INNOVISION JEE MAIN 2017 SOLUTION(CHEMISTRY)
61.
61.
1
1 gram of a carbonate (M2CO3) treatment with excess HCl products 0.01186 mole of CO2. The molar
mass of M2CO3 in g mol–1 is:
(a) 84.3
(b) 118.6
(c) 11.86
(d) 1186
M2CO3 + HCl → CO2
1gm
excess
L.R.
→
M2CO3
Applying
POAC
on
1×
0.01186 mole
‘C’
1
= 1× 0.01186
M
M = 84.36
62.
62.
∴ (a)
Given C(graphite) O2 (g) → CO2(g); ∆rH0 = –393.5 kJ mol–1
1
H2 (g) + O2 (g) → H2O (l); ∆rH0 = –285.8 kJ mol–1
2
CO2 (g) + 2H2O (l) → CH4 (g) + 2O2 (g) ; ∆rH0 = +890.3 kJ mol–1
Based on the above thermochemical equations, the value of ∆rH0 at 298 K for the reaction
C(graphite) + 2H2 (g) → CH4 (g) will be:
(b) – 74.8 kJ mol–1
(a) + 144.0 kJ mol–1
–1
(c) – 144.0 kJ mol
(d) + 74.8 kJ mol–1
C (s) + O2 (g) → CO2 (g)
…….(i)
H2 (g) +
1
O2 (g) → H2O (l) ……..(ii)
2
CO2 (g) + 2H2O (l) → CH4 (g) + 2O2 (g)
……..(iii)
C (s) + 2H2 (g) → CH4 (g)
∆H = eq.(i) + 2 × eq.(ii) + eq.(iii)
= 393.5 + 2 (–285.8) + 890.3
∆H = –74.8
∴(b)
63.
63.
The freezing point of benzene decreases by 0.450 C when 0.2 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)
(a) 80.4%
(b) 74.6%
(c) 94.6%
(d) 64.6%
(CH3COOH)2
2CH3COOH
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INNOVISION JEE MAIN 2017 SOLUTION(CHEMISTRY)
2
nα
2
n (1 – α)
α

moles at eqm. = n 1 − α + 
2

= n (1 − α / 2 )
i = (1 – α / 2)
∆Tf = i kfm
0.2 60
 α
0.45 = 1 −  × 5.12 ×
20 1000
 2
α = 0.9453 or, 94.6%
64.
64.
∴ (c)
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:
(a) 37.5 kg
(b) 7.5 kg
(c) 10 kg
(d) 15 kg
10
Mass of Hydrogen in a human adult = 75 ×
= 7.5kg
100
If all the 1H1 are replaced by 1H2
∴mass of Hydrogen in a human adult will become = 2 × 7.5kg = 15 kg
∴ gain in wt. = 15 – 7.5 = 7.5kg
65.
65.
∴ (d)
∆U is equal to:
(a) Isobaric work
(c) Isothermal work
(b) Adiabtic work
(d) Isochoric work
1st law of thermodynamics
q + Wsurr. = ∆u
or
q – P∆V = ∆u
during adiabatic process, q = 0
∆u = Wsurr.
for isobaric process
q – P ∆u = ∆u
for isothermal process (ideal gas)
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INNOVISION JEE MAIN 2017 SOLUTION(CHEMISTRY)
∆u = 0,
3
q = P∆V
for isochoric process
P∆V = 0,
q = ∆u
∴ (b)
66.
The formation of which of the following polymers involves hydrolysis reaction?
(a) Bakelite
(b) Nylon 6, 6
(c) Terylene
(d) Nylon 6
66.
∴ (d)
67.
Given:
0
ECl0 / Cl − = 1.36V , ECr
= −0.74V
3+
/ Cr
2
0
Cr2 O72− / Cr 3+
E
67.
0
= 1.33V , EMnO
= 1.51V
−
/ Mn 2+
4
Among the following, the strongest reducing agent is:
(a) Mn2+
(b) Cr3+
(c) Cl–
(d) Cr
Lesser the value of SRP, stronger will be the reducing agent.
∴ Cr will be the stronger reducing agent.
∴ (d)
68.
68.
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 greatly in magnitude.
(a)
69.
In the following reactions, ZnO is respectively acting as a/an:
(I) ZnO + Na2O → Na2ZnO2
(II) ZnO + CO2 → ZnCO3
(a) base and base
(b) acid and acid
(c) acid and base
(d) base and acid
69.
(c)
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INNOVISION JEE MAIN 2017 SOLUTION(CHEMISTRY)
70.
