paper-3 (b.e./b. tech.) jee (main) 2020 - resonance · section – 1 : (maximum marks : 80)...

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7340010333

FOR 2020–21



| JEE MAIN-2020 | DATE : 09-01-2020 (SHIFT-1) | PAPER-3 | OFFICAL | CHEMISTRY

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PAGE # 1

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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. Which of these will produce the highest yield in Friedel Craft reaction ?

ÝhMy Øk¶V vfHkfØ;k esa buesa ls dkSu vf/kdre mRikn nsxk \

(1)

CONH2

(2)

OH

(3)

NH2

(4)

Cl

Ans. (4)

Sol. Aniline and phenol form complex with lewis acid so most reactive among the given compounds for Friedel

Craft reaction is chlrobenzene.

Sol. ,uhfyu rFkk fQukWy yqbZl vEy ds lkFk ladqy cukrs gSA blfy;s fn;s x;s ;kSfxdks es ls Dyksjkscsathu ÝhMy Øk¶V

vfHkfØ;k ds izfr vf/kdre fØ;k'khy gksxkA

2. The correct order of heat of combustion for following alkadienes is :

fuEu ,sYdkMkbUl ds fy, ngu dh Å"ek dk lgh Øe gS &

(A)

(B)

(C)

(1) (b) < (c) < (a) (2) (c) < (b) < (a) (3) (a) < (b) < (c) (4) (a) < (c) < (b)

Ans. (3)

Sol. In isomers of hydrocarbon heat of combustion depends upon their stabilities.

As the stability increases heat of combustion decreases.

Stability order

>

>

Sol. gkbMªksdkcZu ds leko;oh;ksa es ngu dh Å"ek muds LFkkf;Ro ij fuHkZj djrh gSA

LFkkf;Ro c<+us ds lkFk] ngu dh Å"ek es deh gksrh gSA

LFkkf;Ro dk Øe

>

>

3. 'X' melts at low temperature and is a bad conductor of electricity in both liquid and solid state. X is :

(1) Carbon tetrachloride (2) Zinc sulphide (3) Silicon carbide (4) Mercury

'X' fuEu rki ij fi?kyrk gS rFkk nzo rFkk Bksl nksuksa voLFkkvksa esa fo|qr dk dqpkyd gSA X gS %

(1) dkcZu VsVªkDyksjkbM (2) ftad lYQkbM (3) flfydkWu dkckZbM (4) edZjh

Ans. (1)

Sol. CCl4 Non-conductor in solid and liquid phase.

CCl4 Bksl rFkk æo izkoLFkk esa dqpkyd gSA





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4. A chemist has 4 samples of artificial sweetener A,B, C and D. To identify these samples, he performed certain experiments and noted the following observations :

(i) A and D both form blue-violet colour with ninhydrin. (ii) Lassaigne extract of C gives positive AgNO3 test and negative Fe4[Fe(CN)6]3 test. (iii) Lassaigne extract of B and D gives positive sodium nitroprusside test. Based on these observations which option is correct ? (1) A : Alitame ; B : Saccharin ; C : Aspartame; D : Sucralose (2) A : Aspartame; B : Alitame ; C : Saccharin ; D : Sucralose (3*) A : Aspartame; B : Saccharin ; C : Sucralose ; D : Alitame (4) A : Saccharin ; B : Alitame ; C : Sucralose ; D : Aspartame

,d dsfeLV ds ikl Ñf=ke e/kqjdksaa A,B, C rFkk D dk 4 izfrn'kZ gSaA bu izfrn'kksZ dks igpkuus ds fy, mlus dqN iz;ksx fd;s x;s rFkk fuEu izs{k.kksa dks uksV fd;k %

(i) A rFkk D nksuksa fuugkbfMªu ds lkFk uhyk cSaxuh jax nsrs gSA

(ii) C dk ySlsa lkjdÙk (Lassaigne extract) AgNO3 ds lkFk ldkjkRed rFkk Fe4[Fe(CN)6]3 ds lkFk udkjkRed ijh{k.k nsrk gSA

