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JEE Main 2016 Chemistry Answer Key & Solution
C
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JEE MAIN - 2016
PART B - CHEMISTRY
y 15 x 4
15
15 x
g.
y y C x H y x O2 xCO2 H 2O 4 4
ng
31. ( )
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31. A 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20 % O2 by volume for complete combustion. After combustion the gases occupy 330 mL . Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure , the formula of the hydrocarbon is : (1) C3H 6 (2) C3H 8 (3) C4H 8 (4) C4H10
y 20 y 15 x 375 or x 5 4 100 4 y = 12 (which is not possible)
Also,
15 x
80 375 Vfinal 330 100
WRONG QUESTION
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Now,
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32. Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure pi and temperature T1 are conncented through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to T2. The final pressure pf is : T1
T1
Pi,V
(1)
32. (3)
T1
T T pi 1 2 T1 T2
T1
Pi,V
Pi,V
(2)
Pi,V
T 2 pi 1 T1 T2
(3)
T 2 pi 2 T1 T2
(4)
TT 2 pi 1 2 T1 T2
Initial total moles = final total moles
or Pf
P1V P1V Pf V Pf V T1 T1 T1 T2
or
T T 2P1V Pf V Pf V Pf V 1 2 T1 T1 T2 T1T2
2P1V T1T2 2P1T2 T1 V (T1 T2 ) V (T1 T2 )
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JEE Main 2016 Chemistry Answer Key & Solution
33. A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of (h /) (where is wavelength associated with electron wave) is given by :
C (1)
33. (4)
meV
(2)
2meV
(3)
(4)
meV
K.E. of electron = (V.l.) Now,
or,
h mv
h 2m( K .E.)
h 2m.v.e
h 2m.v.e
34. The species in which the N atom is in a state of sp hybridization is : (1) NO2+ (2) NO2– (3) NO3– (4)
NO2
NO2+ is sp hybridized
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34. (1)
2meV
C O2 CO2
H1 393.5 KJ/mol
ng
35. (4)
g.
35. The heats of combustion of carbon and carbon monoxide are – 393.5 and – 283.5 kJ mol–1, respectively. The heat of formation (in kJ) of carbon monoxide per mole is : (1) 110.5 (2) 676.5 (3) – 676.5 (4) – 110.5
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1 CO O2 CO2 H 2 283.5 KJ/mol 2 1 C O2 CO H1 H 2 110 KJ/mol 2
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36. 18 g glucose (C6H12O6) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is : (1) 7.6 (2) 76.0 (3) 752.4 (4) 759.0 36. (3)
The question does not have PA° or temperature. However assuming the PA° = 1 atm we can solve : PA n A P P nB 0
or
P 178 .2 180 99 18 18 760 P
P
99 760 = 752. 4 torr 100
37. The equilibrium constant at 298 K for a reaction A + B C + D is 100. If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in mol L–1) will be : (1) 0.182 (2) 0.818 (3) 1.818 (4) 1.182 37. (3)
A + B C +D Keq = 100 Initial 1 1 1 1 Equilibrium (1 – x) (1 – x) (1 + x) (1 + x) (as Q < Keq) 2 2 (1 + x) / (1 – x) = 100 or (1 + x) / (1 – x) = 10 or x = 9 /11. [D] = 1 + x = 1 + ( 9 / 11) = ( 20 / 11) = 1.818 M
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JEE Main 2016 Chemistry Answer Key & Solution
38. Galvanization is applying a coating of : (1) Pb (2) Cr 38. (4)
(3)
Cu
(4)
Zn
Galvanization is the process of formation of a layer of zinc to prevent corrosion.
C
39. Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be : (1) 6.93 × 10–2 mol min–1 (2) 6.93 × 10–4 mol min–1 –1 (3) 2.66 L min at STP (4) 1.34 × 10–2 mol min–1 2 × t1/2 = 50 or t1/2 = 25 min as in 50 min, concentration reduces to (1 / 4)th of original value. K = (0.693 / 25) min–1 – d[H2O2] / dt = K [0.05] = {(0.693 × 0.05) / 25} = 1.386 × 10–3 H2O2 H2O + ½ O2 d[O2] / dt =– ½ d[H2O2] / dt = (1.386 × 10–3) / 2 = 0.693 × 10–3 = 6.93 × 10–4 mol.min–1
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39. (2)
40. (3)
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g.
