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Friday, 16 October 2015

SOME BASIC CONCEPTS OF CHEMISTRY notes




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 Chemistry is the branch of science that deals with the composition, structure and properties of matter.
       Chemistry is called the science of atoms and molecules.
IMPORTANCE OF CHEMISTRY-- Chemistry plays a central role in science and is often intertwined with other branches of science like physics, biology, geology etc. Chemistry also plays an important role in daily life.
1.Chemical principles are important in diverse areas, such as: weather patterns, functioning of brain and operation of a computer.
2. Chemical industries manufacturing fertilizers, alkalis, acids, salts, dyes, polymers, drugs, soaps, detergents, metals, alloys and other inorganic and organic chemicals, including new materials, contribute in a big way to the national economy.
3.Chemistry plays an important role in meeting human needs for food, health care products and other materials aimed at improving the quality of life. This is exemplified by the large scale production of a variety of fertilizers, improved varieties of pesticides and insecticides. Similarly many life saving drugs such as cisplatin and taxol, are effective in
cancer therapy and AZT (Azidothymidine) used for helping AIDS victims, have been isolated from plant and animal sources or prepared by synthetic methods.
4. With a better understanding of chemical principles it has now become possible to design and synthesize new materials having specific magnetic, electric and optical properties. This has lead to the production of superconducting ceramics, conducting polymers, optical fibres and large scale miniaturization of solid state devices.
5. In recent years chemistry has tackled with a fair degree of success some of the pressing aspects of environmental degradation. Safer alternatives to environmentally hazardous refrigerants like CFCs (chlorofluorocarbons), responsible for ozone depletion in the stratosphere, have been successfully synthesised.
       Understanding of bio-chemical processes, use of enzymes for large-scale production of

 chemicals and synthesis of new exotic materials are some of the intellectual challenges for the future generation of chemists. A developing country like India needs talented and creative chemists for accepting such challenges.
 NATURE OF MATTER
   Anything which has mass and occupies space is called matter.
example, book, pen, pencil, water, air, all living beings etc.
Matter can exist in three physical states viz. solid, liquid and gas.
 The properties of these three physical states can be understood with the help of following table---
  SOME BASIC CONCEPTS OF CHEMISTRY  notes for class 11,

  SOME BASIC CONCEPTS OF CHEMISTRY  notes


    A mixture contains two or more substances present in it (in any ratio) which are called its components. Many of the substances present around you are mixtures. For example, sugar solution in water, air, tea etc.,
  A mixture may be classify as homogeneous or heterogeneous.
 Homogeneous mixture, the components completely mix with each other and its composition is uniform throughout. e.g. Sugar solution, and air .
 Heterogeneous mixtures, the composition is not uniform throughout and sometimes the different components can be observed. E.g. the mixtures of salt and sugar, grains and pulses along with some dirt (often stone) pieces,
   Components of a mixture can be separated by using physical methods such as simple hand picking, filtration, crystallisation, distillation etc.
Pure substance/Substances have fixed composition e.g.Copper, silver, gold, water, glucose e.t.c. Pure substances can be classified as elements and compounds.
  An element consists of only one type of particles. These particles may be atoms or molecules. E.g. Sodium, copper, silver, hydrogen, oxygen etc.
   However, the atoms of different elements are different in nature.
Compound When two or more atoms of different elements combine, the molecule of a compound is obtained. E.g. water, ammonia, carbon dioxide, sugar etc.
 Q. Write the differences between--- (with suitable examples)
       a) homogeneous and heterogeneous mixtures
       b) mixture and compound
 PROPERTIES OF MATTER AND THEIR MEASUREMENT--Every substance has unique or characteristic properties. These properties can be classified into two categories – physical properties and chemical properties.
Physical properties are those properties which can be measured or observed without changing the identity or the composition of the substance. E.g. colour, odour, melting point, boiling point, density etc.
The measurement or observation of chemical properties require a chemical change to occur.   e.g. Burning of Mg-ribbon in air
Chemical properties are characteristic reactions of different substances; these include acidity or basicity, combustibility etc.
Many properties of matter such as length, area, volume, etc., are quantitative in nature.
Metric System were being used in different parts of the world. The metric system which originated in France in late eighteenth century, was more convenient as it was based on the decimal system. The need of a common standard system was being felt by the scientific community. Such a system was established in 1960 so it is in detail.
 The International System of Units (SI)
The International System of Units (in French Le Systeme International dUnités – abbreviated as SI) was established by the 11th General Conference on Weights and Measures (CGPM from Conference Generale des Poids at Measures). The CGPM is an inter governmental treaty organization created by a diplomatic treaty known as Metre Convention which was signed in Paris in 1875.
  SOME BASIC CONCEPTS OF CHEMISTRY  notes

