3. LA PART I

IMPORTANT LONG ANSWERS PART ONE

QUESTION NO. 21
(a) What are isotopes? Write a note on relative abundance and occurrence of isotopes. 
(b) What is mass spectrometer? How is it used to determine the relative atomic masses of isotopes?


Isotopes, their Relative Abundance and Occurrence

Isotopes
     Atoms of the same element which have the same atomic number but different mass numbers due to difference in the number of neutrons are called isotopes of that element.
Examples
(i) Isotopes of Hydrogen: Hydrogen consists of three isotopes which are protium(¹H), deuterium(²H) and tritium(³H). All these isotopes have the same atomic number i.e. one, but they have different mass numbers 1,2, and 3 respectively.
(ii) Isotopes of Carbon: Carbon also consists of three isotopes which are C-12, C-13 and C-14. All these isotopes have the same atomic number i.e. six, but they have different mass numbers 12, 13 and 14 respectively.
Relative Abundance of Isotopes

  • The percentage of each isotope in a mixture of isotopes of an element is called relative abundance. 
  • Different isotopes have their own natural abundance.
  • The relative abundance of isotopes is measured by mass spectrometery.
  • The properties of a particular element mostly correspond to the most abundant isotope of that element. 
  • Natural abundance of some common isotopes is given below;


  • The distribution of isotopes among the elements is varied and complex as it is evident from the above table. 

Occurrence of Isotopes

  • At present more than 280 different isotopes occur in nature. 
  • 40 radioactive isotopes are also included in this number (280)
  • Almost 300 unstable radioactive isotopes of different elements have been produced by the artificial disintegration.
  • The elements with odd atomic number almost never posses more than two stable isotopes.
  • The elements with even atomic number usually have larger number of isotopes. 
  • The isotopes whose mass numbers are multiple of four are particularly abundant. 
  • Out of 280 natural isotopes, 154 isotopes have even atomic number and even mass number. 

Determination of Relative Atomic Masses of Isotopes 
By Mass Spectrometer

Mass Spectrometer
     Mass spectrometer is an instrument which is used to measure the exact masses of different isotopes of an element along with their relative abundance.
Determination of Relative Atomic Masses of Isotopes
   
     Different steps involved in the determination of exact atomic masses and the relative abundances of different isotopes of an element by Dempster's mass spectrometer are given below;
(i) Vapourization: The substance whose analysis for the separation of isotopes is required is converted into vapour state. That pressure of these vapours is kept very low, that is, 10⃜ ⁶ to 10⃜ ⁷ torr.
(ii) Ionization: These vapours are then allowed to enter the ionization chamber where fast moving electrons are thrown upon them. The atoms of isotopic element present in the form of vapours, are ionized. These positively charged ions of isotopes of an element have different masses depending upon the nature of the isotopes present in them.
(iii) Acceleration: The positive ions of each isotope has its own (m/e) value. When a potential difference (E) of 500-2000 volts is applied between perforated accelerating plates, then these positive ions are strongly attracted towards the negative plate. In this way, the ions are accelerated.
(iv) Deflection: The beam of accelerated positive ions is then allowed to pass through a strong magnetic field of strength H. This magnetic field is applied in a direction which is perpendicular to the path of the positive ions. The applied magnetic field helps in the separation of positive ions on the basis of their m/e values. The magnetic field makes the ions to move in a circular path. The ions of definite m/e value move in the form of groups one after the other and fall on electrometer.
(v) Mathematical Explanation: The mathematical relationship between m/e values and deflection in the circular path is
m/e = H²r/E
Where H = strength of magnetic field, E = strength of electric field, r = radius of circular path
     If E is increased, by keeping H constant then radius will increase and positive ion of a particular m/e will fall at a different place as compared to the first place. This can also be done by changing the magnetic field. Smaller the m/e of an isotope, smaller the radius of curvature produced by the magnetic field according to the above equation. Each ion sets up a minute electrical current.
(vi) Electrometer (Ion Collector): Electrometer develops electrical current. The strength of the current measured gives the relative abundance of ions of a definite m/e value.
(vii) Comparison with C-12: The ions of other isotopes having different masses are made to fall on the collector and the current strength is measured. The same experiment is performed with C-12 isotope and the current strength is compared. This comparison allows us to measure the exact mass number of the isotopes.
     In modern spectrographs, each ion strikes a detector, the ionic current is amplified and is fed to the recorder. The recorder makes a graph showing the relative abundance of isotopes plotted against the mass number.

