neet(ug) – 2020synthesis.ac.in/neet2020/pdf/chemistrysolutionneet2020_anskey.pdf · neet-2020...

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Page - 1“Gyan Tower” Near Shivbari Circle, Old Shivbari Road, Bikaner(0151-2206735, 2233735, Mob. : 80030948-91/92/93, web.www.synthesis.ac.in)

91. Identify a molecule which does not exist.

(1) He2 (2) Li2 (3) C2 (4) O2

Ans. (1) He2

Sol : 2 * 22He 1s 1s

Nb = NA b A

0

N NB 0

2

He2 does not exist

92. Find out the solubility of Ni(OH)2 in 0.1M NaOH. Given that the ionic product of Ni(OH)2 is 2×10–15.

(1)2×10–13 M (2) 2×10–8M (3) 1×10–13M (4)1×108M

Ans. (1) 2×10–13 M

Sol : IP = [Ni+2][OH–]2

2×10–15= S × (0.1)2

1513

2

2 10S 2 10

0.1

M

93. Identify the correct statements from the following :

(a) CO2(g) is used as refrigerantg for ice-cream and frozen food.

(b) The structure of C60 contains twelve six carbon rings and twenty five carbon rings

(c) ZSM-5, a type of zeolite, is used to convert alcohols into gasoline.

(d) Co is colourless and odourless gas.

(1) (a), (b) and (c) only (2) (a) and (c) only

(3) (b) and (c) only (4) (c) and (d) only

Ans. (4) (c) and (d) only

Sol : (a)N2 is used as refrigerant for ice-creamand frozen food.

(b)Fullerin (C60) contains 12 pentagon and 20 hexagon rings of carbon.

(c) Correct Statement

(d) Correct Statement

Paper Code : [E1] Test Date : 13.09.2020

NEET(UG) – 2020(Question Paper with Answer & Solution)

CHEMISTRY (WITH SOLUTION)

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94. Hydrolysis of sucrose is given by the following reaction

2Sucrose H O Glu cos e Frucrtose

If the equilibrium constant (Kc) is 2×1013 al 300K. the value of rG(-) at the same temperature will be :

Ans. (1) –8.314J mol–1K–1×300K×ln(2×1013)

(1)–8.314J mol–1K–1×300K×ln(2×1013)

(2) 8.314J mol–1K–1 ×300K×ln(2×1013)

(3) 8.314J mol–1 K–1×300K×ln(3×1013)

(4) –8.314J mol–1K–1×300K×ln(4×1013|

Sol :r cG RT ln K

= -8.314Jmol–1 K–1 × 300K × ln(2×1013)

95. Identify compound X in the following sequence of reaction :

(1) (2) (3) (4)

Ans. (3)

Sol :

CH3

OH

OH

2Cl /Hv

–2HCl

2 H–OH

373 K

–2HCl

–H O2

CHCl2

(x)

CHOCH

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96. Identify the incorrect match.

Name IUPAC offical Name

(a) Unnilunium (i) Mendelvium

(b) Unnitrium (ii) Lawrencium

(c) Unnihexium (iii) Seaborgium

(d) Unununnium (iv) Darmstadtium

(1) (a), (i) (2) (b),(ii) (3) (c) (iii) (4) (d), (iv)

Ans. (4) (d), (iv)

Sol : Uuu= Rg= Roentgenium

97. An element have a body centred cubic (bcc) structure with a cell edge of 288pm. The atomic radius is:

(1)3

288pm4 (2)

2288pm

4 (3)

4288pm

3 (4)

4288pm

2

Ans. (1)3

288pm4

Sol : In BCC unit cell 3 4R

3 3R 288pm

4 4

98. Which of the following set of molecules will have zero dipole moment ?

(1) Ammonia, berylium difluoride, water, 1, 4-dichlorobenzene

(2) Boron trifluoride, hydrogen flurodie, carbon dioxide, 1, 3- dichlorobenzene

(3) Nitrogen trifluoride, beryllium difluoride, water, 1,3- dichlorobenzene

(4) Boron trifluoride, beyrlium difluoride, carbon dioxide, 1,4-dichlorobenzene

Ans. (4) Boron trifluoride, beyrlium difluoride, carbon dioxide, 1,4-dichlorobenzene

