124319178XII ISC Guess Question Paper for 2011 to Print

8/6/2019 124319178XII ISC Guess Question Paper for 2011 to Print

1/25

GUESS QUESTION paper IN CHEMISTRY FOR XII (ISC) BOARD

EXAMINATION MARCH 2011Paper 1

(THEORY)(Three Hours)

(Candidates are allowed additional 15 minutes for only reading the paper. They must

NOT start writing during this time)

------------------------------------------------------------------------------------------------------Answer all questions in Part I and six questions from Part II, choosing two questions

from Section A, two from Section B and two from Section C.All working, including rough work, should be done on the same sheet as, and adjacent to,

the rest of the answer. The intended marks for questions or parts of questions are given in

brackets [ ]. Balanced equations must be given wherever possible and diagrams wherethey are helpful. When solving numerical problems, all essential working must be shown.

In working out problems use the following data: Gas ConstantR = 1.987 cal deg-1

mol-1

= 8.314 JK-1

mol-1

= 0.0821 dm3

atm K-1mol

-1

1 l atm = 1 dm3

atm = 101.3 J. 1 Faraday = 96500 Coulombs.

------------------------------------------------------------------------------------------------------------

PART I (Answerallquestions)

Question 1

a) Fill in the blanks choosing appropriate words given in brackets: [5]

(zero, infinity, twelve, negative, positive, four, six, increases, vant Hoffs

factor, does not change, nucleophilic, no alpha, ammoniacal, alpha)

(1) The ratio of observed value of colligative property to the calculated value of

colligative property is called.............................................

(2) The standard reduction potential value of normal hydrogen electrode is.....................(3) The co-ordination number of Na

+ion in rock salt structure is...........

(4) If the free energy change is ...................the reaction does not proceed in the forward

direction.(5) Cannizzaro reaction is shown by an aldehyde containing................

hydrogen atom.

Ans: (1) vant Hoffs factor (2) zero (3) six (4) positive (5) no alpha

(b) Complete the following statements by selecting the correct alternative from the

choices given: [5]

(1) The pair of sugar which give the same product with excess of phenylhydrazine are:

1

a) starch and cellulose b) cellulose and sucrose

Prepared by:Prof. S.Narayana Iyer,M.Sc,M.Phil


2/25

c) sucrose and glucose d) glucose and fructose

(2) Mesotartaric acid is optically inactive due to the presence of:a) Molecular symmetry b) Molecular asymmetry

c) External compensation d) Asymmetric carbon atom

(3) Which of the following is an amorphous solid ?a) NaCl b) glass c) CaF2 d) graphite

(4) Which of the following is not a carboxylic acid?

a) oxalic acid b) picric acid c) propanoic acid d) formic acid(5) Benzaldehyde treated with alkaline KMnO4 gives:

a) CO2 and H2O b) benzyl alcohol c) salicylic acid d) benzoic acidAns:

(1) d) glucose and fructose (2) a) Molecular symmetry (3) b) glass

(4) b) picric acid (5) d) benzoic acid

(c) Answer the following questions: [5]

(1)Why water cannot be separated completely from ethyl alcohol by fractional

distillation?

Ans.Ethyl alcohol and water (95.4% ethyl alcohol and 4.6% water) form constant boiling

mixture (azeotrope) boiling at 351.1 K. Hence, further water cannot be separated

completely from ethyl alcohol by fractional distillation.

(2) Which solution will allow greater conductance of electricity, 1 M NaCl at 293 K or 1

M NaCl at 323 K and why ?

Ans. 1 M NaCl at 323 K as the ionic mobilities increase with increase in temperature.

(3) For a reaction A + H2O B; r = k [A]. What is its (i) Molecularity (ii) Order?

Ans. (i) Pseudo unimolecular reaction (ii) order = 1

(4) Why it is necessary to use a salt bridge in a galvanic cell?

Ans. To complete the inner circuit and to maintain electrical neutrality of the electrolyticsolutions of the half-cells.

