Jp Xii Physical&inorganic Chemistry (3) - Prev Class Xi Chaps.pdf

PHY./INO. CHEMISTRY

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

NO. 04

Date : 13-04-2015 TEST INFORMATION

DATE : 19.04.2015

SCHOLARSHIP TEST

Syllabus : Complete XI Syllabus

This DPP is to be discussed in the week (13-04-2015 to 18-04-2015)

DPP No. # 04 (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]

1 mole of an ideal gas is expanded isothermally and reversibly from 10L to 100L. Which of the following are correct for the process : (A) T  0 (B) H  0 (C*) heat supplied (q)  0 (D) E  0 1 eksy

vkn'kZ xSl] 10 L ls 100 L rd lerkih; rFkk mRØe.kh; :i ls izlkfjr gksrh gSA izØe ds fy, fuEu esa ls dkSulk dFku lgh gS % (A) T  0 (B) H  0 (C*) nh xbZ Å"ek (q)  0 (D) E  0 Sol.

For an isothermal process, T = E = H = 0.

Sol.

lerkih izØe ds fy,] T = E = H = 0.

2.

The correct figure representing isothermal and adiabatic expansions of an ideal gas from a particular initial state is :

(A)

(B*)

(C)

(D)

,d vkn'kZ xSl dk ,d izkjfEHkd fuf'pr voLFkk ls lerkih; rFkk :)ks"e izlkj.k ds fy, lgh xzkQ gksxk&

(A)

(B*)

(C)

(D)

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3.

Select the correct option(s) : S1 : Specific volume and molar heat capacity are intensive parameters. S2 : Change in internal energy for an ideal gas for an isobaric process is expresed as U = nCV (T2 – T1), where the terms used have their usual meaning. S3 : If the path of an irreversible process is reversed, then both system and surroundings shall be restored to their orginal states. S4 : Less work is done on the gas in adiabatic reversible compression than in isothermal reversible compression, if compression is performed between same initial and final volumes.

lgh fodYi @ fodYiksa dk p;u dhft, % S1 : fof'k"V vk;ru rFkk eksyj Å"ek /kkfjrk] ek=kkLora=k ekiØe (intensive parameters) gSaA S2 : ,d vkn'kZ xSl ds ,d lenkch; izØe ds fy,] vkUrfjd ÅtkZ esa ifjorZu dks U = nCV (T2 – T1) }kjk iznf'kZr fd;k tkrk gS] tgk¡ iz;qDr in viuk lkekU; vFkZ j[krs gSaA S3 : ;fn ,d vuqRØe.kh; izØe ds iFk dks O;qRØfer ¼reverse½ fd;k tkrk gS] rc fudk; rFkk ifjos'k nksuksa viuh okLrfod voLFkk iqu% izkIr dj ysrs gSaA S4 : :)ks"eh; mRØe.kh; lEihM+u esa] lerkih; mRØe.kh; lEihM+u dh rqyuk esa] xSl ij de dk;Z fd;k tkrk gS] ;fn lEihM+u leku izkjfEHkd rFkk vfUre vk;ruksa ds e/; lEiUu djk;k tk,A (A) T T T F 4.

(B) F F T T

(C) F T F T

(D*) T T F F

Consider the general hypothetical reaction

bl dkYifud vfHkfØ;k ij fopkj djks A(s) 2B(g) + 3C(g) If the concentration of C at equilibrium is doubled, then after the equilibrium is re-established, the concentration of B will be ;fn C dh lkUnzrk lkE; ij nqxuh dj nh tk;s rc lkE; LFkkfir gksus ij B dh lkUnzrk gksxh& (A) two times the original value (B) one half of its original value (C*) 1/2 2 times the original value

(D) 2 2 times the original value

(A) vius

okLrfod (original) eku dh nks xquh (B) vius okLrfod (original) eku dh vk/kh (C*) vius okLrfod (original) eku dh 1/2 2 xquh (D) vius okLrfod (original) eku dh 2 2 xquh Sol.

(C) K = [B(g)]2 [C(g)]3 = x2y3. If [C(g)] is doubled i.e. = 2y. Suppose [B(g)] is z. Then K = z2 (2y)3 = x2y3 or z2 =

K = [B(g)]2 [C(g)]3 = x2y3. ;fn [C(g)] dks i.e. = 2y.

or

z=

or

z=

8

x=

2 2

x.

nqxuk djs

ekuk [B(g)] , z gS rc

K = z2 (2y)3 = x2y3 or z2 = 5.

