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FOR 2020–21



| JEE MAIN-2020 | DATE : 08-01-2020 (SHIFT-1) | PAPER-2 | OFFICAL PAPER | CHEMISTRY

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PAGE # 1

7340010333

PART : CHEMISTRY

SECTION – 1 : (Maximum Marks : 80) Straight Objective Type (lh/ks oLrqfu"B izdkj)

This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its answer, out of which Only One is correct.

bl [k.M esa 20 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA

1. The first ionization energy (in kJ/mol) of Na, Mg, Al and Si respectively, are :

Na, Mg, Al rFkk Si dh izFke vk;uu ÅtkZ (kJ/mol esa) Øe'k% gSa % (1) 496, 577, 737, 786 (2) 786, 737, 577, 496 (3) 496, 577, 786, 737 (4) 496, 737, 577, 786 Ans. (4) Sol. Correct order of ionisation energy will be : Na < Al < Mg < Si

vk;uu ÅtkZvksa dk lgh Øe gksxk % Na < Al < Mg < Si

2. The stoichiometry and solublity product of a salt with the solubility curve given below is, respectively :

uhps fn;s x;s oØ ds vk/kkj ij] ,d yo.k dh LVkbfd;ksehVªh ¼jllehdj.kfefr½ rFkk foys;rk xq.kuQy] Øe'k% gS %

[Y]/m

M

1 [X]/mM 2 3

1

2

3

(1) X2Y, 2 × 10–9 M3 (2) XY, 2 × 10–6 M3

(3) XY2, 4 × 10–9 M3 (4) XY2, 1 × 10–9 M3

Ans. (3)

Sol.

[Y]/m

M

1 [X]/mM 2 3

1

2

3

XY2(s) 3 3

–

10 2 10

X (aq) 2Y (aq)

Ksp = [X+] [Y–]2

or ;k, Ksp = 10–3 ×(2 × 10–3)2 or ;k, Ksp = 4 × 10–9 M3

3. The third ionization enthalpy is minimum for :

ftlds fy;s r`rh; vk;uu ,UFkSYih U;wure gS] og gS % (1) Ni (2) Co (3) Mn (4) Fe Ans. (4)

Sol. 2626 s4d3ArFe





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4. The complex that can show fac– and mer– isomers is :

og ladj tks fac- rFkk mer- leko;oh iznf'kZr djrk gS] gS : (1) [CoCl2(en)2] (2) [Pt(NH3)2Cl2] (3) [Co(NH3)4Cl2]+ (4) [Co(NH3)3(NO2)3] Ans. (4) Sol. [Ma3b3] type complex shows facial and meridional isomerism

[Ma3b3] izdkj dk ladqy Qydh;&js[kkaf'kr T;kfefr; leko;ork n'kkZrk gSA

5. The strength of an aqueous NaOH solution is most accurately determined by titrating : (Note : consider that an appropriate indicator is used)

tyh; NaOH foy;u dh lkeF;Z lokZf/kd ;FkkFkZrk ls bl rjg vuqekiu }kjk fudkyh tkrh gS % (uksV: fopkj dhft, fd ,d mi;qDr lalwpd dk mi;ksx fd;k x;k gS)

(1) Aq. NaOH in a volumetric flask and concentrated H2SO4 in a conical flask (2) Aq. NaOH in a burette and aqueous oxalic acid in a conical flask (3) Aq. NaOH in a pipette and aqueous oxalic acid in a burette (4) Aq. NaOH in a burette and concentrated H2SO4 oxalic in a conical flask

(1) tyh; NaOH vk;ruh ¶ykLd esa rFkk lkUnz H2SO4 ,d dkWfudy ¶ykLd esa (2) tyh; NaOH ,d C;wjsV esa rFkk tyh; vkDlSfyd vEy ,d dkWfudy ¶ykLd esa (3) tyh; NaOH ,d fiisV esa rFkk tyh; vkDlSfyd ,sflM ,d C;wjsV esa

