of B

(iv) 6 mol A + 3 mol B

(v) 2 mol A + 4 mol B

A limiting reagent determines the extent of a reaction. It is the reactant that gets consumed during a reaction, thus, causing the reaction to stop and limiting the number of products formed.

(i) As per the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 150 molecules of B will react with 250 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent.

(ii) As per the above reaction, 1 mole of A reacts with 1 mole of B. Thus, 3 moles of A will react with only 3 moles of B. As a result, 1 mole of B will not be consumed. Hence, A is the limiting reagent.

(iii) As per the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 300 atoms of A will combine with all 300 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.

(iv) 1 mole of atom A combines with 1 mole of molecule B. Thus, 3 moles of B will combine with only 3 moles of A. As a result, 3 moles of A will be left as such. Hence, B is the limiting reagent.

(v) In the above reaction, 1 mole of atom A combines with 1 mole of molecule B. Thus, 2 moles of A will combine with only 2 moles of B and the remaining 2 moles of B will be left unreacted. Hence, A is the limiting reagent.

Reactions in solutions

When the reactions are carried out in solutions, the quantity of substance present in its given volume can be expressed in any of the following ways:

Mass per cent or weight per cent (w/w%)
Mole fraction
Molarity
molality

Mass per cent or weight per cent (w/w%)

The mass percentage is calculated as the mass of a component divided by the total mass of the mixture, multiplied by 100%.

Mass percent = mass of solute/ mass of the solution × 100

Practice Problem: A solution is prepared by adding 3 g of substance A to 20 g of water. Calculate the mass per cent of the solute.

Solution: Mass percent = mass of A/ mass of solution × 100

= 3 / 3 + 20 × 100

= 3 / 23 ×100

= 13.04 %

Mole Fraction

It is the ratio of the number of moles of a particular component to the total number of moles of the solution.
Consider a substance A dissolve in substance B and their number of moles are n_Aand n Then, the mole fractions of A and B will be as follows:

Mole Fraction of A = No. of moles of A / No. of moles of the solution

= n_A/ (n_A+ n_B)

Mole Fraction of B = No. of moles of A / No. of moles of the solution

= n_B/ ( n_A+ n_B)

Molarity

Molarity is defined as the number of moles of solute dissolved in 1 litre of the solution.
It is a widely used unit and is denoted by M.

Molarity = No. of moles of solute / volume of solution in litres

Molality

It is defined as the number of moles of solute present in 1 kg of solvent.
It is denoted by m.

Molality = no. of moles of solute /mass of solvent in kg

Practice Problem:

The density of 3 M solution of NaCl is 1.25 g mL^–1. Calculate the molality of the solution.

Solution:

M = 3 mol L^–1

Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g

Mass of 1L solution = 1000 × 1.25 = 1250 g

(since density = 1.25 g mL^–1)

Mass of water in solution = 1250 –75.5 = 1074.5 g