Jee Main 2019 Chemistry April Attempt Shift - 2(12th April, 2019) 494h1j

PAPER-1 (B.E./B. TECH.)

JEE (Main) 2019 COMPUTER BASED TEST (CBT)

Questions & Solutions Date: 12 April, 2019 (SHIFT-2) | TIME: 02.30 P.M. to 05.30 P.M.

Duration: 3 Hours | Max. Marks: 360 SUBJECT: CHEMISTRY

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| JEE (MAIN) 2019 | DATE : 12-04-2019 (SHIFT-2) | PAPER-1 | OFFICIAL PAPER | CHEMISTRY

PART : CHEMISTRY

Ans.

The INCORRECT match in the following is : fuEu esa xyr feyku fdlesa gS? (1) G° > 0, K < 1 (2) G° < 0, K < 1 (3) G° = 0, K = 1 (4) G° < 0, K > 1 (2)

Sol.

G= –RTlnK

1.

G< 0  K > 1 2.

The IUPAC name for the following compound is : CH3

CH

H3C CH3 CH2

(1) 3-methyl-4-(3-methylprop-1-enyl)-1-heptyne (2) 3,5-dimethyl-4-propylhept-1-en-6-yne (3) 3,5-dimethyl-4-propylhept-6-en-1-yne (4) 3-methyl-4-(1-methylprop-2-ynyl)-1-heptene fuEUk ;kSfxd ds fy, IUPAC uke gS %

CH3 CH

H3C CH3 CH2

Ans.

(1) 3-esfFky-4-(3-esfFkyizksi-1-bZfuy)-1-gsIVkbu (2) 3,5-MkbesfFky-4-izksfiygsIV-1-bZu-6-vkbu (3) 3,5-MkbesfFky-4-izksfiygsIV-6-bZu-1-vkbu (4) 3-esfFky-4-(1-esfFkyizksi-2-vk;fuy)-1-gsIVhu (2)

Sol.

1

2

3

4

5

6 7

3,5-Dimethyl-4-propylhept-1-en-6-yne 3,5-MkbZesfFky-4-izksfiygsIV-1-bZu-6-vkbZu

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3.

Among the following, the INCORRECT statement about colloids is : (1) They can scatter light. (2) The range of diameters of colloidal particles is between 1 and 1000 nm. (3) The osmotic pressure of a colloidal solution is of higher order than the true solution at the same concentration. (4) They are larger than small molecules and have high molar mass dksykWbMl ds lEcU/k esa fuEu esa ls dkSulk xyr gS : (1) ;s NksVs v.kqvksa dh rqyuk esa cMs gksrs gS vkSj mudk eksyj nzO;eku mPPk gksrk gS (2) dksykbMh d.kksa ds O;kl dk ijkl 1 rFkk 1000 nm ds chp gksrk gSA (3) ,d gh lkanzrk ij dksykbMh foy;u dk ijklj.k nkc okLrfod foy;u ds nkc dh rqyuk esa mPPkrj eku dk

gksrk gSA (4) ;s izdk'k dks izdh.kZ dj ldrs gSA Ans. Sol.

(3) The osmotic pressure of a colloidal solution is of lower order than the true solution at the same concentration.

,d gh lkanzrk ij dksykbMh foy;u dk ijklj.k nkc okLrfod foy;u ds nkc dh rqyuk esa fuEurj eku dk gksrk gSA 4.

The correct name of the following polymer is : H3C CH3 n

(1) Polyisobutane (3) Polyisobutylene

(2) Polytert-butylene (4) Polyisoprene

fuEu cgqyd dk lgh uke gS: H3C CH3 n

Ans.

(1) ikyhvkblksC;wVsu (3) ikyhvkblksC;wVkbyhu (3)

(2) ikyhVVZ-C;wVkbyhu (4) ikyhvkblksizhu H3C CH3

Sol.

Polymerisation CH3–C=CH2 

n

CH3 5.

NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation 2N2O5 (g) 4NO2(g) + O2 (g). The initial concentration of N2O5 is 3.00 mol L–1 and it is 2.75 mol L–1 after 30 minutes. The rate of formation of NO2 is : ,d vfHkfØ;k ds fy, vko';d NO2 dks CCl4 esa N2O5 ds vi?kVu }kjk mRiUUk djrs gS] tSlk fd uhps lehdj.k

esa

gS]

Ans.