(iii) B rFkk D dk ySlsa lkjdÙk lksfM;e ukbVªksiqzlkbM ds lkFk ldkjkRed ijh{k.k nsrk gSA bu izs{k.kksa ds vk/kkj ij dkSu lk fodYi lgh gS\

(1) A : ,syhVse ; B : lSdjhu ; C : ,sLijVse ; D : lqØkykst

(2) A : ,sLijVse; B : ,syhVse ; C : lSdjhu ; D : lqØkykst

(3*) A : ,sLijVse; B : lSdjhu ; C : lqØkykst ; D : ,syhVse

(4) A : lSdjhu ; B : ,syhVse ; C : lqØkykst ; D : ,sLijVse Ans. (3)

Sol. A – Aspartame

B – Saccharine

C – Sucralose

O

D – Alitame





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(i) A & D give positive test with ninhydrin because both have free carboxylic and amine groups. (ii) C form precipitate with AgNO3 in the lassaigne extract of the sugar because it has chlorine atoms.

(iii) B & D give positive test with sodium nitroprusside because both have sulphur atoms.

Sol. A – ,sLikVsZe

B – lSdjhu

C – lqØkykst

O

D – ,sfyVse

(i) A vkSj D fuugkbMªhu ds lkFk ldkjkRed ijh{k.k nsrs gSa] D;ksafd nksuksa es isIVkbM cU/k ik;s tkrs gSA

(ii) C 'kdZjk ds yslkus fu"d"kZ eas, AgNO3 ds lkFk vo{ksi nsrk gS] D;ksafd bles Dyksjhu ijek.kq gksrs gSA

(iii) B vkSj D lksfM;e ukbVªksizqlkbM ds lkFk ldkjkRed ijh{k.k nsrk gSA D;ksafd nksuksa es lYQj ijek.kq gksrs gSA

5. The electronic configurations of bivalent europium and trivalent cerium are :

(atomic number : Xe = 54, Ce = 58, Eu = 63)

(1) [Xe] 4f4 and [Xe] 4f9 (2) [Xe] 4f7 and [Xe] 4f1

(3) [Xe] 4f7 6s2 and [Xe] 4f2 6s2 (4) [Xe] 4f2 and [Xe] 4f7

f}la;kstd ;wjksfi;e rFkk f=kla;kstd lhfj;e ds bysDVªkWfud foU;kl gSa %

(ijek.kq la[;k : Xe = 54, Ce = 58, Eu = 63)

(1) [Xe] 4f4 rFkk [Xe] 4f9 (2) [Xe] 4f7 rFkk [Xe] 4f1

(3) [Xe] 4f7 6s2 rFkk [Xe] 4f2 6s2 (4) [Xe] 4f2 rFkk [Xe] 4f7

Ans. (2)

Sol. Eu2+ : [Xe]4f7

Ce3+ : [Xe]4f1





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6. Complex X of composition of Cr(H2O)6Cln has a spin only magnetic moment of 3.83 BM. It reacts with

AgNO3 and shows geometrical isomerism. The IUPAC nomenclature of X is :

(1) Dichloridotetraaqa chromium(IV) chloride dehydrate

(2) Hexaaqua chromium(III)chloride

(3) Tetraaquadichlorido chromium(III) chloride dehydrate

(4) Tetraaquadichlorido chromium(IV) chloride dihydrate

Cr(H2O)6Cln la?kVu ds ladqy X dk fLiu ek=k dk pqEcdh; vk?kw.kZ 3.83 BM gSA ;g AgNO3 ds lkFk vfHkfØ;k djrk

gS vkSj T;kferh; leko;ork iznf'kZr djrk gSA X dk IUPAC uke gS %

(1) MkbDyksjkbMkVsVªk,sDok Øksfe;e(IV) DyksjkbM MkbgkbMªsV

(2) gsDlk,DokØksfe;e(III) DyksjkbM

(3) VsVªk,sDokMkbDyksjkbMks Øksfe;e(III) DyksjkbM MkbgkbMªsV

(4) VsVªk,sDokMkbDyksjkbMks Øksfe;e(IV) DyksjkbM MkbgkbMªsV Ans. (3)

Sol. Cr(H2O)6Cln (complex)spin = 3.83 B.M.

