Jp Xii Physical&inorganic Chemistry (5) - Prev Class Xi Chaps + Solution & Colligative Properties +solid State.pdf

PHY./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2017 Course : VIJETA (JP)

NO. 07 & 08

Date : 27-04-2015 TEST IN FORMATI ON

DATE : 17.05.2015

PART TEST (PT) - 01 (ADVANCED)

Syllabus : Solution & Colligative Properties, Solid State, Thermodynamics, Ionic equilibrium, s-Block Element. (All ChemInfos and Handouts till date)

This DPP is to be discussed in the week (27-04-2015 to 02-05-2015)

DPP No. # 07 (JEE-MAIN) Total Marks : 60

Max. Time : 40 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.20

1.

(3 marks, 2 min.)

[60, 40]

Consider the reaction CaCO3(s) CaO(s) + CO2(g) in a closed container at equilibrium. What would be the effect of addition of CaCO 3 on the equilibrium concentration of CO2 ?

lkE; ij ,d can fudk; es fuEu vfHkfØ;k dk voyksdu dhft, CaCO3(s) CO2 dh lkE; lkUnzrk (A) Increases (A) o`f) gksxh 2.

ij

CaO(s) + CO2(g) CaCO3 dks feykus dk D;k izHkko gksxk \ (B) Unpredictable (C) Decreases (B) dgk ugha tk ldrk gS (C) de gksxh

(D*) Remains unaffected (D*) vizHkkfor jgsxh

A complex organic compound contains 21 atoms of carbon per molecule. The percentage of carbon in the compound is 70%. If 36 gram of the substance is dissolved in 100 mL of water then the molarity of the solution will be– ,d dkcZfud ladqy ;kSfxd dkcZu ds 21 ijek.kq izfreksy j[krk gSA ;kSfxd esa dkcZu dk 70% mifLFkr gSA ;fn 36 xzke

inkFkZ (substance) dks 100 mL ikuh es feyk;k tk;s rc foy;u dh eksyjrk gksxh– Sol.

(A*) 1 (B) 0.1 (C) 1.43 One mole of compound contains 12 × 21 = 252 g carbon ,d eksy ;kSfxd 12 × 21 = 252 g dkcZu gS Molar weight = 252 ×



eksyj Hkkj = 252 × Molarity =



eksyjrk =

(D) 0.143

100 = 360 70

100 = 360 70

36 1000  =1 360 100 36 1000  =1 360 100

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3.

In the reaction CS2 () + 3O2 (g)  CO2 (g) + 2SO2 (g) H = 265 kcal The enthalpies of formation of CO 2 and SO2 are both negative and are in the ratio 4 : 3. The enthalpy of formation of CS2 is +26kcal/mol. Calculate the enthalpy of formation of SO 2. CS2 () + 3O2 (g)  CO2 (g) + 2SO2 (g) H = 265 kcal

mijksDr vfHkfØ;k esa] CO2 rFkk SO2 nksuksa ds laHkou dh ,UFksYih _.kkRed gS rFkk 4 : 3 ds vuqikr esa gSA CS2 ds laHkou dh ,UFksYih +26kcal/mol gSA SO2 ds laHkou dh ,UFksYih dh x.kuk dhft;s \ Sol.

4.

(A) – 90 kcal/mol (B) – 52 kcal/mol (C) – 78 kcal/mol Cs2 () + 3O2 (g) Co2 (g) + 2So2 + 2so2 H = Let Hf (CO2 , g) = 4 x and Hf (SO2 , g) = 3x Hreaction = Hf (CO2 ,g) = 2 Hf (SO2 .g) – Hf (CS2 ) – 265 = 4 x + 6x –26 x = –23.9 Hf (SO2,g) = 3x = 71.7 kcal / mol. For the reaction at 25°C, X2O4()  2XO2(g) H = 2.1 Kcal and S = 20 cal K–1. The reaction would be : (A*) spontaneous (B) non-spontaneous (C) at equilibrium

(D*) – 71.7 kcal/mol

(D) unpredictable

vfHkfØ;k X2O4()  2XO2(g) ds fy, H = 2.1 Kcal rFkk S = 20 cal K–1 gSA vfHkfØ;k % (A*) Lor% gksxh (B) vLor% gksxh (C) lkE; ij gksxh (D) ds fy, dgk ugha tk ldrk gS 25°C ij Sol.

G = H – TS = 2.1 × 103 – 20 × 298 < 0.

5.