 Definitions of SI Base Units
Unit of length—metre-- The metre is the length of the path traveled by light in vacuum during a time interval of 1/299 792 458 of a second.
Unit of mass-- kilogram -- it is equal to the mass of the international prototype of the kilogram.
Unit of time—second-- The second is the duration of  9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom.
Unit of electric current--ampere --The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed  1 metre apart in vacuum, would produce between these conductors a force equal to 2 × 10–7 newton per metre of length.
Unit of thermodynamic temperature-- kelvin --The kelvin, unit of thermodynamic temperature, is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water.
(Triple point- Temperature at which all three physical states of  water can exist i.e. 00C)
Unit of amount of substance-- mole --1. The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12; its symbol is “mol.”
2. When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.
Unit of luminous intensity-- candela --The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540 × 1012 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian.
 Mass and Weight-- Mass of a substance is the amount of matter present in it while weight is the force exerted by gravity on an object. The mass of a substance is constant whereas its weight may vary from one place to another
  due to change in gravity.
 The mass of a substance can be determined very accurately by using an analytical balance.
Maintaining the National Standards of Measurement
Each modern industrialized country including India has a National Metrology Institute (NMI) which maintains standards of measurements. This responsibility has been given to the National Physical Laboratory (NPL), New Delhi.
Derived units—Units derived with the help of base units of measurement.
Volume-- Volume has the units of (length)3. So volume has units of m3 or cm3 or dm3.
A common unit, litre (L) is not an SI unit, is used for measurement of volume of liquids.
1 L = 1000 mL , 1000 cm3 = 1 dm3
 In the laboratory, volume of liquids or solutions can be measured by graduated cylinder, burette, pipette etc. A volumetric flask is used to prepare a known volume of a solution.
Density Density of a substance is its amount of mass per unit volume.
SI unit of density = SI unit of mass/SI unit of volume
                              = kg/m3 or kg m–3
This unit is quite large and a chemist often expresses density in g cm–3.
Temperature--There are three common scales to measure temperature — °C (degree celsius), °F (degree fahrenheit) and K (kelvin). Here, K is the SI unitGenerally, the thermometer with celsius scale are calibrated from 0° to 100° where these two temperatures are the freezing point and the boiling point of water respectively.
The fahrenheit scale is represented between 32° to 212°.
The temperatures on two scales are related to each other by the following relationship:
                                           