QUESTION NO. 22 
(a) What is empirical formula? Write down the steps involved in determination of empirical formula. 
(b) What is stoichiometry? Give its assumptions. Mention two important laws, which help to perform the stoichiometric calculations. 

Empirical Formula and Its Determination

Empirical Formula
     A chemical formula which shows the simplest whole number ratio between the atoms of different elements present in a compound is called empirical formula.
Examples
Steps Involved in Determination of Empirical Formula
     Following steps are involved in determination of empirical formula.
(i) Determination of Percentage Composition
(ii) Finding the Number of Gram Atoms of Each Element
(iii) Determination of the Atomic Ratio of Each Element
(iv) If the atomic ratio is simple whole number, it gives the empirical formula, otherwise multiply with a suitable digit to get the whole number atomic ratio.

Stoichiometry, Its Assumptions and Calculations

Stoichiometry
     The branch of chemistry which deals with the study of quantitative relationship between reactants and products in a balanced chemical equation is called stoichiometry. 
Stoichiometric Amounts
     The amounts of the reactants or the products as given by the balanced chemical equation are called stoichiometric amounts. 
2H2     +     O2         2H2O
2 mole       1 mole         2 moles
4 g            32 g             36 g
Assumptions
     When stoichiometric calculations are performed, we have to assume the following conditions.
(i) All the reactants are completely converted into products.
(ii) No side reaction occurs.
     During stoichiometric calculations, law of conservation of mass and law of definite proportions are obeyed. 
Calculations
(i) Mass-mass Relationship: If we are given mass of one substance, we can calculate the mass of the other substance involved in the chemical reaction.
(ii) Mass-mole Relationship: If we are given the mass of one substance, we can calculate the moles of the other substance and vice versa.
(iii) Mass-volume Relationship: If we are given mass of one substance, we can calculate the volume of the other substance and vice versa. 
(iv) Similarly, mole-mole calculations can also be performed. 

QUESTION NO. 23 
(a) What is a limiting reactant? How does it control the quantity of the product formed? Explain with examples. 
(b) What is the difference between actual yield and theoretical yield? Why actual yield is less than theoretical yield? 


Limiting Reactant

Limiting Reactant
     A reactant which is used or consumed earlier due to to its lesser amount and controls the amount of product formed in a chemical reaction is called limiting reactant.
Examples
(i) If 2 moles of H2 and 2 moles of O2 are allowed to react then only 2 moles H2O is produced.
2H2 + O2 → 2H2O
Here H2 consumes completely and one mole of O2 is left behind unreacted so H2 is limiting reactant and limits the amount of H2O up to 2 moles.
(ii) If 1 mole of C and 1 mole of O2 are allowed to react then only 1 mole CO2 is produced.
C + O→  CO2
Here O2 consumes completely and one mole of C is left behind unreacted so O2 is limiting reactant and limits the amount of CO2 up to 1 mole.
(ii) During burning O2 is in excess and a combustile material (coal, candle, paper etc) is in smaller quantity. Combustile material is fully consumed and controls the amount of product so coal, candle or paper is limiting reactant.
(iii) If we have 20 slices and 9 kababs, we can only make 9 sandwiches as kababs are limiting reactant.

Actual and Theoretical Yield

Yield
     The amount of products obtained in a chemical reaction is called yield.
Difference Between Actual and Theoretical Yield
Actual Yield is Less Than Theoretical Yield
    Actual yield is less than theoretical yield due to following reasons.
1. Practically inexperienced worker has many shortcomings and cannot get expected yield.
2. Mechanical loss during experimentation e.g., filtration, separation by distillation, or by a separating funnel, washing, drying and crystallization if not properly carried out, decrease the yield.
3. Some of the reactants might take part in a competing side reaction.
4. The reaction might be reversible.
5. The reactants might be impure.
6. Sometimes the reaction conditions are not suitably maintained.

QUESTION NO. 24 
(a) What is general gas equation? Derive its various forms.
(b) State Dalton's law of partial pressure. Also give its applications.

