'Some basic concepts of chemistry' is the most fundamental chapter of complete chemistry. It gives information about the atomic number and mass number of elements. In any chemical reaction, it is important for us to know about the amount and number of reactants that will consume and the products that will produce, thus for estimating all these calculations, we use the laws of chemical combinations.
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In this section, you will study about the important topics of the chapter, overview, formulae and some important tips and guidelines for the preparation of the chapter at the best.
Nature of Matter
Matter is anything that occupies space and has mass. It exists in three physical states—solid, liquid, and gas and can undergo physical or chemical changes based on its properties. Understanding nature of matter is inevitable to decipher the mechanisms of various chemical reactions.
Laws of Chemical Combination
Dalton's Atomic Theory
Mole Concept
Mole concept is fundamental topic for entire physical chemistry. The mole is a standard scientific unit used to measure the amount of substance, defined as containing 6.022×1023 entities (Avogadro's number), simplifying chemical calculations.
Empirical Formula and Molecular Formula
Stoichiometry and Stoichiometric Calculations
Some Important Concentration Terms
Some basic concepts of chemistry are the most basic chapter of chemistry. It has various important concepts that you need to have greater insights for a better understanding of the whole of the chemistry. This article will help you to give important insights into this chapter and will also guide you in some important tips and guidelines.
Nature of Matter
Everything in the universe that has some mass and occupy some space is known as matter. Matter exists in three different physical forms i.e, solid, liquid and gas.
Property | Solid | Liquid | Gas |
Tightness | Very tightly packed | Tightly packed | Loosely packed |
Intermolecular space | Minimum | Intermediate | Maximum |
Force of attraction | Maximum | Intermediate | Minimum |
Kinetic Energy | Minimum | Intermediate | Maximum |
Density | Maximum | Intermediate | Minimum |
Volume | Fixed | Fixed | Variable |
Shape | Fixed | Variable | Variable |
Compressibility factor | Minimum | Intermediate | Maximum |
At the macroscopic level, the matter can be classified into two categories i.e, mixtures and pure compounds as shown in the figure.
Mixtures are those substances in which two or more components are mixed. Mixtures are further classified as homogeneous and heterogeneous mixtures. Homogeneous mixtures are the one in which components are present in a fixed ratio and the properties of this kind of mixture are the same throughout, for example, solution of sugar in water. But heterogeneous mixtures are those in which the components are not mixed in a definite ratio and properties of the mixture vary at different positions of the mixture, for example, sand in water.
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Laws of Chemical Combination
The combination of elements to form some new product follows three basic laws as shown in the figure
John Dalton proposed a theory about matter and atoms in which he proposed the following postulates:
(i) Matter consists of small individual particles known as 'atoms'.
(ii) All atoms of any particular element have the same properties and same mass but atoms of different elements have different properties and mass.
(iii) Compounds are formed when atoms of different elements are combined in a fixed ratio.
(iv) Chemical reactions involve the reorganization of atoms.
A mole is the standard unit to measure the number of particles like atoms or molecules in a given sample. Mathematically, one mole is equal to 6.022 x 1023. In other words, it can be said that it is equal to as many particles as there are atoms in 12g of the carbon-12 isotope. For example, if we have one mole of oxygen gas, that means we have 6.022 x 1023 molecules of O2.
The empirical formula is the simplest whole-number ratio of atoms present in any particular molecule and the molecular formula is the actual representation of the number of atoms present in the molecule.
For example glucose i.e. C6H12O6, its molecular formula is C6H12O6 but its empirical formula is CH2O.
Mathematically, the relation between empirical formula mass and molecular formula mass is given as follows:
n= Molecular mass empirical formula mass
where n is the simplest ratio.
Limiting Reagent
The reactant is consumed first in the reaction. When we are dealing with the balanced chemical equation, if the number of moles of reactants is not in the ratio of the stoichiometric coefficient of the balanced chemical equation, then there should be one reactant that should be limiting reactant.
% yield
Sometimes, experimentally, the reaction does not undergo 100% completion because of many factors which are involved in the actual industrial processes. So in such cases, we need the concept of % yield.
It is defined as the ratio of actual moles of product(s) formed to the number of moles that should have been theoretically formed assuming 100% completion of the reaction.
% yield = Actual number of moles formed Theoretical moles that should have formed
This concept helps us to calculate the mass or amount of reactants and products in the given chemical reaction. For calculations, first, we must have a balanced chemical equation, only then can we predict the mass of reactants and products. For example
The chemical reaction is given as follows:2H2+O2→2H2O
Now, this chemical equation is a balanced equation, thus we are able to predict that one-mole oxygen will combine with 2 moles of hydrogen and form 2 moles of water, and thus accordingly, we can calculate the masses of the respective elements.
In this concept, we also study the "limiting reagent". It is the substance that is present in reactants in a smaller amount. In other words, a limiting reagent is a substance that will be completely finished or react in the given chemical reaction.
Related topics
Prescribed Books for the States of Matter
First, you must finish the class XI NCERT textbook and solve each and every example and unsolved question given in it. Then for advanced level preparation like JEE and NEET, you must follow R.C. Mukherjee and O.P. Tandon. You must definitely solve the previous year papers. Meanwhile, in the preparation, you must continuously write the mock tests for the depth of knowledge. Our platform will help you to provide with the variety of questions for deeper knowledge with the help of videos, articles and mock tests.
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Example 1:
A sample of KCl is placed in 50 ml of solvent. What should be the mass (in gm) of the sample for the molarity to be 2M ?
1) (correct) 7.45
2) 7.81
3) 6.81
4) 7
Solution
Number of Moles = molarity x volume
= 2 x 0.05 = 0.1
So, mass = (39 + 35.5) x 0.01g = 7.45 g
Hence, the answer is an option (1).
Example 2:
The amount (in g) of sugar (C12H22O11) required to prepare 2L of its 0.1 M aqueous solution is:
1) 17.1
2) (correct) 68.4
3) 136.8
4) 34.2
Solution
Molarity -Molarity (M) = (Number of moles of solute)/(volume of solution in litres)
It is defined as the number of moles of the solute in 1 litre of the solution.
As we have learned in the mole concept.
The formula of molarity =(n)solute Vsolution ( in lit )
0.1=wt3422
wt(C12H22O11) = 68.4 gram
Hence, the answer is the option (2).
Example 3:
Calculate the molality of a solution containing of Acetic acid in Ethanol if the mass of solute = 10g and the density of Ethanol = 0.789 gmL-1.
1) 0.2112
2) 0.2012
3) 0.1992
4) 0.2002
Solution
We know.
Molality (m) = (number of moles of solute)/(mass of solvent in kg)
Now,
Moles of solute Acetic acid = mass / molar mass
=1060 moles
If Volume of ethanol = 1 L
Formula, Density = mass/volume
So,
Weight of ethanol = 1000 x 0.789 = 789g = 0.789 Kg
Molality = mass of solute mass of solution in kg Molality =1060×7891000=0.2112 m/kg
Hence, the answer is (0.2112 m/kg).
Basic concepts of chemistry tells about various important concepts like the Mole concept and stoichiometry which play an important role in industry-based setups to find the required amount of reagents. Various chemical reactions and their limiting agents can be identified using these concepts. Students are suggested to read the prescribed books above in this article to get a good command on the entire subject.
Number of moles = Weight / Molecular weight
One mole of substance is equal to 6.023 * 1023 units of that substance.
Unit of Molality is Moles per Kilogram.
Molarity, Molality and Normality are the 3 important terms related to concentration.
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