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JEE MAINS 2017 SOLUTIONS
CHEMISTRY
1. The freezing point of benzene decreases by 0.45°C when O.2 g of acetic acid is added to 20 g of
benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid
in benzene will be :
(Kf for benzene=5.12 K kg mol-1)
(1) 94.6%
(2) 64.6%
(3) 80.4%
(4) 74.6%
Ans. 1
Solution
In benzene
2CH3COOH  (CH3COOH)2
1 
i = 1 +  − 1 α
2 
α
i = 1−
Here α is degree of association ∆Tf = iKf m
2
α = 0.945
% degree of association = 94.5%
2. On treatment of 100 mL 0f 0.1 M solution of CoCl3.6H2O with excess AgNO3; 1.2×1022 ions are
precipitated. The complex is :
(1) [Co(H2O)5Cl]Cl2.H2O
(2) [Co(H2O)4Cl2]Cl.2H2O
(3) [Co(H2O)3Cl3].3H2O
(4) [Co(H2O)6]Cl3
Ans. 2
Solution
Molarity × volume(ml)
1000
100 × 0.1
=
= 0.01 mole
1000
Moles of ions precipitated with excess of
1.2 × 1022
AgNO3 =
6.02 × 1023
= 0.02 moles
Number of Cl– present in ionization sphere =
Mole of ion precipitated with exess AgNO3 0.02
=
= 2
mole of complex
0.01
It means 2Cl– ions present in ionization sphere
∴ complex is [Co(H2O)5Cl]Cl2.H2O
Moles of complex =
3. Which of the following compounds will form significant amount of meta product during mononitration reaction ?
(1)
(2)
(3)
(4)
Ans. 4
Solution
(i) Nitration is carried out in presence of concentrated HNO3 + concentrated H2SO4. (ii) Aniline acts as base. In
presence of H2SO4 its protonation takes place and anilinium
ion is formed
(iii) Anilinium ion is strongly deactivating group and meta directing in nature so it give meta nitration product
in significant amount.
4. The products obtained when chlorine] gas reacts with cold and dilute aqueous NaOH are :
(4) Cl- and ClO(1) Cl- and ClO2(2) ClO and ClO3(3) ClO2- and ClO3Ans. 4
Solution
5. Both lithium and magnesium display several similar properties due to the diagonal relationship;
however, the one which is incorrect, is :
(1) nitrates of both Li and Mg yield NO2 and O2 on heating
(2) both form basic carbonates
(3) both form soluble bicarbonates
(4) both form nitrides
Ans 2
Solution
Li form carbonate which is not basic.
6. A water sample has ppm level concentration of following anions
F- =10; SO42- =100; NO3- =50
The anion/ anions that make/ makes the water sample unsuitable for drinking is/ are :
(1) only SO42-
(2) only NO3-
(3) both SO42- and NO3-
(4) only F-
Ans. 4
Solution
NO3- : The maximum limit of nitrate in drinking water is 50 ppm. Excess nitrate in drinking water can cause
disease. Such as methemoglobinemia. SO42- : above 500 ppm of SO42- ion in drinking water causes laxative
effect otherwise at moderate levels it is harmless F– : Above 2ppm concentration of F– in drinking water cause
brown mottling of teeth.
∴ The concentration given in question of SO42- & NO3- in water is suitable for drinking but the concentration of
F– (i.e 10 ppm) make water unsuitable for drinking purpose :
7. The formation of which of the following polymers involves hydrolysis reaction ?
(1) Terylene
(2) Nylon 6
(3) Bakelite
(4) Nylon 6, 6
Ans 2
Solution
Formation of Nylon-6 involves hydrolysis of its monomer (caprolactam) For reaction refer NCERT
BOOK
8. The Tyndall effect is observed only when following conditions are satisfied :
(a) The diameter of the dispersed particles is much smaller than the wavelength of the light used.
(b) The diameter of the dispersed particle is not much smaller than the wavelength of the light
used.
(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in
magnitude.
(d) The refractive indices of the , dispersed phase and dispersion medium differ greatly in
magnitude.
