Title Med-RM_Chem_SP-1_Ch-1_Some Basic Concepts of Chemistry.pdf ✅

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 Chapter Contents Significant Figures Laws of Chemical Combination Average Atomic Mass Mole Concept Concentration Terms Equivalent Concept Percentage Composition Empirical and Molecular Formula Chemical Stoichiometry Matter Mixture Pure substances Heterogeneous Homogeneous Elements Chemical compounds SIGNIFICANT FIGURES It is defined as all the certain digits plus one doubtful digit of that number beginning with the first non-zero digit. All digits are significant except the leading zeros. Zeros to the right of decimal point are significant. In scientific notation a number is written as N × 10x where N is a number with a single non zero digit to the left of the decimal point and x is an integer. Significant numbers in such a case are counted in N only. In addition and subtraction the result should be reported to the same numbers of decimal places as that of the term with the least number of decimal places. In multiplication and division the result should be reported to the same number of significant figures as the least precise term. An exact number (e.g. 3, 5, etc.) in an expression is considered to have an infinite number of significant figures. Example 1 : Express the following numbers to four significant figures. (1) 6.608792 (2) 42.392800 Solution : (1) As the fifth digit 7 is greater than 5, therefore, the result will be expressed as 6.609. (2) It will be expressed as 42.39. The digit 2 is dropped since it is less than 5 (the figure is not rounded off to the next number). Chapter 1 Some Basic Concepts of Chemistry
2 Some Basic Concepts of Chemistry NEET Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 Example 2 : What is the sum of 3.368 kg and 2.02 kg? Solution : 3.368 2.02 5.388  The sum is rounded off to 2 decimal places. Therefore, the sum is 5.39 kg. LAWS OF CHEMICAL COMBINATION Law of Conservation of Mass (Lavoisier 1774) It deals with the mass of reactants and products and states that in a chemical change the total mass of the products is equal to the total mass of the reactants. e.g., C + O CO 2 2  12g + 32g = 44g Example 3 : 10.0 g of CaCO3 on heating gave 4.4 g of CO2 and x g of CaO. Applying law of conservation of mass, calculate the mass of CaO. Solution : According to the law, Mass of reactants = mass of product, here 3 2 x 10 g 4.4 g CaCO CO CaO   Hence, x = 10 g – 4.4 g = 5.6 g which is mass of CaO. Law of Constant Composition (Proust 1799) A chemical compound always contains same elements combined together in same proportion by mass, e.g., H2 O prepared from any source contains H and O in the ratio of 1 : 8 by mass. Example 4 : Copper oxide was prepared by two different methods. In one case, 1.75 g of the metal gave 2.19 g of oxide. In the second case, 1.14 g of the metal gave 1.43 g of the oxide, show that the given data illustrate the law of constant proportions. Solution : In case I, mass of copper = 1.75 g Mass of copper oxide = 2.19 g % of copper in the oxide Mass of copper 100 Mass of copper oxide   1.75 100 79.9% 2.19    % of oxygen = 100 – 79.9 = 20.1% In case II, mass of copper = 1.14 g Mass of copper oxide = 1.43 g % of copper in the oxide 1.14 100 79.7% 1.43   % of oxygen = 100 – 79.7 = 20.3% Thus, copper oxide prepared by any of the given methods contain copper and oxygen in the same proportion by mass (within the experimental error). Hence, it proves the law of constant proportions.
