1.3 SIGNIFICANT FIGURES figures in a measurement. This important remark makes most of the followingAs discussed above, every measurement observations clear:involves errors. Thus, the result of (1) For example, the length 2.308 cm has fourmeasurement should be reported in a way that significant figures. But in different units, theindicates the precision of measurement. same value can be written as 0.02308 m or 23.08Normally, the reported result of measurement is a number that includes all digits in the mm or 23080 µm. number that are known reliably plus the first All these numbers have the same number of digit that is u Reprint 2025-26 4 PHYSICS This shows that the location of decimal point is negative exponent (or power) of 10. In order to of no consequence in determining the number get an approximate idea of the number, we may of significant figures. round off the number a to 1 (for a ≤5) and to 10 The example gives the following rules : (for 5<a ≤10). Then the number can be • All the non-zero digits are significant. expressed approximately as 10b in which the • All the zeros between two non-zero digits exponent (or power) b of 10 is called order of magnitude of the physical quantity. When only are significant, no matter where the an estimate is required, the quantity is of the decimal point is, if at all. order of 10b. For example, the diameter of the • If the number is less than 1, the zero(s) earth (1.28×107m) is of the order of 107m with on the right of decimal point but to the the order of magnitude 7. The diameter of left of the first non-zero digit are not hydrogen atom (1.06 ×10–10m) is of the order of significant. [In 0.00 2308, the underlined 10–10m, with the order of magnitude zeroes are not significant]. –10. Thus, the diameter of the earth is 17 orders • The terminal or trailing zero(s) in a of magnitude larger than the hydrogen atom. It is often customary to write the decimal after number without a decimal point are not the first digit. Now the confusion mentioned in significant. (a) above disappears : [Thus 123 m = 12300 cm = 123000 mm has 4.700 m = 4.700 × 102 cm three significant figures, the trailing zero(s) = 4.700 × 103 mm = 4.700 × 10–3 km being not significant.] However, you can also The power of 10 is irrelevant to the see the next observation. determination of significant figures. However, all • The trailing zero(s) in a number with a zeroes appearing in the base number in the decimal point are significant. scientific notation are significant. Each number [The numbers 3.500 or 0.06900 have four in this case has four significant figures. significant figures each.] Thus, in the scientific notation, no confusion (2) There can be some confusion regarding the arises about the trailing zero(s) in the base trailing zero(s). Suppose a length is reported to number a. They are always significant. be 4.700 m. It is evident that the zeroes here (4) The scientific notation is ideal for reporting are meant to convey the precision of measurement. But if this is not adopted, we use measurement and are, therefore, significant. [If the rules adopted in the preceding example : these were not, it would be superfluous to write • For a number greater than 1, without anythem explicitly, the reported measurement decimal, the trailing zero(s) are notwould have been simply 4.7 m]. Now suppose we change units, then significant. • For a number with a decimal, the trailing4.700 m = 470.0 cm = 4700 mm = 0.004700 km zero(s) are significant. Since the last number has trailing zero(s) in a number with no decimal, we would conclude (5) The digit 0 conventionally put on the left of a erroneously from observation (1) above that the decimal for a number less than 1 (like 0.1250) number has two significant figures, while in is never significant. However, the zeroes at the fact, it has four significant figures and a mere end of such number are significant in a change of units cannot change the number of measurement. significant figures. (6) The multiplying or dividing factors which are (3) To remove such ambiguities in neither rounded numbers nor numbers determining the number of significant representing measured values are exact and figures, the best way is to report every have infinite number of significant digits. For measurement in scientific notation (in the dpower of 10). In this notation, every number is example in r = or s = 2πr, the factor 2 is an expressed as a × 10b, where a is a number 2 between 1 and 10, and b is any positive or exact number and it can be written as 2.0, 2.00 Reprint 2025-26 UNITS AND MEASUREMENT 5 decimal place. The final result should, therefore, t or 2.0000 as required. Similarly, in T = , n is be rounded off to 663.8 g. n Similarly, the difference in length can be an exact number. expressed as : 1.3.1 Rules for Arithmetic Operations with 0.307 m – 0.304 m = 0.003 m = 3 ×10–3 m. Significant Figures Note that we should not use the rule (1) applicable The result of a calculation involving approximate for