(5) Give the relationship between free energy change and EMF of a cell.

Ans.- G = n FEcell

G = Free energy changes n = moles of electrons F = 96500 Coulomb

Ecell = EMF of a cell

2


3/25

(d) Match the following: [5](i) Benedicts reagent a) ammoniacal silver nitrate

(ii) Fehling solution b) Williamsons synthesis

(iii)Tollens reagent c) formaldehyde(iv) heating of alkyl halide d) copper slulphate + sodium

and sodium alkoxide potassium tartrate

(v) urotropine e) copper sulphate + sodium citrateAns:

(i) Benedicts reagent e) copper sulphate + sodium citrate(ii)Fehling solution d) copper slulphate + sodium potassium tartrate

(iii)Tollens reagent a) ammoniacal silver nitrate

(iv) heating of alkyl halideand sodium alkoxide b) Williamsons synthesis(v) urotropine c) formaldehyde

PART IIAnswersix questions choosing two from Section A, two from Section B and two from

Section C.

SECTION A (Answer any two questions)Question 2

(a) An aqueous solution of cane sugar containing 1.72 g in 100 mL begins to freeze

at -0.093oC. The cryoscopic constant of water is 1.86 K mol

-1kg

-1

Calculate the molecular weight of cane sugar. [3]Ans.

o o

f

f

f

T = 0 - (-0.093 ) = 0.093

1000 x 1.86 x 1.721000 k wm = = = 344T W 0.093 x 100

(b) At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of

4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what

would be its concentration? [3]Ans. Here, T= 300 K = 1.52 bar R = 0.083 bar L K

1mol

1

Applying the relation,

= CRT

3


4/25

C =

R T

C =1.52 bar

0.083 bar L K-1 m ol-1

x 300 K

= 0.061 mol

since the volume of the solution is 1 L, the concentration of the solution would be

0.061M.

(c) Give reasons for the following:(i) The specific conductivity of a solution decreases with dilution [2]

(ii) Ionic solids conduct electricity in molten state but not in solid state. [2]

(i) Specific conductivity of a solution depends upon number of ions present in per

unit volume. On dilution, the number of ions per unit volume decreases. Therefore thespecific conductivity also decreases.

(ii) In ionic compounds, electricity is conducted by ions. In solid state, ions are heldtogether by strong electrostatic forces and are not free to move about within the solid.

Hence, ionic solids do not conduct electricity in solid state. However, in molten state or

in solution form, the ions are free to move and can conduct electricity

Question 3(a) (i) Silver crystallises in fcc lattice. If edge length of the cell is 4.07 10

8cm and

density is 10.5 g cm3

, calculate the atomic mass of silver [3]

Ans.It is given that the edge length, a = 4.077 10

8cm

Density, d= 10.5 g cm3

As the lattice is fcc type, the number of atoms per unit cell,z = 4

We also know that, NA = 6.022 1023

mol1

Using the relation:

4


5/25

d =z M

a3 NA

M =d a3NA

z=

10.5 gcm3 x (4.077 x 10-8 cm )3x 6.023 x 1023

4= 107.143

= 107.143 gmol1

Therefore, atomic mass of silver = 107.143 u

(ii) What is the total number of sigma and pi bonds in the following molecules?(a) C2H2 (b) C2H4 [1]

Ans.a) Structure of C2H2 can be represented as:

CH C H

Hence, there are three sigma and two pi-bonds in C2H2.

b) The structure of C2H4 can be represented as:

C

H

C

H

HH

Hence, there are five sigma bonds and one pi-bond in C2H4.

