Some Basic Concepts of Chemistry Study Module Yakeen NEET 2.0 2025 (Alpha) PDF

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Number of Questions Number of Questions 1 2 3 4 5 6 7 8 9 10 11 12 0 1 2 3 4 5 6 2022 Measurements and Significant Figures 1 0 CHAPTER 2021 2020 Law of Chemical Combinations 1 2020 Covid 2019 Atomic and Molecular Masses 0 2018 Years Topics 2017 Mole Concept and Molar Masses 6 Percentage Composition and Empirical & Molecular Formula 1 2016 I 2016 II 2015 2015 Re Topicwise Number of Questions Analysis (2022-2013) Year Wise Number of Questions Analysis (2022-2013) Stoichiometry & Concentration Terms 9 2014 2013 of Chemistry Some Basic Concepts  Material is another very common term used in chemistry. INTRODUCTION However, the term material has a limited meaning, which corresponds to matter having specific uses. Chemistry is the branch of science that studies the composition, properties and interaction of matter. CLASSIFICATION OF MATTER Anything that has mass and occupies space is called matter. For example, book, pen, pencil, water, air, all living beings Matter can be classified in two ways: etc. are composed of matter. You know that they have mass and (i) Physical classification of matter they occupy space. (ii) Chemical classification of matter Substances Pure Substances Mixtures Elements Compounds Heterogeneous Homogeneous Hydrogen (H) Water (H2O) Carbon (C) Carbon dioxide (CO2) Gas Mixtures Solutions Alloys Air Sugar solution Brass Superheated steam Carbonated soda Bronze Suspension Emulsions Solid Sol Aerosols Iron sludge Milk Iron ore Granite Clouds, smoke Liquid concrete Mayonnaise Physical Classification of Matter Mixtures Depending upon the physical state of matter, it can be classified Mixtures are defined as the substances which are made up of two into three states, namely, solid, liquid and gaseous state. or more pure substances. They can posses variable composition Table: Properties of Solids, Liquids and Gases and can be separated into constituent components by some suitable physical means/methods. Properties Solid Liquid Gas For Example: Alloys (Brass, Bronze) (Brass = Copper + Shape Definite Indefinite Indefinite Zinc) (Bronze = Copper + Tin), Water + alcohol, Water + Salt, Volume Definite Definite Indefinite Water + mustard Oil, Water + Sugar, Water + Kerosene. Attraction Force Strongest Moderate Weakest A mixture may be homogeneous or heterogeneous. Examples Sugar, Iron, Water, Milk, Dihydrogen, In a homogeneous mixture, the components completely mix Gold, Wood oil, Mercury Oxygen, Carbon with each other and its composition is uniform throughout. etc. dioxide, etc The components of such a mixture cannot be seen even under a microscope. Some examples of homogeneous mixtures are air,  heat Solid    heat  Liquid    Gas cool cool gasoline, sea water, brass, coloured glass, Alloys, Water + alcohol, These three states of matter are interconvertible by changing Water + Salt, 22 carat Gold, Water + Sugar, etc. temperature and pressure. In heterogeneous mixtures, the composition is not uniform Chemical Classification of Matter throughout. These consist of two or more parts (called phases) The chemical classification of matter is based upon its composition. which have different compositions. At the macroscopic or bulk level, matter can be classified as For Example: Water + Sand, Water + Mustard oil, Milk, mixture or pure substances. Blood, Air, plastic, smoke, petrol etc. 2 Dropper NEET Module-1 Chemistry PW Pure Substances Some Commonly Used Quantities It consist of single type of particles. 1. Mass and Weight Pure substances can be further classified into elements and Mass of a substance is the amount of matter present in it. The SI compounds. unit of mass is kilogram. 1. Element: An element is the simplest form of a pure substance. It is defined as: Weight is the force exerted on an object by the pull of gravity. The simplest form of a pure substance that can neither be 2. Volume decomposed into nor built from simpler substances by ordinary Volume is the amount of space occupied by an object. So in SI physical or chemical methods. For example Zn, B, Si. system, volume has units cubic meter, m3. 2. Compound: A compound is defined as a pure substance that contains two or more than two elements combined together 3. Density in a fixed proportion by mass and that can be break down into Density of a substance is its amount of mass per unit volume. SI its constituent elements by suitable chemical methods. unit of density is kg/m3 Compounds are further classified into two categories. Density: It is of two type 1. Organic Compound: Obtained from living sources. mass For Example: Oils, fats, derivative of hydrocarbon. ™ Absolute density = volume 2. Inorganic Compound: Obtained from non-living sources. ™ Relative density or specific gravity For Example: HCl, H2O, H2SO4, HClO4, HNO3 etc. density of the substance = density of water at 4°C PROPERTIES OF MATTER We know that density of water at 4ºC = 1 g/ml. AND THEIR MEASUREMENT For Gases Physical and Chemical Properties Molar mass Absolute density (mass/volume) = Physical properties are those properties that can be measured Molar volume or observed without changing the identity or composition of the substance. Example: Colour, Odour. Relative Density or Vapour Density Chemical properties are those in which a chemical change in the Vapour density is defined as the density of the gas with respect to substance takes place. Example: pH, Heat of combustion. that of hydrogen gas at the same temperature and pressure. d gas PM gas /RT Expressing a Physical Quantity Vapour density = = d H2 PM H 2 /RT The value of a physical quantity is always expressed in two parts: (i) Numerical value      (ii) Unit Where P is pressure of gas, M = mol. wt. of gas, R is the gas constant, T is the temperature. The International System of Units (SI Units) M gas M gas The scientists have generally agreed to use the International V.D. = = System of