multiplication and division and write 664 g as measured values of quantities (i.e. values with the result in the example of addition and limited number of significant figures) must 3.00 × 10–3 m in the example of subtraction. They reflect the umeasured values. It cannot be more accurate properly. For addition and subtraction, the rule than the original measured values themselves is in terms of decimal places. on which the result is based. In general, the final result should not have more significant 1.3.2 Rounding off the Ufigures than the original data from which it was The result of computation with approximate obtained. Thus, if mass of an object is measured numbers, which contain more than one to be, say, 4.237 g (four significant figures) and uits volume is measured to be 2.51 cm3, then its for rounding off numbers to the appropriate density, by mere arithmetic division, is significant figures are obvious in most cases. A 1.68804780876 g/cm3 upto 11 decimal places. number 2.746 rounded off to three significant It would be clearly absurd and irrelevant to figures is 2.75, while the number 1.743 would record the calculated value of density to such a be 1.74. The rule by convention is that the precision when the measurements on which the preceding digit is raised by 1 if the value is based, have much less precision. The insignificant digit to be dropped (the following rules for arithmetic operations with underlined digit in this case) is more than significant figures ensure that the final result 5, and is left unchanged if the latter is less of a calculation is shown with the precision that than 5. But what if the number is 2.745 in is consistent with the precision of the input which the insignificant digit is 5. Here, themeasured values : convention is that if the preceding digit is(1) In multiplication or division, the final even, the insignificant digit is simplyresult should retain as many significant dropped and, if it is odd, the preceding digitfigures as are there in the original number with the least significant figures. is raised by 1. Then, the number 2.745 rounded Thus, in the example above, density should off to three significant figures becomes 1.74. On be reported to three significant figures. the other hand, the number 2.735 rounded off to three significant figures becomes 1.74 since 4.237g -3 Density = = 1.69 g cm the preceding digit is odd. 3 2.51 cm In any involved or complex multi-step Similarly, if the speed of light is given as calculation, you should retain, in intermediate 3.00 × 108 m s-1 (three significant figure) and steps, one digit more than the significant digits one year (1y = 365.25 d) has 3.1557 × 107 s (five and round off to proper significant figures at the significant figures), the light year is 9.47 × 1015 m end of the calculation. Similarly, a number (three significant figures). known to be within many significant figures, such as in 2.99792458 × 108 m/s for the speed (2) In addition or subtraction, the final result of light in vacuum, is rounded off to anshould retain as many decimal places as are approximate value 3 × 108 m/s , which is oftenthere in the number with the least employed in computations. Finally, rememberdecimal places. that exact numbers that appear in formulae like For example, the sum of the numbers 436.32 g, 227.2 g and 0.301 g by mere arithmetic L addition, is 663.821 g. But the least precise 2 π in T = 2π , have a large (infinite) number measurement (227.2 g) is correct to only one g Reprint 2025-26 6 PHYSICS of significant figures. The value of π = = 16.2 cm ± 0.6 %. 3.1415926.... is known to a large number of significant figures. You may take the value as Similarly, the breadth b may be written as 3.142 or 3.14 for π, with limited number of b = 10.1 ± 0.1 cm significant figures as required in specific = 10.1 cm ± 1 % cases. Then, the error of the product of two (or more)⊳ Example 1.1 Each side of a cube is experimental values, using the combination of measured to be 7.203 m. What are the errors rule, will be total surface area and the volume of the l b = 163.62 cm2 + 1.6% cube to appropriate significant figures? = 163.62 + 2.6 cm2 Answer The number of significant figures in the measured length is 4. The calculated area This leads us to quote the final result as and the volume should therefore be rounded off l b = 164 + 3 cm2 to 4 significant figures. Here 3 cm2 is the uSurface area of the cube = 6(7.203)2 m2 estimation of area of rectangular sheet. = 311.299254 m2 (2) If a set of experimental data is specified = 311.3 m2 to n significant figures, a result obtained by combining the data will also be valid to n Volume of the cube = (7.203)3 m3 significant figures. = 373.714754 m3 However, if data are subtracted, the number of = 373.7 m3 ⊳ significant figures can be reduced. ⊳ Example 1.2 5.74 g of a substance For example, 12.9 g – 7.06 g, both specified to occupies 1.2 cm3. Express its density by three significant figures, cannot properly be keeping the significant figures in view. evaluated as 5.84 g but only as 5.8 g, as u in a different fashion (smallest number ofmeasured mass whereas there are only 2 decimal places rather than the number of significant figures in the measured volume. significant figures in any of the number added Hence the density should be expressed to only or subtracted). 