Chemistry XII Set 2

8/8/2019 Chemistry XII Set 2

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Ramakrishna Mission School::AaloMid Term Examination [25th Sept 2010]

Class XII (Sci)

Subject: Chemistry (Set 2) F. M.: 70Time: 3 hours P. M.: 23

General Instructions1. All questions are compulsory.2. There are 30 questions in total. Questions 1 to 8 carry one mark each, questions 9

18 carry two marks each, questions 19 to 27 carry three marks each and questions 2to 30 carry five marks each.

3. There is no overall choice. However, an internal choice has been provided in onquestion of two marks, one question of three marks and all three questions of fivmarks each. You have to attempt only one of the given choices in such questions.

4. Use of calculators is not permitted.

Section A [1 x 8 = 8]1. Does the instantaneous rate of a chemical reaction change when a portion of th

reacting solution is taken out? Give the reason very briefly. No. Because, Instantaneous rate of a reaction depends on concentration of reacting solutiand when a portion of the reacting solution is taken out, concentration does not change.

2. Define Coordination number Co-ordination Number: is the number of closest neighbor of any constituent particle inlattice.

3. What do you mean by the term Conductivity?Conductivity: It is a measure of the conductance of a solution of 1 cm length and havingcross-sectional are of 1 square centimeter. Its unit is cm-1

4. What do you understand by Doping?Doping: Conductivity of intrinsic semi-conductor is too low to be of practical use. Theconductivity is increased by adding an appropriate amount of suitable impurity. This procis called Doping.

5. What is meant by Threshold Energy of a chemical reaction?Threshold energy: Minimum amount of energy which colliding molecules must possess order to have effective collisions.

6. What is meant by Electrode potential?Electrode Potential: is the potential difference set up between the metal and a solution of ion. It gives the tendency of an electrode to either lose or gain electrons.

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7. Is Molality better than Molarity in expressing concentration of a solution? Indicawhy, very briefly?Molality: Because, molarity changes with temperature (as liquids expand/contract witemperature), while molality does not change with temperature (as mass is independent temperature).

8. Define Activation energy of a reaction.Activation Energy: Excess energy (over and above the threshold energy) which must bsupplied to the reactants to undergo chemical reactions OR

Difference between (Threshold energy)-(Average kinetic energy of reacting molecules) OR

Activation Energy= Threshold energy Average kinetic energy OR

EL= Et - Ek

Section B[2 x 10 = 20]

9. Explain why conductivity of a metal decreases while that of an electrolyte increaswith increase in the temperature.Electrolyte: Conductivity is due to the movement of ions. Average kinetic energy of ionsthe electrolyte increases with increase in temperature. So, conductivity of electrolytincreases with increase in temperature.

Metals: Conductivity is due to mobility of electrons in the valance band/conductance baAs temperature increases, the mobility of electrons in this band is drastically affected ahence conductivity of metals decreases with increases in temperature.

10. Mention two factors that influence the extent of adsorption of a gas on a solid.Any two factors:

Nature of gas

Nature of adsorbent

Specific area of adsorbent

Effect of temperature

Effect of pressure

Activation of adsorbent

11. Define an electrochemical series. Indicate how it could predict the feasibility ofredox reactionElectrochemical series: is the arrangement of elements in the increasing order of its electro potentials. Also called Activity series.

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A Redox reaction is feasible only if the species which has higher electrode potential reduced (or accepts electrons) and the species which has lower electrode potential is oxidized(or loses electrons).

OR

The species to release electrons must have lower electrode potential than the species thaccepts electrons. This can be discerned from the Electrochemical Series. The E/C hel predict Redox reactions.

12. Explain Frenkels defects in lattices.Frenkels defect: arises when an ion is missing from its lattice site and occupies a

interstitial position. The crystal as a whole remains electrically neutral. Therefore, number oanion and cation remain the same. This type of defect occurs in compounds in which catiand anion are of different sizes. Eg: ZnS, AgCl.