A 10% w/w solution of cane sugar has undergone partial inversion according to the reaction Sucrose + water  glucose + fructose The boiling point of the solution is 100.7°C. What fraction of sugar has inverted ? (Kb (H2O) = 1.8 K.kg/mol) xUuk 'kdZjk ds ,d 10% w/w foy;u dk fuEu vfHkfØ;k ds vuqlkj vkaf'kd izkrhiu gksrk gSA

lqØksl + ty  Xywdksl + ÝDVksl foy;u dk DoFkukad 100.7°C gSA 'kdZjk ds izfriu dk fHkUu fdruk gksrk gS \ Sol.

(Kb (H2O) = 1.8 K.kg/mol) (A) 0.5 (B*) 0.2 (C) 0.3 Let degree of dissociation of sucrose be .  Sucrose + water  glucose + fructose i = 1 + (2 – 1)= 1 +  

Sol.

Tb = iKbm



0.7 = (1 + ) × 1.8 ×



1+=



 = 0.2

(D) 0.4

10  1000 342  90

0.7  342  9 = 1.2 1.8  1000

ekuk lqØksl ds fo;kstu ds fo;kstu dh ek=kk  gSA  lqØksl + ty  Xywdksl + ÝDVksl i = 1 + (2 – 1)= 1 +  

Tb = iKbm



0.7 = (1 + ) × 1.8 ×



1+=



 = 0.2

10  1000 342  90

0.7  342  9 = 1.2 1.8  1000

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6.

Sol.

7.

Sol.

The species C2 Lih'kht C2 j[krk gSA (A) Has one  bond and one  bond (C) Has both  bonds (A) ,d  ca/k rFkk ,d  ca/k

(B*) Has both  bond (D) Does not exist (B*) nksuksa  ca/k

(C) nksuksa  ca/k (D) vfLrRo ugha j[krk gSA Double bond in C2 has both  bonds between the two C-atoms because 4 electron are present in  molecular orbitals C2 esa nks dkcZu ijek.kqvksa ds e/; nks ca/k gksrs gSa rFkk nksuksa gh  ca/k D;ksfd 4 bysDVªkWu  vkf.od d{kdks esa mifLFkr gksrs gSa An acid solution of pH = 6 is diluted 1000 times, the pH of the final solution becomes. pH = 6 ds ,d vEy foy;u dks 1000 xquk ruq fd;k x;kA rc ifj.kkeh foy;u dk pH D;k gks (A*) 6.99 (B) 6 (C) 3 (D) 9 pH = 6 means [H+] = 10–6 M. After dilution [H+] 10–6/1000 = 10–9 M pH = 6 dk vFkZ [H+] = 10–6 M. ruqdj.k ds i'pkr~ [H+] 10–6/1000 = 10–9 M  [H+] from H2O cannot be neglected vr% H2O ls izkIr [H+] vk;uksa dks ux.; ugha fd;k tk ldrkA Total

tkrk gSA

dqy [H+] = 10–9 + 10–7 = 10–7 (10–2 + 1) = 10–7 (1.01)

or ;k pH = –log 1.01 x 10–7 = 7 – 0.0043 = 6.9957 8.

Which of the following has highest electron affinity ?

fuEu esa ls fdldh bySDVªkWu ca/kqrk vf/kdre gS \ Sol.

(A*) Li+ (B) Na+ (C) F– (D) Cl– Addition of an electron to F– and Cl– ions will involve replusions therefore their EA is almost zero. On the other hand, addition of an electron to Na+ and Li+ will involve attraction. Since Li+ is much smaller than Na+ ion, therefore, EA of Li+ is higher than that of Na+. F– rFkk Cl– vk;u esa ,d bysDVªkWu ds ;ksx ls izfrd"kZ.k gksrk gS bl izdkj budh EA yxHkx 'kwU; gks tkrh gSA nwljh vkSj Na+ rFkk Li+ esa

,d bysDVªkWu dk ;ksx ls vkd"kZ.k gksrk gSA ;)fi Li+ , Na+ vk;u fd rqyuk esa vf/kd NksVk gS bl izdkj Li+ fd EA Na+ fd rqyuk esa vf/kd gSA 9.