From data of magnetic moment oxidation number of Cr should be +3

Hence complex is Cr(H2O)6Cl3.

Complex shows geometrical isomerism therefore formula of complex is [Cr(H2O)4Cl2]Cl2H2O.

It's IUPAC Name: Tetraaquadichloridochromium(III) chloride dihydrate

Cr(H2O)6Cln (ladqy)pØ.k = 3.83 B.M.

pqEcdh; vk?kw.kZ ds eku ls Cr dk vkWDlhdj.k vad +3 gksuk pkfg,A

bl izdkj ladqy Cr(H2O)6Cl3 gSA

ladqy T;kferh; leko;ork n'kkZrk gSA blfy, ladqy dk lw=k [Cr(H2O)4Cl2]Cl2H2O gSA

bldk IUPAC uke VsVªk,DOkkMkbZDyksjkbMksaØksfe;e (III) DyksjkbM MkbZgkbMªsV gSA

7. The major product Z obtained in the following reaction scheme is

fuEu vfHkfØ;k Ldhe esa izkIr gksus okyk eq[; mRikn Z gS %

NaNO2 + HCl

273-278 K

NH2

Br

X Cu2Br2

Y HNO3

H2SO4 Z

(1)

NO2

Br Br

(2)

Br

Br

O2N

(3)

Br Br

NO2

NO2

(4)

NO2

Br

Br

Ans. (4)

Sol.

NH2

Br

(1) NaNO2 + HCl (2) Cu2Br2

N2+Cl–

Br

Br

Br

(3)HNO3

Conc. H2SO4

Br

Br

NO2





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8. The major product (Y) in the following reactions is :

fuEu vfHkfØ;kvksa esa eq[; mRikn (Y) gS %

H3C–CH—CCH3

CH3

HgSO4, H2SO4

H3O X

(i) C2H5MgBr, H2O

(ii) Conc. H2SO4Y

(1)

CH3–CH–C=CH2

CH2–CH3

CH3

(2)

CH3–C=C–CH3

CH2–CH3

CH3

(3)

CH3–C–CH–CH3

CH2–CH3

CH2

(4)

CH3–CH–C=CH–CH3

CH3

CH3

Ans. (2)

Sol.

H3C–CH—C–CH3

O CH3

(1) EtMgBr/Ether

(2) H3O

CH3–CH—C–CH3

CH2–CH3

CH3 OH H+/

CH3–C=C–CH3

CH2–CH3

CH3

Sol.

H3C–CH—C–CH3

O CH3

(1) EtMgBr/bZFkj

(2) H3O

CH3–CH—C–CH3

CH2–CH3

CH3 OH H+/

CH3–C=C–CH3

CH2–CH3

CH3

9. B has a smaller first ionization enthalpy than Be. Consider the following statements :

(I) It is easier to remove 2p electron than 2s electron

(II) 2p electron of B is more shielded from the nucleus by the inner core of electrons than the 2s electrons

of Be

(III) 2s electron has more penetration power than 2p electron

(IV) atomic radius of B is more than Be

(atomic number B = 5, Be = 4)

The correct statement are :

(1) (I), (III) and (IV) (2) (II), (III) and (IV)

(3) (I), (II) and (IV) (4) (I), (II) and (III)

B dh izFke vk;uu ,UFkSYih Be ls de gSA fuEu dFkuksa ij fopkj dhft,%

(I) 2s bysDVªkWu dh rqyuk esa 2p bysDVªkWu gVkuk vklku gSA

(II) Be ds 2s bysDVªkWuksa dh rqyuk esa B ds 2p bysDVªkWu vkarfjd dksj bysDVªkWuksa }kjk ukfHkd ls T;knk ifjjf{kr gSaA

