chemistry 2006 set 1 sol

Chemistry 2006 Set 1 Close

Subjective Test

(i) All questions are compulsory.(ii) This question paper consists of four sections A, B, C and D. Section A contains 5 questions of one mark each. Section B is of 7 questions of two marks each. Section C is of 12 questions of three marks each and Section D is of 3 questions of five marks each.(iii) There is no overall choice. However, an internal choice has been provided.(iv) Wherever necessary, the diagrams drawn should be neat and properly labelled.

Section A

Question 1 ( 1.0 marks)

A cubic solid is made of two elements X and Y. Atoms Y are at the corners of the cube and X at thebody centre. What is the formula of the compound?

Solution:

The atom at the body centre makes a contribution of 1 to the unit cell, while the atom at the corner

makes a contribution of to the unit cell.

Thus, number of atoms Y per unit cell

= Number of atoms × Contribution per unit cell

= 8 (at the corners) × atoms per unit cell

= 1

Thus, number of atoms X per unit cell

= Number of atoms × contribution per unit cell

= 1 (at the body centre) × 1

= 1

Thus, the formula of the given compound is XY.


Two liquids A and B boil at 145°C and 190°C respectively. Which of them has a higher vapourpressure at 80°C?

Solution:

Since liquid A has the lower boiling point among the two liquids, it has the higher vapour pressure.


For the reaction A → B, the rate of reaction becomes twenty seven times when the concentration ofA is increased three times. What is the order of the reaction?

Solution:

The reaction is

A → B

Let the order of the reaction be ‘n’

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Let the order of the reaction be ‘n’

Then,

Rate = k [A]n

Or, r = k (a)n … (1)

When the concentration of A is increased by three times, the rate of the reaction becomes twentyseven times, i.e.,

27r = k (3a)n … (2)

On dividing equation (2) by equation (1), we get

Or, n = 3

Thus, the order of the given reaction is 3.


Write the IUPAC name of: CH3.CH (Br) CH2CONHCH3.

Solution:

IUPAC name: N-methyl 3-bromobutanamide


Give a chemical test to distinguish between aniline and N-methyl aniline.

Solution:

Aniline and N-methyl aniline can be distinguished by Hinsberg’s method (i.e., reaction with benzenesulphonyl chloride).

Aniline (primary amine) will form N-phenyl benzene sulphonamide, which dissolves in KOH because ofthe presence of an acidic hydrogen on the N-atom.

On the other hand, N-methyl aniline (secondary amine) will form N-methyl-N-phenyl benzenesulphonamide, which does not dissolve in KOH due to the absence of acidic hydrogen on N-atom.

Section B


(a) Draw the structure of XeF2 molecule.

(b) Write the outer electronic configuration of Cr atom (Z = 24).

Solution:

(a)XeF2 has a linear structure.

F − Xe − F

(b) The outer electronic configuration of Cr atom is 3d5 4s1.


What is meant by entropy-driven reaction? How can a reaction with positive changes of enthalpyand entropy be made entropy driven?

Solution:



A reaction that is spontaneous despite being endothermic, i.e., having a positive value forenthalpy are known as entropy-driven reactions.

We know that

∆G = ∆H − T∆S

A reaction is spontaneous if the value of ∆G is negative.

We know that the value of ∆H is positive. Therefore, in order to make the reaction spontaneous,the value of T∆S should be greater than that of enthalpy. As the value of ∆S is positive, this canbe achieved by increasing the value of T. This means that a reaction with positive values ofenthalpy and entropy can be made entropy driven by increasing the temperature of the system.


Write balanced chemical equations for the following reactions:

(i) Ca3P2 + H2O →

(ii) XeF6 + 3 H2O →

Solution:

(i)

(ii)


Write chemical equations of the reactions involved in the manufacture of potassium permanganatefrom pyrolusite ore.

Solution:

Potassium permanganate can be prepared by the fusion of pyrolusite ore (MnO2) with an alkali metal

hydroxide (KOH) and an oxidising agent like KNO3.

The potassium manganate obtained is green in colour, and is oxidised to permanganateelectrolytically, or by passing through chlorine or ozone to obtain permanganate ion.

During electrolysis:


What are enantiomers? Draw the structures of the possible enantiomers of 3-methyl pent-1-ene.

