Jp Xii Physical&inorganic Chemistry (27) - Prev Chaps + Inorg. Chem.pdf

PHY./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2016 NO. 43

Course : VIJETA (JP)

This DPP is to be discussed in the week (28.09.2015 to 03.10.2015)

ANSWERKEY DPP No. # 43 (JEE-MAIN) 1.

(A)

2.

(A)

3.

(C)

4.

(C)

5.

(A)

6.

(C)

7.

(A)

8.

(D)

9.

(B)

10.

(C)

11.

(C)

12.

(B)

13.

(B)

14.

(B)

15.

(C)

16.

(C)

17.

(B)

18.

(C)

19.

(A)

20.

(D)

DPP No. # 43 (JEE-MAIN) Total Marks : 64

Max. Time : 40 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.16 ChemINFO : 4 Questions ('–1' negative marking) Q.17 to Q.20

1.

(3 marks, 2 min.) (4 marks, 2 min.)

[48, 32] [16, 08]

Which one of the following statements about H3BO3 is not correct ? (A*) It is a strong tribasic acid (B) It is a prepared by acidifying an aqueous solution of borax (C) It has a layer structure in which planarBO 3 units are joined by hydrogen bonds (D) It doesnot act as proton donor as it acts as a Lewis acid by accepting hydroxyl ions H3BO3 ds lanHkZ esa fuEu esa ls dkSulk oDrO; lgh ugha gS \ (A*) ;g

,d izcy f=k{kkjh; vEy gSa (B) ;g cksjsDl ds vEyhd`r tyh; foy;u }kjk cuk;k tkrk gS (C) ;g ,d ijr lajpuk gS ftlesa gkbMªkstu ca/k }kjk lery BO3 bdkbZ dks tksMk tkrk gSA (D) ;g izkV s ksu nkrk ds :i esa dk;Z ugha djrk gS]a cfYd gkbMªkfW Dly vk;u dks xzg.k dj yqbl vEy ds :i esa dk;Z djrk gSA Sol.

It is weak monoprotic acid and it does not act as proton donor but removes proton from water.

;g nqcZy ,d izksfVd vEy gS ;g izksVksu nkrk dh rjg dk;Z ughs djrk gSA ysfdu ;g ikuh ls izksVksu fudky nsrk gSA 2.

Which of the two ions from the list given below that have the geometry that is explained by the same hybridization of orbitals, NO2–, NO3–, NH2–, NH4+, SCN– ? (A*) NO2– and NO3– (B) NH4+ and NO3– (C) SCN– and NH2– (D) NO2– and NH2–

fuEufyf[kr vk;uksa esa ls os dkSu nks vk;u gSa ftudh T;kfefr dks leku d{kdksa ds ladj.k ls le>k;k tk ldrk gS]

Sol.

sNO2–, NO3–, NH2–, NH4+, SCN– ? (A*) NO2– rFkk NO3– (B) NH4+ rFkk NO3– NO2–  sp2 NO3–  sp2 NH2–  sp3 NH4+  sp3 SCN sp

(C) SCN– rFkk NH2–

(D) NO2– rFkk NH2–

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3.

Which of the following compounds has the lowest melting point ?

fuEu ;kSfxdksa esa fdldk xyukad lcls de gksrk gS \ Sol. Sol.

(A) CaCl2 (B) CaBr2 (C*) Ca2 (D) CaF2 Covalent character increases, melting point decreases CaF2 > CaCl2 > CaBr2 > CaI2. lgla;kstd y{k.k esa o`f) gksrh CaF2 > CaCl2 > CaBr2 > CaI2 xyukad esa deh gksrh gSA

4.

The radius of hydrogen in ground state is 0.53 A°. The radius of Li2+ in ground state will be – vk| voLFkk esa gkbMªkstu dh f=kT;k 0.53 A° gSA Li2+ dh vk| voLFkk esa f=kT;k gksxh – (A) 1.06 A° (B) 0.265 A° (C*) 0.17 A° (D) 0.53 A°

Sol.

r 

5.