General Gas Equation

General Gas Equation
     This is called an ideal gas equation or general gas equation. This equation shows if we have any quantity of an ideal gas then the product of its pressure and volume is equal to the product of number of moles, general gas constant and absolute temperature.
Derivation of Gas Laws from General Gas Equation
     The following gas laws can be derived from general gas equation.
Notes Prepared By: Prof. Shahbaz Asghar

Dalton's Law of Partial Pressure and Its Applications

Dalton's Law of Partial Pressure
Statement: The total pressure exerted by a mixture of non-reacting gases is equal to the sum of their individual partial pressure. 
Mathematical Form: Let the gases are designated as 1,2 and 3 and their partial pressures are p1, p2 and p3 respectively. The total pressure Pt of the mixture of gases is given by:
Pt = p1 + p2 + p3
Applications of Dalton's Law of Partial Pressure
     Following are the applications of Dalton's Dalton's law of partial pressure.
(i) Collection of Gases Over Water: Some gases are collected over water in the laboratory. The gas during collection gathers water vapours and becomes moist. The pressure exerted by this moist gas is the sum of partial pressures of dry gas and that of water vapours. The partial pressure exerted by the water vapours is constant at a particular temperature and is called aqueous tension
Pmoist = pdry + pw vap
Pmoist = pdry + aqueous tension
pdry = Pmoist - aqueous tension
(ii) Process of Respiration: The process of respiration depends upon the difference in partial pressures. When animals inhale air then oxygen moves into the lungs as the partial pressure of oxygen in air is 159 torr, while the partial pressure of oxygen in the lungs is 116 torr. CO2 produced during respiration moves out in the opposite direction, as its partial pressure is more in the lungs that that in the air. 
(iii) Breathing at Higher Altitudes: At higher altitude, the pilot feels uncomfortable breathing because the partial pressure of oxygen in the un-pressurized cabin is low as compared to 159 torr, where one feels comfortable breathing.
(iv) Breathing in Depth of Sea: Deep sea divers take oxygen mixed with an inert gas say He and adjust the partial pressure of oxygen according to the requirement. Actually, in sea after every 100 feet depth, the diver experiences approximately 3 atm pressure, so normal air cannot be breathed in depth of sea. Moreover, the pressure of N2 increases in depth of sea and it diffuses in the blood. 

QUESTION NO. 25
(a) State and explain Graham's law of diffusion of gases.
(b) Give postulates of kinetic molecular theory of gases (KMT).

Graham's Law of Diffusion of Gases

Statement
    The rate of diffusion or effusion of a gas is inversely proportional to the square root of its density at constant temperature and pressure.
Mathematical Form
        Let us have two gases 1 and 2, having rates of diffusion as r₁ and r₂ and densities as d₁ and d₂ respectively.
        According to Graham's law,
Demonstration of Graham's Law
     This law can be verified in laboratory by noting the rates of diffusion of two gases in a glass tube, when they are allowed to move from opposite ends.
     Two cotton plugs soaked in HCl and NH3 solutions are introduced in the open ends of 100 cm long tube simultaneously. HCl molecules travel a distance of 40.5 cm and NH3 molecules cover 59.5 cm. They produce dense white fumes of NH4Cl at point of junction. So,
                 1.46 = 1.46
Hence, the law is verified.
   
Kinetic Molecular Theory of Gases (KMT)

Definition
     A set of postulates that describe the nature and behaviour of an ideal gas is called kinetic molecular theory of gases.
Fundamental Postulates
1. Every gas consists of a large number of very small particles called molecules. Gases like He, Ne, Ar have mono-atomic molecules.
2. The molecules of a gas move haphazardly, colliding among themselves and with the walls of the container and change their directions.
3. The pressure exerted by a gas is due to the collision of its molecules with the walls of a container. The collisions among the molecules are perfectly elastic.
4. The molecules of a gas are widely separated from one another and there are sufficient empty spaces among them.
5. The molecules of a gas have no forces of attraction for each other.
6. The actual volume of molecules of a gas is negligible as compared to the volume of the gas.
7. The motion imparted to the molecules by gravity is negligible as compared to the effect of the continued collisions between them.
8. The average kinetic energy of the gas molecules varies directly as the absolute temperature of the gas.

QUESTION NO. 26
(a) How volume and pressure are corrected by Van der Waal's equation?
(b) What is plasma? How is it formed? Explain its various types.

