Last modified on 30 June 2009, at 19:46

Solved Question Papers - IIT JEE/PAPER 1

SOLUTIONS TO IIT-JEE 2009 CHEMISTRY: Paper-I

SECTION – I ''Single Correct Choice Type This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), for its answer out of which ONLY ONE is correct.

Note: Questions with (*) mark are from syllabus of class XI.

1. Among the electrolytes Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effective coagulating agent for Sb2S3 sol is
(A) Na2SO4 (B) CaCl2 (C) Al2(SO4)3 (D) NH4Cl                              
Sol.: According to Hardy–Schulze rule, the coagulating power of an ion is directly proportional to the magnitude of its charge.
       So, For negatively charged Sb2S3 sol, the most effective coagulating agent would be Al2(SO4)3 .
Correct choice: (C)
*2. Given that the abundances of isotopes 54Fe, 56Fe and 57Fe are 5%, 90% and 5%, respectively, the atomic mass of Fe is
(A) 55.85 (B) 55.95 (C) 55.75 (D) 56.06
 Sol.: Average atomic mass of an element = ∑AiXi
  where Ai and Xi represents the atomic mass and mole fraction of the component in the mixture.
  Average atomic mass of Fe =(54×5)+(56×90)+(57×5)/100
                            = 55.95 
Correct choice: (B)
*3. The correct acidity order of the following is
  http://upload.wikimedia.org/wikipedia/commons/4/48/JEE_Benzene_Derivatives.jpg
(A) (III) > (IV) > (II) > (I) (B) (IV) > (III) > (I) > (II) (C) (III) > (II) > (I) > (IV) (D) (II) > (III) > (IV) > (I)
 Sol.: Carboxylic acid is stronger acid than phenol. The presence of electron donating methyl group decreases acidic strength while
     presence of electron withdrawing halogen increases acidic strength.
Correct choice: (A)
*4. The IUPAC name of the following compound is
 http://upload.wikimedia.org/wikipedia/commons/e/e0/Chemistry-IITJEE.jpg
(A) 4-Bromo-3-cyanophenol (B) 2-Bromo-5-hydroxybenzonitrile
(C) 2-Cyano-4-hydroxybromobenzene (D) 6-Bromo-3-hydroxybenzonitrile
 Sol.: –CN group is principal functional group.
Correct choice: (B)
*5. The term that corrects for the attractive forces present in a real gas in the van der Waals equation is
(A) nb     (B) an^2/V^2  (C) -an^2/V^2   (D) -nb
 Sol.: The attractive forces in a real gas decreases its pressure relative to an ideal gas.
           Pr = Pi - an^2/V^2 ; Pi = Pr + an^2/V^2
          The term that accounts for decrease in pressure due to attractive forces among molecules of a real gas must be added to the
          real gas pressure (observed) to get the ideal gas pressure.
Correct choice: (B)
6. Among cellulose, poly(vinyl chloride), nylon and natural rubber, the polymer in which the intermolecular force of attraction is weakest is
(A) Nylon (B) Poly(vinyl chloride) (C) Cellulose (D) Natural Rubber
 Sol.: Nylon and cellulose, both have intermolecular hydrogen bonding whereas polyvinyl chloride has dipole-dipole interaction.
       Natural rubber will have London forces which are weakest.
Correct choice: (D)
7. The reaction of P4 with X leads selectively to P4O6. The X is
(A) Dry O2 (B) A mixture of O2 and N2
(C) Moist O2 (D) O2 in the presence of aqueous NaOH
  Sol.: P4 reacts with O2 in limited supply of air (a mixture of O2 and N2) to give P4O6.
      P4 + 3O2 limited supply of air→ P4O6.
Correct choice: (B)
8. The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm. The mole fraction of N2 in air is 0.8.
The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure is
(A) 4.0 × 10–4 (B) 4.0 × 10–5 (C) 5.0 × 10–4 (D) 4.0 × 10–6
 Sol.: According to Henry’s law
     PN2 = KN2 XN2
  Where K is the Henry’s constant (in atm) and
  XN2 is mole fraction of N2.
  PN2 = XN2 PT = 0.8 × 5 atm = 4.0 atm.
  4 atm = 1.0 × 105 atm × XN2
  4 × 10–5 =     nN2         =     nN2 
               nN2 + nH2O       nN2 + 10
 nN2 = 4 × 10-4 moles
Correct choice: (A)