Pure ammonia is placed in a vessel at a temperature where its degree of dissociation () is appreciable. At equilibrium 'kq) veksfu;k dks fdlh rki ij ik=k esa Mkyk tkrk gS] tgk¡ bldh fo;kstu dh ek=kk () i;kZIr gksrh gSA lkE; ij -

n2 z



rLi 

rH  0.17 Å 3

(A*) Kp does not change with pressure (Kp nkc

ds lkFk ifjorZu ugha gksrk) (B)  does not change with pressure ( nkc ds lkFk ifjofrZr ugha gksrk) (C) Concentration of ammonia does not change with pressure ( veksfu;k dh lkUnzrk nkc ds lkFk ifjofrZr ugha gksrh) (D) Concentration of H2 is less than concentration of N2. H2 dh lkUnzrk N2 dh lkUnzrk ls de gSA Sol.

Kp remains constant at a constant tempurature at does not change with change in pressure fu;r rki ij Kp fu;r jgrk gS bldk eku nkc ij fuHkZj ugha djrkA

6.

If 0.224 litre of H2 gas is formed at the cathode, in the electrolysis of water the volume of O2 gas formed at the anode under identical conditions, is ;fn dSFkksM ij 0.224 yhVj H2 xSl mRiUu gksrh gS] ty ds oS|qr vi?kV~u esa leku ifjfLFkfr;ks esa ,uksM ij mRiUu gksus

okys O2 dk vk;ru gksxkA Sol.

(A) 0.224 L eq. of O2 = eq. of H2 V 0.224  5. 6 11 .2

7.

(B) 0.448 L



(D) 1.12 L

V = 0.112 L

In H2  O2 fuel cell a reaction occuring at cathode is : H2  O2 bZa/ku lsy esa] dSFkksM ij lEiUu gksaus okyh vfHkfØ;k gSa & (A*) 2 H2O + O2 + 4 e  4 OH (B) 2 H2 + O2  2 H2 O (l) (C) H+ + OH  H2O

8.

(C*) 0.112 L

(D) H+ + e 

1 H . 2 2

The solubility of calomel (mercurous chloride) in pure water is equal to

dSyksey ¼ejD;wjl DyksjkbM½ dh ikuh esa foys;rk fuEu ds cjkcj gS& Sol.

9.

(A) 1/2 (2Ksp)1/3 (B) 1/2(Ksp)1/2 +2 Hg2Cl2 Hg2 + 2Cl– s 2s Ksp = 4s3 s = (Ksp/4)1/3

(C) Ksp1/2

(D*) (Ksp/4)1/3

(C) K2CO3

(D) Na2 O2

Solvay process is used for the manufacture of

lksYos fof/k fuEu ds fuekZ.k esa iz;qDr gksrh gS& (A) NaOH

(B*) NaCO3 . 10H2O

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Sol.

Solvay process is used in manufacturing of sodium carbonate Na2CO3 . 10H2O lksYos fof/k Na2CO3 . 10H2O dks cukus ds dke vkrh gSA

10.

Enthalpy change for the reaction, 4H(g)  2H2(g) is – 869.6 kJ. The dissociation energy of H–H bond is :

vfHkfØ;k] 4H(g)  2H2(g) ds fy, ,UFkSYih ifjorZu – 869.6 kJ gSA H–H cU/k dh fo;kstu ÅtkZ gS % Sol.

11.

(A) – 434.8 kJ

(B) – 869.6 kJ

4 H(g)  2H2 (g)

H = –869.6 KJ.

4 H2  4H (g)

H = 869.6 KJ.

H2  2H (g)

H =

(C*) + 434.8 kJ

(D) + 217.4 kJ

869.6 = 434.8 KJ. 2

For the reaction N2(g) + O2(g)

2NO(g), the equilibrium constant is K1. The equilibrium constant is K2

for the reaction 2NO(g) + O2(g) 2NO2(g). What is equilibrium constant for the reaction NO 2(g) ½N2(g) + O2(g) ? vfHkfØ;k] N2(g) + O2(g) 2NO(g) ds fy, lkE; fLFkjkad K1 gS rFkk vfHkfØ;k] 2NO(g) + O2(g) 2NO2(g)

ds fy, K2 gSA vfHkfØ;k NO2(g)

½N2(g) + O2(g) ds fy, lkE; fLFkjkad] (B) 1 / (4K1K2) (C*) [1 / K1K2]½

(A) 1 / (2K1K2) Sol.

N2 + O 2

12.

Sol.

