Jp Xii Physical&inorganic Chemistry (4) - Prev Class Xi Chaps + Solution & Colligative Properties +solid State.pdf

PHY./INO. CHEMISTRY

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

NO. 05 & 06

Date : 20-04-2015 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 (20-04-2015 to 25-04-2015)

DPP No. # 05 (JEE-ADVANCED) Total Marks : 64

Max. Time : 38 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.5 Multiple choice objective ('–1' negative marking) Q.6 to Q.11 Comprehension ('–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.19

1.

(3 (4 (3 (8 (4

marks, marks, marks, marks, marks,

2 2 2 6 2

min.) min.) min.) min.) min.)

[15, [16, [09, [08, [16,

10] 08] 06] 06] 08]

4.4 g of an ideal gas (mol.wt = 44) at 300 K and 10 atm is allowed to expand isothermally against a constant external pressure of 1 atm. Taking 1 L-atm = 100 J, heat absorbed by the gas is : 300 K rFkk 10 atm ij 4.4 xzke vkn'kZ xSl (v.kqHkkj = 44) dks 1 atm fu;r ckÐ; nkc ds fo:) lerkih; :i ls izlkfjr

fd;k tkrk gSA 1 L-atm = 100 J ysdj] xSl }kjk vo'kksf"kr Å"ek fuEu gS % (A) 200.5 J Sol.

Vi =

(B) 212.5 J

(C*) 221.4 J

(D) 242.4 J

0.082  300 nRT 4 .4 = × = 0.246 litre 10 P 44

Pi Vi 10  0.246 Vf = P = = 2.46 litres 1 f q = – W = P(Vf – Vi) = 1 (2.46 – 0.246) × 100 = 221.4 J. 2.

In which case van’t Hoff factor is maximum ? (A) KCl, 50% ionised (B) K2SO4, 40% ionised (C) SnCl4, 20% ionised (D*) FeCl3, 30% ionised

fdl ifjfLFkfr esa okUV gkWQ dkjd vf/kdre gS (A) 50% vk;fur KCl (B) 40% vk;fur K2SO4 (C) 20% vk;fur SnCl4 (D*) 30% vk;fur FeCl3 Sol.

3.

i = 1 + (n–1)  (A) For KCl, (B) For K2SO4, (C) For SnCl4, (D) For FeCl3,

i i i i

= 1 + 0.5 = 1.5 = 1 + 2  0.4 = 1.8 = 1 + 4  0.2 = 1.8 = 1 + 3  0.3 = 1.9

Which of the following increases with dilution at a given temperature?

fn;s x, rki ij fuEu esa ls fdlesa ruqrk ds lkFk o`f) gksrh gSA (A*) pH of 103M acetic acid solution (C*) degree of dissociation of 103M acetic acid

(B) pH of 103 M aniline solution (D*) degree of dissociation of 10–3M aniline

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PAGE NO.- 1

(A*) 103M ,lhfVd

(B) 103 M ,uhyhu

vEy dh pH 3 (C*) 10 M ,lhfVd vEy ds fo;kstu dh ek=kk

ds foy;u dh pH (D*) 10 M ,uhyhu ds fo;kstu dh ek=kk –3

Sol.

Degree of dissocation of WA & WB will increase. [H+] in WA and [OH–] in WB will decrease so pH of WA and pOH of WB will increase.

gy %

WA rFkk WB

ds fo;kstu dh ek=kk] esa o`f) gksxhA WA esa [H ] rFkk WB esa [OH–] esa deh gksxh blfy,] WA dh pH rFkk WB dh pOH esa o`f) gksxhA +

4.

Two liquids X and Y are perfectly immiscible. If X and Y have molecular masses in ratio 1:2, the total vapour pressure of a mixture of X and Y prepared in weight ratio 2:3 should be (P X0 = 400 torr, PY0 = 200 torr) nks nzo X rFkk Y iw.kZr;k vfeJ.kh; gS ;fn X rFkk Y dk eksyj nzO;eku vuqikr 1 : 2 gS] rks vuqikr 2 : 3 esa cuk;k x;k X

rFkk Y ds feJ.k dk dqy ok"i nkc fuEu gksuk pkfg,A (PX0 = 400 torr, PY0 = 200 torr) Sol. 5.