(b) (i) Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If

the decomposition is a first order reaction, calculate the rate constant of the reaction. [3]

Ans. For a 1st

order reaction,

1/20.693t =

k

5


6/25

It is given that t1/2 = 60 min

k =0.693

t 1/2

=0.693

60= 0.01155 min-1

or k = 1.925 x 10-4 s-1 (ii) The rate of a reaction quadruples when the temperature changes from 293 K to 313 K.Calculate the energy of activation of the reaction assuming that it does not change with

temperature. [3]

Ans. From Arrhenius equation, we obtain

logk2

k1=

Ea

2.303 R

T2 - T1

T1T2

It is given that k2 = 4k1

T1 = 293K

T2 = 313K

log

k1=

4k1 Ea

2.3038.314x

313 - 293

293 x 313

0.6021 =20 x Ea

2.303 8.314x 293 x 313x

Ea =0.6021 x 2.303 8.314x 293 x 313x

20=52863.33J mol-1

= 52.86kJmol-1

Hence, the required energy of activation is 52.86 kJmol1.

Question 4(a) (i) Although geometries ofNH3 and H2O molecules are distorted tetrahedral, bond

angle in water is less than that of ammonia. Discuss. [2]

6


7/25

Ans. The molecular geometry of NH3 and H2O can be shown as:

The central atom (N) in NH3 has one lone pair and there are three bond pairs. In H2O,there are two lone pairs and two bond pairs.

The two lone pairs present in the oxygen atom ofH2O molecule repels the two bond

pairs. This repulsion is stronger than the repulsion between the lone pair and the threebond pairs on the nitrogen atom.

Since the repulsions on the bond pairs in H2O molecule are greater than that in NH3, the

bond angle in water is less than that of ammonia.

O

HH

N

HH

H

(ii) Calculate the pH of a buffer solution containing 0.01 mole of acetic acid and 0.015

mole of sodium acetate. Ka of acetic acid is 1.75 x 10-5

[3]

Ans.

pH = - log Ka +log[salt]

[acid]

pH = - log 1.75 x 10

-5

+ log

[0.15]

[0.1]

pH = - 4.756962 + 0.1761 = 4.933

b) Give reasons why:

(i) Example 1: Calculate the standard e.m.f. of the cell : Cd, Cd2+

|| Cu2+

,Cu anddetermine the cell reaction. The standard reduction potentials of

Cu2+

, Cu and Cd2+

, Cd are 0.34V and 0.40 volts respectively. Predict the

feasibility of the cell reaction.

Eocell = Standard EMF of the cell = E

oright E

oleft

= [Std. reduction potential of Cu2+

, Cu] [Std. reduction potentials of Cd2+

, Cd]

= EoCu2+, Cu Eo

Cd2+, Cd

= 0.34 V ( 0.4 V)

= + 0.74 Volts.

7

Left hand electrode (oxidation half cell) reaction is


8/25

Cd(s) Cd2+

+ 2e

Right hand electrode (reduction half cell) reaction isCu

2++ 2e Cu(s)

The cell reaction is

Cd(s) + Cu2+

(aq) Cd2+

(aq)+ Cu(s)

Eocell is positive. Therefore the cell reaction is feasible.

(ii) Determine the standard emf of the cell and standard freeenergy change of the cell reaction.

Ans. Zn, Zn2+

|| Ni2+

, Ni. The standard reduction potentials of Zn2+

, Zn and

Ni2+

, Ni half cells are 0.76 V and 0.25 V respectively.

Eocell =E

oR E

oL = 0.25 ( 0.76)

= + 0.51 V Eocell is + ve. Therefore G

o= ve.

Therefore G

o

= n FE

o

celln = 2 electrons

Therefore Go

= 2 96495 0.51 = 97460 Joules = 97.46 kJ.

SECTION B (Answer any two questions)

Question 5 (a). Describe the extraction of Cu from its ore under the following heads:

(i) concentration (ii) Roasting (iii) Reduction (iv) Refining of crude metal. [3]

8

(Ans). The main ores of copper are copper pyrites (CuFeS2) and copper glance (Cu2S).The various steps usually involved are given below:

Concentration: Concentration is done by froth-floatation. The finely powdered ore is

mixed with water and pine oil in large tank. The ore particles are wetted by the oil,whereas the gangue particles are wetted by water. Air is blown through the mixture. As aresult oil froth containing ore particles is formed which floats on the top of the water and

can be skimmed off easily.