2N2O5 (g) 4NO2(g) + O2 (g). N2O5 dh izkjfEHkd 2.75 mol L–1 gSA NO2 ds lEHkou dh nj gksxh : (1) 8.333 × 10–3 mol L–1 min–1 (2) 2.083 × 10–3 mol L–1 min–1 (3) 1.667 × 10–2 mol L–1 min–1 (4) 4.167 × 10–3 mol L–1 min–1 (3)

lkUnzrk 3.00 mol L–1 rFkk 30 feuV ds ckn dh lkUnzrk

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Sol.

dN2O5  dt 



3.0  2.75 0.25 25 5 1     30 30 3000 600 120

1 dN2O5  1 dNO2   2 dt 4 dt

1 1 dNO2    120 2 dt Rate of formation of NO2 (NO2 ds

6.

lEHkou dh nj½ =

dNO2  dt



1 M min1 = 1.667 × 10–2 mol L–1 min–1 60

Heating of 2-chloro-1-phenylbutane with EtOK/EtOH gives X as the major product. Reaction of X with Hg(OAc)2/H2O followed by NaBH4 gives Y as the major product. Y is 2-Dyksjkss -1-QfuyC;wVsu dks EtOK/EtOH ds lkFk xje djus ij X eq[; :Ik esa izkIr gksrk gSA Hg(OAc)2/H2O ds lkFk X dh vfHkfØ;k rRi'pkr~ NaBH4 ds lkFk vfHkfØ;k ls izkIr Y eq[; mRikn gS :

OH (1) Ph

(2) Ph

OH (4) Ph

(3) Ph

OH Ans.

(3)

OH Sol.

Ph

EtOK  

EtOH(E2 elimination)

Cl

HgOAc 2   Ph Ph  NaBH 4

(Markovnikov addition product is major in step - 2) 7.

25 g of an unknown hydrocarbon upon burning produces 88g of CO2 and 9 g of H2O. This unknown hydrocarbon contains : (1) 22 g of carbon and 3g of hydrogen (2) 24 g of carbon and 1g of hydrogen (3) 20 g of carbon and 5g of hydrogen (4) 18 g of carbon and 7g of hydrogen ,d vKkr gkbMªksdkcZu ds 25 g dks tykus ij 88g CO2 rFkk 9 g H2O mRiUUk gksrs gSA bl vKkr gkbMªksdkcZu esa

;s

lfUUkfgr gSA (1) 22 g dkcZu rFkk 3g gkbMªkstu (2) 24 g dkcZu rFkk 1g gkbMªkstu (3) 20 g dkcZu rFkk 5g gkbMªkstu (4) 18 g dkcZu rFkk 7g gkbMªkstu Ans. Sol.

(2) Mass of CO2 = 88g  Mass of C =

12 × 88 = 24g 44

Mass of H2O = 9 g

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 Mass of H = CO2 dk

æO;eku = 88g

 C dk

æO;eku =

H2O dk



8.

2  9 = 1g 18

12 × 88 = 24g 44

æO;eku = 9 g

H dk

æO;eku =

2  9 = 1g 18

An 'Assertion' and a 'Reason' are given below. Choose the correct answer from the following options : Assertion(A) : Vinyl halides do not undergo nucleophilic substitution easily. Reason (R) : Even though the intermediate carbocation is stabilized by loosely held -electrons, the cleavage is difficult because of strong bonding. (1) Both (A) and (R) are correct statements and (R) is the correct explanation of (A) (2) (A) is a correct statement but (R) is a wrong statement. (3) Both (A) and (R) are correct statement but (R) is not the correct explanation of (A). (4) Both (A) and (R) are wrong statements.

,d dFku rFkk ,d dkj.k uhps fn;k x;k gSA fuEu fodYiks esa ls lgh mrj dk pquko dhft, dFku (A) : foukby gSykbM dk ukfHkdjkxh izfrLFkkiu vklkuh ls ugh gksrkA dkj.k (R) : vn`<+ -byDVªkWuksa }kjk e/;orhZ dkcksZdsVk;u ds LFkkf;Ro ds ckotwn Hkh izcy vkca/ku ds dkj.k fonyu dfBu gSA (1) (A) rFkk (R) nksuksa lgh gS rFkk (R), (A) dh lgh O;k[;k gSA (2) (A) lgh gS ijURkq (R) xyr gSA (3) (A)rFkk (R) nksuksa lgh gS ijUrq (R), (A) dh lgh O;[;k ugh gSA (4) (A) rFkk (R) nksukas gh xyr gSaA Ans.