NEET Some Basic Concepts of Chemistry 3 Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 Law of Multiple Proportion (Dalton 1804) When two elements combine to form two or more compounds then the masses of one of which combines with a fixed mass of the other element bears a simple whole number ratio to one another. Example 5 : Hydrogen and oxygen are known to form two or more compounds. The hydrogen content in one of these is 5.93% while in the other it is 11.2%. Show that this data illustrates the law of multiple proportions. Solution : In the first compound, Hydrogen = 5.93% Oxygen = (100 – 5.93)% = 94.07% In the second compound, Hydrogen = 11.2% Oxygen = (100 – 11.2)% = 88.8% In the first compound the number of parts by mass of oxygen that combine with one part by mass of hydrogen 94.07 15.86 parts 5.93   In the second compound the number of part by mass of oxygen that combine with one part by mass of hydrogen 88.8 7.93 parts 11.2   The ratio of masses of oxygen that combine with fixed mass (1 part) by mass of hydrogen is 15.86 : 7.93 or 2 : 1. Since this ratio is a simple whole number ratio, hence the given data illustrates the law of multiple proportions. Gay Lussac’s Law of Combining Volumes It states that at a given temperature and pressure, when the gases combine they do so in volumes which bear a simple ratio to each other and also to the volume of gaseous product e.g., 2 2 (2L) (1L) (1L) H (g) + Cl (g) 2HCl(g)  The ratio of their volumes is 1 : 1 : 2 Avogadro's law Equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules, e.g., 22.4 lit of every gas at STP (standard temperature and pressure) T = 273 K, P = 1 atm contains equal number of molecules, i.e., 6.022 × 1023. EXERCISE 1. Add (0.001 + 0.02) upto correct number of significant figures (1) 0.021 (2) 0.02 (3) 0.003 (4) 0.001 2. The multiple 5 × 0.2 after rounding off will be (1) 1 (2) 1.0 (3) 1.00 (4) 1.000 3. Round off 0.1576 upto one digit after decimal (1) 0.1 (2) 1.6 × 10–1 (3) 0.2 (4) 1.6
4 Some Basic Concepts of Chemistry NEET Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 4. The value of 5.86 3.96 2.86  will be equal to (1) 8 (2) 8.1 (3) 8.11 (4) 8.113 5. The percentage of hydrogen in water and hydrogen peroxide is 11.1 and 5.9 respectively. These figures illustrate (1) Law of multiple proportions (2) Law of conservation of mass (3) Law of constant proportions (4) Law of combining volumes 6. Element X forms five stable oxides with oxygen of formula X2O, XO, X2O3, X2O4, X2O5. The formation of these oxides explains (1) Law of definite proportions (2) Law of partial pressures (3) Law of multiple proportions (4) Law of reciprocal proportions 7. Which of the following represents Avogadro's hypothesis? (1) Gases react together in volumes which bear a simple ratio to one another (2) Equal volumes of all gases under same conditions of temperature and pressure contain equal number of molecules (3) Equal volumes of all gases under same conditions of temperature and pressure contain equal number of atoms (4) The rates of diffusion of gases are inversely proportional to the square root of their densities 8. The law of conservation of mass is valid for all the following, except (1) All chemical reactions (2) Nuclear reactions (3) Endothermic reactions (4) Exothermic reaction 9. Equal volume of different gases at any definite temperature and pressure have (1) Equal atoms (2) Equal masses (3) Equal densities (4) Equal molecules 10. Which of the following pairs of compound illustrate law of multiple proportions? (1) KOH, CsOH (2) H2O, D2O (3) Ethane, benzene (4) KCl, KBr AVERAGE ATOMIC MASS The average mass of various elements are determined by multiplying the atomic mass of each isotope by its fractional abundance and adding the value thus obtained. For example, chlorine contains two types of atoms having relative masses 35 u and 37 u. The relative abundance of these isotopes in nature is in the ratio 3 : 1. Thus atomic mass of chlorine is the average of these different relative masses as described below: Average atomic mass of chlorine (35 3) (37 1) 35.5u 4     Example 6 : Nitrogen occurs in nature in the form of two isotopes with atomic mass 14 and 15 respectively. If average atomic mass of nitrogen is 14.0067, what is the % abundance of the two isotopes? Solution : Let % abundance of N-14 isotope = x Then % abundance of N-15 isotope = (100 – x) The average atomic mass x 14 (100 x)15 100   