(III) 2p bysDVªkWuksa dh rqyuk esa 2s bysDVªkWu dh izos'kh lkeF;Z (penetration power) T;knk gSA

(IV) B dh ijek.kq f=kT;k] Be ls T;knk gSA

(ijek.kq la[;k B = 5, Be = 4)

lgh dFku gS %

(1) (I), (III) rFkk (IV) (2) (II), (III) rFkk (IV)

(3) (I), (II) rFkk (IV) (4) (I), (II) rFkk (III) Ans. (4)

Sol. Theory Based.

lS)kfUrd





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10. Identify (A) in the following reaction sequence :

(A) (i) CH3MgBr

(ii) H+ , H2O

(3) Conc. H2SO4 /

O3/Zn, H2O

Gives Positive Iodoform test

(B)

C

H3C C=O

O

CH3

H

CH3

fuEu vfHkfØ;k vuqØe esa (A) dh igpku dhft, &

(A) (i) CH3MgBr

(ii) H+ , H2O

(3) lkUnz H2SO4/

O3/Zn, H2O

ldkjkRed vkbMksQkeZ

ijh{k.k nsrk gSA

(B)

C

H3C C=O

O

CH3

H

CH3

(1)

O

CH3 CH3

CH3

(2)

O

CH3

CH3

(3)

O CH3

(4)

O

CH3

CH3

Ans. (4)

Sol. (1) CH3MgBr

(2) H3O+

O3

HO O

H,

Zn, H2O

CH=O

C

H3C CH3

O

CH3

11. According to the following diagram, A reduces BO2 when the temperature is :

200 400 600 800 1000 1200 1400 1600

–1200

–1000

–800

–600

A + O2 AO2

T(ºC)

G

º/kJ

mol–

1

B + O2 BO2

(1) < 1200ºC (2) > 1200ºC but < 1400ºC (3) < 1400ºC (4) > 1400ºC





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fuEu fp=k ds vuqlkj A, BO2 dk vip;u djrk gS tc rki gS %

200 400 600 800 1000 1200 1400 1600

–1200

–1000

–800

–600

A + O2 AO2

T(ºC)

G

º/kJ

mol–

1

B + O2 BO2

(1) < 1200ºC (2) > 1200ºC ijUrq < 1400ºC

(3) < 1400ºC (4) > 1400ºC Ans. (4)

Sol. A + BO2 B + AO2

G = –ve Only above 1400°C

dsoy 1400°C ds Åij

12. The acidic, basic and amphoteric oxides, respectively, are :

vEyh;] {kkjh; rFkk mHk;/kehZ vkWDlkbMsa Øe'k% gSa & (1) Na2O, SO3, Al2O3 (2) N2O3, Li2O, Al2O3 (3) Cl2O, CaO, P4O10 (4) MgO, Cl2O, Al2O3

Ans. (2) Sol. Non-metal oxides are acidic in nature alkali metal oxides are basic in nature Al2O3 is amphoteric.

vèkkfRod vkWDlkbM vEyh; izdf̀r ds gksrs gSA {kkjh; èkkrq vkWDlkbM {kkjh; izd̀fr ds gksrs gSA Al2O3 mHk;èkehZ gSA

13. If the magnetic moment of a dioxygen species is 1.73 B.M., it may be :

(1) 2O or

2O (2) O2 or

2O (3) O2 or

2O (4) O2,

2O or

2O

,d MkbZvkWDlhtu Lih'kht dk pqEcdh; vk?kw.kZ 1.73 B.M. gS] ;g gks ldrh gS %

(1) 2O vFkok

2O (2) O2 vFkok 2O (3) O2 vFkok

2O (4) O2, 2O vFkok

2O

Ans. (1)

Sol. O2 = 1s2 *1s22s2*2s22pz22px

2 = 2py2 *2px

1 = *2py1

O2– = 1s2 *1s22s2*2s22pz

22px2 = 2py

2 *2px2 = *2py

1

O2+ = 1s2 *1s22s2*2s22pz

22px2 = 2py

2 *2px1 = *2py

0

14. The Ksp for the following dissociation is 1.6 × 10–5

PbCl2(s) aq

2aq Cl2Pb

Which of the following choices is correct for a mixture of 300 mL 0.134M Pb(NO3)2 and 100 mL 0.4 M NaCl ?