(2)

9.

The pair that has similar atomic radii is : (1) Ti and Hf (2) Mo and W (3) Sc and Ni (4) Mn and Re

og ;qXe ftldh ijekf.od f=kT;k;sa ,d tSlh gS] (1) Ti rFkk Hf (2) Mo rFkk W (3) Sc rFkk Ni (4) Mn rFkk Re Ans.

(2)

Sol.

rMo  rW due to lanthanoid contraction ysUFkSukbM ladqpu dh otg ls

10.

Which one of the following is likely to give a precipitate with AgNO 3 solution? fuEUk esa ls fdldh AgNO3 foy;u ds lkFk vo{ksi nsus dh laHkkouk gS\ (1) CCl4 (2) CH2=CH–Cl (3) (CH3)3CCl (4) CHCl3 (3) (CH3)3CCl gives a precipitate with AgNO3 solution easily because t-butyl halides give stable carbocation.

Ans. Sol.

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11.

Ans. Sol.

The coordination numbers of Co and Al in [Co(Cl)(en)2]Cl and K3[Al(C2O4)3], respectively, are : (en = ethane-1, 2-diamine) [Co(Cl)(en)2]Cl rFkk K3[Al(C2O4)3], esa Co rFkk Al dh milgla;kstu la[;k;sa Øe'k% gS : (en = ,sFksu -1, 2-Mkb,sehu) (1) 3 and 3 (2) 5 and 6 (3) 6 and 6 (4) 5 and 3 (1) 3 rFkk 3 (2) 5 rFkk 6 (3) 6 rFkk 6 (4) 5 rFkk 3 (2) In [Co(en)2Cl]Cl, co-ordination no = 2 × 2 + 1= 5 In Na3 [Al(C2O4)3], co-ordination no = 3 × 2 = 6 [Co(en)2Cl]Cl

esa, leUo; la[;k = 2 × 2 + 1= 5

Na3 [Al(C2O4)3] 12.

esa, leUo; la[;k = 3 × 2 = 6

In the following skew conformation of ethane, H'–C–C–H'' dihedral angle is : ,Fksu ds fuEu fo"keryh; la:i.k esa] H'–C–C–H'' f}ry dks.k gS : H H' H

H

29°

H'' H

Ans. Sol.

(1) 58° (2)

(2) 149°

H H' H

H

(3) 120°

(4) 151°

(3) graphite

(4) C60

(3) xzsQkbV

(4) C60

29°

120°

H'' H

13.

Ans.

The C-C bond length is maximum in : (1) C70 (2) diamond fuEu esa ls fdlesa C-C vkcU/k yEckbZ vf/kdre (1) C70 (2) ghjk (Mk;eaM) (2)

gS?

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Sol.

Longest C–C bond is present in diamond

nh?kZre C–C ca/k ghjk esa mifLFkr gSA Compound (;kSfxd)

C–C bond length (C–C

14.

caèk yEckbZ)

C70

143.5, 138.3 pm

Diamond (ghjk)

154 pm

Graphite (xzsQkbV)

141.5 pm

C60

143.5, 138.3 pm

What will be the major product when m-cresol is reacted with propargyl bromide (HCC–CH2Br) in presence of K2CO3 in acetone? eq[; mRikn D;k gksxk tc m-fØlkWy dks ,lhVksu esa K2CO3 dh mifLFkfr esa izksiftZy czksekbM (HCC–CH2Br) ds

lkFk vfHkd`r fd;k tkrk gSA

OH

OH

(1)

(2)

CH3

CH3 OH

O (4)

(3)

CH3

CH3 Ans.

(4)

Acetone

CH3

Ans. Sol.

SN 2 reaction + CHC–CH2–Br  

K CO

3 2   

Sol.

15.

O–CH2–CCH

OK

OH

CH3

CH3

Which of the given statements is INCORRECT about glycogen? (1) it is present in animal cells. (2) It is a straight chain polymer similar to amylose (3) Only -linkage are present in the molecule. (4) It is present in some yeast and fungi. Xyk;dkstsu ds lEcU/k esa fn;s x;s dFkuksa esa ls dkSulh lgh ugh gS? (1) ;g izk.kh dksf"kdkvksa eas mifLFkr gSA (2) ,feykst dh rjg ;g ,d _tqJa`[ky cgqyd gSA (3) v.kqvksa esa ek=k -ca/kusa mifLFkr gSA (4) ;g dqN ([kehj) rFkk daodks esa mifLFkr gSA (2) Polymeric chain of glycogen is similar to amylopectin (highly branched polymer).