Roasting: The concentrated ore is roasted. Roasting is done in reverberatory furnace.

2CuFeS2 + O2Roasting

Cu2S + 2FeS + SO2

Cu2S and FeS are partially oxidized.

2Cu2S + 3O2 2Cu2O + 2SO22FeS. + 3O2 2FeO + 2SO2Reduction : Takes place in Bassemer converter. It can be tiled in any position. It has a

basic lining inside.The little FeS present in matte is completely oxidized to FeO and then changed into slag

with reaction of silica. Cuprous sulphide reacts with Cu2O to form blister copper.

Cu2S + 2Cu2O 6Cu + SORefining of copper : The impure copper metal is purified by the process of electrolytic


9/25

refining to get pure copper metal.

The impure metal is made the anode and a strip of pure copper cathode. These areimmersed in a copper sulphide solution with a little sulphuric acid added to it. On passing

an electric current the anode begins to dissolve pure copper gets deposited on the cathode.

The impurities either dissolves in the solution or deposited below the anode as anode

mud.

At anode :Cu (Copper) Cu2+

(copper ion)

At cathode : Cu2+

(Copper ion) Cu (Copper atom) deposits on

cathode

Thus pure copper is obtained at the cathode.

(b) Write balanced equations for the following: (i) NaCl is heated with sulphuric acid in

the presence of MnO2. (ii)Chlorine gas is passed into a solution of NaI in water. [2]

Ans. (i)

(ii)

9


10/25

Question 6

(a) Write the IUPAC names of the following coordination compounds:

(i) [Co(NH3)6]Cl3 (ii) [Co(NH3)5Cl]Cl2 (iii) K3[Fe(CN)6] (iv) K3[Fe(C2O4)3]

(v) K2[PdCl4] (vi) [Pt(NH3)2Cl(NH2CH3)]Cl [2]

Ans. (i) Hexaamminecobalt(III) chloride

(ii) Pentaamminechloridocobalt(III) chloride

(iii) Potassium hexacyanoferrate(III)

(iv) Potassium trioxalatoferrate(III)

(v) Potassium tetrachloridopalladate(II)

(vi) Diamminechlorido(methylamine)platinum(II) chloride

(b) Using IUPAC norms write the formulas for the following:

(i) Tetrahydroxozincate(II)

(ii) Potassium tetrachloridopalladate(II)

(iii) Diamminedichloridoplatinum(II)

(iv) Potassium tetracyanonickelate(II)

(v) Pentaamminenitrito-O-cobalt(III)

(vi) Hexaamminecobalt(III) sulphate

(vii) Potassium tri(oxalato)chromate(III)

(viii) Hexaammineplatinum(IV)

(ix) Tetrabromidocuprate(II)(x) Pentaamminenitrito-N-cobalt(III) [1]

Ans. (i) [Zn(OH]2

(ii) K2[PdCl4] (iii) [Pt(NH3)2Cl2] (iv) K2[Ni(CN)4]

(v) [Co(ONO)(NH3)5]2

(vi) [Co(NH3)6]2 (SO4)3 (vii) K3[Cr(C2O4)3](viii) [Pt(NH3)6]4+

(ix) [Cu(Br)4]2

(x) [Co[NO2)(NH3)5]2+

(c) Draw the structures of optical isomers of: (i) [Cr(C2O4)3]3

(ii) [PtCl2(en)2]2+

(iii) [Cr(NH3)2Cl2(en)]+

[1]

Ans. (i) [Cr(C2O4)3]3

10


11/25

(ii) [PtCl2(en)2]2+

(iii) [Cr(NH3)2Cl2(en)]+

(d) Explain the bonding in coordination compounds in terms of Werners postulates. [1]

11


12/25

Ans Werners postulates explain the bonding in coordination compounds as follows:

(i) A metal exhibits two types of valencies namely, primary and secondary valencies.Primary valencies are satisfied by negative ions while secondary valencies are satisfied

by both negative and neutral ions.