70.
4
Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?
(a)
(b)
(c)
(d)
To be a reducing sugar oxygen present on anomeric carbon must have hydrogen attached.
∴ (d)
71.
The major product obtained in the following reaction is:
(a) C6H5CH = CHC6H5
(c) (–)C6H5CH(OtBu)CH2C6H5
71.
(b) (+)C6H5CH(OtBu)CH2C6H5
(d) (±)C6H5CH(OtBu)CH2C6H5
The reaction is elimination reaction of halides in the presence of base.
∴ (a)
72.
Which of the following species is not paramagnetic?
(a) CO
(b) O2
(c) B2
(d) NO
72.
MOT of
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INNOVISION JEE MAIN 2017 SOLUTION(CHEMISTRY)
*
*
CO = σ 1s 2 σ 1s 2 σ 2 s 2 σ 2 s 2
5
π 2 px 2
σ 2 pz 2
2
π 2 py
*
*
*
O2 = σ 1s 2 σ 1s 2 σ 2 s 2 σ 2 s 2 σ 2 pz
π 2 px 2 π 2 px1
2
*
π 2 py 2 π 2 py1
*
*
B2 = σ 1s 2 σ 1s 2 σ 2 s 2 σ 2 s 2
π 2 px1
π 2 py1
*
π 2 px1
π 2 px 2
2
NO = σ 1s 2 σ 1s 2 σ 2 s 2 σ 2 s 2
2
pz
σ
*
π 2 py 2
π 2 py 0
*
*
Hence CO does not contain the unpaired e–.
∴ (a)
73.
On treatment of 100 mL of 0.1 M solution of CoCl3.6H2O with excess AgNO3; 1.2 × 1022 ions are
precipitated. The complex is:
(b) [Co(H2O)6]Cl3
(a) [Co(H2O)3Cl3].3H2O
(c) [Co(H2O)5Cl]Cl2.H2O
(d) [Co(H2O)4Cl2]Cl.2H2O
73.
moles of ions precipitated =
1.2 × 10 22
= 0.02 moles
6.023 ×10 23
∴ moles of Cl– precipitated = moles of Ag+ precipitated = 0.01 moles
Moles of CoCl3.6H2O =
100 × 0.1
= 0.01 moles
1000
Hence 1 Cl– will be outside the co-ordination sphere, compound will be (C.N. of Co = 6)
∴ (d)
74.
74.
pKa of a week acid (HA) and pKb of a week base (BOH) are 3.2 and 3.4, respectively. The pH of their
salt (AB) solution is:
(a) 6.9
(b) 7.0
(c) 1.0
(d) 7.2
pH of salt of ωA + ωB
pH = 7 +
1
1
pK a – pK b
2
2
=7+
1
1
× 3.2 –
× 3.4 = 6.9
2
2
∴ (a)
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INNOVISION JEE MAIN 2017 SOLUTION(CHEMISTRY)
75.
75.
6
The increasing order of the reactivity of the following halides for the SN1 reaction is:
CH 3CHCH 2CH 3
CH 3CH 2CH 2Cl
p − H 3CO − C6 H 4 − CH 2Cl
|
(II)
(III)
Cl
(I)
(a) (II) < (I) < (III)
(b) (I) < (III) < (II)
(c) (II) < (III) < (I)
(d) (III) < (II) < (I)
⊕
CH 3 − CH − CH 2 − CH 3
20
⊕
CH 3CH 2CH 2
10
Reactivity of halides for SN1 ∝ stability of carbocation generated
Hence III > I > II
∴ (a)
76.
76.
Both lithium and magnesium display several similar properties due to the diagonal relationship;
however, the one which is incorrect, is:
(a) both from soluble bicarbonates
(b) both from nitrides
(c) nitrates of both Li and Mg yield NO2
(d) both from basic carbonates
(d)
77.
The correct sequence of reagents for the following conversion will be:
(a) CH3MgBr, H+/CH3OH, [Ag(NH3)2]+OH–
(b) CH3MgBr, [Ag(NH3)2]+OH–, H+/CH3OH
(c) Ag(NH3)2]+OH–, CH3MgBr, H+/CH3OH
(d) Ag(NH3)2]+OH–, H+/CH3OH, CH3MgBr
O
O
O
H3C
Ag(NH3)2+ OH-
H+ / CH3OH
Tollens reagent
77.
CHO
COO-
3 CH3 MgBr
COOCH3
O- + MgBr
OH
H3O+
O- +MgBr
OH
∴ (d)
78.
