IIT-JEE 2009 Chemistry Paper - I Questions and Solutions

Advertisements

PART I :CHEMISTRYPAPER -ISECTION –IStraight ObjectiveType

---

This section contains8 multiple choice questions. Each question has 4 choices (A), (B), (C)and (D), out of which ONLY ONE iscorrect.

---

For the benefit of 11th/12th Studyingstudents, we have (*) marked the questions which are from 11thsyllabus.You are advised to solve these questions in 100minutes.

*1. Given that the abundances of isotopes54_Fe,56_Fe and57_Fe are 5%, 90% and 5%respectively, the atomic mass of Feis
(A) 55.85 (B)55.95
(C) 55.75 (D)56.05
Key.(B)
Sol. The atomic mass of an element is the average mass numberof all its naturally occurring isotopes,the
averaging being done on the basis of their respectiveabundances. Thus
At. Mass of Fe = 5* 54 +90* 56 +5 *57/100=55.95
Hence(B)

*2. Theterm that corrects for the attractive forces present in a real gas inthe van der Waals equationis
(A) nb(B)an²/v²
(c)-an²/v²(D)–nb
Sol. From Vanderwaals’ equation for n moles of a real gas
The pressure correction factoran²/v²
accounts for the loss of pressure due to inward pull. ‘a’ isvanderwaals’
constant for a gas and it is the measure of the strength ofthe vanderwaals’ intermolecularattraction.
Hence
key(B)

3. Among the electrolytesNa₂SO₄,CaCl₂,Al₂(SO₄)₃and NH₄Cl, the most effective coagulating agentfor
Sb2S2 sol is
(A)Na₂SO₄(B)CaCl₂
(C)Al₂(SO₄)3(D)NH₄Cl
Key.(C)
Sol. Sb2S3 is a negatively chargedcolloid.
∴ cation is responsible for its coagulation andthe most effective isAl₂[SO₄]3
1 mol can furnish 2 ×Al3+ion
i.e. 6 moles of +ve charges which is the maximum in the givenchoices
Hence(C)

4. The Henry’s law constant for the solubility ofN₂ gas in water at 298 K is 1.0 × 105 atm. Themole fraction
of N₂ in air is 0.8.The number of moles of N₂ fromair dissolved in 10 moles of water at 298 K and 5atm
(A) 4.0 × 10–4 (B) 4.0 ×10–5
(C) 5.0 × 10–4 (D) 4.0 ×10–6
Key.(A)
Sol.

This means 1 mole of solution will contain 4 × 10–5 mole ofN₂ and 1 – 4 × 10–5 ≈ 1 molewater.
Thus, 1 mole of water is association with 4 × 10–5 moleN₂
∴ 10 moles of water is associated with 4 × 10–5moleN₂
Hence(A)

5. The reaction of P4 with X leads selectively toP40₆. The Xis
(A) Dry O₂ (B) Amixture of O₂ andN₂
(C) Moist O₂ (D)O₂ in the presence of aqueousNaOH
Key.(B)
Sol. P4 gives P40₆when oxygen is in limited supply. Hence a mixture ofO₂ and N₂ is mostsuitable


*6. The correct acidity order of the followingis

(A) (III) >(IV) > (II) > (I) (B) (IV) > (III)> (I) >(II)
(C) (III) > (II) > (I) >(IV) (D) (II) > (III) > (IV) >(I)
Key.(A)
Sol.
Carboxylic acids are more acidic than phenol. EWG exertsacid-strengthening effect while EDGexerts
acid-weakening effect. The –I effect of Cl is dominating overits +R effect.
∴(A)

7. Among cellulose, poly(vinyl chloride), nylon and naturalrubber, the polymer in which theintermolecular
force of attraction is weakestis
(A) Nylon (B) Poly(vinylchloride)
(C) Cellulose (D) NaturalRubber
Key.(D)
Sol. Nylonand cellulose are Fibres. PVC is a thermoplastic while Natural Rubber is an elastomer.The
intermolecular foces of attraction can be gradedas
Elastomer < Thermoplastic <Fibres.