(In modern terminology, the primary valency corresponds to the oxidation number of themetal ion, whereas the secondary valency refers to the coordination number of the metal

ion.

(ii) A metal ion has a definite number of secondary valencies around the central atom.Also, these valencies project in a specific direction in the space assigned to the definite

geometry of the coordination compound.

(iii) Primary valencies are usually ionizable, while secondary valencies are non-ionizable.

Question 7

(a) Describe the preparation of potassium permanganate from pyrolusite. How does the

acidified permanganate solution react with (i) iron(II) ions (ii) SO2 and (iii) oxalic acid?Write the ionic equations for the reactions. [3]

Ans.Potassium permanganate can be prepared from pyrolusite (MnO2). The ore is fused

with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as

KNO3 or KClO4, to give K2MnO4.

2MnO2 + 4KOHheat

2K2MnO4 + 2H2O

(green)The green mass can be extracted with water and then oxidized either electrolytically or by

passing chlorine/ozone into the solution.Electrolytic oxidation

K2MnO4 2K+

+ MnO2

H2O H+

+ OH-

At anode, manganate ions are oxidized to permanganate ions.

MnO2 MnO4-+ 4e

-

Green Purple

Oxidation by chlorine

2K2MnO4 + Cl2 2KMnO4 + 2KCl

2MnO42- + Cl2 2MnO4- + 2Cl-

Oxidation by ozone

12


13/25

2K2MnO4 + O3 + H2O 2KMnO4 + 2KOH + O2

2MnO42-

+ O3 + H2O 2MnO4-+ 2OH

-+ O2

(i) Acidified KMnO4 solution oxidizes Fe (II) ions to Fe (III) ions i.e., ferrous ions to

ferric ions.

(ii) Acidified potassium permanganate oxidizes SO2 to sulphuric acid.

(iii) Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide.

(b) Why is H2O a liquid and H2S a gas? [2]Ans. H2O has oxygen as the central atom. Oxygen has smaller size and higher

electronegativity as compared to sulphur. Therefore, there is extensive hydrogen bonding

in H2O, which is absent in H2S. Molecules of H2S are held together only by weak van derWaals forces of attraction.

Hence, H2O exists as a liquid while H2S as a solid.

SECTION C (Answer any two questions)

Question 8(a) How will you convert?

(i) Benzene into aniline. [2]

13


14/25

Ans.

(ii) Ethanal into But-2-enal [2]Ans. On treatment with dilute alkali, ethanal gives 3-hydroxybutanal which on heating

produces but-2-enal.

(iii) Benzaldehyde to Benzophenone. [2]Ans.

14

(b) An aromatic compound A on treatment with aqueous ammonia and heating forms


15/25

compound B which on heating with Br2 and KOH forms a compound Cof molecular

formula C6H7N. Write the structures and IUPAC names of compounds A, B and C. [3]

Ans. It is given that compound C having the molecular formula, C6H7N is formed by

heating compound B with Br2 and KOH. This is a Hoffmann bromamide degradation

reaction. Therefore, compound B is an amide and compound C is an amine. The only

amine having the molecular formula, C6H7N is aniline, (C6H5NH2).

Therefore, compound B (from which C is formed) must be benzamide,

(C6H5CONH2).

Further, benzamide is formed by heating compound A with aqueous ammonia.

Therefore, compound A must be benzoic acid.

The given reactions can be explained with the help of the following equations:

15


16/25

(c) Write chemical reaction of aniline with benzoyl chloride and write the name of the

product obtained.

Ans.

Question 9

(a) Give the equations of reactions for the preparation of phenol from cumene. [2]Ans. To prepare phenol, cumene is first oxidized in the presence of air of cumene hydro-

peroxide.

16


17/25

Then, cumene hydroxide is treated with dilute acid to prepare phenol and acetone as by-

products.