Xykbdkstu dh cgqydh J`¡[kyk ,ekbyksisDVhu ¼vR;f/kd 'kkf[kr cgqyd½ ds le:i gksrh gS

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16.

Ans. Sol.

The molar solubility of Cd(OH)2 is 1.84 × 10–5 M in water. The expected solubility of Cd(OH)2 in a buffer solution of pH = 12 is : ty esa Cd(OH)2 dh eksyj foys;rk 1.84 × 10–5 M gSA pH = 12 ds ,d cQj foy;u esa Cd(OH)2 dh lEHkkfor foys;rk gksxh : (1) 2.49 × 10–10 M (2) 6.23 × 10–11 M 2.49 (4) 1.84 × 10–9 M (3)  10– 9 M 1.84 (1) Ksp = 4s3 = 4 × (1.84 × 10–5)3 = 24.9 × 10–15 Ksp = 2.49 × 10–14 Cd2+ (aq) + 2OH– (aq)

Cd(OH)2

10–2

s 2.49 × 10–14 = s(10–2)2 or

;k, s = 2.49 × 10–10 M

Ans.

The temporary hardness of a water sample is due to compound. X. Boiling this sample converts X to compound Y. X and Y, respectively, are : (1) Mg(HCO3)2 and MgCO3 (2) Ca(HCO3)2 and CaO (3) Mg(HCO3)2 and Mg(OH)2 (4) Ca(HCO3)2 and Ca(OH)2 ty izfrn'kZ dh vLFkk;h dBksjrk ;kSfxd X ds dkj.k gSA bl izfrn'kZ dks mckyus ij X cnydj ;kSfxd Y gks tkrk gSA X rFkk Y Øe'k% gS : (1) Mg(HCO3)2 rFkk MgCO3 (2) Ca(HCO3)2 rFkk CaO (3) Mg(HCO3)2 rFkk Mg(OH)2 (4) Ca(HCO3)2 rFkk Ca(OH)2 (3)

Sol.

2Mg(HCO3)2  2MgCO3 + H2O + CO2

17.



–

CO3–2 +H2O  HCO3 + OH

–

–

Mg+2 + 2OH  Mg(OH)2 X : Mg(HCO3)2 Y : Mg(OH)2

Ans.

The INCORRECT statement is : (1) LiNO3 decomposes on heating to give LiNO2 and O2. (2) Lithium is the strongest reducing agent among the alkali metals. (3) LiCl crystallises from aqueous solution as LiCl.2H2O (4) Lithium is least reactive with water among the alkali metals. xyr dFku gS : (1) LiNO3 xje djus ij vi?kfVr gksdj LiNO2 rFkk O2 nsrk gSA (2) {kkj /kkrqvksa esa yhfFk;e izcyre vipk;h deZd gS A (3) LiCl tyh; foy;u ls LiCl.2H2O ds :Ik esa fØLVfyr gksrk gSA (4) {kkj /kkrqvksa esa fyfFk;e ty ds lkFk lcls de vfHkfØ;k'khy gSA (1)

Sol.

4 LiNO3  2 Li2O + 4NO2+ O2

18.

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19.

Benzene diazonium chloride on reaction with aniline in the presence of dilute hydrochloric acid gives :

ruq gkbMªksDyksfjd vEy dh mifLFkfr esa csaTk+hu Mkbtksfu;e DyksjkbM] ,fuyhu ds lkFk vfHkfØ;k djds nsrk gS %

N=N

(1)

NH2 NH2

(2)

N=N–NH

(3)

N=N (4) Ans.

Sol. 20.

Ans. Sol.