The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are:
(b) Cl– and ClO–
(a) ClO2– and ClO3–
(c) Cl– and ClO2–
(d) ClO– and ClO3–
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INNOVISION JEE MAIN 2017 SOLUTION(CHEMISTRY)
78.
7
Cl2 + 2NaOH (cold & dilute) → NaCl + NaClO + H2O
∴ (b)
79.
Which of the following compounds will form significant amount of meta product during mononitration reaction?
(a)
(b)
(c)
(d)
79.
∴ (b)
80.
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:
(a) Zero
(b) Two
(c) Four
(d) Six
80.
∴ (c)
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INNOVISION JEE MAIN 2017 SOLUTION(CHEMISTRY)
81.
81.
8
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 = 8.314 J mol–1 K–1)
(a) 12
(b) 6
(c) 4
(d) 8
A1 = A2
Ea1 = Ea2 + 10
T = 300
K1 = A1e
− Ea1 / RT
K 2 = A2 e
− Ea2 / RT
K1
( E − E ) / RT
= e a2 a1
K2
 K  Ea − Ea1
ln  1  = 2
RT
 K2 
=
−10 × 103
8.314 × 300
K 
10000
ln  2  =
=4
 K1  8.314 × 300
∴ (c)
82.
82.
Which of the following molecules is least resonance stabilized?
(a)
(b)
(c)
(d)
The remaining molecules are resonance stabilized continuously in cyclic system with 6 electrons
each and hence aromatic. While the third molecule is having cross-conjugation.
∴ (c)
83.
83.
The group having isoelectronic species is:
(a) O–, F–, Na, Mg+
(c) O–, F–, Na, Mg2+
(b) O2–, F–, Na, Mg2+
(d) O2–, F–, Na+, Mg2+
where all the species have 10 electrons each.
∴ (d)
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INNOVISION JEE MAIN 2017 SOLUTION(CHEMISTRY)
84.
84.
9
The radius of the second Bohr orbit for hydrogen atom is:
(Planck’s constant h = 6.6262 × 10–34 Js; mass of electron = 9.1091 × 10–31 kg; charge of electron e =
1.60210 × 1019 C; permittivity of vaccum ϵ0 = 8.854185 × 10–12 kg–1m–3A2)
(a) 4.76 Å
(b) 0.529 Å
(c) 2.12 Å
(d) 1.65 Å
2
n
r = 0.529 ×
Å , n = 2; z = 1
z
r = 2.12 Å
∴ (c)
85.
85.
The major product obtained in the following reaction is:
(a)
(b)
(c)
(d)
DIBAL-H can reduce Esters and Carboxylic acids to Aldehydes
∴ (a)
86.
Which of the following reactions is an example of a redox reaction?
(a) XeF2 + PF5 → [XeF]+ PF6–
(b) XeF6 + H2O → XeOF4 + 2HF
(c) XeF6 + 2H2O → XeO2F2 + 4HF
(d) XeF4 + O2F2 → XeF6 + O2
86.
XeF 4 + O2 F2 
→ XeF6 + O2
+4
+1 −1
+6
0
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INNOVISION JEE MAIN 2017 SOLUTION(CHEMISTRY)
10
∴ (d)
87.
A metal crystallizes 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:
(a) 2 2 a
(b) 2 a
a
(c)
(d) 2a
2
87.
ln FCC
face diagonal = a 2
hence, closest distance x =
x=
88.
88.
a 2
2
a
2
∴ (c)
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
(a) HCOONa
(b) CH3COONa
(c) Na2C2O4
(d) C6H5COONa
Na2C2O4 + conc. H2SO4 → Na2SO4 + CO + CO2 + H2O
Na2C2O4 + CaCl2 → CaC2O4 (white) + 2NaCl
5CaC2O4 + 2KMnO4 (purple) + 8H2SO4 → K2SO4 + 5CaSO4 + 2MnSO4 +
10CO2 + 8H2O
∴ (c)
89.
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:
(a) both SO42– and NO3–
(b) only F–
2–
(c) only SO4
(d) only NO3–
89.
max. limit of NO3– = 50ppm.
max. limit of SO42– = 500ppm.
max. limit of F– = 2 ppm.
∴ (b)
90.
Which of the following, upon treatment with tert-BuONa followed by addition of bromine water,
fails to decolourize the colour of bromine?
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INNOVISION JEE MAIN 2017 SOLUTION(CHEMISTRY)
90.
(a)
(b)
(c)
(d)
11
option (a), (b) and (c) will produce alkenes upon treatment with tert-BuONa while option (d) will not
produce alkene.
∴ (d)
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