2 significant figures. (3) The relative error of a value of number 5.74 −3 specified to significant figures depends not Density = g cm 1.2 only on n but also on the number itself. = 4.8 g cm--3 . ⊳ For example, the accuracy in measurement of mass 1.02 g is ± 0.01 g whereas another 1.3.3 Rules for Determining the U in the Results of Arithmetic The relative error in 1.02 g is Calculations = (± 0.01/1.02) × 100 % = ± 1%The rules for determining the u Similarly, the relative error in 9.89 g iserror in the number/measured quantity in = (± 0.01/9.89) × 100 %arithmetic operations can be understood from = ± 0.1 %the following examples. Finally, remember that intermediate results in(1) If the length and breadth of a thin rectangular sheet are measured, using a metre a multi-step computation should be scale as 16.2 cm and, 10.1 cm respectively, there calculated to one more significant figure in are three significant figures in each every measurement than the number of measurement. It means that the length l may digits in the least precise measurement. be written as These should be justified by the data and then l = 16.2 ± 0.1 cm the arithmetic operations may be carried out; Reprint 2025-26 UNITS AND MEASUREMENT 7 otherwise rounding errors can build up. For mass, one dimension in length, and –2 example, the reciprocal of 9.58, calculated (after dimensions in time. The dimensions in all other rounding off) to the same number of significant base quantities are zero. figures (three) is 0.104, but the reciprocal of Note that in this type of representation, the magnitudes are not considered. It is the quality0.104 calculated to three significant figures is of the type of the physical quantity that enters. 9.62. However, if we had written 1/9.58 = 0.1044 Thus, a change in velocity, initial velocity, and then taken the reciprocal to three significant average velocity, final velocity, and speed are figures, we would have retrieved the original all equivalent in this context. Since all these value of 9.58. quantities can be expressed as length/time, This example justifies the idea to retain one their dimensions are [L]/[T] or [L T–1]. more extra digit (than the number of digits in the least precise measurement) in intermediate 1.5 DIMENSIONAL FORMULAE AND steps of the complex multi-step calculations in DIMENSIONAL EQUATIONS order to avoid additional errors in the process The expression which shows how and which of of rounding off the numbers. the base quantities represent the dimensions of a physical quantity is called the dimensional

1.8 Mole concept and Molar Masses molecules Atoms and molecules are extremely small 1 mol of sodium chloride = 6.022 ×1023 formula in size and their numbers in even a small units of sodium chloride amount of any substance is really very large. Having defined the mole, it is easier toTo handle such large numbers, a unit of convenient magnitude is required. know the mass of one mole of a substance Just as we denote one dozen for 12 items, or the constituent entities. The mass of one score for 20 items, gross for 144 items, we mole of a substance in grams is called its use the idea of mole to count entities at the molar mass. The molar mass in grams is microscopic level (i.e., atoms, molecules, numerically equal to atomic/molecular/ particles, electrons, ions, etc). formula mass in u. In SI system, mole (symbol, mol) was Molar mass of water = 18.02 g mol–1introduced as seventh base quantity for the amount of a substance. Molar mass of sodium chloride = 58.5 g mol–1 The mole, symbol mol, is the SI unit of 1.9 Percentage Compositionamount of substance. One mole contains exactly 6.02214076 × 1023 elementary entities. So far, we were dealing with the number of This number is the fixed numerical value of entities present in a given sample. But many the Avogadro constant, NA, when expressed a time, information regarding the percentage in the unit mol–1 and is called the Avogadro of a particular element present in a compound number. The amount of substance, symbol is required. Suppose, an unknown or new n, of a system is a measure of the number of compound is given to you, the first questionspecified elementary entities. An elementary entity may be an atom, a molecule, an ion, an electron, any other particle or specified group of particles. It may be emphasised that the mole of a substance always contains the same number of entities, no matter what the substance may be. In order to determine this number precisely, the mass of a carbon–12 atom was determined by a mass spectrometer and 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 12 C 1 .992648  10 23 g /12 Catom  6 .0221367  1023 