*8. The IUPAC name of the following compoundis

(A) 4-Bromo-3-cyanophenol (B)2-Bromo-5-hydroxybenzonitrile
(C) 2-Cyano-4-hydroxybromobenzene (D)6-Bromo-3-hydroxybenzonitrile
Key.(B)
Sol. The priority orderis:
–CN > Br >–OH
So the compoundis
2-bromo 5-hydroxy benzonitrile

SECTION –II Multiple Correct AnswerType

---

This section contains 4multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C)and(D), out of which ONE OR MORE is/arecorrect.

---

9. The correct statement(s) regarding defects in solidsis(are)
(A) Frenkel defect is usually favoured by a very smalldifference in the sizes of cation andanion
(B) Frenkel defect is a disolocationdefect
(C) Trapping of an electron in the lattice leads to theformation of F-centre
(D) Schottky defects have no effect on the physicalproperties of solids
Key. (B,C)
Sol. Frenkel defect is usually favoured by a very largedifference in the sizes of cation and anion. In Scottydefect pairs of cations andanions left the crystal lattice as a result density Decreases. Hence band ccorrect.

10. The compound(s) that exhibit(s) geometrical isomerismis(are)
(A) [Pt(en)Cl₂] (B)[Pt(en)2]Cl₂
(C)[Pt(en)2Cl₂]Cl₂(D)[Pt(NH3)2Cl₂]
Key(C,D)
SOL:


*11. The compound(s) formed upon combustion of sodium metalin excess air is(are)

(A)Na₂O₂(B)Na₂O
(C) NaO₂ (D)NaOH
Key. (A,B)



*12. The correct statement(s) about the compound H3C(HO)HC-CH= CH – CH(OH)CH3(X)is(are)
(A) The total number of stereoisomers possible for X is6
(B) The total number of diastereomers possible for X is3
(C) If the stereochemistry about the double bond in X istrans, the number of enantiomers possible for Xis
4
(D) If the stereochemistry about the double bond in X is cis,the number of enantiomers possible for X is2
Key. (A,D)

SECTION –III
Linked ComprehensionType

---

Thissection contains 2 paragraphs. Based upon each paragraph, 3 multiplechoice questions have tobe< answered. Each question has 4choices (A), (B), (C) and (D) out of which ONLY ONE iscorrect.

---

Paragraph forQuestions Nos. 13 to15
p-Amino-N, N-dimethylaniline is added to a strongly acidicsolution of X. The resulting solution is treated witha few drops of aqueous solutionof Y to yield blue coloration due to the formation of methylene blue. Treatmentof the aqueous solution of Y with thereagent potassium hexacyanoferrate(II) leads to the formation of an intenseblue precipitate. The precipitatedissolves on excess addition of the reagent. Similarly, treatment of thesolution of Y with the solution of potassium hexacyanoferrate (III) leads to abrown coloration due to the formation ofZ.

13. The compound Xis


14. The compound Yis
(A) MgCl₂ (B)FeCl₂
(C) FeCl₃ (D)ZnCl₂
Key(C)
Sol.

Theprecipitate dissolves in the large excess of the reagent producing anintense blue solution.


15. The compound Zis
(A)Mg₂[Fe(CN)₆](B)Fe[Fe(CN)₆]
(C)Fe4[Fe(CN)₆]₃(D)K₂Zn₃[Fe(CN)₆]₂

Key(B)
Sol.

Paragraph forQuestions Nos. 16 to18
A carbonyl compound P, which gives positive iodoform test,undergoes reaction with MeMgBr followed dehydration to give an olefin Q.Ozonolysis of Q leads to a dicarbonyl compound R, which undergoesintramolecualr aldol reaction to give predominantlyS.



16.The structures of the carbonyl compound Pis


17. The structures of the products Q and R, respectively, are
KeyB


KeyA

18.The Structure of product S,is


KeyB

SECTION –IVMatrix MatchType

---

Thissection contains 2 questions. Each questioncontains statements given in two columns, which have to be matched. The statements in Column Iare labeled A, B, C and D, while the statements in Column II are labelled p, q, r, sand t. Any given statement in Column I can have correct matching with ONE OR MOREstatement(s) in Column II. The appropriate bubbles corresponding to the answers to thesequestions have to be darkened as illustrated in the followingexample: If the correct matches are A –p, s and t; B – q and r; C – p and q; and D – s and t;then the correct darkening ofbubbles will look like the following:

---

19.Match each of the diatomic molecules in Column I with itsproperty/properties in ColumnII


20. Match each of the compounds in Column I with itscharacteristic reaction in ColumnII.