(b) Predict the products of the following reactions: [2]

(i)

(ii)

OC2H5

+ HBr

17


18/25

(iii)

OC2H5Conc.H2SO4

Conc.HNO3

(iv)

(CH3)3C-OC2H5

HI

Ans. (i)

(ii)

18


19/25

(iii)

(iv)

(c) Describe a method for the identification of primary, secondary and tertiary amines.

Also write chemical equations of the reactions involved. [3]

Ans. Primary, secondary and tertiary amines can be identified and distinguished byHinsbergs test. In this test, the amines are allowed to react with Hinsbergs reagent,

benzenesulphonyl chloride (C6H5SO2Cl). The three types of amines react differently with

Hinsbergs reagent. Therefore, they can be easily identified using Hinsbergs reagent.

Primary amines react with benzenesulphonyl chloride to form N-alkylbenzenesulphonyl

amide which is soluble in alkali.

Due to the presence of a strong electron-withdrawing sulphonyl group in the

sulphonamide, the Hatom attached to nitrogen can be easily released as proton. So, it isacidic and dissolves in alkali.

Secondary amines react with Hinsbergs reagent to give a sulphonamide which is

insoluble in alkali.

19


20/25

There is no Hatom attached to the N-atom in the sulphonamide. Therefore, it is not

acidic and insoluble in alkali.

On the other hand, tertiary amines do not react with Hinsbergs reagent at all.

(d) Give the structures of A, B and C in the following reactions: [3](i)

CH3CH2INaCN

AOH-

partial hydrolysisB

NaOH,Br2C

(ii)

C6H5N2ClCuCN

AH2O/H

+

BNH3

C(iii)

CH3CH2BrKCN

A

LiAlH4B

HNO2

0oCC

(iv)

C6H5NO2Fe/HCl

ANaNO2 + HCl

273KB

H2O/H+

C(v)

CH3COOHNH3

ANaOBr

BNaNO2/HCl

C

20


21/25

(vi)

C6H5NO2 Fe/HCl A HNO2

B273K

C6H5OH C

Ans. (i)

CH3CH2INaCN

CH3CH2CN

propanenitrile

(A)

partial hydrolysis

OH-

CH3 - CH2 - C - NH2

||O

(C)

NaOH + Br2

CH3 - CH2 - NH2

ethanamine

(B)

ethyl iodide propanamide

(ii)

21


22/25

(iii)

(iv)

(v)

22


23/25

(vi)

C6H5NO2Fe/HCl

C6H5NH2

Aniline (A)

HNO2

273KC6H5-N2

+Cl-

Benzenediazonium chloride (B)

C6H5OH

N=N

p-Hydroxyazobenzen (C)

Question 10

(a)How will you convert ? [3]1) acetaldehyde to formaldehyde

2) benzene to phenol

3) phenol to benzoic acid.

23


24/25

(b) Classify the following as addition and condensation polymers: Terylene, Bakelite,

Polyvinyl chloride, Polythene. [2]Ans. Addition polymers: Polyvinyl chloride, polythene

Condensation polymers: Terylene, bakelite

(c) An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilutesulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with

chromic acid produced (B). (C) on dehydration gives but-1-ene.Write equations for thereactions involved. [5]Ans. An organic compound A with molecular formula C8H16O2 gives a carboxylic acid

(B) and an alcohol (C) on hydrolysis with dilute sulphuric acid. Thus, compound A must

be an ester. Further, alcohol C gives acid B on oxidation with chromic acid. Thus, B and

C must contain equal number of carbon atoms.Since compound A contains a total of 8 carbon atoms, each of B and C contain 4 carbon

atoms.Again, on dehydration, alcohol C gives but-1-ene. Therefore, C is of straight chain

and hence, it is butan-1-ol.On oxidation, Butan-1-ol gives butanoic acid. Hence, acid B is butanoic acid.

Hence, the ester with molecular formula C8H16O2 is butylbutanoate.

24


25/25

All the given reactions can be explained by the following equations.

25

Documents

124319178XII ISC Guess Question Paper for 2011 to Print