(A) 300 torr (B) 466.7 torr (C*) 600 torr For immiscible solution = PT = PA0 + PB0 = 400 + 200 = 600.

(D) 700 torr

A mixture of two immiscible liquids at a constant pressure of 1 atm boils at a temperature (A) equal to the normal boiling point of more volatile liquid (B) equal to the mean of the normal boiling points of the two liquids (C) greater than the normal boiling point of either of the liquid (D*) smaller than the normal boiling point of either of the liquid. 1 atm fu;r nkc ij nks vfeJ.kh; nzoksa ds ,d feJ.k dk DoFkukad % (A) vf/kd

ok"i'khy nzo ds lkekU; DoFkukad ds cjkcj gksrk gSA (B) nks nzoksa ds lkekU; DoFkukadksa ds ek/; ds cjkcj gksrk gSA (C) nksuksa nzoksa ds lkekU; DoFkukad esa ls fdlh ,d ls vf/kd gksrk gSA (D*) nksuksa nzoksa ds lkekU; DoFkukad esa ls fdlh ,d ls de gksrk gSA Sol.

The vapour pressure of a mixture of two immiscible liquids is the sum of their vapour pressures in the pure states, independent of their relative amounts. Hence, B.Pt. of the mixture will be less than that of either of the liquids, remaining constant throughout.

6.

I. Orbital angular momentum of the electron having n = 5 and having value of the azimuthal quantum number as lowest for this principle quantum number is

h . 

I. n = 5 rFkk bl eq[; Dok.Ve la[;k ds fy, f}xa'kh Dok.Ve la[;k dk eku U;wure ysus ij d{kh; dks.kh; laosx

h 

gksrk gSA

II. If n = 3,  = 0, m = 0, for the last valence shell electron, then the possible atomic number may be 12 or 13. II. ;fn vafre la;ksth dks'k bysDVªkWu ds fy, n = 3,  = 0, m = 0 gS rks laHkkfor ijek.kq Øekad 12 ;k 13 gksxkA III. Total spin of electrons for the atom 25Mn is ± III.ijek.kq

25Mn

ds fy, bysDVªkWuksa dk dqy pØ.k ±

7 . 2

7 gSA 2

IV. Magnetic moment of inert gas is 0 IV. vfØ; xSl dk pqEcdh; vk?kw.kZ 'kwU; gSA Which of the above statement (s) is/are false.

fuEu es ls dkSulk@dkSuls dFku vlR; gS &

Sol.

(A*) I, II and III (B) II and III Only (A*) I, II vkSj III (B) II vkSj III I. For n = 5 lowest value of l = 0 

orbtail angular momentum =

(C) I and IV only (C) I vkSj IV

0(0  1)

(D) None of these (D) buesa ls dksbZ ugha

h =0 2

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II

n = 3, l = 0, m = 0 coresspond to 3s  electronic configuration can be 1s22s22p6 3s1 or 1s22s22p6 3s2 there fore atomic number may be 11 or 12. 5 2 25Mn – [Ar]3d 4s there are 5 unpaired electrons  total spin = ± 5/2 As intert gas has no unpaired electron hence, it’s magnetic moment will be zero. n = 5 ds fy, l = 0 lcls U;wure eku gSA

III

Sol.

IV I.

h

d{kd dks.kh; laosx = 0(0  1) 2 = 0 n = 3, l = 0, m = 0 ls lEcfU/kr 3s d{kd gSA  blfy, bysDVªkWfud foU;kl 

II

1s22s22p6 3s1 1s22s22p6 3s2

;k blfy, ijek.kq Øekad 11 ;k 12 gks ldrk gSA III

– [Ar]3d5 4s2 ij 5 v;qfXer e– gSA

25Mn

;gka

dqy pØ.k = ± 5/2 vfØ; xSl esa dksbZ v;qfXer e– ugha gSA blfy, bldk pqEcdh; vk?kw.kZ 'kwU; gSA 

IV 7.