H2N

(1) + N2

NH2

N=N

NH2

In comparison to boron, beryllium has : (1) greater nuclear charge and lesser first ionization enthalpy. (2) lesser nuclear charge and lesser first ionization enthalpy (3) lesser nuclear charge and greater first ionization enthalpy (4) greater nuclear charge and greater first ionization enthalpy cksjku dh rqyuk esa csjhfy;e j[krk gS : (1) mPPkrj ukfHkdh; vkos'k rFkk fuEUrj izFke vk;uu ,sUFkSYih (2) fuEurj ukfHkdh; vkos'k rFkk mPPkrj izFke vk;uu ,sUFkSYih (3) fuEurj ukfHkdh; vkos'k rFkk fuEUkrj izFke vk;uu ,sUFkSYih (4) mPPkrj ukfHkdh; vkos'k rFkk mPPkrj izFke vk;uu ,sUFkSYih (3) Be has lesser nuclear charge, greater first ionization enthalpy in comparison to B. Be, B dh

21.

dil HCl

+

rqyuk esa U;wu ukfHkdh; vkos'k] vf/kd izFke vk;uu ,sUFkSYih j[krk gSA

The ratio of number of atoms present in a simple cubic, body centered cubic and face centered cubic structure are respectively :

ljy ?kuh;] var% dsfUnzr ?kuh; rFkk Qyd dsfUnzr ?kuh; lajpuk esa mifLFkr ijek.kqvksa dh la[;k dk vuqikr Øe'k% gS %] gksxk :

Ans.

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

Sol.

In sc esa, Z = 8 ×

1 =1Z=1 8 1 + 1 = 2 Z = 2 In bcc esa Z = 8 × 8 1 1 = 4 Z = 4  In fcc esa, Z = 8 × + 6 × 2 8

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22.

The decreasing order of electrical conductivity of the following aqueous solution is : 0.1 M Formic acid (A) 0.1 M Acetic acid (B) 0.1 M Benzoic acid (C)

fuEUk tyh; foy;uksa dh fo|`rh; pkydrk dk ?kVrk Øe gSA 0.1 M QkfeZd ,flM (A) 0.1 M ,flfVd ,flM (B) 0.1 M csUtksbd ,flM (C)

Ans.

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

Sol.

Since acidic strength follows order HCOOH > C6H5COOH > CH3COOH

pwafd vEyh; lkeF;Z fuEu Øe dk vuqlj.k djrh gS HCOOH > C6H5COOH > CH3COOH 23.

Ans. Sol.

In which one of the following equilibria, KP KC? fuEu fdl ,d lkE; esa KP KC gS? N2(g) + O2(g) (1) 2NO(g) (2) 2C(s) + O2(g) 2CO(g) NO(g) + SO3(g) (3) NO2(g) + SO2(g) (4) 2HI(g) H2(g) + I2(g) (2) Kp = Kc(RT)n ng  0 for 2C(s) + O2(g)

2CO(g)

Kp = Kc(RT)n 2C(s) + O2(g)

2CO(g) ds

fy, ng  0

ng  1 24.

The correct statement is : (1) The blistered appearance of copper during the metallurgical process is due to the evolution of CO 2 (2) The Hall-Heroult process is used for the production of aluminium and iron (3) leaching of bauxite using concentrated NaOH solution gives sodium aluminate and sodium silicate. (4) pig iron obtained from cast iron.

lgh dFku gS % (1) /kkfRod izØe ds chp dkWij dk fCyLVMZ :Ik CO2 ds fuxZeu ds dkj.k gksrk gSA (2) ,yqehfu; rFkk vk;ju ds mRiknu ds fy, gky-gsjkYV izØe iz;qDr gksrk gSA (3) lkUnz NaOH foy;u dk iz;ksx djrs gq;s] ckDlkbM dk fu{kkyu lksfM;e ,yqehusV rFkk lksfM;e flyhdsV nsrk gSA (4) dkLV vk;ju (
(3) Leaching of bauxite using concentrated NaOH solution gives sodium aluminate and sodium silicate.

lkUnz NaOH foy;u dk iz;ksx djrs gq;s] ckDlkbM dk fu{kkyu lksfM;e ,yqehusV rFkk lksfM;e flyhdsV nsrk gSA

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| JEE (MAIN) 2019 | DATE : 12-04-2019 (SHIFT-2) | PAPER-1 | OFFICIAL PAPER | CHEMISTRY

Ans.

Thermal decomposition of a Mn compound (X) at 513K results in compound Y, MnO 2 and a gaseous product. MnO2 reacts with NaCl and concentrated H2SO4 to give a pungent gas Z. X, Y and Z, respectively are : (1) KMnO4, K2MnO4 and Cl2 (2) K2MnO4, KMnO4 and SO2 (3) K3MnO4, K2MnO4 and Cl2 (4) K2MnO4, KMnO4 and Cl2 513K ij ,d Mn ;kSfxd (X) ds rkih; vi?kVu ls ;kSfxd Y, MnO2 rFkk ,d xSlh; mRikn izkIRk gksrk gSA NaCl RkFkk lkUnz H2SO4 ls MnO2 vfHkfØ;k djds ,d rh[kh xSl Z nsrk gSA X, Y rFkk Z Øe'k% gS : (1) KMnO4, K2MnO4 rFkk Cl2 (2) K2MnO4, KMnO4 rFkk SO2 (3) K3MnO4, K2MnO4 rFkk Cl2 (4) K2MnO4, KMnO4 rFkk Cl2 (1)

Sol.