Units abbreviated as SI units. M H2 2 Table: Seven base units of SI system Mgas = 2 V.D. Physical Quantity Symbol for Name of Symbol Relative density can be calculated w.r.t. to other gases also. quantity Unit 4. Temperature Length l Metre m There are three common scales to measure temperature: Mass m Kilogram kg Time t Second s 1. The SI scale or Kelvin scale measured in Kelvin (K) Thermodynamic T Kelvin K 2. Celsius scale measured in degree Celsius (°C). Temperature 3. Fahrenheit scale measured in degrees Fahrenheit (°F) Electric current I Ampere A (i) Conversion of celsius to Fahrenheit is Amount of Substance n Mole mol 9 °F = ( °C ) + 32° Luminous Intensity Iv Candela cd 5 (ii) Conversion of Fahrenheit to celsius Some Important Units 5 ™ 1Å = 10–10 m; 1 fm = 10–15 m 9 °C = [°F − 32°] 1nm = 10–9 m; 1 mm = 10–6 m (iii) Conversion of kelvin to celsius 1pm = 10–12 m; 1 mm = 10–3 m Kelvin temperature (K) = °C + 273.15 P W Some Basic Concepts of Chemistry 3  UNCERTAINTY IN MEASUREMENT Example 2: Classify the following as pure substances or mixtures. Also separate the pure substances into elements Significant Figures: The uncertainty in the experimental or and compounds and divide mixture, into homogeneous and the calculated values is indicated by mentioning the number of significant figures. heterogeneous categories: (i) Graphite (ii) Milk Significant figures are those meaningful digits which are known with certainty. The uncertainty is indicated by writing the (iii) Air (iv) Oxygen certain digits and the last uncertain digit. (v) 22 carat gold (vi) Iodized table salt Rules for Determining the Number of Significant Figures: (vii) Wood (viii) Cloud 1. All non-zero digits are significant. For example, 3.132 has Sol. Element: (i), (iv) four significant figures. Homogeneous Mixture : (iii), (v) 2. Zeros between two non zero digits are significant. For Heterogeneous Mixture : (ii), (vi), (vii), (viii) example, 3.01 has three significant figures. Example 3: How many significant figure are there in each 3. The zeros preceding to the first non-zero number (i.e. to the of the following numbers: left of the first non-zero number) are not significant. Such (i) 1.00 × 106 (ii) 0.00010 (iii) π zeros indicate the position of decimal point. For example, 0.324 has three significant figures. Sol. (i) Three  (ii) Two  (iii) An infinite number 4. All zeros at the end or to the right of a number are significant provided they are on the right side of the decimal point. For example, 0.0200 has three significant figures. 5. Exponetial form: N × 10n, Where N shows the significant figure. Concept Application E.g., 1.86 × 104 has three significant figure. 1. How many significant figures should be present in the 6. Rounding off the uncertain digit: answer to the following calculations? (i) If the left most digit to be rounded off is more than 5, the 0.02856 × 298.15 × 0.112 preceding number is increased by one. 0.5785 E.g., 2.16 is rounded to 2.2 (1) 4 (2) 3 (ii) If the left most digit to be rounded off is less than 5, the (3) 2 (4) 1 preceding number is retained. E.g., 2.14 is rounded off to 2.1 (iii) If the left most digit to be rounded off is equal to 5, LAWS OF CHEMICAL COMBINATIONS the preceding number is not changed if it is even and The combination of elements to form compounds is governed by increased by one if it is odd. the following five basic laws. E.g., 3.25 is rounded off to 3.2 2.35 is rounded off to 2.4 Law of Conservation of Mass/Law of Indestructibility of Matter Accuracy and Precision Accuracy is a measure of the difference between the experimental Given by – Lavoisier value or the average value of a set of measurements and the true Tested by – Landolt value. According to law of conservation of mass in all physical & Precision refers to closeness of two or more measurements chemical changes total mass of the system remains constant. of the same quantity that agree with one another. In a physical or chemical change, mass can neither be created nor be destroyed. Train Your Brain i.e. Total mass of the reactant = Total mass of the product. This relationship holds good when reactants are completely converted into products. Example 1: Which of the following mixture(s) are homogeneous? If reactants are not completely consume then the relationship Tap water, Air, Soil, Smoke will be: Sol. Tap water, Air. Total mass of reactant = Total mass of product  + Mass of unreacted reactant 4 Dropper NEET Module-1 Chemistry PW  In CH4, 12 parts by weight of carbon combine with 4 parts Key Note by weight of hydrogen. In H2O, 2 parts by weight of hydrogen P Nuclear reactions are exception to law of conservation combine with 16 parts by weights of oxygen. Thus the weights of mass. In nuclear reaction mass + energy is conserved. of C and O which combine with fixed weight of hydrogen (say 4 parts of weight) are 12 and 32, i.e. they are in the ratio 12 : 32 P According to the modern views, the law of conservation or 3 : 8. of mass is not always valid. The law hold good only in case of such chemical reactions where there is no Now in CO2, 12 parts by weight of carbon combine directly evolution of heat or light. with 32 parts by weight of oxygen, i.e. they combine directly in P During chemical processes, the loss of mass is negligible. the ratio 12 : 32 or 3 : 8 that is the same as the first ratio. But in nuclear reactions, tremendous amount of energy is evolved. Consequently, the change in mass is quite Gay Lussac’s Law of Gaseous Volumes significant. Thus, it is clear that the law of conservation Given by → Gay Lussac: He observed that when gases combine of mass and law of conservation of energy are two ways or are produced in a chemical reaction they do so in a simple of looking at the same law. ratio by volume provided all gases are at same temperature and P Thus, combining the two we get general law known as pressure. law of conservation of mass energy. It states that, Mass Example: 2H2(g) + O2(g) → 2H2O(g) and energy are inter convertible. But the total sum of mass and energy of the system remains constant. 