A pre-weighed vessel was filled with oxygen at NTP and weighed. It was then evacuated, filled with SO3 at the same temperature and pressure and again weighed. The weight of oxygen will be– (A) Same as that of SO3 (B) 4 times that of SO3 (C) 2.5 times that of SO 3 (D) 0.25 times that of SO3 igys ,d [kkyh ik=k dk Hkkj eki dj NTP ij mls vkDlhtu ls Hkjdj Hkkj ekirs gSA vc bl ik=k dks fuokZfrr djds mlh

leku rki vkSj nkc ij SO3 ls Hkjdj nksckjk Hkkj ekirs gSA vkDlhtu dk Hkkj gksxk (A) SO3 ds leku (B*) SO3 dk 4 xquk (C) SO3 dk 2.5 xquk (D) SO3 dk 0.25 xquk Sol.

Since T & P are same & vessel is also same so at constant P, V, & T no of males in both the cases will be same tc dh rki vkSj nkc leku vkSj ik=k Hkh leku gS blfy, fu;r P, V, & T ij eksyks dh la[;k Hkh leku gksxh

WSO3 80 WO2 =

=

WO 2 32

2 WSO 3 5 = 0.4 WSO3

10.

Which will give maximum [H+] ion concentration by dissolving equal moles in water fuEu es ls fdldks ikuh es ?kksyus ij [H+] vk;u dh lkUnzrk vf/kdre gksxhA ;fn lcds eksy cjkcj (A*) LiOH (B) NaOH (C) KOH (D) RbOH

gks&

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PAGE NO.- 3

Ans.

 (A) LiOH is least basic  [OH ]  min imum

 [H ]  max imum  [OH ]  (A) LiOH lcls

11.

de {kkjh; gSA

 [H ] 

A chloride dissolves appreciably in cold water. When placed on a platinum wire in Bunsen flame, no distinctive colour is noticed. The cation is– ,d DyksjkbM B.Ms ikuh es ?kqy tkrk gSA tc bls IysVhue ¼platinum½ ds rkj ij j[kdj cqUlsu Tokyk esa xeZ djrs gS rks

dksbZ jax izkIr ugh gskrk gSA og /kuk;u gS& Ans.

(A) Ca2+ (B*) Mg2+ (C) Ba2+ (D) Pb2+ (B) PbCl2 is insoluble in cold water. Ca2+ & Ba2+ give bricke red & apple green colours respectively to the Bunsen flame (B) PbCl2 B.Ms ikuh esa vfoys; gSA Ca2+ & Ba2+ cqUlsu

12.

Ans.

13.

Tokyk ds lkFk Øe'k% bZaV tSlk yky vkSj lsc tSlk gjk (bricke red & apple green colours ) jax nsrk gSA

The solvent which is better to be used during ebullioscopic, measurement has– DoFkkukadekih (ebullioscopic) vuqekiu ds nkSjku dkSuls foyk;d dk mi;ksx djuk vPNk gS (A*) High Kb (B) Low Kb (C) Kb = Kf (D) Kf < Kb (A) Higher the value of Kb , more will be Tb . Hence, it can be measured more accurately (A) Kb dk eku T;knk gksxk rc Tb dk eku Hkh T;knk gksxk blfy, ;g vf/kd lgh ekik tk ldrk

gSA

Standard molar enthalpy of formation of CO 2 is equal to (A) zero (B) the standard molar enthalpy of combustion of gaseous carbon (C) the sum of standard molar enthalpies of CO & O 2 (D*) the standard molar enthalpy of combustion of C(s) CO2 dh ekud eksyj ,UFkSYih fdlds cjkcj gS (A) 'kwU; (B) xSlh;

dkcZu ds ngu dh ekud eksyj ,UFkSYih ds cjkcj gS (C) CO vkSj & O2 dh ekud eksyj ,UFkSYih ds ;ksx ds cjkcj gS (D*) C(s) ds ngu dh ekud eksyj ,UFksYih ds cjkcj gS Ans.

(D)

14.

The most unsymmetrical crystal system is: (A) Cubic (B) Hexagonal

(C*) Triclinic

fuEufyf[kr esa ls lokZf/kd vlefer fØLVy fudk; gS & (A) ?kuh; (Cubic) (B) "kVdks.kh; (Hexagonal) (C*) f=kurk{k (Triclinic) Sol.

For triclinic a  b  c &       90°

15.

The number of crystal systems known are :

(D) Orthorhombic (D) fo"keyEck{k (Orthorhombic)

dqy fdrus izdkj ds fØLVyh; fudk; Kkr gS % (A*) 7 16.