Sol : (i)

F|

B/ \

F FRµ 0 (ii) F Be F (µR =0 )

(iii) O = C = O (µR =O) (iv)

Cl

Cl

µR = 0

99. On electrolysis of dil. Sulphuric acid using platinum (pt) electrode, the product obtained at anode will be:

(1) Hydrogen gas (2) Oxygen gas (3) H2S gas (4) SO2

Ans. (2) Oxygen gas

Sol : 2

2 4 4H SO 2H SO

At cathode : 22H 2e H

At anode: [(OH–)] ,(SO4)–2]

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24SOOH

SOP SOP OH(-) will be discharged preferrably

24SOOH

DP DP

2 2

12OH 2e O H O

2

100. Reaction between acetone and methylmagnesium chloride followed by hydrolysis will give :

(1) Isopropyl alcohol

(2) Sec. butyl alcohol

(3) Tert. Butyl alcohol

(4) Isobutyl alcohol

Ans. (3) Tert. Butyl alcohol

Sol :

Cl

tert. Butylalcohol

OH

CH C–CH + CH –M –Cl3 3– g3 CH C–CH3 3–

(CH ) C–OH + Mg3 3

CH3

HOH

O–Mg–Cl

H–OH

101. Which of the following oxoacid of sulphur has –O–O– linkage ?

(1) H2SO4, sulphurous acid (2) H2SO4, sulphuric acid

(3) H2S2O8, peroxdisulphuric acid (4) H2S2O7, pyrosulphuric acid

Ans. (3) H2S2O8, peroxdisulphuric acid

Sol : (1)

O||

H O S O H (2)

O||

H O S O H||O

(3)

O O|| ||

H O S O O S O H|| ||O O

(4)

O O|| ||

H O S O S O H|| ||O O

102. Which of the following amine will give the carbylamine test :

(1) (2)

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(3) (4)

Ans. (1) 4.90 BM

Sol : Primary amine give carbylamine test.

103. The calculated spin only magnetic moment of Cr2+

(1) 3.87 BM (2) 4.90 BM (3) 5.92 BM (4)2.84BM

Ans. (2) 4.90 BM

Sol : 2 2 2 6 2 6 4 0Cr 1s 2s 2p 3s 3p 3d 4s

3d4= n =4

mµ n n 2 BM 4 4 2 BM 24 B.M. = 4.90BM

104. The correct option for free expansion of an ideal gas under adiabatic conditionis :

(1) q =0, T=0 and W=0 (2) q =0, T <0 and W > 0

(3) q < 0, T =0 and W =0 (4) q > 0, T> 0 and W> 0

Ans. (1) q =0, T=0 and W=0

Sol : Adiabatic process q =0

Free expansion w=0

No internal change will take place (because E =q +W)

T=0. for ideal gas.

105. The freezing point depression constant (Kf) of benzene is 5.12K kg mol–1. The freezing point depression

for the solution of molality 0.078 m containinga non-electrolyte solute in benzene is (rounded off upto

two decimal places) :

(1)0.20K (2) 0.80K (3) 0.40K (4) 0.60 K

Ans. (3) 0.40K

Sol :f fT iK m

i = 1, kf = 5.12 k kg mol–1, m=0.078m

fT 1 5.12 0.078K 0.40K

106. The number of Faradays (F) required to produce 20g of calcium from molten CaCl2 (Atomic mass of Ca

=40g mol–1) is:

(1)1 (2) 2 (3) 3 (4)4

Ans. (1) 1

Sol : Gram eq. of electricity = Gm eq. of Ca

gm

F

qQ

EW

gm

F

qQ z

Aw F

20gmQ 2 1 F

40gm / mol

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107. Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as :

(1) Aldolcondensation

(2) Cannizzaro’s reaction

(3) Cross Cannizzaro’s reaction

(4) Cross Aldol condensation

Ans. (4) Cross Aldol condensation

Sol : Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as cross Aldol

condensation.

108. Paper chromatography is an example of :

(1) Adsorption chromatography

(2) Partition chromatography

(3) Thin layer chromatography

(4) Column chromatography

Ans. (2) Partition chromatography

Sol : (i) Adsorption chromatographyColumn Chromatography

Thin layer Chromatography (T.L.C.)