100 ml 50 ml 100 ml 2 volumes 1 volume 2 volumes vol of H2 : vol of O2 : vol of steam Law of Definite Proportions Given by → Joseph Proust: A chemical compound always 2 : 1 : 2 contains same elements combined together in same proportion by Avogadro Law mass. i.e, chemical compound has a fixed composition & it does not depends on the method of its preparation or the source from Avogadro proposed that equal volumes of gases at the same which it has been obtained. temperature and pressure should contain equal number of molecules. Example: Carbon dioxide can be produced by different methods such as burning of carbon, heating lime stone etc. It Example: 22.4 L of every gas at STP (Standard temperature has been observed that each sample of CO2 contains carbon and and Pressure, i.e., T = 273 K, P = 1 atm) contains equal number of oxygen combined in the ratio 3:8 by mass. This means that the molecules, which is equal to 6.022 × 1023. composition of a compound always remain the same irrespective of the method by which it is prepared. Law of Multiple Proportions Train Your Brain Given by → Dalton: According to this law, if two elements can combine to form more than one compound, the masses of one Example 4: 10 g of CaCO3 on heating gives 4.4 g of CO2 element that combine with a fixed mass of the other element, are then determine weight of produced CaO in quintal. in the ratio of smallest whole numbers. (1) 5.6 × 10–5 quintal (2) 2.8 × 10–5 quintal Example: Carbon (C) can combine with oxygen (O) to form (3) 5.6 × 10–8 quintal (4) 2.8 × 10–8 quintal more than one compound, namely CO, CO2. Here ratio of masses Sol. (1) Total mass of reactant = 10 g of O that combine with fixed mass of C is 16:32 or 1:2. Mass of CO2 = 4.4 g Law of Reciprocal Proportion Mass of produced CaO = x Given by → Richter: The ratio of the weights of two elements According to law of conservation of mass A and B that combine separately with fixed weight of the third 10 = 4.4 + x element C is either the same or some simple multiple of this ratio of the weights in which A and B combine directly with each other. 10 – 4.4 = x Example: The elements C and O combine separately with the x = 5.6 g third element H to form CH4 and H2O and they combine directly ∴ 1 quintal = 100 kg with each other to form CO2 as shown in the below figure. ∴ 1 Kg = 1000 g H Kg = 5.6 g × = 5.6 × 10–3 × Kg CH4 1000 H2O 1 = 5.6 × 10–3 × quintal = 5.6 × 10–5 quintal C O 100 CO2 P W Some Basic Concepts of Chemistry 5  Example 5: For the gaseous reaction H2 + Cl2 → 2HCl ATOMIC AND MOLECULAR MASSES If 40 ml of hydrogen completely reacts with chlorine then Atomic Mass Unit find out the required volume of chlorine and volume of 1 produced HCl? It is defined as exactly th of the mass of a carbon-12 atom. It 12 Sol. According to Gay Lussac’s Law: is represented as amu. [Now a new symbol ‘u’ called unified mass H2 + Cl2 → 2HCl is used.] 12 1 1 ml of H2 will react will 1 ml of Cl2 and 2 ml of HCl Mass= of 1 amu × = 1.67 × 10–24 g will be produced. 6.022 × 10 23 12 ∴ 40 ml of H2 will react with 40 ml of Cl2 and 80 ml Today, ‘amu’ has been replaced by ‘u’ which is known as of HCl will produce. unified mass. Required vol. of Cl2 = 40 ml Average Atomic Mass Produced vol. of HCl = 80 ml. When we take into account the existence of the isotopes and their relative abundance (Percent occurrence), the average atomic mass of that element is calculated. Average atomic mass of an element is the sum of the masses Concept Application of its isotopes each multiplied by its natural abundance. Mathematically, average atomic mass of X (Ax) a x + a x +..... + an xn 2. Element ‘X’ forms five stable oxides with oxygen of = 1 1 2 2 100 formula X2O, XO, X2O3, X2O4, X2O5. The formation a1 = atomic mass; x1 % occurrence in nature of these oxides explains: (1) Law of definite proportions Key Note (2) Law of partial pressure P Relative atomic mass is nothing but the number of (3) Law of multiple proportions nucleons present in the atom. (4) Law of reciprocal proportions DALTON’S ATOMIC THEORY Train Your Brain The assumption of Dalton’s Atomic theory are: Example 6: Naturally occuring chlorine is 75% Cl35 which 1. Matter consists of indivisible atoms. has an atomic mass of 35 amu and 25% Cl37 which has a mass of 37 amu. Calculate the average atomic mass of 2. All the atoms of a given element have identical properties chlorine: including identical mass. Atoms of different elements differ (1) 35.5 amu (2) 36.5 amu in mass. (3) 71 amu (4) 72 amu 3. Compounds are formed when atoms of different elements Sol. (1) Average atomic mass combine in a fixed ratio. % of Ι isotope × its atoms mass 4. Chemical reactions involve reorganisation of atoms. These + % of ΙΙ isotope × its atomic mass are neither created nor destroyed in a chemical reaction. = 100 Dalton’s theory could explain the laws of chemical combination. 75 × 35 + 25 × 37 = = 35.5 amu The main failures of Dalton’s atomic theory are: 100 1. It failed to explain how atoms of different elements differ Example 7: Indium (atomic mass = 114.82) has two from each other i.e., did not tell anything about structure of naturally occurring isotopes, the predominant one form has the atom. isotopic mass 114.9041 and abundance of 95.72%. Which of the following isotopic mass is most likely for the other 2. It does not explain how and why atoms of different element isotope? combine with each other to form compound. (1) 112.94 (2) 115.90 3. It failed to explain the nature of forces present between (3) 113.90 (4) 114.90 different atoms in a molecule. Sol. (1) Let atomic weight of other isotope is ‘M’ 4. It fails to explain Gay Lussac’s law of combining volumes. 114.9041× 95.72 + M × 4.28 114.82 = 5. It did not make any difference between ultimate particle of 100 an element that takes part in reaction (atoms) and ultimate M = 112.94 particle that has independent existence (molecules). 