13. How many atoms are there in a unit cell when a metal crystallizes in a :a. FCC structure b. BCC structure.

(a) FCC structure: four atoms/unit cell (b) BCC structure: two atoms/unit cell

14. Explain Brownian movement.Colloids have a dispersion medium and a dispersion phase. The molecules of the medium constantly colliding with the particles of the phase. These impacts are however random aunbalanced. These impacts give a random, zigzag motion to the colloidal particles, giving stirring effect, not allowing the phase particles to settle. Thus Brownian motion is responsi

for the stability of colloids. If there was no Brownian motion, the phase particles would hagradually settled due to gravity.

15. Explain the statement: the conductivity of a metal decreases while that of anelectrolyte increases with increase in the temperature. (1) Energy: Only those collisions of reactants will give products which possess energigreater than threshold energy.

(2) Orientation: Colliding molecules should have proper orientation so that old bonds m break and new bonds are formed.

16. Give any two distinguishing features between Physisorption and Chemisorption.Physisorption Chemisorption

(1) Low enthalpy of adsorption. (1) High enthalpy of adsorption

(2) Force between molecules of adsorbate andadsorbent is weak Vanderwaals force.

(2) Force is strong chemical force similar tochemical bonds.

(3) Reversible. (3) Irreversible.

(4) Activation energy not needed. (4) High activation energy needed.

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(5) Forms multi-molecular layer. (5) Mono-molecular layer.

(6) Rate of adsorption increases with increasein pressure of adsorbate.

(6) Rate of adsorption decreases as pressure of adsorbate increases.

17. Silver is deposited on a metallic vessel whose surface area is 900 cm2 by passing acurrent of 0.5 amp for 2 hours. Calculate the weight of silver deposited. Use thfollowing data: Atomic number of Ag = 108 a.m.u. F = 96500 C; density of Ag = 10.5

g/cm 2

Ag++e- Ag

Quantity of electricity passed,

Now, 96,500c would deposit 108g of Ag.

Let thickness of Ag deposited = x cm

i.e.

Or

Differentiate between multimolecular & macromolecular colloids.Multimolecular colloids Macromolecular colloids

(1) Aggregates of small molecules (1) Aggregate of large size molecules

(2) Molecular mass is low (2) Molecular mass is high

(3) They have lyophilic character (3) They are lyophobic in nature

(4) Atoms/molecules held by weakVanderwaals force. Hence colloidalarrangement is rigid.

(4) Colloidal arrangement is flexible and can takeany shape.

18. What is the role of cyrolite in the metallurgy of Aluminum?Aluminum is generally extracted from Bauxite (Alumina). Purified alumina is subject

to Electrolysis for extracting Aluminum. For this, purified Aluminum is dissolved in molcryolite. (Na3AlFe). Molten cryolite decreases melting point of fused Alumina to 1173K,

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simultaneously increasing its electrical conductivity, thus enabling the electrolytic extractiof the Al from Alumina.

Section C[3 x 9 = 27]

19. a. Define Standard Electrode Potential of a cell.Standard Electrode Potential: When the concentration of all species involved in a hacell is unity, than the electrode potential is called Standard Electrode Potential.

b. How is the Standard Electrode Potential of a cell related to:i. Equilibrium constant

ii. Gibbs free energy change.(i)

Where: n = number of electrons gained during electrode reactionMn+= Molar concentration of ions

E = Standard Electrode Potential

E = Electrode Potential

(ii)

Where: = Standard Gibbs Energy for reaction

n = number of electrons gained at electrode reaction

E = Standard Electrode Potential

F = Faradays constant of electricity

20. An element X with an atomic mass of 60 g/mol has density of 6.23 g/cm3. If thedge length of unit cell is 400 pm, identify the type of cubic unit cell. Calculate radiof an atom of this element.