(D) 1 / (K1K2)

2NO K1 .... (i)

2NO + O2 NO2

dk eku D;k gksxk \

2NO2 K2

1 N + O2 2 2

K=

.... (ii)

1 k 1.k 2

The freezing point depression constant for water is – 1.86ºC m–1. If 5.00 g Na2SO4 is dissolved in 45.0 g H 2O, the freezing point is changed by – 3.82ºC. Calculate the van’t Hoff factor for Na 2SO4. ty dk fgekad voueu fLFkjkad – 1.86ºC m–1 gSA ;fn 5.00 g Na2SO4 dks 45.0 g H2O esa ?kksyk tkrk gS rks fgekad – 3.82ºC ls ifjofrZr (A) 2.05 Kf = –186° cm –1 Tf = i × Kf . m 3.82 = i × 1.86 ×

gks tkrk gSA Na2SO4 ds fy, okWUV gkWQ xq.kd dh x.kuk dhft,A (B*) 2.63

(C) 3.11

(D) 0.381

5  1000 142  45

i = 2.63 13.

Of the following complex ions, which is diamagnetic in nature ?

fuEufyf[kr ladqy vk;uksa esa ls dkSulk izfrpqEcdh; izd`fr dk gS \ Sol.

(A) [NiCl4]2– [Ni(CN)4]2– Ni2+ = 3d84s°

(B*) [Ni(CN)4]2–

(C) [CuCl4]2–

(D) [CoF6]3–

Diamagnetic (izfrpqEcdh;)

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14.

Standard electrode potential for Sn4+ / Sn2+ couple is + 0.15 V and that for the Cr3+ / Cr couple is – 0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be : Sn4+ / Sn2+ ;qXe ds fy, ekud bysDVªksM foHko + 0.15 V gS rFkk Cr3+ / Cr ;qXed ds fy;s ekud bysDVªksM foHko – 0.74 V gSaA bu nksuksa (A) +1.19 V

Sol.

;qXeksa dks ekud voLFkk esa tksM+dj ,d lsy rS;kj gksrk gSA lsy foHko gksxk % (B*) +0.89 V

(C) +0.18 V

(D) +1.83 V

E oSn4  / Sn 2  = +0.15 V E oCr 3  / Cr = – 0.74 V E°cell = E°C –E°A = 0.15 – (–0.74) = 0.89 V

15.

X, Y and Z are elements in the same short period. The oxide of X is a giant molecule, the oxide of Y is a simple molecule and the oxide of Z is ionic. The arrangement of the elements in order of increasing atomic number would be leku y?kqvkorZ esa X, Y rFkk Z rRo gSaA X dk vkWDlkbM ,d o`gn~ v.kq gS] Y dk vkWDlkbM ,d lkekU; v.kq gSa rFkk Z dk

vkWDlkbM vk;fud gSaA ijek.kq Øekad dh o`f) ds Øe es rRo dk foU;kl fuEu gSA (A) X, Y, Z 16.

(B) X, Z, Y

(C*) Z, X, Y

(D) Y, Z, X

Which of the following anions is not easily removed from aqueous solutions by precipitation?

fuEu esa ls dkSulk _.kk;u] vo{ksi.k }kjk tyh; foy;u ls vklkuh ls ugha fudkyk tkrk gS & Sol.

(A) Cl  (B) SO42  (C*) NO3  Because all the nitrates are much soluble in water.

(D) CO3 2

D;ksafd lHkh ukbVªsV ty esa vf/kd foys; gSA

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ChemINFO-1.20

Special Concentration Units I. Volume Strength for H2O2

Daily Self-Study Dosage for mastering Chemistry

Volume Strength for H2O2 : Volume strength (or volumetric conc.) of H2O2 is represented as 10V , 20 V , 30 V etc. 20V H2O2 means one litre of this sample of H2O2 on decomposition gives 20L of O2 gas at STP. Decomposition of H2O2 is given as : H2O2  H2O +

1 O 2 2

1 × 22.4 L O2 at STP 2 = 11.2 L O2 at STP

1 mole = 34g

For 20 L O 2 , we should decompose atleast Ex.