Solution:

Out of current syllabus


Write the reactions and the conditions involved in the conversion of:

(a) Propene to 1-Propanol

(b) Phenol to Salicylic acid

Solution:

(a) Propene to 1-propanol

Diborane reacts with propene to produce tri-n-propylborane, which upon subsequent oxidation withalkaline H2O2 gives 1-propanol. The chemical equations involved in the reaction are as follows:



2 2

(b) Phenol to salicylic acid

When phenol is treated with sodium metal, sodium phenoxide is formed. When sodium phenoxide isheated with carbon dioxide at 400 K, under a pressure of 4 − 7 atmospheres, sodium salicylate isproduced. Sodium salicylate, on acidification, produces salicylic acid (2-hydroxybenoic acid). Thechemical equations involved in the reaction are as follows:


Write the structures of monomers used in the preparation of:

(a) Teflon

(b) PMMA

Or

(a) How does vulcanisation change the character of natural rubber?

(b) Why are the numbers 66 and 6 put in the names of nylon-66 and nylon-6?

Solution:

(a) Teflon

Monomer used is tetrafluoroethene.

Structure:

(b) PMMA



(b) PMMA

Monomer used is methyl 2-methyl propenoate

Structure:

Or

(a) The rubber obtained by the vulcanisation of natural rubber has excellent elasticity over a longrange of temperature. This is because heating rubber with sulphur causes cross-linking of polymerchains through disulphide bonds. This prevents the tearing of the polymers on stretching. Thus,vulcanised rubber has low water-absorption tendency, and is resistant to the action of organicsolvents and oxidising agents.

(b) Nylon-66 is derived from two monomers, hexamethylenediamine and adipic acid. Each of thesemonomeric units contains six carbon atoms. Hence, the two ‘6’s in nylon-66 represent the numbersof carbon atoms present in these monomers (adipic acid and hexamethylenediamine).

Nylon-6 is derived from the monomer, caprolactam. Caprolactam contains six carbon atoms. Thus,the number 6 in nylon-6 represents the number of carbon atoms present in this monomer.

Section C


State Heisenberg’s uncertainty principle. An electron has a velocity of 50 ms−1 accurate up to

99.99%. Calculate the uncertainty in locating its position? (Mass of electron = 9.1 ×10−31 kg, h = 6.6

× 10 −34 Js).

Solution:

Heisenberg’s uncertainty principle states that it is impossible to simultaneously measure the positionand momentum of a small particle with absolute accuracy or certainty

The product of the uncertainties in the position (∆x) and momentum (∆p) is always equal to or

greater than , where h is Planck’s constant, i.e.,

Where, m is the mass of the particle and ∆v is the uncertainty in velocity

In the given numerical,

Uncertainty in speed (∆v) = 99.99% of 50 m s−1

Mass of electron (m) = 9.1 × 10−31 kg

h = 6.6 × 10−34 Js

On applying the uncertainty principle, we get



∴ Uncertainty in locating the position of electron = 1.16 × 10−6 m


An element has a body-centred cubic structure with a cell edge of 288 pm. The density of the

element is 7.2 g cm−3 Calculate the number of atoms present in 208 g of the element.

Solution:

Cell edge (a) = 288 pm

Volume of unit cell = a3

= (288 pm)3

= (288 × 10−10 cm)3

= 2.389 × 10−23 cm3

Volume of 208 g of the element =

= 28.89 cm3

Number of unit cells =

= 12.09 × 1023

In a bcc structure, the number of atoms per unit cell = 2

∴ Number of atoms in 208 g of the given element = 2 × 12.09 × 1023

= 24.18 × 1023


(a) Why is the vapour pressure of a solution of glucose in water lower than that of water?

(b) A 6.90 M solution of KOH in water contains 30% by mass of KOH. Calculate the density of theKOH solution?

[Molar mass of KOH = 56 g mol −1]

Solution:

(a) Vapour pressure of a solution decreases when solute is added to it. When a non-volatile solute(glucose) is added to water, a part of the water surface is occupied by glucose molecules. Thus, theeffective surface area for the vaporisation of water molecules gets decreased. As a result, thevapour pressure of glucose solution is lower than that of water.

(b) Given, molar mass of KOH = 56 g mol−1

30% by mass KOH solution means that 30 g of KOH is present in 100 g of the solution.

6.90 M solution contains 6.90 mol of KOH in 1000 cm3 of the solution.

1 mol of KOH = 56 g

6.90 mol of KOH = (56 × 6.90) g

= 386.4 g

∴ 386.4 g of KOH is present =

= 1288 g of solution

Density of KOH solution =



= 1.288 g cm−3

Hence, the density of the given solution is 1.288 g cm−3


Answer the following in brief:

(a) Which of the two isomers of butane is more stable at 25°C and why?