Volume and Pressure Correction By Van der Waals's Equation

     Van der Waals pointed out that both volume and pressure factors in ideal gas equation needed correction to make it applicable to the real gases.
Volume Correction
(i) Compression of a Gas: When a gas is compressed, the molecules are pushed so close together that the repulsive forces operate between them. When pressure is increased further, it is opposed by the molecules themselves. Actually, the molecules have definite volume, no doubt, very small as compared to the vessel, but it is not negligible.
(ii) Van der Waals' Postulate: Van der Waals postulated that the actual volume of molecules can no longer be neglected in a highly compressed gas. If the effective volume of the molecules per mole of a gas is represented by b, then the volume available to gas molecules is the volume of the vessel minus the volume of gas molecules.
Vfree = Vvessel - b (where Vfree is the volume available to gas molecules)
(iii) Excluded Volume "b": The volume of a gas which is occupied by 1 mole of gas molecules in highly compressed state, but not in the liquid state, is called excluded volume or effective volume or incompressible  volume (b). It is a constant and characteristic of a gas. It value depends upon the size of the molecules. It is also a Van der Waals constant. It is not equal to the actual volume of gas molecules. In fact, it is four times the actual volume of molecules.
b = 4Vm (where Vm is actual volume of one mole of gas molecules. 
Pressure Correction
(i) Attraction Between Molecules: A molecule in the interior of a gas is attracted by other molecules on all sides, so these attractive forces are cancelled out. However, when a molecule strikes the wall of a container, it experiences a force of attraction towards the other molecules in the gas. The decreases the force of its impact on the wall.
(ii) Pressure on the Wall of Container: Consider a molecule "A" which is unable to create pressure on the wall due to the pressure of attractive forces due to "B" type molecule. Let the observed pressure on the wall of the container is "P". The pressure is less than the actual pressure Pi by an amount P'. So,
P = Pi - P'
Where Pi is the true kinetic pressure, if the forces of attraction would have been absent. P' is the amount of pressure lessened due to attractive forces. Ideal pressure Pi is;
Pi = P + P'     ..... (i)
It is suggested that a part of the pressure "P" for one mole of a gas used up against inter-molecular attractions should decrease as volume increases.
(iii) Value of P': The value of P' in terms of a constant "a" which accounts for the attractive forces and the volume V of vessel can be written as:
P' = a/V² ..... (ii) 
Greater the attractive forces among the gas molecules, smaller the volume of vessel, greater the value of lessened pressure P'. "a" is a coefficient of attraction or attraction per unit volume. It has a constant value for a particular real gas.
(iv) Value of Pi: Putting the value of P' from equation (ii) into (i)
Pi = P + a/V²
(v) For One Mole of a Gas:
(vi) For "n" Moles of a Gas
     This is called van der Waals's equation. "a" and "b" are called van der Waals's constants.
(vii) Common and SI Units of "a" and "b"

Plasma, Its Formation and Types

Plasma

  • Plasma is a hot ionized gas consisting of approximately equal numbers of positively charged ions and negatively charged electrons. 
  • Plasma is often called "fourth state of matter", the other three being solid, liquid and gas. 
  • Plasma was identified by the English scientist William Crookes in 1879. 
  • Plasma is estimated to constitute more than 99 percent of the visible universe. 
  • Although, naturally occurring plasma is rare on earth, there are many man-made examples.  
  • It occurs only in lightning discharges and in artificial devices like fluorescent lights, neon signs etc. 
  • It is everywhere in our space environment. All the stars that shine are all plasma. 
 How is Plasma Formed?
     When more heat is supplied, the atoms are molecules may be ionized. An electron may gain enough energy to escape its atom. This atom loses one electron and develops a net positive charge. It becomes and ion. In a sufficiently heated gas, ionization happens many times, creating clouds of free electrons and ions. However, all the atoms are not necessarily ionized, and some of them may remain completely intact with no net charge. This ionized gas mixture, consisting of ions, electrons and neutral atoms is called plasma. 
     It means that plasma is a distinct state of matter containing a significant number of electrically charged particles, a number sufficient to affect its electrical properties and behaviour.
Types of Plasma
     There are two types of plasma, artificial plasma and natural plasma.
(i) Artificial Plasma: Artificial plasma can be created by ionization of a gas, as in neon signs. Plasma at low temperatures is hard to maintain outside a vacuum, low temperature plasma reacts rapidly with any molecule it encounters. This aspect makes this material, very useful and hard to use.
(ii) Natural Plasma: Natural plasma exits only at very high temperatures, or low temperature vacuums. Natural plasma does not breakdown or react rapidly, but is extremely hot (over 20,000⁰C minimum). Their energy is so high that they vaporize any material they touch.
QUESTION NO. 27
(a) Define hydrogen bonding. Explain hydrogen bonding in H2O, NH3 and HF, how is it helpful in explaining the structure of ice?
(b) What is vapour pressure of a liquid? Describe measurement of vapour pressure by manometric method.