The kelvin scale is related to celsius scale as follows :
                                           K = °C + 273.15
Note—Temperature below 0 °C (i.e. negative values) are possible in Celsius scale but in Kelvin scale, negative temperature is not possible.
 UNCERTAINTY IN MEASUREMENT
The study of chemistry, deals with experimental data as well as theoretical calculations. There are meaningful ways to handle the numbers conveniently and present the data realistically with certainty to the maximum possible extent. These ideas are discussed below in detail.
 Scientific Notation
As chemistry is the study of atoms and molecules which have extremely low masses and are present in extremely large numbers, a chemist has to deal with numbers very large (Avogadro’s no.) as well as very small (mass of a H atom). It is very difficult to count numbers involving so many zeros and it offers a real challenge to do simple mathematical operations of addition, subtraction, multiplication or division with such numbers.
This problem is solved by using scientific notation, i.e., exponential notation for such numbers.
 In which any number can be represented in the form N × 10n (Where n is an exponent having positive or negative values and N can vary between 1 to 10).
 e.g.  We can write 232.508 as 2.32508 X102 in scientific notation. Similarly, 0.00016 can be written as 1.6 X 10–4.
Now, for performing mathematical operations on numbers expressed in scientific notations, the following points are to be kept in mind.
Multiplication and Division
These two operations follow the same rules which are there for exponential numbers, i.e.-----(cw)
Addition and Subtraction
For these two operations, first the numbers are written in such a way that they have same exponent. After that, the coefficient are added or subtracted.
Significant Figures -- Every experimental measurement has some amount of uncertainty associated with it. Every one would like the results to be precise and accurate.
Precision refers to the closeness of various measurements for the same quantity.
Accuracy is the agreement of a particular value to the true value of the result. e.g. if the true value for a result is 2.00 g. Precision and accucacy clearly understood from the data given in Table
                        1                    2                Average (g)                  Remark
Student A     1.95              1.93                    1.940              values are precise as they are close to each
                                                                                                                            other but are not accurate.
Student B      1.94             2.05                    1.995                      neither precise nor accurate
Student C      2.01             1.99                    2.000                    These values are both precise and accurate.
       The uncertainty in the experimental or the calculated values is indicated by mentioning the number of significant figures. Significant figures are meaningful digits which are known with certainty.
    The uncertainty is indicated by writing the certain digits and the last uncertain digit. E.g. If we write a result as 11.2 mL, we say the 11 is certain and 2 is uncertain and the uncertainty would be +1 or -1 in the last digit.
  There are certain rules for determining the number of significant figures.
 (1) All non-zero digits are significant.e.g. in 285 cm-- three and in 0.25 mL -- two S.F.
(2) Zeros preceding to first non-zero digit are not significant. E.g in 0.03-- one and in 0.0052 -- two significant figures.
(3) Zeros between two non-zero digits are significant.e.g. in 2.005 -- four significant figures.
(4) Zeros at the end or right of a number are significant provided they are on the right side of the decimal point. E.g. in 0.200 g --three significant figures.
But, in 100 -- one significant figure, but 100. -- three significant figures and 100.0 – four significant figures. Such numbers are better represented in scientific notation. We can express the number 100 as 1×102 – one significant figure, 1.0×102 – two significant figures and 1.00×102 – three significant figures.
(5) Counting numbers of objects, for example, 2 balls or 20 eggs, have infinite significant figures as these are exact numbers and can be represented by writing infinite number of zeros after placing a decimal i.e., 2 = 2.000000 or 20 = 20.000000
   In numbers written in scientific notation, all digits are significant e.g., 4.01×102 has three significant figures, and 8.256 × 10–3 has four significant figures.
Addition and Subtraction of Significant Figures
The result cannot have more digits to the right of the decimal point than either of the original numbers.
     12.11 + 18.0 + 1.012 = 31.122
Here, 18.0 has only one digit after the decimal point so the result should be reported as 31.1.
Multiplication and Division of Significant Figures
2.5×1.25 = 3.125  Since 2.5 has two significant figures, the result should be 3.1.
  The following points should be keep in mind for rounding off the numbers to write the result to the required number of significant figures
1. If the rightmost digit to be removed is more than 5, the preceding number is increased by one.
  E.g. 1.386 If we have to remove 6, we have to round it to 1.39
2. If the rightmost digit to be removed is less than 5, the preceding number is not changed. e.g.   
   4.334 if 4 is to be removed, then the result is rounded up to 4.33.
3. If the rightmost digit to be removed is 5, then the preceding number is not changed if it is an even number but it is increased by one if it is an odd number. E.g. if 6.35 is to be rounded by removing 5, we have to increase 3 to 4 giving 6.4 as the result. However, if 6.25 is to be rounded off it is rounded off to 6.2.
 Dimensional Analysis During calculations generally there is a need to convert units from one system to other. This is called factor label method or unit factor method or dimensional analysis.  Examples----( C.W.)
 LAWS OF CHEMICAL COMBINATIONS
The combination of elements to form compounds is governed by the following five basic laws.
 Law of Conservation of Mass (Given by Antoine Lavoisier in 1789).
    It states that matter (mass) can neither be created nor destroyed.
               C + O2 ---à CO2
             12g + 32g  =    44g
 Law of Definite Proportions (Given by, a French chemist, Joseph Proust.)
   He stated that a given compound always contains exactly the same proportion of elements by
weight. (Proust worked with two samples of cupric carbonate one of which was of natural origin and the other was synthetic one. He found that the composition of elements present in it was same)
e.g. If we collect water from different sources or prepare in lab, It always has H & O in fix ratio by mass or by volume or by no. of atoms.
 Law of Multiple Proportions (Given by Dalton in 1803.)
According to this law, if two elements combine together and form two or more than two compounds, the masses of one element that combine with a fixed mass of the other element, are in the simple ratio of small whole numbers.
 e.g. N and O combine together and form five oxides N2O, NO, N2O3, NO2, and N2O