(4) tyh; NaOH ,d C;wjsV esa rFkk lkUnz H2SO4 ,d dkWfudy ¶ykLd esa Ans. (2) Sol. Oxalic acid is a primary standard solution while H2SO4 is a secondary standard solution.

vkWDtsfyd vEy izkFkfed ekud foy;u gS tcfd H2SO4 f}rh;d ekud foy;u gSA

6. A graph of vapour pressure and temperature for three different liquids X, Y and Z is shown below :

rhu vyx&vyx nzoksa X, Y ,oa Z ds fy, ok"i nkc rFkk rki ds chp ,d xzkQ uhps fn;k x;k gS %

Vap

our

pres

sure

Temperature

X Y 800

500

400

200

293 313 333 353

Z

(m

m H

g)

The following inferences are made : (1) X has higher intermolecular interactions compared to Y. (2) X has lower intermolecular interactions compared to Y. (3) Z has lower intermolecular interactions compared to Y. The correct inference(s) is/are : (1) (C) (2) (B) (3) (A) and (C) (4) A

fuEu fu"d"kZ fudkys x;s %

(1) Y dh rqyuk esa X dk varjkv.kqd vU;ksU; fØ;k mPprj gSA

(2) Y dh rqyuk esa X dk varjkv.kqd vU;ksU; fØ;k fuEurj gSA

(3) Y dh rqyuk esa Z dk varjkv.kqd vU;ksU; fØ;k fuEurj gSA

lgh fu"d"kZ@fu"d"ks± gS@gSa % (1) (C) (2) (B) (3) (A) rFkk (C) (4) A

Ans. (2) Sol. At a particular temperature as intermolecular force of attraction increases vapour pressure decreases.

,d fuf'pr rki ij vUrjkv.kqd vkd"kZ.k cy esa o`f) ds lkFk ok"i nkc esa deh vkrh gSA





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7. The rate of a certain biochemical reaction at physiological temperature (T) occurs 106 times faster with enzyme than without. The change in the activation energy upon adding enzyme is :

,d tSo&jklk;fud vfHkfØ;k dh nj 'kjhj fØ;kRed rki (T) ij fcuk ,Utkbe dh rqyuk esa ,Utkbe }kjk 106 xquk rst gksrh gSA ,Utkbe ds feykus ij lfØ;.k ÅtkZ esa ifjorZu gS %

(1) + 6(2.303)RT (2) + 6RT (3) – 6RT (4) – 6(2.303) RT Ans. (4) Sol. K = Ae–E/RT ……………….(1)

106k = RT/E– CAe ……………….(2)

1equation

2equation 106 = RT/EE Ce

6 ln 10 = RT/EE C

6303.2RT

EE C

or ;k, E–EC = 2.303 × 6RT or ;k, RT6303.2–EEE Ca

8. Among the gases (a) – (e), the gases that cause greenhouse effect are : (a) CO2 (b) H2O (c) CFCs (d) O2 (e) O3 (1) (a), (c), (d) and (e) (2) (a), (b), (c) and (e) (3) (a), (b), (c) and (d) (4) (a) and (d)

xSlksa (a) – (e) esa, xSlsa tks xzhugkml izHkko iSnk djrh gSa] gksaxh %

(a) CO2 (b) H2O (c) CFCs (d) O2 (e) O3

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

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

Ans. (2) Sol. CO2, H2O vapours CFC's and O3 are green house gases.

CO2, H2O ok"i CFC gS rFkk O3 xzhu gkÅl xSls gS

9. When gypsum is heated to 393 K, it forms : (1) Anhydrous CaSO4 (2) CaSO4 · 0.5 H2O (3) CaSO4 · 5 H2O (4) Dead burnt plaster

393K ij ftIle dks xeZ djus ij izkIr gksrk gS&

(1) futZy CaSO4 (2) CaSO4 · 0.5 H2O

(3) CaSO4 · 5 H2O (4) e`r tyk gqvk IykLVj

Ans. (2)

Sol. CaSO4 . 2H2O 393 K CaSO4 .