Hydrogen Bonding and Its Examples

Hydrogen Bonding
     The electrostatic force of attraction between hydrogen atom (bonded to a small highly electronegative atom) and the electronegative atom (F, O, N) of another molecule is called hydrogen bonding.
Examples
(i) Water (H2O): In case of H2O, oxygen is a more electronegative element as compared to hydrogen, so water is a polar molecule. Hence, there will be dipole-dipole interactions between partial positively charged hydrogen partial negatively charged oxygen atoms. Actually, hydrogen bonding is something more than dipole-dipole interaction. Firstly, oxygen atom has two lone pairs. Secondly hydrogen has sufficient partial positive charge. Both the hydrogen atoms of water molecule create strong electrical field due to their small sizes. The oxygen atom of the other molecule links to form a coordinate covalent bond with hydrogen using one of its lone pairs of electrons. Thus, loose bond formed is definitely stronger than simple dipole-dipole interaction. Because of the small size of hydrogen atom, it can take part in this type of bonding. This bonding acts as a bridge between two electronegative oxygen atoms.



Vapour Pressure and Its Measurement



QUESTION NO. 28
(a) What are liquid crystals? Give their uses in daily life.
(b) Write down the types of solids. Describe the properties of ionic and molecular solids.

Liquid Crystal and Their Uses

Liquid Crystals
(i) Definition: The turbid liquid phase of a solid that exists in between the melting and clearing temperature is called liquid crystal.
Crystal ⇋ Liquid Crystal ⇋ Liquid
(ii) Discovery: In 1888, Frederick Reinitzer, an Austrian botanist discovered the liquid crystals. He was studying an organic compound Cholesteryl Benzoate. This compound turns milky liquid at 145⁰C and becomes a clear liquid at 179⁰C. When the substance is cooled, the reverse process occurs. This turbid liquid state was called liquid crystal.
(iii) Characteristics: Liquid crystals have both properties of liquids and crystals (solids). Liquid like properties include viscosity, surface tension, and fluidity etc. Crystal like properties include optical properties and molecules have some orderly arrangement. We can say that the properties of liquid crystals are intermediate between those of crystals and isotopic liquids. A crystalline solid may be isotopic or anisotropic but liquid crystals are always anisotropic.
(iv) Types: Those substances which make the liquid crystals are often composed of long rod like molecules. In the normal liquid phase, these molecules are oriented in random directions. In liquid crystalline phase, they develop some ordering of molecules. Depending upon the nature of ordering, liquid crystals can be divided into; Nematic, Smectic and Cholesteric.
Uses of Liquid Crystals
(i) As Temperature Sensor: Like solid crystals, liquid crystals can diffract light. When one of the wavelengths of white light is reflected from a liquid crystal, it appears coloured. As the temperature changes, the distances between the layers of the molecules of liquid crystals change. Therefore, the reflected light changes accordingly. Thus liquid crystals can be used as temperature sensors.
(ii) To Find Potential Failure/As Room Thermometers: Liquid crystals are used to find the point of potential failure in electrical circuits. Room thermometers also contain liquid crystals with a suitable temperature range. As the temperature changes, figures show up in different colours.
(iii) Medical Diagnosis: Liquid crystals are used to locate veins, arteries, infections and tumors. The reason is that these parts of the body are warmer than the surrounding tissues. Specialists can use the techniques of skin thermography to detect the blockages in veins and arteries. When a layer of liquid crystal is painted on the surface of a breast, a tumor shows up as a hot area which is coloured blue. This technique has been successful in the early diagnosis of breast cancer.
(iv) Electrical Devices: Liquid crystals are used in the display of electrical devices such as digital watches, calculators and laptop computers. These devices operate due to the fact that temperature, pressure and electromagnetic fields easily affect the weak bonds, which hold molecules together in liquid crystals.
(v) Solvents in Chromatography: In chromatographic separation, liquids crystals are used as solvents.
(vi) Oscillograph and TV Displays: Oscillograph and TV displays also use liquid crystal screens.