1

1

1 2 x 8

1 2 x 8

1

1 8

x=

2 2

x.

For the process H2O() (1 bar, 373 K)  H2O(g) (1 bar, 373 K), the correct set of thermodynamic parameters is:

izØe H2O() (1 bar, 373 K)  H2O(g) (1 bar, 373 K) ds fy,, Å"ekxfrdh ekiØeksa dk lgh lewg gSA Sol. 6.

(A*) G = 0, S = + ve (C) G = + ve, S = 0 Definition.

(B) G = 0, S = – ve (D) G = – ve, S = + ve

XeF6 + SiO2  XeO3 (x) + SiF4 (y) The hybridisation states of central atoms in the products (x) and (y) are respectively : (A) sp3 and sp3d (B) sp2 and sp3 (C*) both sp3 (D) both sp 3d XeF6 + SiO2  XeO3 (x) + SiF4 (y) mRikn (x) rFkk (y) esa dsUnzh; ijek.kqvksa ds ladj.k (A) sp3 rFkk sp3d

(B) sp2 rFkk sp3

dh voLFkk Øe'k% D;k gS \ (C*) nksuksa esa sp3

(D) nksuksa

esa sp3d

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Sp3 ,

Sol.

7.

Sp3

Which of the following salts undergoes anionic hydrolysis?

fuEu esa ls dkSulk yo.k dk _.k&vk;fud ty vi?kVu n'kkZrk gSA (A) CuSO4 8.

(B) NH4Cl

(C) FeCl3

(D*) Na2CO3

Blood is buffered with CO2 and HCO3–. What is the ratio of the base concentration to the acid (i.e. CO2(aq.) plus H2CO 3) concentration to maintain the pH of blood at 7.4 ? The first dissociation constant of H2CO3 (H2CO3 H+ + HCO3–) is 4.2 × 10–7 where the H2CO3 is assumed to include CO2(aq.) i.e.,dissolved CO2. ( log 2 = 0.3, log 3 = 0.48, log 7 = 0.85, antilog1.06 = 11.5 ) jDr CO2 rFkk HCO3– ds lkFk ,d cQj foy;u gSA jDr dh pH 7.4 ij O;ofLFkr j[kus ds fy, {kkj dh lkanzrk o vEy

dh lkanzrk (i.e. CO2(tyh;) + H2CO3) dk vuqikr D;k gksuk pkfg, \ H2CO3 dk izFke fo;kstu fLFkjkad (H2CO3 H++ HCO3–) = 4.2 × 10–7 gSA tgk¡ H2CO3 dks CO2(aq.) ;qDr ekuk tkrk gSA (i.e., ?kqfyr CO2) ( log 2 = 0.3, log 3 = 0.48, log 7 = 0.85, antilog1.03 = 10.7) (A*) 10.7 (B) 1.8 (C) 10

[B]  [ A]

log

[B] = 1.03  [ A]

(D) 12

[B] = 10.7 [A ]

Sol.

7.4 = (7 – log 4.2) + log

9.

The best indicator for the detection of end point in titration of a weak acid & a strong base is: (A) methyl orange (pH 3 to 4) (B) methyl red (pH 5 to 6) (C) bromothymol blue (pH 6 to 7.5) (D*) phenolphthalein (pH 8 to 9.6)

,d nqcZy vEy rFkk ,d izcy {kkj ds vuqekiu esa vfUre fcUnq ds fu/kkZj.k ds fy, dkSulk lwpd lcls vf/kd mi;qDr gSA (A) esfFky vkWjsUt (pH 3 to 4) (B) esfFky jsM (pH 5 to 6) (C) czkseksFkk;kseksy CY;w (pH 6 to 7.5) (D*) fQuksYQFksyhu (pH 8 to 9.6) 10.

Zinc and hydrochloric acid react according to the reaction : Zn(s) + 2HCl(aq.)  ZnCl2(aq.) + H2(g) If 0.3 mole of Zn are added to hydrochloric acid containing 0.52 mole HCl, how many moles of H 2 are produced:

ftad vkSj gkbMªksDyksfjd vEy dh fØ;k fuEu vfHkfØ;k ds vuqlkj gksrh gS Zn(s) + 2HCl(aq.)  ZnCl2(aq.) + H2(g) ;fn HCl ds 0.52 eksy esa Zn ds 0.3 eksy dks feyk fn;k tk;s] rks H2 ds (A*) 0.26 (B) 0.52 (C) 0.14 Sol.