(B) 8

(D) 4

Tetragonal crystal system has the following unit cell dimensions: (A) a = b = c and  =  =  = 90° (B*) a = b  c and  =  =  = 90° (C) a  b  c and  =  =  = 90° (D) a = b  c and  =  = 90°,  = 120° f}leyEck{k (tetragonal) fØLVy fudk;] fuEufyf[kr esa ls dkSulh bdkbZ dks"Bhdk foek;sa j[krk (A) a = b = c

rFkk  =  =  = 90° (C) a  b  c rFkk  =  =  = 90° Sol.

(C) 6

gSA

(B*) a = b  c

rFkk  =  =  = 90° (D) a = b  c rFkk  =  = 90°,  = 120°

a  b  c &  =  =  = 90° the crystal system is orthorhombic Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 4

17.

Which has no axis of rotation of symmetry ? (A) Hexagonal (B) Orthorhombic

fuEu esa ls fdlesas leku :i ls ?kw.kZu dk dksbZ v{k (A) "kV~dks.kh; (B) fo"keyEck{k Sol.

(C) Cubic ugha gS ?

(D*) Triclinic

(C) ?kuh;

(D*) f=kurk{k

In triclinic unit sell.

f=kurk{k ,dd dksf"Vdk esa a b c      18.

Three elements P, Q and R crystallize in a cubic solid lattice. The P atoms occupy the corners, Q atoms the cube centres and R atoms the edges. The formula of the compound is: rhu rRo P, Q rFkk R ?kuh; Bksl tkyd ds :i esa fØLVyhd`r gksrh gSA P ijek.kq dksuksa ij] Q ijek.kq ?ku ds dsUnz ij

rFkk R ijek.kq fdukjksa ij mifLFkr gSA ;kSfxd dk lw=k D;k gS & (A) PQR

(B) PQR2

(C*) PQR3

(D) PQ3R.

1 1 = 1 ; Q = 1 = 1 ; R = 12  = 3 ; formula = PQR 3 8 4

Sol.

P=8

19.

A compound alloy of gold and copper crystallises in a cubic lattice in which the gold atoms occupy the lattice points at the corner of a cube and the copper atoms occupy the centres of each of the cube faces. Hence compound alloy has formula :

xksYM rFkk dkWij ds feJ/kkrq ;kSfxd ?kuh; tkyd ds :i esa fØLVyhd`r gksrh gSA ftlesa xksYM ijek.kq ?ku ds tkyd fcUnq] dksuksa ij rFkk dkWij ijek.kq ?ku ds izR;sd Qyd dsUnz ij fLFkr gksrs gSA blizdkj feJ/kkrq ;kSfxd dk lw=k gS & (A) AuCu

(B) Au3Cu 1 =1 8

Cu = 6 

(C) Au2Cu

(D*) AuCu3

1 = 3 formula AuCu3 2

Sol.

Au = 8 

20.

Column A describe nature of bonding and Column B the solid having that type of bonding :

Correct matching of A and B is in alternate : dkWye A cU/k dh izd`fr rFkk dkWye B Bksl esa cU/k

A rFkk B dk I (A) P (C) Q

ds izdkj dks crkrk gSA

lgh feyku fuEu fodYi esa gSA II Q P

III R R

IV S S

(B*) (D)

I Q S

II S P

III P Q

IV R R

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PAGE NO.- 5

DPP No. # 08 (JEE-ADVANCED) Total Marks : 77

Max. Time : 45 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.7 Multiple choice objective ('–1' negative marking) Q.08 to Q.11 Integer type Questions ('–1' negative marking) Q.12 to Q.14 Match the Following (no negative marking) Q.15 ChemINFO : 4 Questions ('–1' negative marking) Q.16 to Q.20

1.

(3 (4 (4 (8 (4

marks, 2 min.) marks, 2 min.) marks 3 min.) marks, 6 min.) marks, 2 min.)

[21, [16, [12, [08, [20,

14] 08] 09] 06] 08]

In a face centred cubic arrangement of A and B atoms whose A atoms are at the corner of the unit cell and B atoms at the face centres. One of the A atom Is missing from one corner in unit cell. The simplest formula of the compound is A rFkk B ijek.kqvksa dh Qyd dsUnzh; ?kuh; O;oLFkk esa] A ijek.kq bdkbZ dks"Bhdk ds dksuksa ij rFkk B ijek.kq Qyd ds dsUnz

ij mifLFkr gSA bdkbZ dks"Bhdk ds ,d dksus ls ,d A ijek.kq vuqifLFkr gS rc] ;kSfxd dk ljyre lw=k gS& (A) A7B3 Sol.