1

2H2O

10. As per Hardy–Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order :

gkMhZ&lqYls la:i.k ds vuqlkj] Qsfjd gkbMªkWDlkbM lkWy ds fy, fuEu dk Å.kZu eku bl Øe esa gS %

(1) K3[Fe(CN)6] < K2CrO4 < KBr = KNO3 = AlCl3 (2) K3[Fe(CN)6] < K2CrO4 < AlCl3 < KBr < KNO3 (3) K3[Fe(CN)6] > AlCl3 > K2CrO4 > KBr > KNO3 (4) AlCl3 > K3[Fe(CN)6] > K2CrO4 > KBr = KNO3 Ans. (1)

Sol. According to hardy-schultz rule, Coagulation value or flocculation value powernCoagulatio

1

gkMhZ&'kqYt fu;e ds vuqlkj] laxq.ku eku vFkok Å.kZu eku {kerkÅ.kZu

1





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11. Which of the following statement is not true for glucose ?

Xywdksl ds fy, dkSu lk dFku lR; ugha gS \

(1) Glucose exists in two crystalline forms and (2) Glucose reacts with hydroxylamine to form oxime. (3) Glucose gives Schiff's test for aldehyde. (4) The pentaacetate of glucose does not react with hydroxylamine to give oxime.

(1) Xywdksl nks fØLVyh; :iksa rFkk esa feyrk gSA

(2) Xywdksl] gkbMªkfDly,sehu ds lkFk vfHkfØ;k djds vkWDlkbe cukrk gSA

(3) Xywdksl] ,fYMgkbM ds fy, f'kQ+ ijh{k.k nsrk gSA (4) Xywdksl dk isUVk,slhVsV vkWDlkbe cukus ds fy, gkbMªk WfDly,sehu ls vfHkfØ;k ugha djrkA

Ans. (3) Sol. Open chain form of glucose is very very small, hence does not gives Schiff's test.

Xywdksl dh [kqyh J`a[kyk cgqr de ek=kk esa gksrh gS] blfy, ;g f'kQ&ijh{k.k ugha nsrk gSA

12. Arrange the following compounds in increasing order of C–OH bond length : Methanol, phenol, p-ethoxyphenol

(1) methanol < phenol < p-ethoxyphenol (2) phenol < methanol < p-ethoxyphenol (3) phenol < p-ethoxyphenol < methanol (4) methanol < p-ethoxyphenol < phenol

fuEu ;kSfxdksa dks C–OH vkcU/k yEckbZ ds c<+rs Øe esa O;ofLFkr dhft, % esFksukWy] QhukWy] p-,FkkDlhQhukWy (1) esFksukWy < QhukWy < p-,FkkDlhQhukWy (2) QhukWy < esFksukWy < p-,FkkDlhQhukWy

(3) QhukWy < p-,FkkDlhQhukWy < esFksukWy (4) esFksukWy < p-,FkkDlhQhukWy < QhukWy Ans. (3) Sol. There is not any resonance in CH3–OH. Resonance is poor in p-Ethoxyphenol than phenol. so C-OH

bond length order is: CH3–OH > p-ethoxyphenol (p-EtO-C6H4–OH) Ph–OH

CH3–OH esa dksbZ vuqukn ugha gksrk gS p-,FkkDlhQhukWy esa fQukWy dh rqyuk esa nqcZy vuqukn ik;k tkrk gSA vr% C-OH caU/k yEckbZ dk Øe % CH3–OH > p-,FkkDlhQhukWy (p-EtO-C6H4–OH) > Ph–OH

13. For the Balmer series in the spectrum of H atom,

22

21

Hn

1

n

1R , the correct statements among (I) to

(IV) are : (I) As wavelength decreases, the lines in the series converge (II) The integer n1 is equal to 2 (III) The lines of longest wavelength corresponds to n2 = 3 (IV) The ionization energy of hydrogen can be calculated from wave number of these lines

gkbMªkstu ijek.kq ds LisDVªe esa ckej Js.kh ds fy, %

22

21

Hn

1

n

1R , (I) - (IV) esa lgh dFku gSa %

(I) tSls&tSls rjaxnS/;Z ?kVrh gS] Js.kh esa ykbusa vfHklfjr djrh gSA

(II) iw.kk±d n1 2 ds cjkcj gSA

(III) nh?kZre rjaxnS/;Z dh ykbusa n2 = 3 ds vuq:i gksrh gSaA (IV) bu ykbuksa dh rjax la[;k ls gkbMªkstu ds vk;uu ÅtkZ dh x.kuk dh tk ldrh gSA