LR  HCl, so Mole of H2 = LR  HCl,

11.

fdrus eksy cusxsa % (D) 0.30

Mole of HCl 0.52 = = 0.26 2 2

blfy, H2 ds eksy =

HCl 2

=

0.52 = 0.26 2

Property of the alkaline earth metals that increases with their atomic number is : (A) ionisation energy (B*) solubility of their hydroxides (C) solubility of their sulphates (D) electronegativity

ijek.oh; la[;k c<+us ds lkFk {kkjh; e`nk /kkrqvksa ds fdl xq.kksa esa o`f) gksrh gS % (A) vk;uu ÅtkZ (B*) muds gkbMªksDlkWbM dh foys;rk (C) muds lYQsV dh foys;rk (D) fo|qr _.kkRedrk Sol.

For alkaline earth metal hydroxides, down the group the basic strength increases due to increase in metallic character. So down the group, hydroxides easily get dissolved producing OH – ions. So down the group with increasing atomic number, the solubility of their hydroxides increase. 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

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g y-

{kkjh; e`nk /kkrq ds gkbMªksDlkbMksa ds fy,] oxZ esa uhps tkus ij /kkfRod y{k.k c<+us ds dkj.k {kkjh; lkeF;Z c<+rk gSA vr% oxZ eas uhps tkus ij gkbMªksDlkbM vklkuh ls foys; gksdj OH– vk;u cuk ysrs gSaA blfy, oxZ esa uhps tkus ij ijek.kq Øekad c<+us ij] buds gkbMªkWDlkbMksa dh foys;rk c<+rh gSaA

12.

Hydrogen ion concentration in mol/L in a solution of pH = 5.4 will be pH = 5.4 ds ,d foy;u esa gkbMªkstu vk;uksa dh lkanzrk ¼eksy@yhVj esa½ gksxh % (A) 3.98 × 108 (B) 3.88 × 106 (C) 3.68 × 10–6 (D*) 3.98 × 10–6 + + –6 pH = – log[H ]  [H ] = antilog (–pH) = antilog (–5. 4) = 3.98 × 10

Sol. 13.

Bond angle of 109º 28' is found in : 109º 28' dk ca/k dks.k ik;k tkrk gS % (A) NH3 (B) H2O

(C) CH3+

(D*) NH4+

Sol.

14.

Which one of the following represents acid-base reaction according to Lewis definition, but not according to Bronsted- Lowry definition ? fuEu esa ls dkSu yqbZl ifjHkk"kk ds vuqlkj vEy {kkj vfHkfØ;k dks iznf'kZr djrk gSA ysfdu czksULVsM-yksjh ifjHkk"kk ds vuqlkj

iznf'kZr ugha djrk gS \

Sol.

(A) MgCl2(s)  Mg2+(aq) + 2Cl– (aq) (C*) NH3(g) + HBr(g)  NH4Br (s) H3O+(aq) NH3(g) + HBr(g)  NH4Br(s)

(B) CN– + H2O()  HCN (aq) + OH– (aq) (D) Fe(H2O)63+ (aq) + H2O  [Fe(H2O) 5OH]2+(aq) +

15.

A 0.20 molar solution of a weak monoprotic acid, HA, has a pH of 3.0. The ionization constant of the acid is : ,d nqcZy izksfVd vEy HA dk 0.20 eksyj foy;u 3.0 dh pH j[krk gSA vEy dk vk;uu fu;rkad gS& (A) 5.0 × 10–7 (B) 2.0 × 10–7 (C*) 5.0 × 10–6 (D) 5.0 × 10–3

Sol.

[H+] =

16.

The vapour pressure of a pure liquid A is 40 mmHg at 310 K. The vapour pressure of this liquid in a solution with liquid B is 32 mmHg. Mole fraction of A in the solution, if it obeys Raoult’s law is : 310 K ij 'kq) nzo A dk ok"inkc 40 mmHg gSA bl nzo dk] nzo B ds lkFk foy;u esa ok"i nkc 32 mmHg gSA ;fn foy;u

Ka = 5 × 10–6.

Ka  C

jkmYV fu;e dk ikyu djrk gS rks foy;u esa A dh eksy fHkUu gS % (A*) 0.8 Sol.

(B) 0.5

(D) 0.4

pA = XA p°A 32 = XA 40

17.