(ii) Partition chromatography – Paper chromatography

109. An increase in the concentration of the reactant of a reaction leads to change on :

(1) activation energy (2) heat of reaction (3) threshold energy (4) collision frequency

Ans. (4) collision frequency

Sol : By increasing concentration of the reactants, no. of reactant molecules per unit volume increase, hence

no. of collisions per unit time (collision frequency) increases.

110. A mixture of N2 and Ar gases in a cylinder contain 7g of N2 and 8g of Ar. If the total pressure of t mixture

of the gases in the cylinder is 27 bar partial pressure of N2 is :

(1) 9 bar (2) 12 bar (3) 15 bar (4)18 bar

Ans. (3) 15 bar

Sol :2

2

N

N total

total

nP P

n

2N

7 1n mol mol 0.25mol

28 4

Ar

8 1n mol mol 0.20mol

40 5

ntotal

= 0.45 mol

2N

0.25P 27

0.45 bar =15 bar

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111. Identify the correct statement from the following :

(1) Wrought iron is impure iron with 4%carbon

(2) Blister copper has bistered appearance due to evolution of CO2

(3) Vapour phase refining is carried out Nickel by Van Arkel method.

(4) Pig iron can be moulded into a variety shapes.

Ans. (4) Pig iron can be moulded into a variety shapes.

Sol : (1) Wrought iron has 0-0.2% carbon

(2) Blister copper has blistered appearance due to evolution of SO2

(3) Vapour phase refining of Ni is carried out by Mond’s process. Van Arkel method is used for Ti.

(4) Pig iron can be moulded into variety of shapes

112. A tertiary butyl carbocation is more stable the secondary butyl carbocation because of which the following:

(1) – I effect of – CH3 groups

(2) + R effect of – CH3 groups

(3) – R effect of – CH3 groups

(4) Hyperconjugation

Ans. (4) Hyperconjugation

Sol : A tertiary butyl carbocation is more stable than secondary butyl carbocation because of hyper conjugation

(H.C. effect).

113. Which of the following is a cationic detergent :

(1) Sodium lauryl sulphate

(2) Sodium stearate

(3) Cetyltrimethyl ammonium bromide

(4) Sodium dodecylbenzene sulphonate

Ans. (3) Cetyltrimethyl ammonium bromide

Sol : Cationic detergent = celyltrimethyl ammonium bromide.

114. Elimination reaction of 2-Bromo-pentane to form pent-2-ene is :

(a) -Elimination reaction

(b) Follows Zaitsev rule

(c) Dehydrohalogenation reaction

(d) Dehydration reaction

(1) (a), (b) , (c) (2) (a) , (c), (d)

(3) (b), (c), (d) (4) (a), (b), (d)

Ans. (1) (a), (b) , (c)

Sol : Elimination reaction of 2-Bromopentane to form pent-2-ene is

H–C–C–C–C–C–H CH –CH –CH=CH–CH3 2 3

Pent–2–eneH

H

H

H

H

H

Br

H

H

H

It is type of B-Elimination reaction and follow Zaitsev rule and dehydrohalogenation reaction.

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115. The mixture which shows positive deviation from Reoult’s law is :

(1) Ethanol + Acetone (2) Benzen + Toluene

(3) Acetone+ Chloroform (4) Chloroethane + Bromethane

Ans. (1) Ethanol + Acetone

Sol : (1) Ethanol + Acetone =+ve deviation from Roult’s law

(2) Benzene + Toluene = Ideal Solution

(3)Acetone + CHCl3 = –ve deviation

(4)C2H5–Cl + C2H5Br = Ideal solution

116. Which of the following is the correct order of increasing field strength of ligands to form coordination

compounds :

(1) 22 4SCN F C O CN (2)

22 4SCN F CN C O

(3) 22 4F SCN C O CN (4) 2

2 4CN C O SCN F

Ans. (1) 22 4SCN F C O CN

Sol : Spectrochemical series

2

2 4SCN F Cr O CN

117. Which of the following is a basic amino acid :

(1) Serine (2) Alanine

(3) Tyrosine (4) Lysine

Ans. (4) Lysine

Sol : Basic amino acid is lysine.

118. HCl was passed through a solution of CaCl2 , MgCl2 NaCl. Which of the followign compound (s) crystallise

(s) ?