 KMnO4   K2MnO4 + MnO2 + O2 (g)

25.

(X)

(Y)

MnO2 + 2NaCl + 2H2SO4   MnSO4 + Na2SO4 + Cl2 +H2O (Z) 26.

Consider the following reactions : Ag2O ppt  A

Hg2++/H+

B

NaBH4

C

ZnCl2 Turbidity within 5 conc. HCl minutes

'A' is :

fuEu vfHkfØ;kvksa ij fopkj dhft, %

A

Ag2O  Hg2++/H+

Ans.

ppt B

NaBH4

C

ZnCl2 5 feuV conc. HCl

'A' gS : (1) CHCH (3) CH3–CCH (3)

(2) CH3–CC–CH3 (4) CH2=CH2

O Sol.

esa vkfoyrk

1 Hg2

 CH3—C—CH3 CH3–CCH (X)  H

OH 3  ZnCl 2 2  NaBH4  (Lucas test in 5   CH3—CH—CH3  HCl

min.) 2º alcohols give Lucas test with ZnCl2,HCl in 5 min. 2º ,Ydksgky ZnCl2,HCl ds lkFk 5 feuV esa Y;qdkl ijh{k.k 27.

nsrs gSA

The primary pollutant that leads to photochemical smog is : (1) acrolein (2) sulphur dioxide (3) nitrogen oxides (4) ozone

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| JEE (MAIN) 2019 | DATE : 12-04-2019 (SHIFT-2) | PAPER-1 | OFFICIAL PAPER | CHEMISTRY

izkFkfed iznw"kd tks izdk'kjlk;fud /kwedqgk iSnk djrk gS] gS : (2) lYQj MkbvkWDlkbM vkWDlkbMsa (4) vkstksu

Ans. 28.

(1) ,Øksyhu (3) ukbVªkstu (3)

Among the following the energy of 2s orbital is lowest in :

fuEu esa] 2s d{kd dh ÅtkZ fdlesa fuEure gS? Ans. Sol.

(1) Na (2) Li (3) H (4) K (4) Energies of the orbitals in the same subshell decrease with increase in the atomic number

leku midks'k esa d{kdksa dh ÅtkZ, ijek.kq Øekad esa o`f) ds lkFk ?kVrh gSA E2s(H) > E2s(Li) > E2s(Na) > E2s(K)SS 29.

The compound used in the treatment of lead poisoning is : (1) D-penicillamine (2) EDTA (3) desferrioxime B (4) Cis-platin

ysM fo"kfDrRkk ds mipkj esa iz;qDr ;kSfxd gS % (1) D-isuhflykekbu (3) MslQsjhvkDlkbe B

(2) EDTA (4) fll-IysfVu

Ans. Sol.

(2) EDTA is used in the treatment of lead poisoning. EDTA, ysM ds fo"k ds mipkj esa ç;qDr fd;k tkrk gSA

30.

A solution is prepared by dissolving 0.6 g of urea (molar mass = 60g mol–1) and 1.8g of glucose (molar mass = 180 g mol–1) in 100 mL of water at 27°C. The osmotic pressure of the solution is : (R = 0.08206 L atm K–1 mol–1) 27°C ij] ,d foy;u dks 100 mL ty esa 0.6 g ;wfj;k (eksyj nzO;eku = 60g mol–1) rFkk 1.8g Xywdkst (eksyj nzO;eku = 180 g mol–1) ?kksydj rS;kj fd;k x;kA foy;u dk ijklj.k nkc gksx : (R = 0.08206 L atm K–1 mol–1) (1) 8.2 atm (2) 1. 64 atm (3) 2.46 atm (4) 4.92 atm (4)  = CRT  0.6 1.8  1000   (n  n )  1000  =  1 2  RT =  60  180   100   0.082  300 V      = [(0.01 + 0.01) × 10] × 0.082 × 300 = 4.92 atm

Ans. Sol.

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