Relationship between volume strength and Molarity for H2O2. Assume ‘V’ volume H2O2. It means

At STP 1 Lit H2O2 Soln  V Lit O2 



34 × 20 g H2O2 11.2

Molarity of H2O2 (M) =

34 V g H O (by equation). 11 .2 2 2

or

Molarity =

Volume strength of H2O2 11.2

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice.

17.

Find the volume strength of H2O2 solution prepared by mixing of 250 mL of

3 M M H2O2 & 750 mL of H O 2 2 2 2

solution : (A) 1.5 V 18.

(C) 5.6 V

(D) 11.2 V

A fresh H2O2 solution is labelled 11.2 V. This solution has the same concentration as a solution which is : (A) 3.4% (wt/wt)

19.

(B*) 8.4 V

(B) 3.4% (vol/vol)

(C*) 3.4% (wt/vol)

(D) None of these

Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation: 2H2O2(aq)  2H2O() + O2(g) under conditions where 1 mole of gas occupies 24 dm3. 100 cm3 of XM solution of H2O2 produces 3 dm3 of O2. Thus X is................ (A*) 2.5

20.

(B) 1

(C) 0.5

(D) 0.25

(C) 1.5 % (w/v)

(D*) All of these

A 16.8 V H 2O2 solution is equivalent to : (A) 5.1 % (w/v) H 2O2

(B) 51 g/L H 2O2

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ChemINFO-1.20

Special Concentration Units

Daily Self-Study Dosage for mastering Chemistry

I. H2O2dk

vk;ru lkeF;Z

H2O2 dk vk;ru lkeF;Z : H2O2 ds vk;ru lkeF;Z ¼;k vk;ruh lkUnzrk½ dks 10V , 20 V , 30 V bR;kfn ls iznf'kZr fd;k tkrk gSA 20V H2O2 dk vFkZ gS % H2O2 ds bl uewus dk 1 yhVj fo;ksftr gksus ij S.T.P. ij O2 xSl dk 20 yhVj nsrk gSA H2O2 dk fo;kstu fuEu izdkj ls fn;k tkrk gSA H2O2  H2O +

1 O 2 2

1 eksy

O2

ds

= 34g

= O2 ds 11.2 L S.T.P. ij

20 L O2 ds Ex.

1 × 22.4 L S.T.P ij 2

fy, gesa de ls de

34 × 20 g H2O2 dks fo?kfVr djuk iM+sxkA 11.2

H2O2 ds

fy, vk;ru lkeF;Z o eksyjrk esa lEcU/kA ekuk ‘V’ vk;ru H2O2 gS blls rkRi;Z gS % STP ij 1 Lit H2O2 Soln  V Lit O2  H2O2 dh

eksyjrk (M) =

34 V g H O (by equation). 11 .2 2 2

or

eksyjrk =

H2 O2 11.2

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice.

17.

3 M M H2O2 foy;u ds 250 mL rFkk H O foy;u ds 750 mL dks feyk dj cuk;s x;s H2O2 foy;u dk vk;ru lkeF;Z 2 2 2 2

Kkr dhft, % (A) 1.5 V

(B*) 8.4 V

(C) 5.6 V

(D) 11.2 V

18.

,d rktk H2O2 foy;u dks 11.2 V ls yscy fd;k x;k gSA bl foy;u dh lkUnzrk] ml foy;u dh lkUnzrk ds cjkcj gksxh] ftlesa % (A) 3.4% (Hkkj/Hkkj) (B) 3.4% (vk;ru/vk;ru) (C*) 3.4% (Hkkj/vk;ru) (D) buesa ls dksbZ ugha

19.

tyh; foy;u esa gkbMªkstu ijvkWDlkbM xeZ djus ij fuEu lehdj.k ds vuql kj fo;ksftr gksdj vkWDlhtu nsrk gSA 2H2O2(aq)  2H2O() + O2(g) 1 eks y X

xSl 24 dm 3 xzg .k djrh gSA bu ifjfLFkr;ksa esa X M H2O2 foy;u ds 100 cm3 3 dm 3 O2 nsrs gSA bl izdkj

gS ................

(A*) 2.5 20.

(B) 1

(C) 0.5

(D) 0.25

(C) 1.5 % (w/v)

(D*) mijksDr

,d 16.8 V H2O2 foy;u fdlds lerqY ; gS % (A) 5.1 % (w/v) H 2O2

(B) 51 g/L H 2O2

lHkhA

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