6 Dropper NEET Module-1 Chemistry PW  Volume Number of moles = Concept Application Molar volume Molar Mass 3. An element, X has the following isotopic composition, The mass of 1 mol of a substance in grams is called its molar mass. 200X : 90% For Eg: Molar mass of Na2CO3 = 2 × 23 + 12 + 3 × 16 = 106 gm/mol 199X : 8.0% Mass-Mole-Number Relationship 302X : 2.0% Mass The weighted average atomic mass of the naturally- Number of moles = occurring element X is closed to Molar massin g mol−1 (1) 200 amu (2) 201 amu (3) 202 amu (4) 199 amu PERCENTAGE PURITY Percentage purity is the percentage of a pure compound in an Gram Atomic Mass impure sample. When numerical value of atomic mass of an element is expressed mass of pure compound in sample in grams then the value becomes gram atomic mass (GAM) % purity = × 100 total mass of impure sample GAM = Mass of 1 gram atom = Mass of 1 mole atoms = Mass of NA atoms = Mass of 6.022 × 1023 atoms For eg: A 150.0g sample of an iron ore contains 88.2g of pure iron. Molecular Mass   Its % purity: It is the sum of atomic masses of the elements present in a 88.2 molecule. It is obtained by multiplying the atomic mass of each = ×100 = 58.8 % 150.0 element by the number of its atoms and adding them together. Formula Mass In ionic compounds we use formula mass instead of molecular mass. Formula mass of an ionic compound is the sum of the Train Your Brain atomic masses of all atoms in a formula unit of compound. Example 8: Which of the following contains the greatest Key Note number of atoms? (1) 1.0 g of butane (C4H10) Equivalent Mass (E.M.) (2) 1.0 g of nitrogen (N2) Atomic mass (3) 1.0 g of silver (Ag) P E.M. of an element = Valency (4) 1.0 g of water (H2O) Molecular mass P E.M. of an acid = 1 Basicity Sol. (1) No. of atom of (C4H10) = 58 × 14 Na Molecular mass P E.M. of a base = 1 Acidity (2) No. of atom of (N2) = 28 × 2 Na 1 (3) No. of atom of (Ag) = × Na MOLE CONCEPT AND MOLAR MASSES 108 1 ‘Mole’ was introduced as the seventh base quantity for the amount (4) No. of atom of water (H2O) = × 3 Na 18 of substance in SI system. Hence greatest No. of atoms = C4H10 One mole is the amount of a substance that contains as many entities (atoms, molecules or other particles) as there are atoms in Example 9: The number of sodium atoms in 2 moles of exactly 12 gm (or 0.012 kg) of 12C isotope. sodium ferrocyanide (Na4[Fe(CN)6]) is: From mass spectrometer we found that there are 6.023 × 1023 (1) 12 × 1023 (2) 26 × 1023 atoms present in 12 gm of 12C isotope. This number is known as (3) 34 × 1023 (4) 48 × 1023 avogadro constant (NA = 6.023 × 1023). Sol. (4) 1 mole of Na4[Fe(CN)6] contains 4 mole of Na Mole Concept in Gaseous Reaction So, 2 moles contains: Molar volume is related to volume of one mole of gaseous = 8 × NA atoms of sodium substance. The volume occupied by 1 mol of a gaseous substance = 8 × 6.023 × 1023 is called molar volume. 1 mole occupies 22.414 L or 22414 mL at = 48 × 1023 STP i.e., 273 K and 1 atm. P W Some Basic Concepts of Chemistry 7  (b) Molecular formula: It shows the exact number of different Example 10: 5.6 litre of oxygen at STP contains: types of atoms present in a molecule of a compound. (1) 6.02 × 1023 atoms (2) 3.01 × 1023 atoms eg., MF of benzene is C6H6 (3) 1.505 × 1023 atoms (4) 0.7525 × 1023 atoms Sol. (2) 22.4 L of oxygen at STP contains 6.02 × 1023 Determination of Chemical Formula molecules. 5.6 L of oxygen at STP contains 1.505 × (a) Determination of Empirical Formula: 1023 molecules which contains 3.01 × 1023 atoms of Step (I): Determination of percentage of each element oxygen. Step (II): Determination of mole ratio Example 11: The molecular mass of H2SO4 is 98 amu. Step (III): Making it whole number ratio Calculate the number of moles of each element in 294 g of H2SO4. Step (IV) : Simplest whole number ratio Sol. Gram molecular mass of H2SO4 = 98 gm (b) Determination of Molecular Formula 294 MF = (EF) × n; Moles of H2SO4 = = 3 moles 98 Where n is a simple whole number. H2SO4 H S O Molecular formula = n × Empirical formula One molecule 2 atom one atom 4 atom Key Note 1 × NA 2 × NA 1 × NA 4 × NA atoms atoms atoms Molecular weight n= ∴ 1 mole 2 mole 1 mole 4 mole Empirical weight ∴ 3 mole 6 mole 3 mole 12 mole Train Your Brain Concept Application Example 12: Phosgene, a poisonous gas used during World war-I, contains 12.1% C, 16.2% O and 71.7% Cl by mass. 4. Which of the following contains the largest number of What is the empirical formula of phosgene. oxygen atoms? 1.0 g of O atoms, 1.0 g of O2, 1.0 g of ozone O3. (1) COCl2 (2) COCl (1) O2 (3) CHCl3 (4) C2O2Cl4 (2) O3 Sol. (1) (3) O atom (4) All have the same number of oxygen atoms Simple 5. The total number of g-molecules of SO2Cl2 in 13.5 g of % of A.mu Relative Sim- whole sulphuryl chloride is Element Symbol ele- of ele- no. of plest no. (1) 0.1 (2) 0.2 ment ment atoms ratio atomic ratio (3) 0.3 (4) 0.4 12.1 1.01 = 1.01 =1 Carbon C 12.1 12 1 PERCENTAGE COMPOSITION 12 1.01 We know that according to law of definite proportions any sample 16.2 1.01 = 1.01 =1 of a pure compound always possess constant ratio with their Oxygen O 16.2 16 16 1.01 1 combining elements in terms of mass. Mass percentage of an element 71.7 2.02 = 2.02 =2 Mass of that element in the compound Chlorine Cl 71.7 35.5 35.5 1.01 2 × 100 Molar mass of the compound Then empirical formula = COCl2 CHEMICAL FORMULA Example 13: 1.615 g of anhydrous ZnSO4 was left in moist It is of two types: air. After a few days its weight was found to be 2.875 g. (a) Empirical formula: It represent the simplest whole number What is the molecular formula of hydrated salt? ratio of various atoms present in a compound. eg. EF of (At. masses: Zn = 65.5, S = 32, O = 16, H = 1) benzene (C6H6) is CH. 8 Dropper NEET Module-1 Chemistry PW  Chemical Equation and Balanced Chemical Equation Sol. Molecular mass of anhydrous ZnSO4 Chemical Reaction: It is a process in which two or more than two = 65.5 + 32 + 4 × 16 = 161.5 g substances interact with each other where old bonds are broken and So, 1.615 g of anhydrous ZnSO4 combine with water new bonds are formed. = 2.875 – 1.1615 = 1.260 g 1.615 g of anhydrous ZnSO4 combine with water = 1.260 g Chemical equation is a scientific method of representing a 161.5 g of anhydrous ZnSO4 combine with chemical change in terms of symbols and formula of reactants 1.260 and products involved in it. = × 161.5 = 126g 1.615 e.g., Zn + H2SO4 → ZnSO4 + H2 126 No. of moles of water = =7 However, a balanced chemical equation gives us a lot of quantitative 18 information, mainly the molar ratio in which reactants combine Hence, Formula is ZnSO4. 