(C) 0.2



XA =

32 = 0.8. 40

The azeotropic mixture of water (B.P. = 100ºC) and HCI (B.P. = 86ºC) boils at about 120ºC During fractional distillation of this mixture it is possible to obtain: (A) pure HCI (B) pure H2O (C) pure H2O as well as pure HCI (D*) neither pure H2O nor HCI ty (B.P. = 100ºC) rFkk HCI (B.P. = 86ºC) dk fLFkjdokaFkh(azeotropic) feJ.k yxHkx 120ºC ij mcyrk gSA bl feJ.k

ds izekih vklou ds nkSjku fuEu dks izkIr djuk laEHko gS : (A) 'kq) HCI dks (B) 'kq) H2O dks (C) 'kq) H2O ds lkFk–lkFk 'kq) HCI dks (D*) 'kq) H2O rFkk 'kq) HCI nksuks dksbZ ughaA 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

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Sol.

From an azeotropic mixture the components cannot be separated at boiling point because it is a constant temperature boiling mixture.

,d fLFkjdokaFkh feJ.k ds ?kVdks A vo;oksa dks doFkukad fcUnq ij i`Fkd ugha fd;k tk ldrk gSA D;ksafd ;g ,d fu;r rki do;u feJ.k gksrk gSSA+ 18.

Which is the correct statement for positive deviation of solution. (A) The energy released in solvation exceeds the energy used in breaking up interactions / intermolecular forces. (B*) Solute-solvent interaction is weaker than their own interactions. (C) Volume would contract during formation of solution. (D) Boiling point of solution will be higher than expected.

foy;u ds /kukRed fopyu ds fy, fuEu esa ls lgh dFku dkSulk gSA (A) foyk;du esa eqDr ÅtkZ] vUrZfØ;k @ vUrjkf.od cyksa dks rksM+us esa iz;qDr ÅtkZ ls vf/kd gksrh gSA (B*) foys;&foyk;d vUrZfØ;k ¼interaction½ buds Loa; dh vUrZfØ;k ¼interaction½ dh rqyuk esa nqcZy gksrh gSA (C) foy;u fuekZ.k ij vk;ru esa ladqpu gksrk gSA (D) foy;u dk DoFkukad vuqekfur eku ls mPp gksxkA Sol. Sol.

H = +ve, V = +ve (expansion) and A – B interaction is relatively weaker than A–A and B–B interactions. Pobs. > Pideal so, (B.P.)obs. > (B.P.)ideal H = +ve, V = +ve (izlkj) rFkk A – B vUrZfØ;k] A–A rFkk B–B vUrZfØ;k dh rqyuk esa nqcZy gksrh gSA Pizsf{kr > Pvkn'kZ

19.

vr%,

(B.P.)izsf{kr > (B.P.)vkn'kZ

An azeotropic solution of two liquids has boiling point lower than that of either of the liquids when it (A) Shows a negative deviation from Raoult's law (B) Shows no deviation from Raoult's law (C*) Shows positive deviation from Raoult's law (D) Is saturated

nks nzOkksa ds ,d fLFkj DokFkh foy;u dk DoFkukad nksuksa nzoksa ls de gksxk tc % (A) ;g jkmYV fu;e ls _.kkRed fopyu n'kkZrk gS (B) ;g jkmYV fu;e ls dksbZ fopyu ugha n'kkZrk gS (C*) ;g jkmYV fu;e ls /kukRed fopyu n'kkZrk gS (D) ;g lar`IRk gSA 20.

For an ideal solution of A and B liquid, when x A = 0.2, total vapor pressure over solution was 560 torr and partial pressure of A was 80 torr. What will be the composition of vapor when x A = 0.8 ? A rFkk B æo ds ,d vkn'kZ foy;u ds fy, tc xA = 0.2 gS rc foy;u ij dqy ok"i nkc 560 Vksj Fkk rFkk A dk vkaf'kd

nkc 80 Vksj FkkA ok"i dk la?kVu D;k gksxk] tc xA = 0.8 gS ? 1 7 PA = 80 torr (A) yA = Sol.

xA = 0.2  PA° =

(B*) yA =

8 11

(C) yB =

3 7

(D) yB =

9 11

PA 80  = 400 torr x A 0.2

PT = 560 torr PB = 560 – 80 = 480 torr  PB° =

480 = 600 torr 0 .8

When xA = 0.8 PT = 400 × 0.8 + 600 × 0.2 = 320 + 120 = 440 torr  yA =

320 8  440 11

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