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Chemistry 2008 Set 3 Close Subjective Test These are additional set 3 questions. Question 1 ( 1.0 marks) What type of substances exhibits antiferromagnetism? Solution: Antiferromagnetism is exhibited by substances which have domain structure similar to ferromagnetic substances, but are oppositely oriented (thereby cancelling out each other’s magnetic moment), e.g., MnO. Schematic alignment of magnetic moments in antiferromagnetic substances Question 2 ( 1.0 marks) Express the relation between conductivity and molar conductivity of a solution. Solution: The relation between conductivity and molar conductivity of a solution is Where, c = Concentration of the solution Question 3 ( 1.0 marks) Which has a higher enthalpy of adsorption, physisorptions or chemisorptions? Solution: Chemisorptions have higher enthalpy of adsorption than physisorptions. This is because physisorptions involve weak Van der Waal’s forces, while chemisorptions involve strong chemical bonds. Question 10 ( 2.0 marks) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10 -3 S cm -1 ? Solution: Given, Conductivity, κ = 0.146 × 10 -3 S cm -1 Resistance, R = 1500 Ω Cell constant = κ × R = 0.146 × 10 -3 × 1500 = 0.219 cm -1 Question 19 ( 3.0 marks) How would you account for the following? 2/4/2011 Subjective Test Paper - Chemistry - Meri… …meritnation.com/…/KSFLbZQ7rODN7… 1/4

Chemistry 2008 Set 3 Sol

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Page 1: Chemistry 2008 Set 3 Sol

Chemistry 2008 Set 3 Close

Subjective Test

These are additional set 3 questions.

Question 1 ( 1.0 marks)

What type of substances exhibits antiferromagnetism?

Solution:

Antiferromagnetism is exhibited by substances which have domain structure similar to ferromagneticsubstances, but are oppositely oriented (thereby cancelling out each other’s magnetic moment),e.g., MnO.

Schematic alignment of magnetic moments in antiferromagnetic substances

Question 2 ( 1.0 marks)

Express the relation between conductivity and molar conductivity of a solution.

Solution:

The relation between conductivity and molar conductivity of a solution is

Where, c = Concentration of the solution

Question 3 ( 1.0 marks)

Which has a higher enthalpy of adsorption, physisorptions or chemisorptions?

Solution:

Chemisorptions have higher enthalpy of adsorption than physisorptions. This is becausephysisorptions involve weak Van der Waal’s forces, while chemisorptions involve strong chemicalbonds.

Question 10 ( 2.0 marks)

The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the

cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10−3 S cm−1?

Solution:

Given,

Conductivity, κ = 0.146 × 10−3 S cm−1

Resistance, R = 1500 Ω

∴Cell constant = κ × R

= 0.146 × 10−3 × 1500

= 0.219 cm−1

Question 19 ( 3.0 marks)

How would you account for the following?

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Page 2: Chemistry 2008 Set 3 Sol

How would you account for the following?

(i) Frenkel defects are not found in alkali metal halides.

(ii) Schottky defects lower the density of related solids.

(iii) Impurity doped silicon is a semiconductor.

OR

Explain the following properties giving suitable examples:

(i) Ferromagnetism

(ii) Paramagnetism

(iii) Ferrimagnetism

Solution:

(i) Frenkel defects are shown by ionic solids having large differences in the sizes of ions. Solids such

as ZnS, AgCl show these defects due to the small size of Zn2+ and Ag+ ions, and the large size ofanions. Alkali metals are not so small so as to show these defects. Hence, Frenkel defects are notfound in alkali metal halides.

(ii) Schottky defects are basically vacancy defects in ionic solids. In these defects, lattice sitesbecome vacant. As a result, the density of the substance decreases.

(iii) Silicon is an intrinsic semi-conductor in which conductivity is very low. To increase its conductivity,silicon is doped with an appropriate amount of suitable impurity. When doped with electron-richimpurities such as P or As, n-type semi-conductor is obtained, and when doped with electron-deficient impurities, p-type semi-conductor is obtained. In n-type semiconductor, negatively chargedelectron is responsible for increasing conductivity, and in p-type semiconductor, electron hole isresponsible for increasing conductivity.