7H2O. and the molar ratio in which products are formed. Features of a Balanced Chemical Equation Concept Application (a) It contains the same number of atoms of each element on both sides of equation. (POAC) (b) It should follow law of charge conservation on either side. 6. Calculate the molecular formula of compound which contains 20% Ca and 80% Br molecular weight of (c) Physical states of all the reagents/reactants should be included compound is 200. in brackets. (1) Ca1/2Br (2) CaBr2 (d) All reagents/reactants should be written in their standard (3) CaBr (4) Ca2Br forms (Molecular, Atomic,Solid etc.) 7. A compound of X and Y has an equal mass of them. If (e) The coefficients give relative molar ratios of each reagent/ their atomic weights are 30 and 20 respectively. The reactant. molecular formula of that compound (it’s mol wt. is 120) could be: Balanced Chemical Equation (1) X2Y2 (2) X3Y3 (3) X2Y3 (4) X3Y2 Volume of gaseous Mass of reactant FORMULAS reactant ÷m Mass of 1 atom of that element ola L Relative Atomic Mass = rm Mole of each ÷2 2.4 Mass of 1 atom of C − 12 isotope ass reactants Atomic Mass = Relative atomic mass × 1 amu Σin=1 atomic mass of isotope × % abundance Average atomic mass = Mole/Coefficient ratio 100 for each reactants If a Molecular mass = sum of atomic masses of all atoms present in a molecule e ll ra re sam tio are Mass of element io a = × 100 rat not Mass % of Element If all sam e Molecular mass Molecular formula = Empirical Formula × n Stoichiometric Non-Stoichiometric STOICHIOMETRY AND STOICHIOMETRIC Stoichiometry deals with the calculation of masses and CALCULATIONS sometimes volumes of the reactants and products in a reaction. The coefficients of reactants and products in a balanced chemical Stoichiometric calculations can be done equation is called the stoichiometric coefficients. by using following methods Steps: 1. Write the balanced chemical equation. 2. See the number of moles of various reactants that take part in the reaction and products formed. Mass-Mass Mass-volume Mole-Mole Mole-Volume analysis analysis analysis analysis 3. Calculate the number of moles or amount of substance formed. P W Some Basic Concepts of Chemistry 9 Interpretation of balanced chemical equations: We can write, Once we get a balanced chemical equation then we can interpret a Moles of A reacted moles of B reacted chemical equation by following ways: = a b ™ Mass - mass analysis moles of C formed moles of D formed = = ™ Mass - volume analysis c d ™ Mole - mole analysis Key Note Number of moles (n) P In fact mass-mass and mass-vol analysis are also interpreted in terms of mole-mole analysis. You can see in the following Mu chart also. M yM ltip Di ly yb M vid by M ipl e N by by lt Mu A e N vid A Di Mass Divide by NA/MM Number of in molecules grams (W) Multiply by NA/MM (N) MM: Molecular mass NA: Avogadro's No. Fig.: Relation between Mass, No. of moles and No. of molecules Mass-mass Analysis In the following reaction  According to stoichiometry  2KClO3 → 2KCl+ 3O 2   Mass − mass ratio : 2×122.5 2×74.5 3×32  of the reaction  Mass of KClO3 2 ×122.5 Mass of KClO3 2 ×122.5 Fig.: Relation of mole in terms of mass, number and volume or = , = Mass of KCl 2 × 74.5 Mass of O 2 3 × 32 Mass–Volume Analysis Considering decomposition of KClO3 Train Your Brain 2KClO3 → 2KCl + 3O2 Example 14: 367.5 gram KClO3 (M = 122.5) when heated Mass–Volume ratio: 2 × 122.5 g : 2 × 74.5 g : 3 × 22.4 litre at how many gram of KCl and oxygen is produced. STP We can use two relation for volume of oxygen Sol. Balance chemical equation for heating of KClO3 is Mass of KClO3 2 ×122.5 ...(i) = 2KClO3 → 2KCl + 3O2 volume of O 2 at STP 3 × 22.4 lt Mass-mass ratio: 2 × 122.5 g : 2 × 74.5 g : 3 × 32 g Mass of KCl 2 × 74.5 ...(ii) = mass of KClO3 2 ×122.5 367.5 122.5 volume of O 2 at STP 3 × 22.4 lt = ⇒ = mass of KCl 2 × 74.5 W 74.5 Mole-Mole Analysis W = 3 × 74.5 = 223.5 g This is very much important for quantitative analysis point of Mass of KClO3 2 ×122.5 367.5 2 ×122.5 view. Consider again the decomposition of KClO3. = ⇒ = Mass of O 2 3 × 32 W 3 × 32 2KClO3 → 2KCl+ 3O2 W = 144 g In very first step of mole-mole analysis, you should read the Example 15: 367.5 g KClO3 (M = 122.5) when heated, balanced chemical equation like 2 moles KClO3 on decomposition how many litre of oxygen gas is produced at STP gives us 2 moles KCl and 3 moles O2 and from the stoichiometry of reaction we can write Sol. mass of KClO3 2 ×122.5 367.5 2 ×122.5 Moles of KClO3 Moles of KCl Moles of O 2 = ⇒ = = = volume of O 2 at STP 3 × 22.4 lt V 3 × 22.4 lt 2 2 3 Now, For any general balanced chemical equation like: V = 3 × 3 × 11.2 ⇒ V = 100.8 lt   a A + b B → c C + d D 10 NEET (XI) Module-1 PW  Concept Application From Step III Mole of HCl Moles of CO 2 produced = 2 1 8. The equation: 4 3 ∴ mole of CO2 produced = = 2 2Al(s) + O2(g) → Al2O3(s) show that 2 2 ∴ volume of CO2 produced at S.T.P. = 2 × 22.4 = 44.8 lt. 3 Example 17: If 0.5 mol of BaCl2 is mixed with 0.2 mole (1) 2 moles of Al reacts with moles of O2 to produce 7 2 of Na3PO4, the maximum amount of Ba3(PO4)2 that can be moles of Al2O3 formed is : 2 Sol. 0.10 mol 3 (2) 2g of Al reacts with g of O2 to produce 1 mole 3 BaCl2 + 2 Na3PO4 → 6 NaCl + Ba3(PO4)2 of Al O 2 2 3 molar ratio 3 2 6 1 3 initial moles 0.5 0.2 0 0 (3) 2g mole of Al reacts with litres of O2 to produce 1 mole of Al2O3 2 Limiting reagent is Na3PO4 hence it would be consumed, and the yield would be decided by it initial 3 moles. (4) 2 moles of Al reacts with moles of O2 to produce 1 mole of Al2O3 2 2 moles of Na3PO4 give 1 mole of Ba3 (PO4)2 , ∴ 0.2 moles of Na3PO4 would give 0.1 mole of of Ba3(PO4)2 Limiting Reagent ™ The reactant that gets consumed during the reaction & limits the amount of product formed is known as the limiting reagent. ™ Limiting reagent is present in least stoichiometric amount and Concept Application therefore, controls amount of product. 