atoms/mol Fig. 1.11 One mole of various substances Reprint 2025-26 Some Basic Concepts of Chemistry 19 you would ask is: what is its formula or what 1.9.1 Empirical Formula for Molecular are its constituents and in what ratio are they Formula present in the given compound? For known An empirical formula represents the simplest compounds also, such information provides a whole number ratio of various atoms present check whether the given sample contains the in a compound, whereas, the molecular same percentage of elements as present in a formula shows the exact number of different pure sample. In other words, one can check types of atoms present in a molecule of a the purity of a given sample by analysing this compound. data. If the mass per cent of various elements Let us understand it by taking the example present in a compound is known, its empirical of water (H2O). Since water contains hydrogen formula can be determined. Molecular formula and oxygen, the percentage composition of can further be obtained if the molar mass is both these elements can be calculated as known. The following example illustrates follows: this sequence. Mass % of an element = mass of that element in the compound × 100 molar mass of the compound Problem 1.2 A compound contains 4.07% hydrogen,Molar mass of water = 18.02 g 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are itsMass % of hydrogen = empirical and molecular formulas? = 11.18 Solution 16.00Mass % of oxygen = × 100 Step 1. Conversion of mass per cent 18.02 to grams = 88.79 Since we are having mass per cent, it is Let us take one more example. What is the convenient to use 100 g of the compound percentage of carbon, hydrogen and oxygen as the starting material. Thus, in the in ethanol? 100 g sample of the above compound, 4.07g hydrogen, 24.27g carbon andMolecular formula of ethanol is: C2H5OH 71.65g chlorine are present.Molar mass of ethanol is: Step 2. Convert into number moles of(2×12.01 + 6×1.008 + 16.00) g = 46.068 g each element Mass per cent of carbon Divide the masses obtained above by 24.02g = × 100 = 52.14% respective atomic masses of various 46.068g elements. This gives the number of Mass per cent of hydrogen moles of constituent elements in the compound 6.048g = × 100 = 13.13% 46.068g 4.07 g Moles of hydrogen = = 4.04 1.008gMass per cent of oxygen 16.00g = × 100 = 34.73% 24.27 g Moles of carbon = = 2 .021 46.068g 12 .01 g After understanding the calculation of 71.65g per cent of mass, let us now see what Moles of chlorine = = 2 .021 35 .453 ginformation can be obtained from the per cent composition data. Reprint 2025-26 20 chemistry equation of a given reaction. Let us consider Step 3. Divide each of the mole values the combustion of methane. A balanced obtained above by the smallest number equation for this reaction is as given below: amongst them CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g) Since 2.021 is smallest value, division Here, methane and dioxygen are called by it gives a ratio of 2:1:1 for H:C:Cl. reactants and carbon dioxide and water are In case the ratios are not whole numbers, called products. Note that all the reactants then they may be converted into whole and the products are gases in the above number by multiplying by the suitable reaction and this has been indicated by coefficient. letter (g) in the brackets next to its formula. Step 4. Write down the empirical Similarly, in case of solids and liquids, (s) and formula by mentioning the numbers (l) are written respectively. after writing the symbols of respective The coefficients 2 for O2 and H2O are elements called stoichiometric coefficients. Similarly CH2Cl is, thus, the empirical formula the coefficient for CH4 and CO2 is one in each of the above compound. case. They represent the number of molecules (and moles as well) taking part in the reaction Step 5. Writing molecular formula or formed in the reaction. (a) Determine empirical formula mass by Thus, according to the above chemical adding the atomic masses of various reaction, atoms present in the empirical formula. For CH2Cl, empirical formula mass is • One mole of CH4(g) reacts with two moles 12.01 + (2 ×1.008) + 35.453 of O2(g) to give one mole of CO2(g) and = 49.48 g two moles of H2O(g) (b) Divide Molar mass by empirical • One molecule of CH4(g) reacts with formula mass 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 = 2 = (n) (g) to give 22.7 L of CO2 (g) and 45.4 L of (c) Multiply empirical formula by n H2O(g) obtained above to get the molecular • 16 g of CH4 (g) reacts with 2×32 g of O2 formula (g) to give 44 g of CO2 (g) and 2×18 g of Empirical formula = CH2Cl, n = 2. Hence H2O (g). molecular formula is C2H4Cl2. From these relationships, the given data can be interconverted as follows: 1.10 Stoichiometry and mass Stoichiometric Calculations The word ‘stoichiometry’ is derived from two Greek words — stoicheion (meaning, Mass = Density element) and metron (meaning, measure). Volume Stoichiometry, thus, deals with the