(1) (I), (III), (IV) (2) (I), (II), (IV) (3) (II), (III), (IV) (4) (I), (II), (III) Ans. (4) Sol. Theory based.

Sol. lS)kfUrd





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14. The predominant intermolecular forces present in ethyl acetate, a liquid, are :

nzo ,fFky ,slhVsV esa mifLFkr izeq[k varjkv.kqd cy gSa %

(1) Dipole-dipole and hydrogen bonding

(2) London dispersion, dipole-dipole and hydrogen bonding

(3) London dispersion and dipole-dipole

(4) Hydrogen bonding and London dispersion

(1) f}/kqzo&f}/kqzo rFkk gkbMªkstu vkcU/k

(2) yUMu ifj{ksi.k] f}/kqzo&f}/kqzo rFkk gkbMªkstu vkcU/k

(3) yUMu ifj{ksi.k rFkk f}/kqzo&f}/kqzo

(4) gkbMªkstu vkcU/k rFkk yUMu ifj{ksi.k

Ans. (3)

Sol. Ethyl acetate is polar molecule so dipole-dipole interaction will be present in it.

,fFky ,flVsV /kzqoh; v.kq gS blfy, ;gk¡ f}/kzqo&f}/kzqo vU;ksU; fØ;k ik;h tkrh gSA

15. The most suitable reagent for the given conversion is :

fn;s x;s :ikUrj.k ds fy, lokZf/kd mi;qDr vfHkdeZd gS %

CONH2

CN

C=O

HO2C

CH3

?

CONH2

CN

COCH3

HOH2C

(1) H2/Pd (2) NaBH4 (3) LiAlH4 (4) B2H6 Ans. (4) Sol. B2H6 is very selective reducing agent and usually used to reduce acid to alcohol.

B2H6 cgqr vf/kd p;ukRed izo`fÙk dk vipk;d gS rFkk ;g lkekU;r% vEy dks ,Ydksgy esa vip;u ds fy, iz;qDr gksrk gSA

16. The number of bonds between sulphur and oxygen atoms in S2O82– and the number of bonds between

sulphur and sulphur atoms in rhombic sulphur, respectively, are : (1) 8 and 8 (2) 8 and 6 (3) 4 and 6 (4) 4 and 8

S2O82– esa lYQj rFkk vkWDlhtu ijek.kqvksa ds chp vkcU/kksa dh la[;k rFkk fo"keyEck{k lYQj esa lYQj ijek.kqvksa

rFkk lYQj ds chp vkcU/kksa dh la[;k Øe'k% gSa % (1) 8 rFkk 8 (2) 8 rFkk 6 (3) 4 rFkk 6 (4) 4 rFkk 8

Ans. (1)

Sol. S2O82–

–O–S–O–O–S–O–

O O

O O

S8

S

SS

SS

S

S

S





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17. The major product of the following reaction is :

fuEufyf[kr vfHkfØ;k dk eq[; mRikn gS %

OH

H3C H3C

H3C

2 4dil H SO

(1)

OH CH3 H3C

H3C

OH

OH

(2) OH

CH3 H3C

CH3

(3)

OH

CH3 H3C

CH3

(4)

HO CH3 H3C

H3C OH OH

Ans. (3)

Sol. OH 2 4dil H SO

CH2

+ 1

2 3 4 5

6

1

2 3

4

5

6

H2O

OH

18. The decreasing order of reactivity towards dehydrohalogenation (E1) reaction of the following compounds

is :

fuEu ;kSfxdksa ds fMkgMªksgSykstsus'ku (E1) vfHkfØ;k ds izfr vfHkfØ;k'khyrk dk ?kVrk Øe gS %

(A) Cl

(B)

Cl

(C)

Cl

(D)

Cl

(1) B > D > C > A (2) D > B > C > A (3) B > A > D > C (4) B > D > A > C

Ans. (2)

Sol. E1 reaction proceeds via carbocation formation, therefore greater the stability of carbocation, faster the

E1 reaction.