9. 0.5 mole of H2SO4 is mixed with 0.2 mole of Ca(OH)2. ™ The remaining or leftout reactant is called the excess reagent. The maximum number of moles of CaSO4 formed is: ™ If we are dealing with balanced chemical equation, then (1) 0.2 (2) 0.5 if number of moles of reactants are not in the ratio of stoichiometric coefficient of balanced chemical equation, then (3) 0.4 (4) 1.5 there should be one reactant which should be limiting reactant. How to Find Limiting Reagent CONCENTRATION TERMS Step-I Divide the given moles of reactant by the respective stoichiometric coefficient of that reactant. The concentration of a solution or the amount of substance present Step-II See that for which reactant, this division comes out to in its given volume can be expressed in any of the following ways. be minimum. The reactant having minimum value is 1. Mass percent or weight percent (w/w%) limiting reagent for you. 2. Mole fraction Step-III Now, if we have found limiting reagent, then our focus 3. Molarity should be on limiting reagent to find the amount of the 4. Molality product. 5. Normality Mass Percent Train Your Brain It is obtained by using the following relation: Mass of solute Mass percent = × 100 Example 16: 6 moles of Na2CO3 and 4 moles of HCl are Mass of solution made to react. Find the volume of CO2 gas produced at STP. Mole Fraction The reaction is Na2 CO3 + 2HCl → 2 NaCl + CO2 + H2O It is no. of moles of a certain component to the total no. of moles Sol. From Step I & II of the solution. Na2 CO3 HCl No.of molesof A nA 6 4 = 6= 2 ( division is minimum ) Mole fraction of A = = No.of molesof solution nA + nB 1 2 ∴ HCl is limiting reagent Mole fraction is a pure number. It will remain independent of temperature changes. P W Some Basic Concepts of Chemistry 11 Molarity It is defined as the number of moles of the solute in 1 litre of the solution. It is denoted by M Train Your Brain No.of moles of solute Example 18: 255 g of an aqueous solution contains 5 g of Molarity (M) = Volume of solution in litre urea. What is the concentration of the solution in terms of molality. (Mol. wt. of urea = 60) Molarity is the unit that depends upon temperature. It varies inversely with temperature. Sol. Mass of urea = 5 g Molecular mass of urea = 60 g Mathematically: Molarity decreases as temperature increases. Number of moles of urea = 0.083 1 1 Molarity ∝ ∝ Mass of solvent = (255 – 5) = 250 g temperature volume ∴ Molality of the solution If a particular solution having volume V1 and molarity = M1 is Number of moles of solute diluted to V2 mL then = = ×1000 Mass of solvent in gram M1V1 = M2V2 0.083 = × 1000 = 0.332 m M2 : Resultant molarity 250 Example 19: 0.5 g of a substance is dissolved in 25 g of a If a solution having volume V1 and molarity M1 is mixed with solvent. Calculate the percentage amount of the substance another solution of same solute having volume V2 & molarity M2 in the solution. then M1V1 + M2V2 = MR (V1 + V2) Sol. Mass of substance = 0.5 g M1V1 M 2 V2 Mass of solvent = 25 g MR = Resultant molarity = ∴ Percentage of the substance (w/w) V1 V2 0.5 Molality = × 100 = 1.96 0.5 + 25 It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by m. Concept Application No. of moles of solute Thus, Molality(m) = Mass of solvent in kg Molality is independent of temperature changes. 10. What approximate volume of 0.40 M Ba(OH)2 must be There are other terms also used to express concentration of added to 50.0 mL of 0.30 M NaOH to get a solution in solution which the molarity of the OH– ions is 0.50 M? (1) 33 mL (2) 66 mL Normality (N) (3) 133 mL (4) 100 mL It is the number of gram equivalent of a solute dissolved per litre of the solution. Equivalent Weight (E) No. of gram equivalents of solute Normality ( N ) = Molecular weight Vol. of solution in litres E= n-factor Mass of solute n-Factor Gram equivalent = Equivalent weight Species n-factor Mass of solute in gram Element Valency Normality = Equivalent weight in gram × vol.of soltution in litres Ion Charge on ion Normality equation: N1V1 = N2 V2 Acid Basicity (no. of moles H+ ion released by 1 mole of acid) Formality Base Acidity (no. of moles OH– ion released by 1 mole Wt. of ionic solute of base) Formality = Formula Wt. of solute × Vol. of solution in lit. Salt Total + ve or – ve charge produced by 1 mole of salt 12 Dropper NEET Module-1 Chemistry PW Degree of Hardness of Water Actual yield % yield of product = mass of CaCO3 Theoritical maximum yield Hardness of water = × 106 For eg: During a chemical reaction, 0.5g of product is formed. Total mass of water The max. calculated yield is 1.6g. The percent yield of the reaction Yield of Product will be: In many cases, the actual yield of a product is less than the Actual yield 0.5 = theoretical maximum yield. The percentage yield of the product × 100= ×100 = 31.25 % Theoretical yield 1.6 is thus defined as. Train Your Brain Example 20: What is the degree of hardness of a sample of water containing 48 mg of MgSO4 (molecular mass 120) per kg of water? (1) 10 ppm (2) 20 ppm (3) 30 ppm (4) 40 ppm Sol. (4) Degree of hardness is no. of parts of calcium carbonate or equivalent to various calcium and magnesium salts present in ppm. 48 mg of MgSO4 present in 103g of water (Given) So, 106g water will contain = 48000 mg of MgSO4 = 48 g of MgSO4 1 mole MgSO4 = 1 mole of CaCO3  120 g of MgSO4 = 100 g of CaCO3 48 ×100 ∴ 48 g of MgSO4 = = 40 g of CaCO3 120 ∴ Hardness of water = 40 ppm Example 