Given: M= 600g/mol P= 6.23 g/cm3 NA= 6.023x1023/mol

We know that:

i.e.

or

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It is an FCC crystal lattice.

Radius = 141.4 pm

21.a. Differentiate between Average rate & Instantaneous rate of a reaction.

Average rate of reaction =

Instantaneous rate of reaction = Average rate of reaction when t is infinitely small.

b. Why is the use of instantaneous rate of reaction preferred over average rate oreaction?Except for a Zero-order reaction, for all reactions, the rate of reaction continuousldecreases with time. As a consequence, the Average rate of reaction loses itssignificance. Whereas, if we consider Instantaneous rate of reaction, it is not over period of time, but at any given instant of time. If the reactants are active, and threaction is going on, there will always be a definitely value for this. So, isgenerally preferred over .

22. What is Raolts Law? What are its limitations?Raolts Law: At any temperature, the partial vapor pressure of any volatile component osolution is equal to the product of vapor pressure of the pure component and mole fractioncomponent in solution.

Limitations:

(1) It is applicable only to dilute solutions.

(2) It is applicable to solutions of only non-electrolytes.

Or Define Kohlrauschs Law. How will you calculate m of Ba(OH)2 with the help of this law? Use the following data: m of NaOH, NaCl & BaCl2 are 2.481 x 10-2,1.265 x 10-2 & 2.800 x 10-2 Sm2/mol respectively.Kohlrauschs Law: At infinite dilution, when dissociation is complete, each ion makesdefinite contribution towards molar conductivity of the electrolyte, irrespective of the natof the other ion with which it is associated. From Kohlrauschs Law, we know that molconductance at infinite dilution for a salt is the sum of individual conductance from the ioof the electrolyte.

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23. Explain:a. Desorption

Desorption: Removal of adsorbate particles from the solid surface is calledDesorption. It is the reverse of Adsorption. It can occur on heating.

b. PeptizationPeptization: Process of conversion of a freshly prepared precipitate into a colloid badding a suitable electrolyte is called Peptization.

c. Shape selective Catalysis.Shape selective catalyst: is a catalyst whose catalytic action depends upon itsporestimulation and molecular sizes of reactants and productions. Eg. Zeolite.

24. Explain the conduction of electricity on the basis of Band Theory.Band model: Metal lattices have an extremely large number of atoms. Atomic orbitals metal atoms form a large number of molecular orbitals which are so close in energy to eaother that they form a band. If the band is partially filled of if it overlaps with a higher eneunoccupied band, then electrons can easily flow under an applied electric field. This is seas conductivity. The outermost filled energy band is called the Valence band. The next bain which electrons can move is called Conduction band. The space between these 2 banrepresents energies forbidden to the electron called Forbidden zone.

Metals: Conduction band is very close to valance band. So, easy movement of electronsthere into the conduction band. Therefore metals are good conductors.

Semi-metals/semi-conductors: Very small gap between conduction band and valance banAt temperature nearing absolute zero, conduction band would be empty. Electrons wouoccupy their lowest possible energy levels. The material would become a perfect insulatBut, at ordinary temperatures, some electrons get thermally excited from valance band conduction band and hence material becomes a conductor.

Insulators: Energy gap (forbidden zones) is very large. So, electrons for valence band ca jump to the conduction band. Hence, there is no conductivity.

25. Differentiate between amorphous & crystalline solids:a. Melting point b. Enthalpy of fusion.c. Anisotropy

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Amorphous solids Crystalline solids

Melting Point: They do not have sharpmelting point. They have amelting range, which softensover a scale of temperature

values.

They have a sharp andcharacteristic melting point.

Enthalpy of fusion:-

Amorphous solids: Crystalline Solids:

* They do not have definite enthalpy of fusion. * They have definite and characteristicEnthalpy of fusion.