Types and Properties of Solids

Types of Solids
     There are four types of crystalline solids depending upon the type of bond present in them.
(i) Ionic Solids
(ii) Covalent Solids
(iii) Molecular Solids
(iv) Metallic Solids

Properties of Ionic Solids
(i) Definition: The crystalline solids in which the particles forming the crystal are positively and negatively charged ions which are held together by strong electrostatic forces of attraction (ionic bond) are called ionic solids. The crystals of NaCl, KBr etc. are ionic solids.
(ii) Physical State: The cations and anions are arranged in a well defined pattern, so they are crystalline solids at room temperature. Under ordinary conditions of temperature and pressure they never exist in the form of liquids or gases.    
(iii) Hardness, Volatility, Melting and Boiling Points: Ionic crystals are very stable compounds. Very high energy is required to separate the cations and anions from each other against the forces of attraction. That's why ionic solids are very hard, have low volatility and high melting and boiling points.
(iv) Nature of Ionic Solids: Ionic solids do not exist as individual neutral independent molecules. The cations and anions attract each other and these forces are non-directional. The close packing of the ions enables them to occupy minimum space. A crystal lattice is developed when the ions arrange themselves systematically in an alternate manner.
(v) Radius Ratio: The structure of ionic crystals depends upon the radius ratio of cations and anions.
Radius ratio = r+/r-
In NaCl; Na⁺ = 95 pm, Cl⁻ = 181 pm
Radius ratio = 95/181 = 0.525
NaCl and CsF have the same geometry because the radius ratio in both the cases is the same.
Properties of Molecular Solids
(i) Definition: Those solid substances in which the particles forming the crystals are polar or non-polar molecules or atoms of a substance are called molecular solids. Ice, sugar, Iodine (I2) and phosphorus (P4) are examples of molecular solids.
(ii) Regular Arrangement: X-rays analysis has shown the regular arrangement of atoms in the constituent molecules of these solids and we get the exact positions of all the atoms.
(iii) Softness: The forces, which hold the molecules together in the molecular crystals, are very weak. So they are soft and easily compressible.
(iv) Volatility, Melting and Boiling Points: They are mostly volatile and have low melting and boiling points.
(v) Conduction, Solubility and Density: They are bad conductors of electricity, have low densities and sometimes transparent to light. Polar molecular crystals are mostly soluble in polar solvents, while non-polar molecular crystals are usually soluble in non-polar solvents.


QUESTION NO. 29
(a) Explain Millikan's oil drop experiment to determine the charge on an electron.
(b) Give the postulates of Bohr's atomic model. Also writes the defects in this model.


QUESTION NO. 30
(a) Derive the equation for the radius of nth orbit of hydrogen atom using Bohr's atomic model.
(b) What are quantum numbers? Briefly describe their types.


QUESTION NO. 31
(a) Define electron affinity. How is it changed in periodic table? What factors are responsible for its variation?
(b) Write down the main postulates of VSEPR theory and explain the structure of ammonia on the basis of this theory.


QUESTION NO. 32
(a) Define hybridization process and explain the bonding and structure of methane, ethene and ethyne.
(b)  Describe the important points of molecular orbital theory (MOT). Explain the para-magnetic nature of O2 according to this theory.


QUESTION NO. 33
(a) State first law of thermodynamics. How does it explain (i) qv = ∆E  (ii) qp = ∆H.
(b) Define enthalpy of a reaction. How is it measured with glass calorimeter?


QUESTION NO. 34
(a) State and explain Hess's law of constant heat summation with an example.
(b) State and explain Le-Chatelier's principle.


QUESTION NO. 35
(a) What is common ion effect? Give its applications.
(b) Write a note on synthesis of ammonia gas by Haber's process, keeping in mind the application of chemical equilibrium in industry.


QUESTION NO. 36
(a) State different forms of Raoult's law. How this law can help us to understand the ideality of a solution?
(b) Give graphical explanation for elevation of boiling point of a solution. Describe one method to determine the boiling point elevation of a solution.


QUESTION NO. 37
(a) Balance the following equations by oxidation number method.
(i) Zn + HNO3 Zn(NO3)2 + NO + H2O
(ii) MnO2 + HCl  MnCl2 + H2O + Cl2
(b) Describe the construction and working of standard hydrogen electrode. (SHE)


QUESTION NO. 38
(a) What is order of reaction? Describe two methods for finding order of reaction.
(b) What is Arrhenius equation? How can it be used to calculate the energy of a reaction?