(1) Q = Ksp (2) Q > Ksp (3) Not enough data provided (4) Q < Ksp





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fuEu fo;kstu ds fy;s Ksp dk eku 1.6 × 10–5 gS]

PbCl2(s) aq

2aq Cl2Pb

0.134M Pb(NO3)2 ds 300 mL rFkk 0.4 M NaCl ds 100 mL dks feykdj cuk;s x;s feJ.k ds fy, fuEu eas ls dkSu lk fodYi lgh gS \

(1) Q = Ksp (2) Q > Ksp

(3) i;kZIr vk¡dM+k miyC/k ugha (4) Q < Ksp

Ans. (2) Sol. Q = [Pb2+][Cl–]2

2300 0.134 100 0.4

400 400

23 0.1340.1

4

= 0.105 × 10–2 =1.005 ×10–3

Q > Ksp

15. If enthalpy of atomisation for Br2(I) is x kJ/mol and bond enthalpy for Br2 is y kJ/mol the relation between them : (1) is x = y (2) is x < y (3) does not exist (4) is x > y

;fn Br2(I) ds fy, d.ku ,UFkSYih (enthalpy of atomisation) x kJ/mol gks rFkk Br2 ds fy, vkcU/k ,UFkSYih y kJ/mol gks] rks muds chp lEcU?k & (1) x = y gksxk (2) x < y gksxk (3) curk ugha gSA (4) x > y gksxk Ans. (4)

Sol.

Br2()

Br2(g)

HVap.

Hatomisation = x kJ/mole 2Br(g)

Bond energy = y kJ/mole

Br2()

Br2(g)

HVap.

ijek.oh;dj.k = x kJ/mole 2Br(g)

caèk ÅtkZ = y kJ/mole

Hatomisation = Hvap + Bond energy Hence x > y

Hijek.oh;dj.k = Hvap + caèk ÅtkZ bl izdkj x > y

16. [Pd(F)(Cl)(Br)(I)]2– has n number of geometrical isomers. Then, the spin-only magnetic moment and crystal field stabilisation energy [CFSE] of [Fe(CN)6]n–6, respectively, are :

[Note : Ignore the pairing energy]

(1) 5.92 BM and 0 (2) 0 BM and –2.4 0

(3) 1.73 BM and –2.0 0 (4) 2.84 BM and –1.6 0

[Pd(F)(Cl)(Br)(I)]2– ds T;kferh; leko;oksa dh la[;k n gSA rc [Fe(CN)6]n–6 dk fLiu ek=k pqEcdh; vk?kw.kZ rFkk fØLVy {ks=k LFkk;hdj.k ÅtkZ [CFSE] Øe'k% gSa%

[uksV : ;qXeu ÅtkZ dks NksM+ nhft,A]

(1) 5.92 BM rFkk 0 (2) 0 BM rFkk –2.4 0

(3) 1.73 BM rFkk –2.0 0 (4) 2.84 BM rFkk –1.6 0

Ans. (3)





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Sol. Number of Geometrical Isomers in square planar [PdFClBrI]2– are = 3 Hence, n = 3 [Fe(CN)6]3–

Fe3+ = 3d5 , According to CFT configuration is 221 00

2g gt e

2n n

=1.73 B.M.

0 2 0–0.4 0.6g egCFSE nt n

= –0.40 × 5 = –2.00

Sol. oxZ leryh; [PdFClBrI]2– esa T;kfefr; leko;oh;ksa dh la[;k = 3 gSA blfy;s, n = 3

[Fe(CN)6]3–

Fe3+ = 3d5 , CFT ds vuqlkj foU;kl 221 00

2g gt e

2n n

=1.73 B.M.