Given [n-butane (∆fH 0 = −120 kJ mol−1) and iso-butane (∆fH

0 = −130 kJ mol−1)]

(b) For the change H2O (l) →H2O (g), predict the sign of ∆S.

(c) For the reaction N2(g) + 3H2 (g) → 2NH3 (g), predict whether work is done by the system or on

the system and why?

Solution:

(a) Given,

For n-butane, ∆fH0 = −120 kJ mol−1

For iso-butane, (∆fH0 = −130 kJ mol−1

The process of formation of both the isomers is exothermic, but ∆fH0 for iso-butene is more

negative than ∆fH0 for n-butane. This means that more energy is released during the formation

of isobutene. Therefore n-butane is more stable than iso-butane.

(b) Entropy is the measure of the degree of randomness or disorder in a system. The greater thedisorder, the higher is the entropy. Water in gaseous state has more disordered structure thanwater in liquid state. Therefore the entropy of water in gaseous state is higher than that ofwater in liquid state. Hence, for the change H2O(l) → H2O(g), the sign of ∆S is positive.

(c) In the given reaction, the volume of the system decreases. Therefore, work is done on thesystem.


The rate of a particular reaction triples when temperature changes from 50°C to 100°C. Calculatethe activation energy of the reaction.

[log 3 = 0.4771; R = 8.314 J K −1 mol−I]

Solution:

Energy of activation (Ea) is related to rate constant by Arrhenius equation.

It is given that the rate of reaction triples when temperature changes from 50°C to 100° C, i.e.,

when temperature is increased by 50°C and

= 913.51 J K−1 mol−1

∴ Ea = 913.51 J K−1 mol−1


(a) How can a colloidal solution and true solution of the same colour be distinguished from eachother?



(b) List four applications of adsorption.

Or

Explain the following observations:

(a) Lyophilic colloid is more stable than lyophobic colloid.

(b) Coagulation takes place when sodium chloride solution is added to a colloidal solution of ferrichydroxide.

(c) Sky appears blue in colour.

Solution:

(a) A colloidal solution and a true solution of the same colour can be distinguished from each otherby Tyndall effect. True solution does not show Tyndall effect, while colloidal solution shows Tyndalleffect.

(b) Four applications of adsorption are as follows:

(i) To create a very high vacuum, charcoal is used for adsorbing the remaining traces of air from avessel evacuated by a vacuum pump.

(ii) Chromatographic analysis of compounds is based upon the phenomenon of adsorption.

(iii) Silica and aluminium gels are used as adsorbents for removing moisture and controlling humidity.

(iv) Gas mask is a device which consists of activated charcoal or mixture of adsorbents. It is generallyused for breathing in coal mines and to adsorb poisonous gas such as CH4, CO, etc.

Or

(a) Lyophobic colloids can be easily precipitated (or coagulated) on the addition of small amounts ofelectrolytes, by heating or by shaking; hence, they are not stable.

On the other hand, lyophilic colloids are not easily coagulated. This is because the dispersed phasein a lyophilic colloid adsorbs the solvent and is distributed evenly in the medium. Thus, lyophiliccolloid is more stable than lyophobic colloid.

(b) Ferric hydroxide is a positively charged colloid. When an excess of electrolyte (e.g., NaCl solution)is added to the colloidal solution of ferric hydroxide, colloidal particles get precipitated. This is

because of the fact that colloids interact with the ions carrying the charge opposite (Cl−) to the onepresent on them. This causes neutralisation, which leads to their coagulation.

(c) Dust particles along with water suspended in air scatter blue light, which then reaches our eyes,thereby making the sky appear blue in colour.


Name the chief ore of silver. Describe with chemical equations the extraction of silver from this ore.

Solution:

Argentite, which is a sulphide ore, is the chief source of silver.

It is extracted by the process of leaching with a dilute solution of sodium cyanide.

Silver dissolves and forms a complex, argentocyanide. It is further treated with scrap zinc, whichdisplaces silver from the complex. This is called displacement method or hydrometallurgy. Thechemical reactions involved are given below.


(a) Using valence bond theory predict the geometry and magnetic behaviour of [Cr (NH3)6]3+ ion [Cr

= 24].