21: For the reaction, A + 2B → C + 2D The correct statement is (1) Equivalents of A = 2 × Equivalents of B (2) Moles of A reacted = Moles of D formed (3) Equivalents of B = Equivalents of C (4) Moles of B reacted = 2 × Moles of D formed Sol. (3) A + 2B → C + 2D Number of = Number of = Number of = Number of eq. of A   eq. of B   eq. of C   eq. of D Moles of A reacted = 2 × moles of D formed Moles of B reacted = moles of D formed Concept Application 11. Potassium chlorate decomposes upon slight heating in the presence of a catalyst, according to the reaction below: 2KClO3(s) → 2KCl(s) + 3O2(g) In a certain experiment, 40.0g of KClO3 is heated until it completely decomposes and the collected oxygen gas actual yield is found to be 14.9g. What is the theoretical yield of oxygen gas and percent yield of the reaction respectively? (1) 15.7; 94.9 (2) 13.6; 74.5 (3) 12.5; 87.6 (4) 17.4; 68.4 P W Some Basic Concepts of Chemistry 13  Short Notes Various Relations of Moles Important Formulae of Concentration Terms w2 W2 × 1000 1. w/w % = × 100 7. M = w1 + w 2 M 2 × V(ml) w 2. w/v % = w2 × 100 % × d × 10 Vsol 8. M = w Molar mass of solute v2 W2 ×1000 3. v/v % = × 100 =9. m Moles of the solute = v1 + v 2 mass of solvent(kg) M 2 × W1 ( g ) w2 4. ppm = × 106 x 2 × 1000 w1 + w 2 10. m = x1 × Mw1 ( g ) w2 5. ppb = × 109  1 M (g)  w1 + w 2 = 11. d M + 2   m 1000  n2 n1 = 6. X 2 = ; X1 n1 + n 2 n1 + n 2 12. M = M 1000dX 2 x1M1 + x 2 M 2 ( g ) Conc. Terms Formula Moles of component nA Mole Fraction Mole fraction = x = Total number of moles present in solution A n A + n B + n C..... mass of solute Parts per million ppm of a solute in solution = × 106 mass of solution mass of CaCO3 Degree of hardness of water Hardness of water = × 106 Total mass of water Mass of solute Mass Percentage %w/w = × 100 Mass of solution 14 Dropper NEET Module-1 Chemistry PW  Mass of solute = Mass by volume Percentage %w/v × 100 Volume of solution Volume of solute ( cm3 ) Volume by volume percentage %v/v = × 100 Volume of solution ( cm3 ) Moles of solute Molarity Molarity, (M) = Volume of solution (L) Moles of solute Molality Molality, (m) = Mass of solvent (Kg) Some Important Relations 10 × %w/v M= M solute 10 × %w/w × d M= M solute %w/v = % w/w × d X solute 1000 m= × X solvent M solvent 1000 × M m= 1000d − M × M solute Where, M = Molarity, m = Molality d = Density M1/Msolvent = Molar mass of solvent M2/Msolute = Molar mass of solute Xsolute = Mole fraction of solute Xsolvent = Mole fraction of solvent P W Some Basic Concepts of Chemistry 15  Solved Examples 1. It takes 2.56 × 10–3 equivalent of KOH to neutralize 0.1254g 5. The chloride of a metal (M) contains 65.5% of chlorine. H2XO4. The number of neutrons in X is 100 ml of the vapour of the chloride of the metal at STP (1) 16 (2) 8 (3) 7 (4) 32 weight 0.72g. the molecular formula of the metal chloride is (1) MCl3 (2) MCl (3) MCl2 (4) MCl4 0.1254 Sol. (1) Molecular mass of chloride of metal = weight of 22,400 ml Sol. (4) Mole= of H 2 XO 4 = [M x Atomic mass of x] M x + 66 0.72 × 22, 400 vapour of metal at STP = = 161.28g 100 'n' factor of H2XO4 = 2 [H2XO4 is dibasic acid] 100g of metal chloride contains = 65.5 g chloride 0.1254 65.5 × 161.28 161.28g metal chloride contains = = 105.6g ∴ 2.56 × 10–3 = ×2 100 M X + 66 Therefore, the number of mole of chlorine atoms per mole Mx= 31.96 g/mol = 32 g/mol of metal chloride = 105.6/35.5 = 3 2. How many grams of sodium bicarbonate are required to Hence the molecular formula of metal chloride is MCl3 neutralize 10.0 ml of 0.902 M vinegar? 6. Gaseous mixture of propane and butane of volume 3 litre on complete combustion produces 11.0 litre CO2 under standard (1) 8.4g (2) 1.5g (3) 0.75g (4) 1.07g conditions of temperature and pressure. The ratio of volume Sol. (3) NaHCO3 + CH 3 COOH  CH 3 COONa + CO 2 + H 2 O of butane to propane is Vinegar (1) 1 : 2 (2) 2 : 1 (3) 3 : 2 (4) 3 : 1 Sol. (2) C3H8 + 5O2 → 3CO2 + 4H2O 10 × 0.902 x litres of propane produce 3x litre of CO2 Equivalent of acid = 1000 C4H10 + 6.5O2 → 4CO2 + 5H2O Equivalent of NaHCO3 = 9.02 × 10–3 (3 – x) litres of butane produce 4(3 – x) lit of CO2 3x + 4(3 – x) = 11 Amount of NaHCO3 = 9.02 ×10−3 × 84 = 0.758 3x + 12 – 4x = 11 3. A sample of hard water contains 244 ppm of HCO3− ions. 12 – x = 11 x = 1 litre What is the minimum mass of CaO required to remove Volume of butane : propane = 2 : 1 HCO3− ions completely from 1 kg of such water sample 7. In an experiment, 50 ml of 0.1 M solution of a salt (1) 56 mg (2) 112 mg (3) 168 mg (4) 244 mg reacted with 25 ml of 0.1 M solution of sodium sulphite. The half equation for the oxidation of sulphite ion: Sol. (2) Mass of HCO3– in 1 kg or 106 mg water = 244 mg 2− SO3(aq) + H 2O  − → SO 24(aq) + + 2H (aq) +2e 244 Millimoles of HCO3– = =4 If the oxidation number of the metal in the salt was 3, what 61 would be the new oxidation number of the metal? Ca(HCO3) + CaO → CaCO3 + H2O + CO2 + 2e– (1) 0 (2) 1 (3) 2 (4) 4 millimoles of CaO = 2 Sol. (3) SO32– get oxidised and its ‘n’ factor is 2 The metal must have been reduced mass of CaO = 56 × 2 = 112 mg Applying the law of equivalence 4. 