 Oxide of N      
reacting mass of N   
reacting mass of O  
fix mass of N 
reacting mass of O with  fix mass of N 
Ratio of reacting mass of O with  fix mass of N 
N2O
28g
16g
14g
8g
1
NO
14g
16g
14g
16g
2
N2O3
28g
48g
14g
24g
3
NO2
14g
32g
14g
32g
4
N2O5
28g
80g
14g
40g
5

Gay Lussac’s Law of Gaseous Volumes (Given by Gay Lussac in 1808.)
   According to this law when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.  E.g.
                H2(g) + Cl2(g)   ---à   2HCl(g)
               1V           1V                     2V
    All reactants and products have simple ratio 1:1:2.
Note--Gay-Lussac’s discovery of integer ratio in volume relationship is actually the law of definite proportions by volume.
 Avogadro Law (In 1811, Given by Avogadro)
  According to this law equal volumes of gases at the same temperature and pressure should contain equal number of molecules.
Avogadro made a distinction between atoms and molecules Avogadro could explain the result of chemical reactions by considering the molecules to be polyatomic. Two volumes of hydrogen react with One volume of oxygen to give Two volumes of water vapour
 DALTON’S ATOMIC THEORY--In 1808, Dalton published ‘A New System
of Chemical Philosophy’ in which he proposed the following :
1. Matter consists of indivisible atoms.
2. All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.
3. Compounds are formed when atoms of different elements combine in a fixed ratio.
4. Chemical reactions involve reorganization of atoms. These are neither created nor destroyed in a chemical reaction.
     Dalton’s theory could explain the laws of chemical combination.
 ATOMIC AND MOLECULAR MASSES
 Today, we use mass spectrometry technique for determining the atomic masses.
   Since 1961 C-12 (isotope of a carbon) taken as standard to calculate relative masses of other elements.
     In this system, 12C is assigned a mass of exactly 12 atomic mass unit (amu) and masses of all other atoms are given relative to this standard.
   One atomic mass unit is defined as a mass exactly equal to one twelfth the mass of one carbon - 12 atom. And 1 amu = 1.66056×10–24 g. Mass of an atom of hydrogen = 1.6736×10–24 g.    
   Thus, in terms of amu, the mass of hydrogen atom =1.6736×10–24g/1.66056×10–24g=1.0078 amu
                                                                                                                                        = 1.0080 amu Today, ‘amu’ has been replaced by ‘u’ which is known as unified mass.
 Average Atomic Mass--Many naturally occurring elements exist as more than one isotope. When we take into account the existence of these isotopes and their relative abundance (per cent occurrence),the average atomic mass of that element can be calculated. For example, (C.W.)
 Molecular Mass
Molecular mass is the sum of atomic masses of the elements present in a molecule. E.g. (c.w.)
Q.  Calculate molecular mass of glucose (C6H12O6) molecule.
 Formula Mass---In crystalline substances e.g. sodium chloride do not contain discrete molecules as their constituent units. In such compounds, positive (sodium) and negative (chloride) entities are
arranged in a three-dimensional structure, so for ease of calculations , simple ratio of these entities taken as formula unit and its mass known as formula mass. e.g. (c.w.)
Thus, formula mass of sodium chloride = atomic mass of sodium + atomic mass of chlorine
                                                                = 23.0 u + 35.5 u = 58.5 u
 MOLE CONCEPT AND MOLAR MASSES
One mole is the amount of a substance that contains as many particles or entities (i.e. atoms, molecules or ions) as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope.
It is very important to note that the mole of a substance always contain the same number of entities, no matter what the substance may be.
   This number of entities in 1 mol is given a separate name and symbol. It is known as ‘Avogadro constant’, denoted by NA in honour of Amedeo Avogadro.
   The mass of one mole of a substance in grams is called its molar mass.
The molar mass is numerically equal to atomic/molecular/ formula mass in u but expressed in grams in place of u. e.g. Molecular mass of water=18.02u and Molar mass of water = 18.02 g
              (mass of 1- water molecule)                        (mass of 6.02x1023 - water molecules)
PERCENTAGE COMPOSITION—This determination is important to check the purity of a given sample. Let us consider the example of water (H2O). Water contains hydrogen and oxygen, the percentage composition of both these elements can be calculated as :
      Mass % of an element = mass of that element in the compound × 100
                                                         molar mass of the compound
    Molar mass of water = 18.02 g
   Mass % of hydrogen = 2× 1.008 × 100
                                                  18.02
                                         = 11.18
Mass % of oxygen = 16.00 × 100
                                        18.02
                               = 88.79        (other examples C.W.)
 Empirical Formula for Molecular Formula—
            An empirical formula represents the simplest whole number ratio of various atoms present in a compound. E.g. CH is the empirical formula of benzene.
   The molecular formula shows the exact number of different types of atoms present in a molecule of a compound. E.g. C6H6 is the molecular formula of benzene.
Determination of the Empirical Formula and Molecular Formula---With the help of mass per cent of various elements present in a compound, its empirical formula can be determined. Molecular formula can further be obtained if the molar mass is known. Example--
Q. A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas ?
Solution-----
Name of element
Percentage of elements
Step-1Conversion of mass per cent to grams.
Step 2. number moles of each element