26. Mention the principle on which the following metallurgical operations are based:a. Froth floatation process

Froth Flotation: This works on the principle of difference in the wetting properties othe ore and gangue particles with water and oil. It is used when the ore particles ar preferentially wetted by oil and gangue particles are wetted by water. It is useful foconcentration of all sulfide ores like galena (PbS), Zinc blende (ZnS), Pyrites(CuFeS2)

b. LevigationLevigation: Powdered ore is dropped into a hydraulic classifier from a hopper. A powerful jet of water is sprayed over the falling ore. Lighter particles of gangue arcarried upwards and are removed from an outlet at the top. Heavier ore particleswhich are free of physical impurities, settle down to the bottom.

c. Zone refining.Zone Refining: This process is based on the principle that impurities are moresoluble in the molten state if the metal than in its solid state. Hence, in this procesthe mixture containing the metal and its impurities is solidified. Crystals of very pumetal are formed, leaving the impurities in the molten part of the metal. The methois used for refining silicon, germanium, etc. for semiconductors.

27. Explain why the following phenomena occur:a. A beam of light passing through a colloidal solution has a visible path.

Due to Tyndall Effect. Since diameter of colloidal particles is comparable towavelength of visible light, light gets scattered and its path is visible.

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Amorphous solids Crystalline solids

Anisotropy: They are isotropic; have same physical properties in alldirections.

They areanisotropic and havedifferent physical properties indifferent directions.


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b. Passing an electric current through a colloidal solution removes colloida particles from it.Colloidal particles are electrically charged. They get discharged and accumulate netheir respective electrodes.

c. FeOH sol coagulates on addition of a solution of Potassium Sulphate.Peptization occurs.

Section D[5 x 3 = 15]

28. What are micelles? How do they differ from a normal colloidal solution? Explain wan example.Micelles: are substances which behave as an electrolyte at lowconcentration and as a colloidat high concentration. Generally they are associated colloids.

Micelles may contain as many as 100 molecules or more.

Eg: Soap.C 17H 35COONa Upon dissolving in water, it gives Na+ and C17H35COO- stearateions. The stearate ion is hydrophilic. If any oily dirt is present in water, say on a cloth dippin water, the Na+ ion enters into the oil, while the stearate ions stick out of the oil in the waterlike bristles. This leads to Micelle formation.

Every oil droplet is surrounded by number of suchmicelles. Gradually when the concentration reaches a critical limit, the electrically charghydrophilic heads start ripping off the oil droplets away from the cloth and start floatingwater, forming emulsions. Thus, at low concentrations, they behave as normal electrolyt but at higher concentration, they become emulsions, behaving as colloidal particles.

Or Show that time required for completion of three-fourths of a 1st order reaction is twicethe time required for completion of half reaction.

For aION- reaction:

Or

When t = t 1/2 ; x = a/2

----------

When t = t 3/4 ; x = 3a/4

s ----------

Now from and :

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29.a. Derive the Nernst equation for the following cells:

CuCu2+(0.13M) Ag+ (1 x 10-4M) AgUse the following data: E(Cu2+Cu) = +0.34V &E(Ag2+Ag) = +0.80V

To calculate the Nernst Equation for:

Cu Cu 2+ (o.13M) Ag + (1x10-4M) Ag

Given: E (Cu2+ Cu) = +0.34V

E (Ag2 Ag) = +0.80V

The Electrode reactions are:

The cell reaction is:

b. Can we store CuSO4 in an iron vessel or not? Explain. Use the following dataReduction potential of Cu2+Cu & Fe2+Fe are +0.34V & -0.44Vrespectively.

Reduction potential of Cu2+ Cu is +0.34V

Reduction Potential of Fe2+ Fe is 0.44V

This means Cu2+will be reduced to Cu and Fe will be oxidized toFe2+. i.e.the vessel will dissolve.

It is not possible to store CuSO4 in an iron vessel.

c. Explain how rusting of iron can be visualized as setting up of anelectrochemical cell.