(1) Both MgCl2 and CaCl2 (2) Only NaCl

(3) Only MgCl2 (4) NaCl,MgCl2 and CaCl2

Ans. (2) Only NaCl

Sol : To obtain pure sodiumchloride. The crude Salt is dissolved inminimum amount of water and filtered to

remove insoluble impurities. the Solution is then saturated with hydrogen chloride gas. Crystals of pure

sodium chloride separate out.Calciumand magneisum chloride. Beingmore soluble than sodium chloride,

remain in solution.

119. Which of the following is a natural polylmer :

(1) cis-1, 4-polyisoprene

(2) poly (Butadiene-styrene)

(3) polybutadiene

(4) poly (Butadiene-acrylonitrile)

Ans. (1) cis-1, 4-polyisoprene

Sol : Natural polymer is cis 1, 4 polyisoprene.

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120. Which of the following is not correct but carbon monoxide ?

(1) It forms carboxyhaemoglobin.

(2) It reduces oxygen carrying ability of Blood

(3) The carboxyhaemoglobin (haermogiobound to CO) is less stable the oxyhaemoglobin.

(4) It is produced due to incomplete combustion.

Ans. (3) The carboxyhaemoglobin (haermogiobound to CO) is less stable the oxyhaemoglobin.

Sol : Carbomyhaemoglobin is more stable than oxyhaemoglobin because CO is strong field ligand w.r.t. O2

(Co-ordination chemistry)

121. Sucrose on hydrolysis gives :

(1) -D-Glucose + -D-Fructose (2) -D-Glucose + -D-Glucose

(3) -D-Glucose + -D-Fructose (4) -D-Fructose + -D-Fructose

Ans. (3) -D-Glucose + -D-Fructose

Sol :

122. The following metal ion activates many enzyme participates in the oxidation of glucose to produce ATP

and with Na, is responsible for transmission of nerve signals :

(1) Iron (2) Copper

(3) Calcium (4) Potassium

Ans. (4) Potassium

Sol : Potassium ion is responsible with Na ion for transmission of Nerve signals.

123. Which one of the following has maximum number of atoms ?

(1) 1g of Ag(s) [Atomic mass of Ag =108] (2) 1g of Mg(s) [Atomic mass of Mg = 24]

(3)1g of O2(g)[Atopmic mass of O =16] (4) 1gof Li(s) [Atomic massof Li = 7]

Ans. (4) 1gof Li(s) [Atomic massof Li = 7]

Sol : Natom =qgm atom × NA

(1) For 1 gm of Ag Natoms= A

1N

108

(2)1 gm of Mg atoms A

1N N

24

(3)1 gm of O2 atoms A

1N N

16

(4) 1 gm of Li atoms A

1N N ....maximum

7

124. The number of protons, neutrons and electron 17571 Lu , respectively, are :

(1)71, 104 and 71 (2)104, 71 and 71 (3) 71, 71 and 104 (4) 175, 104 and 71

Ans. (1) 71, 104 and 71

Sol : 17571Lu No. of photons = Z= 71

Mass No.=175

No. of neutrons = Mass no. –No. of protons

= 175 – 71 = 104

Pr otonseN N 71

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125. What is the change inoxidation number of carbon in the following reaction ?

4 2 4CH g 4Cl g CCl l 4HCl g

(1) +4 to +4 (2) 0 to +4 (3) (4) 0 to –4

Ans. (3) –4 to +4

Sol : In CH4 O.N. of C–

x 1 4 O x 4

In CCl4 O. No. of C–

x 4 0 x 4

126. Identify the correct statements.

(1)Cr2+(d4) is stronger reducing agent than Fe2+ (d6 ) in water.

(2) The transitionmetalsand their compound are known for their catelytic activity due to their ability to

adopt multiple oxidation states and to formcomplexes.

(3)Interstitial compounds are those that areformed when small atoms like H,C or N are trapped inside

the cyrstal latice ofmetals

(4) The oxidation state of chromium in 24CrO and 2

2 7Cr O are not the same

Ans. (4) The oxidation state of chromium in 24CrO and 2

2 7Cr O are not the same

Sol : In 2

4CrO

O. N. of Cr :

x 8 2 x 6

In 22 7Cr O O. N. of Cr. :

2x –14 = –2 x 6

O. N. of Cr in 2

4CrO

and 2

7CrO

are same.