calculation of masses (sometimes volumes also) of the 1.10.1 Limiting Reagent reactants and the products involved in a Many a time, reactions are carried out with chemical reaction. Before understanding how the amounts of reactants that are different to calculate the amounts of reactants required than the amounts as required by a balanced or the products produced in a chemical chemical reaction. In such situations, one reaction, let us study what information reactant is in more amount than the amount is available from the balanced chemical required by balanced chemical reaction. The Reprint 2025-26 Some Basic Concepts of Chemistry 21 reactant which is present in the least amount important to understand as how the amount gets consumed after sometime and after that of substance is expressed when it is present in further reaction does not take place whatever the solution. The concentration of a solution be the amount of the other reactant. Hence, or the amount of substance present in its the reactant, which gets consumed first, given volume can be expressed in any of the limits the amount of product formed and is, following ways. therefore, called the limiting reagent. 1. Mass per cent or weight per cent (w/w %) In performing stoichiometric calculations, 2. Mole fraction this aspect is also to be kept in mind. 3. Molarity 1.10.2 Reactions in Solutions 4. Molality A majority of reactions in the laboratories Let us now study each one of them in are carried out in solutions. Therefore, it is detail. Balancing a chemical equation According to the law of conservation of mass, a balanced chemical equation has the same number of atoms of each element on both sides of the equation. Many chemical equations can be balanced by trial and error. Let us take the reactions of a few metals and non-metals with oxygen to give oxides 4 Fe(s) + 3O2(g) → 2Fe2O3(s) (a) balanced equation 2 Mg(s) + O2(g) → 2MgO(s) (b) balanced equation P4(s) + O2 (g) → P4O10(s) (c) unbalanced equation Equations (a) and (b) are balanced, since there are same number of metal and oxygen atoms on each side of the equations. However equation (c) is not balanced. In this equation, phosphorus atoms are balanced but not the oxygen atoms. To balance it, we must place the coefficient 5 on the left of oxygen on the left side of the equation to balance the oxygen atoms appearing on the right side of the equation. P4(s) + 5O2(g) → P4O10(s) balanced equation Now, let us take combustion of propane, C3H8. This equation can be balanced in steps. Step 1 Write down the correct formulas of reactants and products. Here, propane and oxygen are reactants, and carbon dioxide and water are products. C3H8(g) + O2(g) → CO2 (g) + H2O(l) unbalanced equation Step 2 Balance the number of C atoms: Since 3 carbon atoms are in the reactant, therefore, three CO2 molecules are required on the right side. C3H8 (g) + O2 (g) → 3CO2 (g) + H2O (l) Step 3 Balance the number of H atoms: on the left there are 8 hydrogen atoms in the reactants however, each molecule of water has two hydrogen atoms, so four molecules of water will be required for eight hydrogen atoms on the right side. C3H8 (g) +O2 (g) → 3CO2 (g)+4H2O (l) Step 4 Balance the number of O atoms: There are 10 oxygen atoms on the right side (3 × 2 = 6 in CO2 and 4×1= 4 in water). Therefore, five O2 molecules are needed to supply the required 10 CO2 and 4×1= 4 in water). Therefore, five O2 molecules are needed to supply the required 10 oxygen atoms. C3H8 (g) +5O2 (g) → 3CO2 (g) + 4H2O (l) Step 5 Verify that the number of atoms of each element is balanced in the final equation. The equation shows three carbon atoms, eight hydrogen atoms, and 10 oxygen atoms on each side. All equations that have correct formulas for all reactants and products can be balanced. Always remember that subscripts in formulas of reactants and products cannot be changed to balance an equation. Reprint 2025-26 22 chemistry Problem 1.3 the limiting reagent in the production of NH3 in this situation. Calculate the amount of water (g) produced by the combustion of 16 g Solution of methane. A balanced equation for the above Solution reaction is written as follows : The balanced equation for the combustion of methane is : CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g) Calculation of moles : (i) 16 g of CH4 corresponds to one mole. Number of moles of N2 (ii) From the above equation, 1 mol of 1000 g N 2 1 mol N 2 × 2 × CH4 (g) gives 2 mol of H2O (g). = 50 .0 kg N 1 kg N 2 28 .0 g N 2 2 mol of water (H2O) = 2×(2+16) = 17.86×102 mol = 2×18 = 36 g Number of moles of H2 H 2 O ⇒18g 1000 g H 2 1 mol H 2 = 1 1 mol H2O = 18 g H2O × = 10 .00 kg H 2 × 1mol H 2 O 1 kg H 2 2 .016 g H 2 18g H 2 O = 4.96 × 103 mol Hence, 2 mol H2O× 1mol H 2 O According to the above equation, 1 mol N2 (g) requires 3 mol H2 (g), for the = 2×18 g H2O = 36 g H2O reaction. Hence, for 17.86×102 mol of Problem 1.4 N2, the moles of H2 (g) required would be How many moles of methane are 2 3 mol H 2 g 2  required to