A=7

(B) AB3 1 7 = ; 8 8

B=6

(C*) A7B24

(D) A2B3

1 =3 2

Formula = A 7 / 8 B3 or A7B24 2.

A compound is formed by elements A and B. This crystallises in the cubic structure where the A atoms are at the corners of the cube and B atoms are at the body centres. The simplest formula of the compound is : ,d ;kSfxd dks A rFkk B rRoksa }kjk cuk;k tkrk gSA ;g ?kuh; lajpuk esa fØLVyhd`r gksrh gS tgk¡ A ijek.kq ?ku ds dksuks

ij rFkk B ijek.kq dk; dsUnz ij mifLFkr gSA ;kSfxd dk ljyre lw=k fuEu gS : (A) A8B4 Sol.

(B) AB6

(C*) AB

In simple cubic contribution of one corner =

(D) A6B

1 8

total corner = 8 effective no. of atom per unit cell =

1 ×8=1 8

No. of bondy center in simple cubic = 1 No. of atom in body center = 1 AB Sol.

1

,d dksus ds fy, ljy ?kuh; ;ksxnku = 8 dqy dksus = 8 izfr ,dd dksf"Vdk ds fy, izHkkoh la[;k =

1 ×8=1 8

ljy ?kuh; esa dk; dsUnzdksa dh la[;k = 1 dk; dsUnz esa ijek.kq dh la[;k = 1 3.

Let the solubilities of AgCl in H2O, 0.01 M CaCl2, 0.01 M NaCl & 0.05 M AgNO3 be S1, S2, S3 & S4 respectively what is the correct relationship between these quantities . Neglect any complexation. ekuk AgCl dh foys;rk H2O, 0.01 M CaCl2, 0.01 M NaCl ,oe~ 0.05 M AgNO3 esa Øe'k% S1, S2, S3 , S4 gSa] rks bu

ek=kkvksa esa lgh laca/k D;k gSA fdlh Hkh rjg dh tfVyrk dks ux.; ekfu;s (A) S1 > S2 > S3 > S4

(B) S1 > S2 = S3 > S4

(C*) S1 > S3 > S2 > S4 (D) S4 > S2 > S3 > S1

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4.

Sol.

Two mole of an ideal gas is heated at constant pressure of one atomosphere from 27ºC to 127ºC. If Cv,m = (20 + 10–2 T) JK–1 mol–1, then q and U for the process are respectively : ,d vkn'kZ xSl ds nks eksy ,d ok;qe.My ds fLFkj nkc ij 27ºC ls 127ºC rd xeZ fd;s tkrs gSaA ;fn Cv,m= (20+10–2 T) JK–1 mol–1 gS] rc vfHkfØ;k ds fy;s q vkSj U Øe'k% gS % (A*) 6362.8 J, 4700 J (B) 3037.2 J, 4700 J (C) 7062.8, 5400 J (A) w = – nR T = –2 × 8.314 × 100 = –1662.8 J U = n

400

400





C V, m dT = 2 ×

300

 5.

300

(D) 3181.4 J, 2350 J

 10 2 400 2 – 300 2 (20 + 10–2 T) dT = 220  100  2 

  = 4700 J 

4700 = q – 1662.8 q = 6362.8 J.

For a reversible adiabatic process with Ti > Tƒ on an ideal gas, which of the following statement is INCORRECT: PiVi

=

Pƒ Vƒ

(A) Work is done by the gas.

(B) T i

(C*) H = 0

(D) W = nCVT

Tƒ

,d vkn'kZ xSl ij Ti > Tƒ okys ,d mÙØe.kh; :)ks"e çØe ds fy,] fuEu esa ls dkSulk dFku vlR; gS\ (A) xSl

Sol.

PiVi

(B) T i

}kjk dk;Z fd;k x;k gSA

(C*) H = 0 Reversible adiabatic process. So P iVi = PƒVƒ

=

Pƒ Vƒ Tƒ

(D) W = nCVT

Pƒ Vƒ Pi Vi = Tƒ (Valid for all processes on an ideal gas) Ti

Ti > Tƒ. So internal energy of the gas decreases during the process. Also the process is adiabatic. So q = 0. Therefore, from first law, work is done by the gas. Also the process is adiabatic. So q = 0. Therefore, from first law, W = E = nCVT. mRØe.kh; :)ks"e izØe] vr% PiVi = PƒVƒ Pƒ Vƒ Pi Vi Ti = Tƒ (vkn'kZ

xSl ds fy, lHkh izØeksa ij ykxw)

Ti > Tƒ. vr%

izØe ds nkSjku xSl dh vkUrfjd ÅtkZ ?kVrh gSA ;|fi izØe :)ks"e gSA vr% q = 0 blfy, izFke fu;e ds vuqlkj] xSl }kjk dk;Z fd;k x;k gSA ;|fi izØe :)ks"e gSA blfy, q = 0, izFke fu;e ds vuqlkj] W = E = nCVT. 6.