Step 3. Divide the mole value by the smallest number

C
24.27%
24.27g
24.27/12 = 2.0225
2.0225/2.018  = 1
H
4.07%
4.07g
4.07/1= 4.07
4.07/2.018 =2
Cl
71.65%
71.65g
71.65/35.5 = 2.018
2.018/2.018 = 1
In step- 3 we get the ratio of different elements in the compound.
   C:H:Cl = 1:2:1
(Note-In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.)
Step 4. Write empirical formula with the help of ratio of elements.
     C1H2Cl1 or CH2Cl, is the empirical formula of the above compound.
Step 5. Writing molecular formula
Determine empirical formula mass as— (For CH2Cl, empirical formula mass is)
(a)      12 + (1x2) + 35.5 = 49.5
(b)    Divide Molar mass by empirical formula mass
                     =            Molar mass                        =           98.96 g   = 2 = (n)
                              Empirical formula mass                      49.48 g
 (c) Multiply empirical formula by n obtained above to get the molecular formula
Empirical formula = CH2Cl, n = 2. Hence molecular formula is C2H4Cl2.
 STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS
 We know chemical equation is maximum informative, when it is written in the form of balanced chemical equation. It gives quantitative relationship between the various reactants and products in terms of moles, masses, molecules and volumes. This is called stoichiometry. (The word ‘stoichiometry’ is derived from two Greek words - stoicheion (meaning element) and metron (meaning measure) i.e. measurement of an element.). The coefficients of the balanced chemical equation are known as stoichiometric coefficients. (examples—C.W.)
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.
 Limiting Reagent—The reactant which gets consumed first or limits the amount of product formed is known as limiting reagent.  (Examples- c.w.)
 Reactions in Solutions--A majority of reactions in the laboratories are carried out in solutions. The concentration of a solution or the amount of substance present in its given volume can be expressed in any of the following ways.
1. Mass per cent or weight per cent (w/w %)                  2. Mole fraction
3. Molarity                                                                       4. Molality
1. Mass per cent--It is obtained by using the following relation:
                       Mass per cent (w/w%) =   Mass of solute x 100
                                                                    Mass of solution
2. Mole Fraction (X)
It is the ratio of number of moles of a particular component to the total number of moles of the
solution. If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are nA and nB respectively; then the mole fractions of A and B are given as
                     Mole fraction of A (XA)   =              No.of moles of A    =        nA
                                                                      No.of moles of solution         nA + nB
  Similarly we can calculate the mole fraction of B (XB).
3. Molarity (M) -- It is defined as the number of moles of the solute in 1 litre of the solution. Thus,
            Molarity (M)         =       No. of moles of solute
                                                Volume of solution in litres
   Molarity on dilution can be calculated by using the general formula
                        M1V1 = M2V2
4. Molality(m)-- It is defined as the number of moles of solute present in 1 kg of solvent.
               Molality (m)    =   No. of moles of solute
                                              Mass of solvent in kg

Exercise-1
1.1 Calculate the molecular 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.




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 percent 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___ (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  and Mass of dioxygen are as respectively---
(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 ----àAB2
    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) -----à2NH3 (g)
(i) Calculate the mass of ammonia produced if 2.00 x 103 g dinitrogen reacts with 1.00 x 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 ten 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
1.28 Which one of the following will have 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?
(i)  0.02856 x 298.15 x  0.112    =
                    0.5785
 (ii) 5 x 5.364 =
(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) -------à 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) ------à 2H2O (l) + MnCl2(aq) + Cl2 (g)
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