(4) (a) and (c)
(1) (b) and (c)
(2) (a) and (d)
(3) (b) and (d)
Ans 3
Sollution: Refer NCERT Book
9. pKa of a weak acid (HA) and pKb of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of
their salt (AB) solution is :
(4) 7.0
(1) 1.0
(2) 7.2
(3) 6.9
Ans. 3
Solution
Given
pKa (HA) = 3.2
pKb (BOH) = 3.4
As given salt is of weak acid and weak bas
10. The major product obtained in the following reaction is :
(1)
(2)
(3)
(4)
Ans. 3
Solution
DIBAL – H is electrophilic reducing agent reduces cynide, esters, lactone, amide, carboxylic acid into
corresponding Aldehyde (partial reduction)
11. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution ?
(1)
Ans. 2
Solution
(2)
(3)
(4)
(1) Ester in presence of Aqueous KOH solution give SNAE reaction so following reaction takes place
(2) In above compound in presence of Aq. KOH (SNAE) reaction takes place & ∝-Hydroxy carbonyl compound
is formed which give ⊕ve Tollen's test So this compound behave as reducing sugar in an aqueous KOH
solution.
12. The correct sequence of reagents for the following conversion will be :
(1) [Ag(NH3)2]+OH-, CH3MgBr, H+/CH3OH
(3) CH3MgBr, H+/CH3OH, [Ag(NH3)2]+OHAns. 2
Solution
(2) [Ag(NH3)2]+OH-, H+/CH3OH, CH3MgBr
(4) CH3MgBr, [Ag(NH3)2]+OH-, H+/CH3OH
13. Which of the following species is not paramagnetic ?
(1) B2
(2) NO
(3) CO
(4) O2
Ans 3
Soluiton
NO ⇒ has one unpaired electron is present in π* molecular orbital.
CO ⇒ has no unpaired electron
O2 ⇒ Has two unpaired electrons are present in π* molecular orbitals.
B2 ⇒ has two unpaired electrons are present in π bonding molecular orbitals.
14. Which of the following, upon treatment with tert-BuONa followed by addition of bromine water,
fails to decolourize the colour of bromine ?
(1)
Ans. 2
Solution
(2)
(3)
(4)
15. Which of the following reactions is an example of a redox reaction ?
(1) XeF6 + 2H2O → XeO2F2 + 4HF
(2) XeF4 + O2F2 → XeF6 + O2
(4) XeF6 + H2O → XeOF4 + 2HF
(3) XeF2 + PF5 → [XeF]+ PF6Ans 2
Solution
In the reaction XeF4 + O2F2 → XeF6 + O2 oxidation number of xenon is increasing from 4 to 6 while
that oxygen is decreaing from +2 to 0.
16. ∆U is equal to :
(4) Adiabatic work
(1) Isothermal work
(2) Isochoric work
(3) Isobaric work
Ans 4
Solution
Sol. From 1st law :
∆U = q + w
For adiabatic process :
q=0
∴ ∆U = w
∴ Work involve in adiabatic process is at the expense of change in internal energy of the system.
17. Which of the following molecules is least resonance stabilized ?
(1)
(2)
(3)
(4)
Ans 1
Solution
This is non aromatic so least stable
18. The increasing order of the reactivity of the following halides for the SN1 reaction is :
(1) (II) < (III) < (I)
Ans. 3
(2) (III) < (II) < (I)
(3) (II) < (I) < (III)
Solution.
For any S 1 reaction reactivity is decided by ease of dissociation of alkyl halide
N
(4) (I) < (III) < (II)
Higher the stability of R⊕ (carbocation) higher would be reactivity of SN1 reaction.
Since stability of cation follows order.
Hence correct order is
II < I < III
19. 1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The
molar mass of M2CO3 in g mol-1 is:
(4) 118.6
(1) 11.86
(2) 1186
(3) 84.3
Ans 3
Solution
20. Sodium salt of an organic acid ’X’ produces effervescence with conc. H2SO4. ’X’ reacts with the
acidified aqueous CaCl2 solution to give a white precipitate which decolourises acidic solution of
KmnO4. ’X’ is :
(4) CH3COONa
(1) Na2C2O4
(2) C6H5COONa
(3) HCOONa
21. The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%);
Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person
would gain if all 1H atoms are replaced by 2H atoms is:
(1) 10 kg
(2) 15 kg
(3) 37.5 kg
(4) 7.5 kg
Ans 4
Solution
Since H is 10% and 1H atoms are replaced by 2H atoms then weight of hydrogen is just doubled.