127. For the reaction 22Cl g Cl g , the correct option is :

(1)r rH 0 and S 0 (2)

r rH 0 and S 0

(3)r rH 0 and S 0 (4)

r rH 0 and S 0

Ans. (4)r rH 0 and S 0

Sol : Cl Cl Cl Cl

Reaction will be exothermic rH Ve rH 0

2 moles of gaseous atoms converts into 1 mol of molecular gas gaseous moles are descreasing.

entropy will decreases rS ve

rS O

128. Measuring Zeta potential is useful indetermining which property of colloidal solution ?

(1) Viscosity

(2) Solubility

(3) Stability of the colloidal particles

(4) Size of the colloidal particles

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Ans. (3) Stability of the colloidal particles

Sol : Zeta potential represents stability of colloidal particles in solution

129. Urea reacts with ater to form A which decompose to formB. B when passes through Cu2+ (aq.) deep

blue colour solutionC is form what is the formula of C from the following ?

(1)CuSO4 (2) [Cu(NH3)4]2+ (3) Cu(OH)2 (4) CuCO3 . Cu(OH)2

Ans. (2) [Cu(NH3)4]2+

Sol :

H—OH

NH —C—NH2 2

||O

OH—H

(H CO ) +2NH2 3 3

H O+CO +2NH2 2 3

(B)

223 3 4

Cu NH Cu NH

C

130. Match the following andidentify the correct option.

(a) CO(g) + H2(g) (i) Me(HCO3)2 +Ca(HCO3)2

(b) Temporary hardness of water (ii) An electron deficient hydrogen

(c) B2H6 (iii) Synthesis gas

(d) H2O2 (iv) Non-planar structure

(a) (b) (c) (d)

(1) (i) (ii) (iii) (iv)

(2) (ii) (i) (iv) (iii)

(3) (iii) (iv) (i) (ii)

(4) (iv) (iii) (ii) (i)

Ans. (1) (a) (b) (c) (d)

(i) (ii) (iii) (iv)

Sol : (a)2CO H Synthesisgas

(b)Temporary hardness of water is due to bicarbonates of Mg and Ca

(c) B2H6 is an e– deifficient hydride

Dimer of BH3

(d) H2O2 has openbook [Non-Planar)structure

H

O

O

H

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131. Match the following :

Oxide Nature

(a) CO (i) Basic

(b) BaO (ii) Neutral

(c) Al2O3 (iii) Acidic

(d) Cl2O7 (iv) Amphoteric

Which of the following is correct option ?

(a) (b) (c) (d)

(1) (i) (ii) (iii) (iv)

(2) (ii) (i) (iv) (iii)

(3) (iii) (iv) (i) (ii)

(4) (iv) (iii) (ii) (i)

Ans. (2) (a) (b) (c) (d)

(ii) (i) (iv) (iii)

Sol : CO = Neutroal

BaO =Basic

Al2O3=Amphoteric

Cl2O7=acidic

132. The rate constant for a first order reaction 4.606×10–3 S–1. The time required to red 2.0g of the

reactant to 0.2g is :

(1) 100s (2) 200s (3) 500s (4)1000s

Ans. (3) 500s

Sol : For a I order reaction :

0

t

a2.303K log

t a

4.606×10–3sec–1=2.303 2

logt 0.2

3

2.303 log10t sec

4.606 10

= 500 sec

133. An alkene on ozonolysis gives methanal as one of the product. Its structure is :

(1) (2) (3) (4)

Ans. (3)

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Sol :

O

CH –CH=CH2 2 CH –CHO2CH –CH CH2 2

O /Zn3O—O + HCHO

Methanol

134. Which of the following alkane cannot be made in good yield by Wurtz reaction :

(1) n-Hexane

(2) 2, 3-Dimethylbutane

(3) n-Heptane

(4) n-Butane

Ans. (3) n-Heptane

Sol : Symmetrical alkanes are formed in good yield in Wurtz reaction. Since the heptane alkane would be

asymmetrical, it will not have a good yield.

135. Anisole on cleavage with HI gives :

(1) (2) (3) (4)

Ans. (1)

Sol :

OCH3 OH

H–I+ CH I3

Anisole

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