E1 vfHkfØ;k dkcZ/kuk;u ds fuekZ.k ls lEiUu gksrh gS bl izdkj dkcZu/kuk;u dk vf/kd LFkkf;Ro] E1 vfHkfØ;k dks

rhoz djrk gSA





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19. The major products A and B in the following reactions are :

fuEu vfHkfØ;kvksa esa eq[; mRikn A rFkk B gSa %

CN

heat

Peroxide ekÅ"

ijkWDlkb M [A]

[A] + [B]

(1) A =

CN

and rFkk B =

CN

(2) A =

CN

and rFkk B =

CN

(3) A =

CN

and rFkk B = CN

(4) A =

CN

and rFkk B =

CN

Ans. (2)

Sol.

CN

heat

PeroxideekÅ"

ijkWDlkb M CN

[A]

CN

+

CN

[A] [B] [A] would be more stable radical and undergoes radical addition to form [B].

[A] vf/kd LFkk;h ewyd gksuk pkfg, rFkk ewyd ;ksxkRed vfHkfØ;k }kjk [B] dk fuekZ.k djrk gSA

20. A flask contains a mixture of isohexane and 3-methylpentane. One of the liquids boils at 63ºC while the

other boils at 60ºC. What is the best way to seperate the two liquids and which one will be distilled out

first ?

,d ¶ykLd esa vkblksgsDlsu rFkk 3-esfFkyisUVsu dk feJ.k gSA bu nzoksa esa ,d 63ºC ij mcyrk gS tcfd nwljk 60ºC

ij mcyrk gSA bu nks nzoksa dks iF̀kd djus dk lcls ljy mik; D;k gS rFkk buesa dkSu loZizFke vklfor gksxk \

(1) simple distillation, isohexane (2) simple distillation, 3-methylpentane

(3) fractional distillation, 3-methylpentane (4) fractional distillation, isohexane

(1) lk/kkj.k vklou, vkblksgsDlsu (2) lk/kkj.k vklou, 3-esfFkyisaVsu

(3) izHkkth vklou, 3-esfFkyisaVsu (4) izHkkth vklou, vkblksgsDlsu

Ans. (4)

Sol. Liquid having lower boiling point comes out first in fractional distillation. Fractional distillation is generally

used if boiling point difference is small.(Boiling point of 3-methylpentane = 63°C, isohexane = 60°C)

fuEu DoFkukad fcUnq okyk nzOk izHkkth vklou esa igys iF̀kd gksrk gSA dFkukad fcUnq esa vUrj de gksus ij izHkkth

vklou fof/k dk mi;ksx fd;k tkrk gSA (DoFkukad fcUnq 3-esfFkyisaVsu = 63°C, vkblksgsDlsu = 60°C)





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SECTION – 2 : (Maximum Marks : 20)

This section contains FIVE (05) questions. The answer to each question is NUMERICAL VALUE with two digit integer and decimal upto one digit.

If the numerical value has more than two decimal places truncate/round-off the value upto TWO decimal places. Full Marks : +4 If ONLY the correct option is chosen. Zero Marks : 0 In all other cases

[kaM 2 ¼vf/kdre vad% 20)

bl [kaM esa ik¡p (05) iz'u gSA izR;sd iz'u dk mÙkj la[;kRed eku (NUMERICAL VALUE) gSa] tks f}&vadh; iw.kkZad rFkk n'keyo ,dy&vadu eas gSA

;fn la[;kRed eku esa nks ls vf/kd n’'keyo LFkku gS ] rks la[;kRed eku dks n'keyo ds nks LFkkuksa rd VªadsV@jkmaM vkWQ (truncate/round-off) djsaA

vadu ;kstuk :

iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSA 'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA

21. The magnitude of work done by a gas that undergoes a reversible expansion along the path ABC shown in the figure is ___________ .

ml xSl ds }kjk] tks fp=k esa fn[kk;s x;s ABC iFk ds vuqlkj mRØe.kh; izlkj.k djrh gS] fd;s x;s dk;Z dk ifjek.k gksxk ___________ .