Which of the following are isostructural ?

fuEu esa ls dkSulh Lih'kht lelajpukRed gS \ NO3–, (I) (A*) (I) and (IV) (A*) (I) rFkk (IV)

Sol.

(I)

CO32–, (II) (B) (II) and (V) (B) (II) rFkk (V)

(II)

ClO3– , (III)

SO3, (IV) (C) (III) and (IV) (C) (III) rFkk (IV)

XeO3 (V) (D) (IV) and (V) (D) (IV) rFkk (V)

(III)

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PAGE NO.- 7

(IV)

7.

(V)

For a real gas with very large value of molar volume, which of the following equation can most suitably be applied:

vR;f/kd eksyj vk;ru okyh ,d okLrfod xSl ds fy, fuEu esa ls dkSulh lehdj.k lokZf/kd mi;qDr :i ls iz;q Dr dh tk ldrh gS % a (A) Z = 1 – V RT m Sol.

(B*) PVm = RT

(C) Z = 1 +

Pb RT

a (D) PVm – RT = V m

For very large value of molar volume (Vm) a Vm and b can be neglected, so gas behaves as Ideal

 PVm = RT

vR;f/kd eksyj vk;ru (Vm) ds fy, a Vm

rFkk b dks ux.; ekuk tk ldrk gS vr% xSl vkn'kZ O;ogkj djrh gSA

 PVm = RT 8.*

Sol.

When the gas is an ideal one and process is isothermal, then the correct relation is/are : ;fn ,d xSl] ,d vkn'kZ xSl gS rFkk izØe lerkih; gS] rc lgh lEcU/k gS@gSa : (A*) P1V1 = P2V2 (B*) E = 0 (C) W = 0 (D*) H = 0 The value of work done will not be zero.

fd;s x;s dk;Z dk eku 'kwU; ugha gksxkA 9*.

Critical temperature for a particular gas is – 177°C then for which of the following case value of compressibility factor of the gas may be more than unity. ,d fuf'pr xSl ds fy, ØkfUrd rkieku – 177°C gS rks fuEu esa ls fdl xSl ds fy, lEihMh;rk xq.kkad(z) dk eku bdkbZ

ls vf/kd gksrk gSA (A) at 0°C and 0.01 atm (C*) at 60°C and 0.01 atm 10*.

Sol.

gy %

(B*) at 0°C and 2000 atm (D*) at 60°C and 10 atm

Solubility of AgCN will be greater in which of the following solutions as compared to distilled water? vklqr ty dh rqyuk esa AgCN dh foys;rk fuEu esa ls fdl foy;u esa vf/kd gksxh \ (A) 0.5 M AgNO3 (B*) 0.5 M HNO3 (C*) 5 M KCN (D) All of above mijksDr lHkh (A) Common ion effect. In (B) and (C), solubility equilirbium of AgCN will be pushed forward due to reaction of H+ with CN– (in B) and complexation of Ag+ with CN– (in C) respectively. (A) levk;u izHkko (B) o (C) esa] (B) esa CN– ds

lkFk H+ dh vfHkfØ;k ds dkj.k AgCN ds foys;rk lkE; dks vxz fn'kk esa ys tkrs gS (C) esa] CN– ds lkFk Ag ds ladqy ds dkj.k AgCN ds foys;rk lkE; dks vxz fn'kk esa ys tkrs gSA +

11*.

Select incorrect statement : (A) Na+ and K+ ions are responsible for maintaining isotonic property of inside and outside of the cell of organism. (B) Aquatic species are more comfortable in lake present at sea level in compare to lake present at high altitude. (C*) Solubility of N2 decrease in presence of He when oxygen cylinder is utilised by Scuba divers. (D*) The KH value of CO 2 is higher than KH of N2. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 8

xyr dFku@dFkuksa dk p;u dhft;sA (A) Na+ rFkk K+ vk;u] tSo dksf’kdk dh vkUrfjd rFkk ckg~; leijkljh xq.k ds fy, mÙkjnk;h gksrs gSA (B) tyh; iztkrh;kW leqnzh LRkj ij mifLFkr >hy esa mPp LFkyksa ij mifLFkr >hy dh rqyuk esa vf/kd lgt eglw l djrs gSA (C*) leqnzh xksrk[kksjksa }kjk iz;qDr fd;s tkus okys vkWDlhtu flys.Mj esa He dh mifLFkr esa N2 dh foys;rk ?kVrh gSA (D*) CO2 dk KH eku] N2 lsa vf/kd gksrk gSA 12.