(b) Write IUPAC name of:

[Pt (NH 3)2Cl2]

Solution:

(a) The oxidation number of chromium (Cr) in [Cr (NH3)6]3+ ion is +3. Electronic configuration of Cr is



3 6

[Ar] 3d54s1.

Since the resulting complex [Cr(NH3)6]3+ involves d2sp3 hybridisation, it is octahedral in shape. The

presence of three unpaired electrons in the complex makes the complex paramagnetic.

(b) IUPAC name of [Pt (NH3)2Cl2] is diamminedichloridoplatinum (II).


(a) The isotope of decays in 14 steps, by loss of 8 α-particles and 6 β- particles. What are the

mass number and atomic number of the end product?

(b) Write the equation for the complete reaction (D, α).

(c) How is the nuclear binding energy related to the stability of the nucleus?

Solution:

(a) The decaying of the isotope of by the loss of 8 α-particles and 6 β-particles can be

represented as

Thus, the mass number and atomic number of the end product are 203 and 82 respectively.

(b) The complete equation for the given reaction is

(c) Nuclear binding energy gives an idea of the relative stability of the nuclide. Larger the value ofbinding energy per nucleon, larger will be its relative stability.


(a) Describe the following giving suitable examples:

(i) Cannizzaro reaction

(ii) Aldol condensation

(b) Give a chemical test to distinguish between ethanal and propanal.

Solution:

(a) (i) Cannizarro reaction

Aldehydes having no α-hydrogen atoms undergo self oxidation and reduction (disproportionation)reactions on treatment with a concentrated alkali. In such reactions, one molecule of aldehyde getsoxidised to form an acid and the other molecule of aldehyde gets reduced to form an alcohol.

For example, two molecules of formaldehyde, in the presence of concentrated KOH, producemethanol and potassium formate.

(ii) Aldol condensation

In the presence of a dilute alkali, aldehydes and ketones (having at least one α-hydrogen atom)produce β-hydroxyl aldehyde (aldol) and β-hydroxyl ketone respectively. The aldol and ketol thenreadily lose water to give α, β-unsaturated carboxyl compounds. Such reactions are called aldolcondensation.



condensation.

For example, in the presence of dil. NaOH, ethanal (aldehyde) produces 3-hydroxybutanal (aldol),which then readily loses water to produce but-2-enal.

(b) Ethanal contains the CH3CO− group linked to H, but propanal does not contain the CH3CO-

group linked to H or C. Hence, ethanal reacts with I2/NaOH to give yellow precipitate of iodoform

(positive iodoform test) and propanal does not reacts with I2/NaOH


Account for the following:

(a) Electrophilic substitution in case of aromatic amines takes place more readily than benzene.

(b) CH3CONH2 is a weaker base than CH3CH2NH2.

(c) Nitro compounds have higher boiling points than hydrocarbons having almost same molecularmass.

Solution:

(a) The benzene ring in aniline is highly activated. This is because of the presence of a lone pair ofelectrons on the nitrogen atom. The displacement of the lone pair on nitrogen towards the benzenering facilitates the electrophilic attack on the ring. Thus, due to the presence of −NH2 group,

electrophilic substitution takes place more readily in aromatic amines than benzene.

(b) In CH3CONH2, the electron pair on nitrogen is withdrawn towards carboxyl group, as shown by

the following resonance structures.

Thus, the lone pair of electrons is not available for donation.

On the other hand, because of the +I effect of the C2H5 group, the lone pair of electrons can be

easily donated in CH3CH2NH2. This is why CH3CONH2 is a weaker base than CH3CH2NH2.

(c) Nitro group are highly polar in nature, and therefore, there is a strong attraction between thenitrogen and oxygen of nitro groups. As a result, a large amount of energy is required to separatethem. Hence, compounds containing nitro groups have high boiling points in comparison withhydrocarbons of the same molecular mass.


Define the following and give one example of each:

(a) Tranquillisers

(b) Mordant

(c) Hybrid rocket propellants

Solution:

(a) Tranquillisers are the neurologically active drugs. They are the class of chemical compounds usedfor the treatment of stress, anxiety, and mild or severe mental diseases. They are essentialcomponents of sleeping pills.



(b) Out of current syllabus

(c) Out of current syllabus

Section D


(a) Explain why electrolysis of aqueous solution of NaCl gives H2 at cathode and Cl2 at anode. Write

overall reaction.