Weight gain will be 75 x .1 = 7.5 kg
22. The major product obtained in the following reaction is :
(1) (-)C6H5CH(OtBu)CH2C6H5
(2) (1)C6H5CH(OtBu)CH2C6H5
(4) (+)C6H5CH(OtBu)CH2C6H5
(3) C6H5CH = CHC6H5
Ans 3
Solution
Elimination reaction is highly favoured by
(a) Bulkier base
(b) High temperature
The above reaction is biomolecular ellimination reaction giving C6H5CH = CHC6H5 as major product.
23. Given
C(graphite) + O2(g) → CO2(g) ;
∆rH°= -393.5 kJ mol-1
H2(g) + ½ O2(g) → H2O(l);
∆rHo = -285.8 kJ mol-1
CO2(g) +2H2O(l) →CH4(g) +2O2(g) ;
∆rH° = +890.3 kJ mol-1
Based on the above thermochemical equations, the value of ∆rH° at 298 K for the reaction
C(graphite) +2H2(g) → CH4(g) will be :
(1) -144.0 kJ mol-1
(2) +74.8 kJ mol-1
(3) +144.0 kJ mol-1
(4) -74.8 kJ mol-1
Ans 4
Solution
C(graphite) + O2(g) → CO2(g) ;
∆rH°= -393.5 kJ mol-1
H2(g) + ½ O2(g) → H2O(l);
∆rHo = -285.8 kJ mol-1
CO2(g) +2H2O(l) →CH4(g) +2O2(g) ;
∆rH° = +890.3 kJ mol-1
multiply second reaction by 2 and add all the 3 reactions we will get
C(graphite) +2H2(g) → CH4(g) ∆rH will be -74.8 kJ mol-1
24. In the following reactions ZnO is respectively acting as a/an :
(a) ZnO+Na2O → Na2ZnO2
(b) ZnO+CO2 → ZnCO3
(1) acid and base
(2) base and acid
(3) base and base
Ans 1
Solution
(a) ZnO+Na2O → Na2ZnO2
here ZnO acts as an acid
(b) ZnO+CO2 → ZnCO3
here ZnO acts as an base
ZnO is an amphoteric oxide but in given reaction.
25. The radius of the second Bohr orbit for hydrogen atom is :
(Planck’s Const. H= 6.6262×10-34 Js;
mass of electr0n=9.1091×10-31 kg;
charge of electron e = 1.60210×10-19 C;
permittivity of vacuum
∈0 = 8.854185×10-12 kg-1·m-3 A2)
(1) 2.12 Å
(2) 1.65 Å
(8) 4.76 Å
1
2
3
(4) acid and acid
(4) 0.529 Å
Ans. 1
Solution
Radius of nth Bohr orbit in H-atom
= 0.53 n2Å
Radius of II Bohr orbit = 0.53 ×(2)2
= 2.12 Å
26. Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds
that of R2 by 10 kJ mol-1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300
K,/then ln(k2/k1) is equal to :
(R = 8.814 J mol-1 K-1)
(4) 6
(1) 4
(2) 8
(3) 12
Ans 1
Solution
From arrhenius equation,
k = Ae
Ea
RT
−
k1 = Ae
−
Ea 1
RT
−
Ea 2
RT
k2 = Ae
1
2
dividing 2 by 1 and taking ln
ln
k2 1000
=
k1 30 R = 4
27. A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is ’a’, the
closest approach between two atoms in metallic crystal will be :
(2) 2a
(1) a / 2
(4) 2a
(3) 2 2a
Ans 1
Solution
In FCC unit cell atoms are in contant along face
diagonal
So,
2a = 4R
closest distance (2R) = 2a / 2 = a / 2
28. The group having isoelectronic species is :
(1) O-, F-,Na+,Mg2+
(2) O2-, F-, Na+, Mg2+
Ans 2
Solution
ions
O–2 F–
Na+ Mg+2
10
10
10
10
29. Given
(3) O-,F-,Na, Mg+
(Among the following, the strongest reducing agent is :
(1) Cl(2) Cr
(3) Mn2+
Ans 2
Solution
Reducing property increases as electrode potential E0 value decreases.
(4) O2- ,F-,Na, Mg2+
(4) Cr3+
30. 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The
number of possible stereoisomers for the product is :
(4) Two
(1) Four
(2) Six
(3) Zero
Anti markownikov product (4 stereo isomers possible because of 2 chiral centre. The molecule is
nonsymmetric)