How many B—O—B bond are present in the structure of borax ? cksjsDl dh lajpuk esa fdrus B—O—B ca/k mifLFkr gS \

Ans.

5

13.

Certain amount of a gas confined in a piston-filled cylinder is heated from 27ºC to 127ºC keeping its pressure constant and the gas expanded against a constant pressure doing 4.157 kJ of work on surroundings. Find the number of moles of gas present in the cylinder ?

,d xSl dh fuf'pr ek=kk dks] ,d fiLVu ;qDr flys.Mj esa j[kk tkrk gS] bl Hkjs gq;s flys.Mj dks bldk nkc fu;r j[krs gq, 27ºC ls 127ºC rd xeZ fd;k tkrk gS rFkk xSl fu;r nkc ds fo:) izlkfjr gksdj] ifjos'k ij 4.157 kJ dk dk;Z djrh gS flys.Mj esa] mifLFkr xSl ds eksyksa dh la[;k Kkr dhft;s \ Ans. Sol.

5 W = – PV W = – nRT

14.

or

n=

4.157  103 =5 8.314  100

How many of the following orbitals have x–y plane as nodal pane ? fuEu esa ls fdrus d{kd uksMy ry ds :i esa x–y ry j[krs gS\ (a) 3s

(b) 2pz

(c) 3px

(h) 5pz

(i) 3dyz

(j) 4 dz2

(d) 3 dx 2 – y 2

(e) 4dxy

(f) 3dxz

(g) 4p y

Ans. Sol.

4 (b), (f), (h), (i)

15.

1.22 g of a monobasic acid is dissolved in 100 g of benzene. Boiling point of solution increases by 0.13ºC with respect to pure benzene. Find the molecular mass of acid in benzene solvent (in u). Report your answer after dividing it by 100 and (Round it off to nearest integer). 1.22 g ,dy {kkjh; vEy 100 g csUthu esa ?kksyk tkrk gS rc bl foy;u dk DoFkukad ‘kq) csUthu ls 0.13ºC c<+ tkrk gSA

csUthu foyk;d esa vEy dk vkf.od nzO;eku (u esa) Kkr fdft,A vkidk mÙkj 100 ls Hkkx nsdj nhft;sA (fudVorhZ iw.kkZd es)a Ans. Sol.

2 Tb = Kb × m 0.13 = 2.6 ×

1.22 / M 0.1

 M = 244 u  Reported answer (iz'ukuqlkj

mÙkj) = 2

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PAGE NO.- 9

16.

Column-I and Column-II contains four entries each. Entries of Column-I are to be matched with some entries of Column-II. One or more than one entries of Column-I may have the matching with the same entries of Column-II. Column-I Column-II (Bravais Lattice(s)) [Crystal system] (A) Primitive, face centered, body centered, end centered (p) Cubic (B) Primitive, face centered, body centered (q) Orthorhombic (C) Primitive, body centered (r) Hexagonal (D) Primitive only (s) Tetragonal LrEHk -I rFkk LrEHk-II izR;sd pkj izfof"V;k¡ j[krs gSA LrEHk-I dh izfof"V;ksa dks LrEHk-II dh dqN izfof"V;ksa ds lkFk lqesfyr fd;k

tkrk gSA LrEHk-I dh ,d ;k ,d ls vfèkd izfof"V;k¡ LrEHk-II dh leku izfof"V;ksa ds lkFk lqesfyr gksrh gSA LrEHk -I LrEHk -II (czsfol tkyd) ( fØLVy fudk; ) (A) izhehVho] Qyd dsfUær] dk; dsfUær] vUr% dsfUær (p) ?kuh; (B) izhehVho] Qyd dsfUær] dk; dsfUær (q) fo"ke yEck{k (C) izhehVho] dk; dsfUær (r) "kV~dks.kh; (D) dsoy izhehVho (s) prq"dks.kh; Ans.