Pressure

(Pa)

4 Volume

(m3)

(2, 2) 6 8 10 12

10

8

6

4

BA

Ans. 48.00 to 48.00

Sol. J4861062

1W

22. The volume (in mL) of 0.125 M AgNO3 required to quantitatively precipitate chloride ions in 0.3 g of [Co(NH3)6]Cl3 is _________.

M[Co(NH3)6]Cl3 = 267.46 g/mol MAgNO3 = 169.87 g/mol

[Co(NH3)6]Cl3 ds 0.3 g esa DyksjkbM vk;u dks ek=kkRed :i ls vo{ksfir djus ds fy, 0.125 M AgNO3 dk fdruk vk;ru (mL esa) vko';d gksxk _________.

M[Co(NH3)6]Cl3 = 267.46 g/mol MAgNO3 = 169.87 g/mol Ans. 26.80 to 27.00

Sol. [ML6]Cl3 + 3AgNO3 3AgCl 0.3 g v ml, 0.125 M

346.267

3.0 = 0.125 × V ×10–3 or, V =

125.046.267

100033.0

= 26.92 ml.





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23. Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve 10 ppm of iron in 100 kg of wheat is ___________.

Atomic weight : Fe = 55.85 ; S = 32.00 ; O = 16.00

Qsjl lYQsV gsIVkgkbMªsV dks vkgkj ds iq"Vhdj.k esa vk;ju ds fy;s iz;ksx fd;k tkrk gSA xsgw¡ ds 100 kg esa vk;ju dk 10 ppm izkIr djus ds fy, yo.k dh ek=kk (xzke esa) gksxh ___________.

ijek.kq nzO;eku : Fe = 55.85 ; S = 32.00 ; O = 16.00 Ans. 4.95 to 4.97

Sol. 10 = 6Massof Fe(ing)10

100 1000

or ;k, mass Fe = 1 g

FeSO4.7H2O (M = 277.85) 55.85 g in 1 mole

1 g — 1

55.85mole

1

55.55× 277.55 g = 4.97 g

24. The number of chiral centres in penicillin is

isfuflyhu esa dkbjy dsUnzksa dh la[;k gSA Ans. 3.00 to 3.00

Sol. S

O

CH3

CH3

COOH

H

*N *

HN

H H

*R—C

|| O

Star marked atoms are chiral centres.

Rkkjkafdr ijek.kq fdjSy dsUnz gS

25. What would be the electrode potential for the given half cell reaction at pH = 5 ?

2H2O O2 + 4H+ + 4e– ; E0red = 1.23 V (R = 8.314 J mol–1 K–1 ; Temp = 298 K ; oxygen under std.

pressure of 1 bar)

pH = 5 ij] nh xbZ v)Z lsy vfHkfØ;k ds fy, bysDVªksM foHko D;k gksxk \

2H2O O2 + 4H+ + 4e– ; E0red = 1.23 V (R = 8.314 J mol–1 K–1 ; Temp = 298 K ; vkWDlhtu ekud ok;qe.Myh;

nkc 1 bar ij) Ans. 1.52 to 1.53 Sol. From the given data

Eop = E°op – 0.059

4 log [H+]4

Eop = –1.23 – 4

0591.0 log [H+]4

= –1.23 + 0.0591 × pH = –1.23 + 0.0591 × 5 = –1.23 + 0.2955 = – 0.9345 V = –0.93 V




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PAPER-2 (B.E./B. TECH.) JEE (Main) 2020 - Resonance · This solution was download from Resonance...

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