produce 22g CO2 (g) after 17 .86  10 mol N 1 mol N 2 g combustion? = 5.36 × 103 mol H2 Solution But we have only 4.96×103 mol H2. According to the chemical equation, Hence, dihydrogen is the limiting CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g) reagent in this case. So, NH3(g) would be formed only from that amount of 44g CO2 (g) is obtained from 16 g CH4 (g). available dihydrogen i.e., 4.96 × 103 mol [∴1 mol CO2(g) is obtained from 1 mol Since 3 mol H2(g) gives 2 mol NH3(g) of CH4(g)] Number of moles of CO2 (g) 2 mol NH 3 g 2 g 4.96 ×103 mol H2 (g) × 3 mol H 2 g = 22 g CO2 (g) ×1 molCO 44 gCO 2 g = 3.30 × 103 mol NH3 (g) = 0.5 mol CO2 (g) 3.30 × 103 mol NH3 (g) is obtained. Hence, 0.5 mol CO2 (g) would be obtained from 0.5 mol CH4 (g) or 0.5 mol If they are to be converted to grams, it of CH4 (g) would be required to produce is done as follows : 22 g CO2 (g). 1 mol NH3 (g) = 17.0 g NH3 (g) Problem 1.5 .0 g NH 3 g 50.0 kg of N2 (g) and 10.0 kg of H2 (g) 3.30 ×103 mol NH3 (g) ×17 3 g are mixed to produce NH3 (g). Calculate 1 mol NH the amount of NH3 (g) formed. Identify Reprint 2025-26 Some Basic Concepts of Chemistry 23 3. Molarity = 3.30×103×17 g NH3 (g) It is the most widely used unit and is denoted = 56.1×103 g NH3 by M. It is defined as the number of moles of = 56.1 kg NH3 the solute in 1 litre of the solution. Thus, No. of moles of solute 1. Mass per cent Molarity (M) = Volume of solution in litres It is obtained by using the following relation: Suppose, we have 1 M solution of a substance, say NaOH, and we want to prepare a 0.2 M solution from it. 1 M NaOH means 1 mol of NaOH present in 1 litre of the solution. For 0.2 M solution, Problem 1.6 we require 0.2 moles of NaOH dissolved in 1 litre solution. A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate Hence, for making 0.2M solution from 1M the mass per cent of the solute. solution, we have to take that volume of 1M NaOH solution, which contains 0.2 mol of NaOH Solution and dilute the solution with water to 1 litre. Now, how much volume of concentrated (1M) NaOH solution be taken, which contains 0.2 moles of NaOH can be calculated as follows: If 1 mol is present in 1L or 1000 mL solution then, 0.2 mol is present in 1000 mL × 0 .2 mol solution 1 mol 2. Mole Fraction = 200 mL solutionIt is the ratio of number of moles of a particular component to the total number Thus, 200 mL of 1M NaOH are taken and enough water is added to dilute it to make it 1 litre.of moles of the solution. If a substance ‘A’ dissolves in substance ‘B’ and their number In fact for such calculations, a general of moles are nA and nB, respectively, then the formula, M1×V1 = M2 × V2 where M and V are molarity and volume, respectively, can be used.mole fractions of A and B are given as: In this case, M1 is equal to 0.2M; V1 = 1000 mL and, M2 = 1.0M; V2 is to be calculated. Substituting the values in the formula: 0.2 M × 1000 mL = 1.0 M × V2 Note that the number of moles of solute (NaOH) was 0.2 in 200 mL and it has remained the same, i.e., 0.2 even after dilution ( in 1000 mL) as we have changed just the amount of solvent (i.e., water) and have not done anything with respect to NaOH. But keep in mind the concentration. Reprint 2025-26 24 chemistry Problem 1.7 Problem 1.8 Calculate the molarity of NaOH in the The density of 3 M solution of NaCl is solution prepared by dissolving its 4 g 1.25 g mL–1. Calculate the molality of in enough water to form 250 mL of the the solution. solution. Solution Solution M = 3 mol L–1 Since molarity (M) 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 No. of moles of solute Molality = Mass of solvent in kg 3 mol = = 2.79 m Note that molarity of a solution depends 1 .0745 kg upon temperature because volume of a Often in a chemistry laboratory, a solution is temperature dependent. solution of a desired concentration is prepared by diluting a solution of known 4. Molality higher concentration. The solution of higher concentration is also known as It is defined as the number of moles of solute stock solution. Note that the molality present in 1 kg of solvent. It is denoted by m. of a solution does not change with temperature since mass remains No. of moles of solute Thus, Molality (m) = unaffected with temperature. Mass of solvent in kg Summary Chemistry, as we understand it today is not a very old discipline. People in ancient India, already had the knowledge of many scientific phenomenon much before the advent of modern science. They applied the knowledge in various walks of life. The study of chemistry is very important as its domain encompasses every sphere of life. Chemists study the properties and structure of substances and the changes undergone by them. All substances contain matter, which can exist in three states – solid, liquid or gas. The constituent particles are held in different ways in these states of matter and they exhibit their characteristic properties. Matter can also be classified into elements, compounds or mixtures. An element