QUESTION NO. 39
Solve the following numerical problems. 
(i) Calculate the number of grams of Al2S3 which can be prepared by the reaction of 20 g of Al and 30 g of sulphur. How much the non-limiting is in excess?
(ii) A mixture of two liquids, hydrazine N2H4 and N2O4 are used as a fuel in rockets. They produce N2 and water vapours. How many grams of N2 gas will be formed by reacting 100 g of N2H4 and 200 g of N2O4? (2N2H4 + N2HO4  → 3N2 + 4H2O)
(iii) The combustion analysis of an organic compound shows it to contain 65.44% carbon, 5.50% hydrogen and 29.06% oxygen. What is the empirical formula of the compound? If the molecular mass of this compound it 110.5/mole. Calculate the molecular formula of the compound.
(iv) A well known ideal gas is enclosed in a container having volume 500 cm³ at S.T.P. Its mass comes out to be 0.72 g. What is the molar mass of this gas?
(v) Mg metal reacts with HCl to give H2 gas. What is the minimum volume of HCl solution (27% by mass) required to produce 12.1 g of H2? The density of HCl solution is 1.14 g/cm³.
Mg(s) + 2HCl(aq MgCl2(aq) + H2(g)
(vi) NH3 gas can be prepared by heating together two solids, NH4Cl and Ca(OH)2. If a mixture containing 100 g of each solid is heated then calculate the number of grams of NH3 produced.
2NH4Cl(s) + Ca(OH)2(s)  CaCl2(s) + 2NH3(g) + 2H2O(l)
(vii) What pressure is exerted by a mixture of 2.00 g of H2 and 8.00 g of N2 at 273k in a 10 dm³ vessel?
(viii) Calculate the mass of 1dm³ of NH3 gas at 30⁰C and 1000mm Hg pressure, considering that NH₃ is behaving ideally.
(ix) 250cm³ of the sample of hydrogen effuses four times as rapidly as 250cm³ of an unknown gas. Calculate the molar mass of unknown gas.
(x) One mole of methane gas is maintained at 300k. Its volume is 250cm³. Calculate the pressure exerted by the gas when the gas is non-ideal.
a = 2.253atm dm⁶ mole⃜ ²  b = 0.0428dm³ mole⃜ ¹
QUESTION NO. 40
Solve the following numerical problems. 
(i) When 2.00 moles of H2 and 1.00 mole of O2 at 100⁰C and 1 torr pressure to produce 2.00 moles of gaseous water, 484.5 kJ of energy are evolved. What are the values of (a) ∆H (b) ∆E for the production of one mole of H2O(g)?
(ii) Neutralization of 100 cm³ of 0.5M NaOH at 25⁰C with 100 cm³ of 0.5M HCl at 25⁰C raised the temperature of the reaction mixture to 28.5⁰C. Find the enthalpy of neutralization. Specific heat of water is 4.2 J/K/g.
(iii) 10.16 g of graphite is burnt in a bomb calorimeter and the temperature rise recorded is 3.8K. Calculate the enthalpy of combustion of graphite, if the heat capacity of the calorimeter (bomb, water, etc) is 86.02 kJ/K.
(iv) Benzoic acid C6H5COOH is a weak mono-basic acid Ka = 6.4 x 10⃜ ⁵ mole/dm³. What is the PH of the buffer containing 7.2 g of sodium benzoate and 0.02 mole benzoic acid?
(v) The solubility of CaF2 in water at 25⁰C is found to be 2.05 x 10⃜ ⁴ mole/dm³. What is the value of Ksp at this temperature?
(vi) N2(g) and H2(g) combine to give NH3(g). The value of Kc in this reaction at 500⁰C is 6.0 x 10⃜ ². Calculate the value of Kp for this reaction.
(vii) What is the percentage ionization of acetic acid in a solution in which 0.1 mole of it has been dissolved per dm³ of the solution?
(viii) Calculate PH of a buffer solution in which 0.11 molar CH3COONa and 0.09 molar acetic acid solutions are present. Ka for CH3COOH is 1.85 x 10⃜ ⁵.
(ix) The solubility of PbF2 at 25⁰C is 0.64 dm⃜  ³. Calculate Ksp of PbF2.
(x) Ca(OH)2 is sparingly soluble compound. Its solubility constant is 6.5 x 10⃜ ⁶. Calculate the solubility of Ca(OH)2.
      

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