(b) Calculate the emf of the cell Zn/Zn2+ (0.1 M) || Cd2+ (0.01 M)/Cd at 298 K, [Given

and

Or

(a) Account for the following:

(i) Alkaline medium inhibits the rusting of iron.

(ii) Iron does not rust even if the zinc coating is broken in a galvanised iron pipe.

(b)

(i) Construct a galvanic cell using the above data.

(ii) For what concentration of Ag + ions will the emf of the cell be zero at 25°C, if the concentration of

Cu2+ is 0.01 M?[log 3.919 = 0.593]

Solution:

(a) At cathode, the following reduction reactions can take place:

A reduction reaction with higher reduction potential is preferred. Therefore, the reaction at thecathode during electrolysis is

This is why electrolysis of aqueous solution of NaCl gives H2 at the cathode.

At anode, the following oxidation reactions can take place:

At anode, the reaction with lower value of E0 is preferred. But, due to overvoltage, oxidation ofchloride ion occurs and chlorine gas is obtained. Hence, the reaction at the anode during electrolysisis

This is why electrolysis of aqueous solution of NaCl gives Cl2 at the anode.

The overall cell reaction is given below.

.



(b) emf of the given cell

Thus, the emf of the give cell is 0.36 V.

OR

(a)

(i) Alkaline medium decreases the availability of ions. In the rusting of iron, ions facilitate the

oxidation of iron by first picking up electrons produced from water, and then, by reducing theoxygen. Thus, alkaline medium inhibits the rusting of iron.

(ii) The layer of zinc on the surface of iron produces a protective layer of basic zinc carbonate[ZnCO3.Zn(OH)2]. This is produced by the reaction between zinc, carbon dioxide and moisture in air.

Also, zinc, being more electropositive, undergoes oxidation first.

Thus, even if the zinc coating is broken in a galvanised iron pipe, iron does not rust because of theprotective layer of basic zinc carbonate.

(b) (i)

(ii) The reaction having more reduction potential occurs at the cathode. Thus,

Cathode reaction:

Anode reaction:

For the given galvanic cell,





Give reasons for the following:

(a) Molten aluminium bromide is poor conductor of electricity.

(b) Nitric oxide becomes brown when released in air.

(c) PCl5 is ionic in nature in the solid state.

(d) Ammonia acts as a ligand.

(e) Sulphur disappears when boiled with an aqueous solution of sodium sulphite.

Or

Name the principal ore of tin or lead. Describe the different steps (along with equations for thereactions) involved in the extraction of the metal from the ore named. Name an important alloy ofeach, tin and lead.

Solution:

(a) Aluminium bromide exists as a dimer, having a coordinate bond. In this structure, there

are no free electrons, and thus, molten aluminium bromide is a poor conductor of electricity.

(b) In the presence of sunlight, colourless nitric oxide is oxidised to nitrogen dioxide. This NO2 is

brown in colour. Thus, nitric oxide becomes brown when released in air.

(c) PCl5 exists as a salt in solid state. It contains a tetrahedral cation and an octahedral

anion . Thus, PCl5 is ionic in nature in the solid state.

(d) Ammonia contains a lone pair of electrons. Thus, it can easily bind with other chemical entities toform complexes, and acts as a good ligand.

(e) Out of current syllabus

Or

Out of current syllabus


(a) State two main differences between globular proteins and fibrous proteins.

(b) Based on their chemical composition, state how are lipids classified? Give one example of eachclass.

Or

(a) ‘Hormones are chemical messengers.’ Explain?

(b) Name the main disease caused due to lack of the vitamin and its source in each of the following:A, B6 and E.

Solution:

(a)The two main differences between globular proteins and fibrous proteins are listed in the giventable.

Globular proteins Fibrous proteins

1. They are cross-linked condensation polymers of acidicand basic amino acids.

1. They are linearcondensation products.

2. They are usually soluble in water. 2. They are generallyinsoluble in water.

(b) Out of current syllabus

OR

(a) Hormones transfer information from one group of cells to a distant tissue or organ. They arecarried to different parts of the body by the blood stream. Thus, hormones are chemical messengers.

(b) The main disease caused due to the lack of the given vitamins and their sources are listed in thegiven table.

Name ofvitamins

Sources Deficiency diseases

Vitamin AFish liver oil, butter,

milkNight blindness



Vitamin B6 Yeast, milk, egg yolk Convulsions

Vitamin E Vegetable oilsLoss of sexual power of reproduction and

muscular power



chemistry 2006 set 1 sol

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