40. For a linear plot of log ( x / m) versus log p in a Freundlich adsorption isotherm, which of the following statements is correct ? (k and n are constats) (1) Both k and 1/n appear in the slope term (2) 1/n appears as the intercept (3) Only 1/n appears as the slope (4) log(1/n) appears as the intercept. Freundlich equation is : (x / m) = kp1/n or log (x / m) = log k + (1 / n) log P slope = (1 / n)
Sc - 4s2 3d1 (631 kJ / mol)
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41. (4)
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41. Which of the following atoms has the highest first ionization energy ? (1) Rb (2) Na (3) K (4)
Sc
Na = 3s1 (496 kJ / mol)
42. Which one of the following ores is best concentrated by froth floatation method? (1) Magnetite (2) Siderite (3) Galena (4) Malachite 42. (3)
Froth floatation process is used for concentration of metal sulphides. Magnetite - Fe3O4 Galena - PbS Siderite - FeCO3 Malachite - CuCO3.Cu(OH)2.
43. Which one of the following statements about water is FALSE? (1) Water is oxidized to oxygen during photosynthesis (2) Water can act both as an acid and as a base (3) There is extensive intramolecular hydrogen bonding in the condensed phase (4) Ice formed by heavy water sinks in normal water. 43. (3)
There is extensive intermolecular hydrogen bonding in the condensed phase where each H2O molecule is surrounded by three neighbouring molecules.
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JEE Main 2016 Chemistry Answer Key & Solution
44. The main oxides formed on combustion of Li, Na and K in excess of air are, respectively : (1) Li2O, Na2O and KO2 (2) LiO2, Na2O2 and K2O (3) Li2O2, Na2O2 and KO2 (4) Li2O, Na2O2 and KO2
C
44. (4)
excess Li Li2O (normal oxide) air
excess Na Na2O2 (peroxide) air
excess
K KO2 (superoxide) air
45. The reaction of zinc with dilute and concentrated nitric acid, respectively, produces : (A) N2O and NO2 (B) NO2 and NO (C) NO and N2O (D) NO2 and N2O 45. (1)
4Zn + 10 HNO3 (dilute) 4Zn(NO3)2 + N2O + 5H2O Zn + 4HNO3(conc.) Zn(NO3)2 + 2NO2 + 2H2O
Orthophosphorus Acid (+ 3 o.s) (H3PO3)
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46. (1)
O
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OH OH
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P
H
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g.
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46. The pair in which phosphorous atoms have a formal oxidation state of +3 is : (1) Orthophosphorous and pyrophosphorous acids (2) Pyrophosphorous and hypophosphoric acids (3) Orthophosphorous and hypophosphoric acids (4) Pyrophosphorous and pyrophosphoric acids
w
Pyrophosphorous acid ( + 3 o.s) (Dibasic acid) (H4P2O5) O
O
P
HO
P
O
H
OH
H
Hypophosphoric acid is H4P2O6 which is tetrabasic with an oxidation state of +4 and pyrophosphoric acid is H4P2O7 which is tetrabasic with an oxidation state of +4.
47.
Which of the following compounds is metallic and ferromagnetic? (1) TiO2 (2) CrO2 (3) VO2 (4)
47. (2)
MnO2
CrO2 is a metallic metal oxide and Ferromagnetic (These have unpaired electrons and strongly attracted in magnetic field).
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JEE Main 2016 Chemistry Answer Key & Solution
48. The pair having the same magnetic moment is : [ At. No. Cr =24, Mn = 25, Fe = 26, Co = 27] (1) [Cr(H2O)6]2+ and [CoCl4]2– (2) (3) [Mn(H2O)6]2+ and [Cr(H2O)6]2+ (4)
[Cr(H2O)6]2+ and [Fe(H2O)6]2+ [CoCl4]2– and[Fe(H2O)6]2+
C 3d
48. (2)
4s
4p
[Cr(H2O)6]2+
sp3d2
[CoCl4]–2
sp3
[Fe(H2O)6]2+
[Mn(H2O)6]2+
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g.