0 2 0–0.4 0.6g egCFSE nt n = –0.40 × 5 = –2.00

17. The de Broglie wavelength of an electron in the 4th Bohr orbit is

pkSFkh cksj d{kk esa ,d bysDVªkWu dh Mh&czksXyh rjaxnS/;Z gksxh %

(1) 4a0 (2) 2a0 (3) 6a0 (4) 8a0

Ans. (4)

Sol. 2 r = n

2 × 2n

Za0 = n

2 × 24

1a0 = 4

= 8 a0

18. The compound that cannot act both as oxidising and reducing agent is :

og ;kSfxd tks mipk;d rFkk vipk;d nksuksa dh rjg dk;Z ugha dj ldrk gS &

(1) HNO2 (2) H3PO4 (3) H2SO3 (4) H2O2

Ans. (2) Sol. As in H3PO4 Phosphorous is present it's maximum oxidation number state hence it cannot act as reducing

agent.

pwafd H3PO4 esa QkWLQksjl bldh mPpre vkWDlhdj.k voLFkk esa mifLFkr gSA blfy, ;g vipk;d ds leku dk;Z ugha dj ldrk gSA

19. The increasing order of basicity for the following intermediates is (from weak to strong)

fuEu e/;ofrZ;ksa ds fy, {kkjh;rk dk c<+rk Øe gS (nqcZy ls izcy) :

CH3

H3C–C

CH3

(i)

H2C=CH–CH2

(ii)

HCC

(iii)

CH3

(iv)

CN

(v)

(1) (iii) < (iv) < (ii) < (i) < (v) (2) (iii) < (i) < (ii) < (iv) < (v) (3) (v) < (iii) < (ii) < (iv) < (i) (4) (v) < (i) < (iv) < (ii) < (iii) Ans. (3) Sol. Basicity is inversely proportional to electronegativity.

Sol. {kkjh;rk] fo|qr_.krk ds O;qRØekuqikrh gksrh gSA





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20. For following reactions

A K700 product

A catalyst

K500 product

it was found that the Ea is decreased by 30 kJ/mol in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assumer pre exponential factor is same) :

(1) 105 kJ/mol (2) 195 kJ/mol (3) 135 kJ/mol (4) 75 kJ/mol

fuEu vfHkfØ;k ds fy,

A K700 mRikn

A mRizsjd

K500 mRikn

;g ik;k x;k gS fd mRizsjd dh mifLFkfr esa Ea , 30 kJ/mol ls ?kV xbZA ;fn nj vifjofrZr jgs rks mRiszfjr vfHkfØ;k ds fy, lafØ;.k ÅtkZ gksxh (eku yhft;s iwoZ pj?kkrkadh xq.kd ogh jgrk gS) :

(1) 105 kJ/mol (2) 195 kJ/mol (3) 135 kJ/mol (4) 75 kJ/mol Ans. (4) Sol. Kcat = K

1

1

Ea

RTAe

=

2

2

Ea

RTAe

1 2

1 2

Ea Ea

T T Ea1 = Ea2 – 30

2Ea 30

500

= 2Ea

700

5Ea2 = 7Ea2 – 210

Ea2 = 210

2 = 105 kJ/mole

Activation energy of the catalysed reaction = 105 – 30 = 75 kJ/mole

mRizsfjr vfHkfØ;k dh lafØ;.k ÅtkZ = 105 – 30 = 75 kJ/mole

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





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7340010333

21. 108 g of silver (molar mass 108 g mol–1) is deposited at cathode from AgNO3(aq) solution by a certain quantity of electricity. The volume (in L) of oxygen gas produced at 273 K and 1 bar pressure from water by the same quantity of electricity is _________________