(A) - (q) ; (B) - (p) ; (C) - (s) ; (D) - (r)

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ChemINFO-1.5

Solid State Crystal Systems

Daily Self-Study Dosage for mastering Chemistry

There is an unlimited number of possible lattices because there is no restriction on the length of lattice translation vector or on the angle  between them. on the basis of geometrical consideration theoritically there can be 32 different combinations of symmetry of crystal. On other hand bravais show there is only seven types of crystal system. (a) Cubic (b) Tetragonal (c) Orthorhombic (d) Rhombohedral (e) Hexagonal (f) Monoclinic (g) Triclinic Seven Crystal system

Crystal System Cubic

Axial Axial angles distances a=b=c α = β = γ = 90°

Tetragonal

a=bc

Examples

Type of Unit cells

Cu, ZnS, KCl

S,BCC,FCC

α = β = γ = 90°

Sn(white tin), SnO2,TiO2

S,BC

Orthorhombic a  b  c

α = β = γ = 90°

Rhombic sulphur, CaCO3

S,BCC,FCC,ECC

Monoclinic

abc

α = γ = 90°;β  90°

Monoclinic sulphur, PbCrO2

S,BCC

Hexagonal

a=bc

α = β = 90°;γ = 120°

Graphite, ZnO

S

Rhombohedral a = b = c

α = β = γ  90°

CaCO3(Calcite), HgS(Cinnabar)

S

Triclinic

α β γ  90°

K2Cr2O7, CuSO4.5H2O

S

Cubic

abc

; Tetragonal

Hexagonal

;

; Orthorhombic

Rhombohedral

; Monoclinic

;

; Triclinic

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 17.

The lattice parameters of a given crystals are a = 8.42 Å, b = 8.42 Å and c = 9.80 Å. There coordinate axes are mutually perpendicular to each other. The crystal is (A*) Tetragonal (B) Ortho rhombic (C) Monoclinic (D) Trigonal

18.

The crystal dimensions a = b  c ;  =  = 900  = 1200 represents (A) Tetragonal (B*) Hexagonal system (C) Monoclinic (D) Orthorhombic system

19.

Which of the following unit cells has least symmetry ? (A) Tetragonal (B) Monoclinic (C*) Triclinic

(D) Cubic

Among the unit cells given below, which has highest symmetry ? (A) Hexagonal (B) Orthorhombic (C*) Cubic

(D) Monoclinic

20.

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PAGE NO.- 11

ChemINFO-1.5

Solid State Crystal Systems

Daily Self-Study Dosage for mastering Chemistry

;gk¡ vla[; la[;k esa lEHko tkyd gksrs gS D;ksafd ;gk¡ muds chp fdlh çdkj ds tkyd :ikUrj.k dh yEckbZ vFkok dks.k ij çfrcU/k ugha gSA T;kferh; voyksdu ds vk/kkj ij lS)kfUrd :i ls ;gk¡ fØLVy dh leferrk ds 32 fofHkUu la;kstu gks ldrs gSA nwljh vksj czsfol fØLVy ra=k ds dsoy lkr çdkj n'kkZrk gSA (a) ?kuh; (b) f}leyack{k (c) fo"keyack{k (d) f=klerk{k (e) "kV~dks.kh; (f) ,durk{k (g) f=kurk{k lkr fØLVy ra=k

;

?kuh;

f}leyack{k

"kV~dks.kh; ra=k

;

;

;

fo"keyack{k

;

f=klerk{k

;

,durk{k

f=kurk{k

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 17.

,d fn, x, fØLVy ds fy, tkyd çkpky a = 8.42 Å, b = 8.42 Å and c = 9.80 Å gSA muds funsZ'kh; v{k ,d nwljs ds vU;ksU; :i ls yEcor gSaA fØLVy fuEu gSA (A*) f}leyack{k (B) fo"keyack{k (C) ,durk{k (D) f=kdks.kh;

18.

fØLVy foek a = b (A) f}leyack{k

19.

20.

 c ;  =  = 900  = 1200 çnf'kZr

(B*) "kV~dks.kh;

ra=k

djrh gSA (C) ,durk{k

(D) fo"keyack{k

fuEu esa ls fdl ,dd dksf"Bdk ds fy, lcls de leferrk gS \ (A) f}leyack{k (B) ,durk{k (C*) f=kurk{k

(D) ?kuh;

fups fn;s x;s ,dd dksf"Bdkvksa esa ls fdlesa lokZf/kd leferrk gS \ (A) "kV~dks.kh; (B) fo"keyEck{k (C*) ?kuh;

(D) ,durk{k

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PAGE NO.- 12