sp3d2
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sp3d2
+
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en
Co
49. (2)
en
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49. Which one of the following complexes shows optical isomerism? (1) [Co(NH3)3Cl3] (2) cis [Co(en)2Cl2]Cl (3) trans [Co(en)2Cl2]Cl (4) [Co(NH3)4Cl2]Cl (en = ethylenediamine)
Cl
Cl
Cl
Cl en
Cis-d -isomer
en
+
Co
en
Cis-l -isomer
50. The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was found to be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm respectively. This water is unsuitable for drinking due to high concentration of : (1) Fluoride (2) Lead (3) Nitrate (4) Iron 50. (3)
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JEE Main 2016 Chemistry Answer Key & Solution
51. The distillation technique most suited for separating glycerol from spent-lye in the soap industry is: (1) Simple distillation (2) Fractional distillation (3) Steam distillation (4) Distillation under reduced pressure
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51. (4)
Glycerol can be separated from spent -lye in soap industry by using distillation under reduced pressure. This method in used to purify liquids having very high b.p. and those which decomposed at or below their boiling point.
52. The product of the reaction given below is : 1.NBS/hv 2. H2O/K2CO3 X
OH
(2)
(3)
OH
Br
H2O/K2CO3
g.
NBS/hv
(4)
co m
(1)
COOH
O
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52. (2)
53. The absolute configuration of
53. (2)
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CH3
(1)
(2R, 3S)
H
H
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CO2H OH Cl
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H H
(2)
(2S, 3R)
(3)
(2S, 3S)
(4)
(2R, 3R)
COOH OH (S) (R)
Cl
CH3
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54. 2-chloro-2-methyl pentane on reaction with sodium methoxide in methanol yields : CH3 C2H5CH C CH3
C2H5CH2C CH2
C (a)
C2H5CH2C
OCH3
(b)
CH3
CH3
(1)
All of these
(2)
(c)
(a) and (c)
(3)
(c) only
CH3
(4)
(a) and (b)
Cl
54. (3)
+ CH CH2 CH3 CH3ONa E2 H CH3
CH3 C
CH3 C
CH CH2 CH3
CH3
CH3 CH CH2
Cl
ng
HOCl –OH + Cl+
+
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55. (2)
g.
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55. The reaction of propene with HOCl(Cl2+ H2O) proceeds through the intermediate : (1) CH3 – CH+ – CH2 – OH (2) CH3 – CH+ – CH2 – Cl (3) CH3 – CH(OH) – CH2+ (4) CH3 – CHCl – CH2+
CH3
+ CH CH2
OH
OH
(intermediate)
CH3 CH CH2
Cl
w
Cl
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56. In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of amine produced are : (1) One mole of NaOH and one mole of Br2 (2) Four moles of NaOH and two moles of Br2 (3) Two moles of NaOH and two moles of Br2 (4) Four moles of NaOH and one mole of Br2 O
56. (4)
R C NH2 + Br2 + 4NaOH R – NH2 + 2KBr + K2CO3 + 2H2O
Hoff mann - Bromamide reaction.
57. Which of the following statements about low density polythene is FALSE ? (1) Its synthesis requires high pressure (2) It is a poor conductor of electricity (3) Its synthesis requires dioxygen or a peroxide initiator as a catalyst. (4) It is used in the manufacture of buckets , dust -bins etc. 57. (4)
Manufacture of buckets, dust-bins etc are done by using High density polythene (HDP).
JEE MAIN - 2016
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JEE Main 2016 Chemistry Answer Key & Solution
Thiol group is present in : (1) Cytosine (2)
Cystine
(3)
Cysteine
(4)
Methionine
C COOH
58. (3)
Cystcine is
H2N
H
CH2 SH
59. Which of the following is an anionic detergent? (1) Sodium stearate (3) Cetyltrimethyl ammonium bromide 59. (2)
(2) (4)
Sodium lauryl sulphate Glyceryl oleate
(CH3)(CH2)10 CH2OSO3– Na+ is sodium lauryl sulphate is anionic reagent.
60. The hottest region of Bunsen flame shown in the figure below is :
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region 4 region 3 region 2
region 1
(2)
region 2
(3)
region 3
(4)
region 4
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(1)
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g.
region 1
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outer Nonluminous flame Hottest Region Inner blue cone
60. (2)
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