,d fuf'pr fo|qr ek=kk }kjk AgNO3(tyh;) ls 108 g flYoj (eksyj nzO;eku 108 g mol–1) dSFkksM+ ij fu{ksfir fd;k x;kA fo|qr dh mlh ek=kk }kjk 273 K rFkk 1 ckj nkc ij cuk;h xbZ vkWDlhtu dk vk;ru (L esa) gksxk _________________

Ans. 5.66 to 5.67

Sol. 1081

108Ag deposit

n mole

–Ag e Ag

1F charge is required to deposit 1 mole of Ag

Ag ds ,d eksy fu{ksi.k ds fy, 1 F vkos'k dh vko';drk gksrh gSA

–2 2

12 2

2

H O O H e

2F charge deposit 1

2 mole

2F vkos'k fu{ksfir djrk gS1

2mole

1F charge will deposit 1

4mole

1F vkos'k fu{ksfir djrk gS 1

4mole

2O

nRTV

P

=1 0.08314 273

4 1

2OV = 5.674 L

22. How much amount of NaCl should be added to 600 g for water ( = 1.00 g/mL) to decrease the freezing point of water to –0.2ºC ? __________. (The freezing point depression constant of water = 2 K kg mol–1)

600g ikuh ( = 1.00 g/mL) esa NaCl dh fdruh ek=kk feyk;h tk;s fd mldk fgekad ?kVdj –0.2ºC gks tk;s ?

__________. (ikuh ds fy, fgekad voueu fLFkjkad = 2 K kg mol–1)

Ans. 1.74 to 1.76

Sol. Tf = 0.2°C

Tf = ikfm

0.2 = 2×2×600

1000

5.58

w

41000

6005.582.0w

1.2 58.5

40

= 1.76g





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PAGE # 12

7340010333

23. The hardness of a water sample containing 10–3 M MgSO4 expressed as CaCO3 equivalents (in ppm) is _____________.

(molar mass of MgSO4 is 120.37 g/mol)

10–3 M MgSO4 okys ty ds izfrn'kZ dh dBksjrk ftldks CaCO3 lerqY; (ppm esa) vfHkO;Dr fd;s tkus ij] gksxh _____________.

(MgSO4 dk eksyj lagfr (molar mass) 120.37 g/mol)

Ans. 100.00 to 100.00

Sol. 10–3 molar MgSO4 10–3 moles of MgSO4 present in 1 L solutions.

10–3 eksyj MgSO4 1 L foy;u esa mifLFkr MgSO4 ds 10–3 eksy

3 4CaCO MgSOn n

3(in term of CaCO )ppm = 310 100

1000

× 106

3(in term of CaCO )ppm = 100 ppm

24. The mass percentage of nitrogen in histamine is _____________

fgLVSfeu esa ukbVªkstu dh nzO;eku izfr'krrk gS _____________

Ans. 37.80 to 38.20

Sol. Structure of Histamine is

N

NH

NH2

Molecular formula of Histamine is C5H9N3

Molecular mass of Histamine is 111

Percentage nitrogen by mass in Histamine = 100111

42 = 37.84%

Sol. fgLVkfeu dh lajpuk

N

NH

NH2

fgLVkfeu dh vkf.od lajpuk C5H9N3

fgLVkfeu dk vkf.od nzO;eku 111

fgLVkfeu es nzO;eku ls ukbVªkstu dk izfr'kr = 100111

42 = 37.84%

25. The molarity of HNO3 in a sample which has density 1.4 g/mL and mass percentage of 63% is __________. (Molecular Weight of HNO3 = 63 )

ml izfrn'kZ esa] ftldk ?kuRo 1.4 g/mL rFkk nzO;eku izfr'krrk 63% dh gks] HNO3 dh eksyjrk gksxh __________.

(HNO3 dk v.kqHkkj = 63 )

Ans. 14.00 to 14.00

Sol. 63% w/w HNO3 solution foy;u

M = 63 1.4

63 100

× 1000 mole/L = 14 mole/L




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7340010333



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