contains particles of only one type, which may be atoms or molecules. The compounds are formed where atoms of two or more elements combine in a fixed ratio to each other. Mixtures occur widely and many of the substances present around us are mixtures. When the properties of a substance are studied, measurement is inherent. The quantification of properties requires a system of measurement and units in which the quantities are to be expressed. Many systems of measurement exist, of which the English Reprint 2025-26 Some Basic Concepts of Chemistry 25 and the Metric Systems are widely used. The scientific community, however, has agreed to have a uniform and common system throughout the world, which is abbreviated as SI units (International System of Units). Since measurements involve recording of data, which are always associated with a certain amount of u is very important. The measurements of quantities in chemistry are spread over a wide range of 10–31 to 10+23. Hence, a convenient system of expressing the numbers in scientific notation is used. The u figures, in which the observations are reported. The dimensional analysis helps to express the measured quantities in different systems of units. Hence, it is possible to interconvert the results from one system of units to another. The combination of different atoms is governed by basic laws of chemical combination — these being the Law of Conservation of Mass, Law of Definite Proportions, Law of Multiple Proportions, Gay Lussac’s Law of Gaseous Volumes and Avogadro Law. All these laws led to the Dalton’s atomic theory, which states that atoms are building blocks of matter. The atomic mass of an element is expressed relative to 12C isotope of carbon, which has an exact value of 12u. Usually, the atomic mass used for an element is the average atomic mass obtained by taking into account the natural abundance of different isotopes of that element. The molecular mass of a molecule is obtained by taking sum of the atomic masses of different atoms present in a molecule. The molecular formula can be calculated by determining the mass per cent of different elements present in a compound and its molecular mass. The number of atoms, molecules or any other particles present in a given system are expressed in the terms of Avogadro constant (6.022 × 1023). This is known as 1 mol of the respective particles or entities. Chemical reactions represent the chemical changes undergone by different elements and compounds. A balanced chemical equation provides a lot of information. The coefficients indicate the molar ratios and the respective number of particles taking part in a particular reaction. The quantitative study of the reactants required or the products formed is called stoichiometry. Using stoichiometric calculations, the amount of one or more reactant(s) required to produce a particular amount of product can be determined and vice-versa. The amount of substance present in a given volume of a solution is expressed in number of ways, e.g., mass per cent, mole fraction, molarity and molality. exerciseS 1.1 Calculate the molar mass of the following: (i) H2O (ii) CO2 (iii) CH4 1.2 Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4). 1.3 Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass. 1.4 Calculate the amount of carbon dioxide that could be produced when (i) 1 mole of carbon is burnt in air. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. (iii) 2 moles of carbon are burnt in 16 g of dioxygen. 1.5 Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1. Reprint 2025-26 26 chemistry 1.6 Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69%. 1.7 How much copper can be obtained from 100 g of copper sulphate (CuSO4)? 1.8 Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively. 1.9 Calculate the atomic mass (average) of chlorine using the following data: % Natural Abundance Molar Mass 35Cl 75.77 34.9689 37Cl 24.23 36.9659 1.10 In three moles of ethane (C2H6), calculate the following: (i) Number of moles of carbon atoms. (ii) Number of moles of hydrogen atoms. (iii) Number of molecules of ethane. 1.11 What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2L? 1.12 If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution? 1.13 Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below: 1Pa = 1N m–2 If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal. 1.14 What is the SI unit of mass? How is it defined? 1.15 Match the following prefixes with their multiples: Prefixes Multiples (i) micro 106 (ii) deca 109 (iii) mega 10–6 (iv) giga 10–15 (v) femto 10 1.16 What do you mean by significant figures? 1.17 A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass). (i) Express this in per cent by mass. (ii) Determine the molality of chloroform in the water sample. 