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Water layer present on the surface of Iron dissolves acidic oxides present in air sucas CO2, SO2,etc. to form acids which dissociate to give H+ ions:

The particular spot which has H+ ions starts losing electrons to formferrous ions.

Oxidation takes place and that spot behaves as anode.

The electrons released at this anodic spot move through the metal to reach anothespot where H+ ions and dissolved oxygen use these electrons. Reduction takes placeand that spot behaves as cathode.

The overall reaction is:

Thus, an electrochemical cell sets up on the surface.

Or Explain the metallurgical extraction of Aluminum from Bauxite.

Extraction of Al from Bauxite involves three steps:(1) Purification of bauxite: Bayers process (Leaching)

Powdered bauxite is treated with hot concentrated solution of sodium hydroxide at 473-52&35 bar pressure. Alumina (Al2O3) dissolves in the solution to form sodium aluminate. This process is called Leaching.

Impurities are insoluble and they are removed by filtration. Soduim aluminate is neutraliz by passing CO2 gas.

We get hydrated alumina again. Freshly prepared alumina crystals are added as seed fo precipitation. The filtered alumina is filtered and heated up to 1473K to get pure alumi

(2) Electrolysis of fused alumina: Hall-Herolt processPurified alumina is dissolved in molten cryolite and put in an iron

tank lined with carbon bricks. This carbon bricked wall acts as the cathode. Carbon rods

used as anode. Molten cryolite increases the conductivity of solution and decreases melti point to 1173K. Fused electrolyte is covered with powdered coke.Page11 of 13


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Aluminum gets collected at the bottom of the tank, while carbon in the anode gets buraway. Thus anodes need to be replaced regularly.

(3) Electrode refining of Aluminum: Hoops process

An iron tank lined with carbon bricks is taken. Into this, three layers of molten liquid a poured. The bottom layer is molten 99% pure alumina obtained from the Hall-Herolt proceThe middle layer is fluorides of sodium, barium, and aluminum. The top layer consists pure molten aluminum. Carbon rods which act as cathode are dipped in the top layer.

When electrolysis takes place, aluminum ions from the middle layer are discharged at tcathode as pure molten aluminum. An equal amount of aluminum ions are displaced from lowest layer into the middle layer and the process continues until the entire lower layerdepleted of aluminum.

30. Calculate the freezing point of 1 molar aqueous solution (density = 1.04 g/L) of KC(Use the following data: K f for water = 1.86 kg/mol; atomic mass of K = 39; Cl = 36)

We know that:

= 2 * 1.86 * 1.0357

= 3.853o

Freezing point of Solution = 0 3.853o

= -3.853o

Or Explain the following with suitable examples:

a. FerromagnetismFerromagnetism: When there is spontaneous alignment of magnetic moments o

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b. ParamagnetismParamagnetism: Substances which have permanent magnetic dipoles and areattracted by a magnetic field are Paramagnetic substances and this property isParamagnetism.

c. FerrimagnetismFerrimagnetism; Where the magnetic movements of domains are aligned in paralleand antiparallel directions in unequal numbers resulting in a net magnetic movementhe substance are Ferrimagnetic and this property is Ferrimagnetism.

d. Anti-ferromagnetismAnti-Ferromagnetism: If the alignment of magnetic moments of domain is in acompensatory way so as to give a net zero magnetic moment because of cancelatioof individual magnetic moments, the substance is an Anti-Ferromagnet and the property is Anti-Ferromagnetism.

e. 12-16 and 13-15 compounds.12-16 & 13-15 Compounds: are types of semi-conductors. Solidbinary compounds prepared by combining elements of Gr-12 and 16 are 12-16 compounds. Eg: ZnS anCdS. Solid binary compounds prepared by combining elements of Gr-13 and 15 ar13-15 compounds. Eg: AlP ad GnAs.

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Chemistry XII Set 2

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