100 ml of each of 0.5 N NaOH, N/5 HCl and N/10 H2SO4 50 × 0.1 × (3 – n) = 25 × 0.1 × 2 are mixed together. The resulting solution will be 8. The chloride of a metal contains 71% chlorine by weight (1) Acidic (2) Neutral (3) Alkaline (4) None and the vapour density of it is 50. The atomic weight of the metal will be Sol. (3) Meq. of NaOH = 100 × 0.5 = 50 (1) 29 (2) 58 (3) 35.5 (4) 71 1 Sol. (1) Molecular weight of metal chloride = 50 × 2 = 100 Meq. of HCl = ×100 = 20 = 20 Let metal chloride be MCln then 5 29 71 1 Eq. of metal = eq. of chloride, or = Meq. of H2SO4 = × 100 = 10 29 E 35.5 10 \ E= Total meq. of acid = 20 + 10 = 30 2 Now a + 35.5n = 100 Total meq. of NaOH = 50 or n.E + 35.5n = 100 meq. of NaOH left = 50 – 30 = 20 n = 2 Thus, solution will be alkaline. Therefore a = 2 × E = 2 × 29/2 = 29 16 Dropper NEET Module-1 Chemistry PW  Exercise-1 (Topicwise) MEASUREMENT AND SIGNIFICANT FIGURES 10. One part of an element A combines with two parts of another 1. Light travels with a speed of 3 × 108 m/sec. The distance element B, 6 parts of element C combines with 4 parts travelled by light in 1 Femto sec is of (B) If A and C combine together the ratio of their weights, (1) 0.03 mm (2) 0.003 mm will be governed by (3) 3 mm (4) 0.0003 mm (1) law of definite proportion 2. Area of nuclear cross-section is measured in “Barn”. It is (2) law of multiple proportion equal to (3) law of reciprocal proportion (1) 10–20 m2 (2) 10–30 m2 (4) law of conservation of mass (3) 10 m–28 2 (4) 10–14 m2 11. The law of conservation of mass holds good for all of the 3. Two students X and Y report the mass of the same substance following except. as 7.0 g and 7.00 g respectively, which of the following statement is correct? (1) All chemical reactions (2) Nuclear reaction (1) Both are equally accurate (3) Endothermic reactions (4) Exothermic reactions (2) X is more accurate than Y 12. The % of copper and oxygen in samples of CuO obtained (3) Y is more accurate than X by different methods were found to be same. This proves (4) Both are inaccurate scientifically the law of 4. The number of significant figures in value of π are (1) Constant Proportion (2) Reciprocal Proportion (1) 1 (2) 2 (3) 3 (4) ∞ (3) Multiple Proportion (4) Conservation of mass. 5. The correctly reported answer of the addition of 29.4406, 13. Two elements X and Y combine in gaseous state to form 3.2 and 2.25 will have significant figures XY in the ratio 1 : 35.5 by mass. The mass of Y that will be (1) 3 (2) 4 (3) 2 (4) 5 required to react with 2 g of X is 6. If an object has a mass of 0.2876 g, then find the mass of nine (1) 7.1 g (2) 3.55 g (3) 71 g (4) 35.5 g such objects. Report the answer to correct significant figures 14. 4.4 g of an oxide of nitrogen gives 2.24 L of nitrogen and (1) 2.5884 g (2) 2.5886 g 60 g of another oxide of nitrogen gives 22.4 L of nitrogen (3) 2.588 g (4) 2.5 g at S.T.P. The data illustrates LAW OF CHEMICAL COMBINATIONS (1) Law of conservation of mass 7. In Habers process, the volume at S.T.P of ammonia relative (2) Law of constant proportions to the total volume of reactants at STP is (3) Law of multiple proportions (1) One fourth (2) One half (4) Law of reciprocal proportions (3) Same (4) Three fourth 8. 6 g of carbon combines with 32 g of sulphur to form CS2, 15. If law of conservation of mass was to hold true, then 20.8 g 12 g of C also combine with 32 g oxygen to form CO2. 10 of BaCl2 on reaction with 9.8 g of H2SO4 will produce 7.3 g g of sulphur combines with 10 g of oxygen to form Sulphur of HCl and BaSO4 equal to dioxide. Which law is illustrated by this? (1) 11.65 g (2) 23.3 g (3) 25.5 g (4) 30.6 g (1) Law of multiple proportions 16. One of the following combinations which illustrates the law (2) Law of constant composition of reciprocal proportions is (3) Law of reciprocal proportions (1) N2O3, N2O4, N2O5 (2) NaCl, NaBr, NaI (4) Gay Lussac’s law 9. Which of the following data illustrates the law of conservation (3) CS2, CO2, SO2 (4) PH3, P2O3, P2O5 of mass? 17. Hydrogen and oxygen combine to form H2O2 and H2O (1) 56 g of C reacts with 32 g of Oxygen to produce 44 g containing 5.93% and 11.2% hydrogen respectively, the data of CO2 illustrates (2) 1.70 g of AgNO3 reacts with 100 ml of 0.1M HCl to (1) Law of conservation of mass produce 1.435 g of AgCl and 0.63 g of HNO3 (2) Law of Constant proportions (3) 12 g of C is heated in vacuum and on cooling, there is no change in mass (3) Law of reciprocal proportions (4) 36 g of S reacts with 16 g of O2 to produce 48 g of SO2 (4) Law of multiple proportions P W Some Basic Concepts of Chemistry 17 18. Two elements X (of mass 16) and Y (of mass 14) combine to 27. If isotopic distribution of C-12 and C-14 is 98% and form compounds A, B and C. The ratio of different masses 2% respectively, then the number of C-14 atoms in 12 g of of Y which combine with a fixed mass of X in A, B and C is carbon is 1 : 3 : 5, if 32 parts by mass of X combines with 84 parts (1) 1.032 × 1022 (2) 3.01 × 1022 by mass of Y in B, then in C, 16 parts by mass of X will (3) 5.88 × 1023 (4) 6.02 × 1023 combine with; 28. 5.6 L of a gas at S.T.P. weights equal to 8 g. The vapour (1) 14 parts by mass of Y (2) 42 parts by mass of Y density of gas is (3) 70 parts by mass of Y (4) 84 parts by mass of Y (1) 32 (2) 16 (3) 8 (4) 40 ATOMIC AND MOLECULAR MASSES 29. One atom of an element weighs 1.8 × 10–22 g, its atomic mass 19. Insulin contains 3.4% sulphur by mass. What will be the is minimum molecular weight of insulin? (1) 29.9 g (2) 18 g (1) 94.117 u (2) 1884 u (3) 108.36 g (4) 154 g (3) 941 u (4) 976 u 30. If H2SO4 ionises as H2SO4 + 2H2O →2H3O+ + SO42–. Then 20. Avogadro’s number is the number of molecules present in total number of ions produced by 0.1 mol H2SO4 will be (1) 1 g of molecule (2) 1 atom of molecule (1) 9.03 × 1021 (2) 3.01 × 1022 (3) 6.02 × 10 22 (4) 1.8 × 1023 (3) gram molecular mass (4) 1 litre of molecule 21. The number of molecules present in one milli litre of a gas 31. Which of the following will not have a mass of 10 g? at STP is known as (1) 0.1 mol CaCO3. (2) 1.51 × 1023 Ca2+ ions (1) Avogadro number (2) Boltzman number (3) 0.16 mol of CO32- ions (4) 7.525 × 1022 Br atom (3) Loschmidt number (4) Universal gas constant th 32. x L of N2 at S.T.P. contains 3 × 1022 molecules. The number 1 of molecules in x/2 L of ozone at S.T.P. will be 22. If the atomic mass unit ‘u’ were defined to be of the 5 (1) 3 × 1022 (2) 1.5 × 1022 mass of an atom of C-12, what would be the atomic weight 21 (3) 1.5 × 10 (4) 1.5 × 1011 of nitrogen in amu or ‘u’ in this state? Atomic weight of N on conventional scale is 14 33. A person adds 1.71 gram of sugar (C12H22O11) in order to (1) 6.77 u (2) 5.834 u sweeten his tea. The number of carbon atoms added are: (mol mass of sugar = 342) (3) 14 u (4) 23 u (1) 3.6 × 1022 (2) 7.2 × 1021 23. A 100 g sample of Haemoglobin on analysis was found to