1.18 Express the following in the scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012 1.19 How many significant figures are present in the following? (i) 0.0025 (ii) 208 (iii) 5005 Reprint 2025-26 Some Basic Concepts of Chemistry 27 (iv) 126,000 (v) 500.0 (vi) 2.0034 1.20 Round up the following upto three significant figures: (i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808 1.21 The following data are obtained when dinitrogen and dioxygen react together to form different compounds: Mass of dinitrogen Mass of dioxygen (i) 14 g 16 g (ii) 14 g 32 g (iii) 28 g 32 g (iv) 28 g 80 g (a) Which law of chemical combination is obeyed by the above experimental data? Give its statement. (b) Fill in the blanks in the following conversions: (i) 1 km = ...................... mm = ...................... pm (ii) 1 mg = ...................... kg = ...................... ng (iii) 1 mL = ...................... L = ...................... dm3 1.22 If the speed of light is 3.0 × 108 m s–1, calculate the distance covered by light in 2.00 ns. 1.23 In a reaction A + B2  AB2 Identify the limiting reagent, if any, in the following reaction mixtures. (i) 300 atoms of A + 200 molecules of B (ii) 2 mol A + 3 mol B (iii) 100 atoms of A + 100 molecules of B (iv) 5 mol A + 2.5 mol B (v) 2.5 mol A + 5 mol B 1.24 Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N2 (g) + H2 (g)  2NH3 (g) (i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen. (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass? 1.25 How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different? 1.26 If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced? 1.27 Convert the following into basic units: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg Reprint 2025-26 28 chemistry 1.28 Which one of the following will have the largest number of atoms? (i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g of Cl2(g) 1.29 Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one). 1.30 What will be the mass of one 12C atom in g? 1.31 How many significant figures should be present in the answer of the following calculations? 0.02856 × 298.15 × 0.112 (i) (ii) 5 × 5.364 0 .5785 (iii) 0.0125 + 0.7864 + 0.0215 1.32 Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes: Isotope Isotopic molar mass Abundance 36Ar 35.96755 g mol–1 0.337% 38Ar 37.96272 g mol–1 0.063% 40Ar 39.9624 g mol–1 99.600% 1.33 Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He. 1.34 A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula. 1.35 Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + CO2(g) + H2O(l) What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl? 1.36 Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction 4 HCl (aq) + MnO2(s) → 2H2O (l) + MnCl2(aq) + Cl2 (g) How many grams of HCl react with 5.0 g of manganese dioxide? Reprint 2025-26 Unit 2 structure of atom The rich diversity of chemical behaviour of different Objectives elementsstructure canof atomsbe tracedof theseto theelements.differences in the internal After studying this unit you will be able to • know about the discovery of The existence of atoms has been proposed since the time electron, proton and neutron and of early Indian and Greek philosophers (400 B.C.) who their characteristics; were of the view that atoms are the fundamental building • describe Thomson, Rutherford blocks of matter. According to them, the continued and Bohr atomic models; subdivisions of matter would ultimately yield atoms which would not be further divisible. The word ‘atom’ • understand the important features has been derived from the Greek word ‘a-tomio’ which of the quantum mechanical model means ‘uncut-able’ or ‘non-divisible’. These earlier ideas of atom; were mere speculations and there was no way to test • u n d e r s t a n d n a t u r e o f them experimentally. These ideas remained dormant for electromagnetic radiation and a very long time and were revived again by scientists in Planck’s quantum theory; the nineteenth century. • explain the photoelectric effect The atomic theory of matter was first proposed and describe features of atomic on a firm scientific basis by John Dalton, a British spectra; school teacher in 1808. His theory, called Dalton’s • state the de Broglie relation and atomic theory, regarded the atom as the ultimate Heisenberg u able to explain the law of conservation of mass, law of • define an atomic orbital in terms constant composition and law of multiple proportion of quantum numbers; very successfully. However, it failed to explain the results • state aufbau principle, Pauli of many experiments, for example, it was known that exclusion principle and Hund’s substances like glass or ebonite when rubbed with silk rule of maximum multiplicity; and or fur get electrically charged. • write the electronic configurations In this unit we start with the experimental observations of atoms. made by scientists towards the end of nineteenth and beginning of twentieth century. These established that atoms are made of sub-atomic particles, i.e., electrons, protons and neutrons — a concept very different from that of Dalton. Reprint 2025-26 30 chemistry

1.1 Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.