OR

(i) Ferromagnetism:The substances that are strongly attracted by a magnetic field are calledferromagnetic substances. Ferromagnetic substances can be permanently magnetised even in theabsence of a magnetic field. Some examples of ferromagnetic substances are iron, cobalt, nickel,gadolinium and CrO2.

When such a substance is placed in a magnetic field, all the domains that act as tiny magnets getoriented in the direction of the magnetic field. As a result, a strong magnetic effect is produced. Thisordering of domains persists even after the removal of the magnetic field. Thus, the ferromagneticsubstance becomes a permanent magnet.

Schematic alignment of magnetic moments in ferromagnetic substances

(ii) Paramagnetism: The substances that are weakly attracted by a magnetic field are called

paramagnetic substances. Some examples of paramagnetic substances are O2, Cu2t, Fe3t and Cr3t.

Paramagnetic substances get magnetised in a magnetic field in the same direction, but losemagnetism when the magnetic field is removed. To undergo paramagnetism, a substance must haveone or more unpaired electrons. This is because the unpaired electrons are attracted by a magneticfield, thereby causing paramagnetism.

(iii) Ferrimagnetism: The substances in which the magnetic moments of the domains are aligned inparallel and anti-parallel directions, in unequal numbers, are said to have ferrimagnetism. Examplesinclude Fe3O4 (magnetite), ferrites such as MgFe2O4 and ZnFe2O4.

Ferrimagnetic substances are weakly attracted by a magnetic field as compared to ferromagneticsubstances. On heating, these substances become paramagnetic.

Schematic alignment of magnetic moments in ferrimagnetic substances

Question 21 ( 3.0 marks)

Explain the basic principles of the following metallurgical operations:

(i) Zone refining

(ii) Vapour phase refining

(iii) Electrolytic refining

Solution:

(i) Zone refining:

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Page 3: Chemistry 2008 Set 3 Sol

This method is based on the principle that impurities are more soluble in the molten state of metal(the melt) than in the solid state. In the process of zone refining, a circular mobile heater is fixed atone end of a rod of impure metal. As the heater moves, the molten zone of the rod also moves alongwith it. As a result, pure metal crystallises out of the melt and the impurities pass on to the adjacentmolten zone. This process is repeated several times, which leads to the segregation of impurities atone end of the rod. Then, the end with the impurities is cut off.

Silicon, boron, gallium, indium etc. can be purified by this process.

(ii) Vapour phase refining:

Vapour phase refining is the process of refining metal by converting it into its volatile compound, andthen, decomposing it to obtain a pure metal. To carry out this process,

(a) the metal should form a volatile compound with an available reagent, and

(b) the volatile compound should be easily decomposable so that the metal can be easily recovered.

Nickel, zirconium and titanium are refined using this method.

(iii) Electrolytic refining:

Electrolytic refining is the process of refining impure metals by using electricity. In this process, animpure metal is made the anode and a strip of pure metal is made the cathode. A solution of asoluble salt of the same metal is taken as the electrolyte. When electric current is passed, metal ionsfrom the electrolyte are deposited at the cathode as pure metal and the impure metal from theanode dissolves into the electrolyte in the form of ions. The impurities present in the impure metalgets collected below the anode. This is known as anode mud.

Question 22 ( 3.0 marks)

Explain what is observed when:

(i) an electrolyte, KCl, is added to hydrated ferric oxide sol

(ii) an electric current is passed through a colloidal solution

(iii) a beam of strong light is passed through a colloidal solution

Solution:

(i) When KCl is added to hydrated ferric oxide sol, it dissociates to give K+ and Cl− ions. Particles offerric oxide sol are positively charged. This neutralises the colloidal particles which then unite to formbigger particles that are consequently precipitated. Thus, they get coagulated in the presence of

negatively charged Cl− ions.

(ii)The colloidal particles are charged and carry either a positive or a negative charge. The dispersionmedium carries an equal and opposite charge. This makes the entire system neutral. Under theinfluence of an electric current, the colloidal particles move towards the oppositely chargedelectrode. When they come in contact with the electrode, they lose their charge and coagulate.

(iii)When a beam of light is passed through a colloidal solution, scattering of light is observed. This isknown as Tyndall effect. This scattering of light illuminates the path of the beam in the colloidalsolution.

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