Which of the following molarity values of ions in a aqueous solution of 5.85 % w/v NaCl, 5.55% w/v CaCl2 and 6% w/v NaOH are correct [Na = 23, Cl = 35.5 , Ca = 40, O = 16 ] 5.85 % w/v NaCl, 5.55% w/v CaCl2 rFkk 6% w/v NaOH ds ,d tyh; foy;u esa fuEu esa ls dkSu ls vk;uksa dh eksyjrk

lgh gSaA [Na = 23, Cl = 35.5 , Ca = 40, O = 16 ] (A*) [Cl–1] = 2M Sol.

(B) [Na+] = 1M

(C*) [Ca2+] = 0.5 M

(D*) [OH–] = 1.5 M

Only single solution have all these means 100 ml solution have 5.85 gm NaCl = 0.1 mole and 5.55 gm CaCl2 = 0.05 mole and 6 gm NaOH = 0.15 mole [Cl–] =

(0.1 0.05  2 ) 1000 =2M 100

[Ca2+] =

0.05  1000 = 0.5 M 100

(0.1 0.15 ) 1000 = 2.5 M 100



[Na+] =



[OH–] = 1.5 M

mijksDr lHkh dsoy ,d gh foy;u j[krs gSa vFkZkr~ 100 ml foy;u 5.85 gm NaCl ds 0.1 mole eksy j[krs gSa rFkk 5.55 gm CaCl2 = 0.05 eksy rFkk 6 gm NaOH = 0.15 eksy [Cl–] =

(0.1 0.05  2 ) 1000 =2M 100

[Na+] =

(0.1 0.15 ) 1000 = 2.5 M 100

[Ca2+] = 8.

0.05  1000 = 0.5 M 100

[OH–] = 1.5 M.

In which of the following cases van't Hoff factor is greater than 3 (i > 3) fuEu esa ls fdl ifjfLFkfr esa] okWUV gkWQ xq.kkad dk eku 3 (i > 3) ls vf/kd gS % (A*) AlCl3 if  = 0.8. (B) BaCl2 if  = 0.9. (C*) Na3PO4 if  = 0.9 (D*) K4[Fe(CN)6] if  = 0.7. 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.- 3

Sol.

(A) i = 1 + ( n – 1)  i = 1 + (4 – 1) ´ 0.8 = 3.4 (B) For BaCl2, i = 1 + ( 3 – 1)  0.9 = 2.8 (C) For Na3PO4, i = 1 + (4 – 1)  0.9 = 3.7 (D) For K4[Fe(CN)6], i = 1 + ( 5–1)  0.7 = 3.8

9.

In which of the following pairs of solutions will the values of the van't Hoff factor be the same? (A) 0.05 M K4 [Fe(CN)6] and 0.10 M FeSO 4. (B*) 0.10 M K4[Fe(CN)6] and 0.05 M FeSO4.(NH4)2SO4. 6H2O. (C) 0.20 M NaCl and 0.10 M BaCl2 (D*) 0.05 M FeSO4.(NH4)2SO4.6H2O and 0.02 M KCl . MgCl2 . 6H2O fuEu esa ls dkSuls foy;u ;qXeksa ds fy, okUVgkWQ dkjd (factor) dk eku leku gksxkA (A) 0.05 M K4 [Fe(CN)6] rFkk 0.10 M FeSO4 (B*) 0.10 M K4[Fe(CN)6] rFkk 0.05 M FeSO4.(NH4)2SO4. 6H2O (C) 0.20 M NaCl rFkk 0.10 M BaCl2

Sol.

10.

(D*) 0.05 M FeSO 4.(NH4)2SO4 . 6H2O rFkk 0.02 M KCl . MgCl2 . 6H2O. Those which have same no of ions after dissociation will have same Van't Hoff factor. K4[Fe(CN)6] & FeSO4.(NH4)2SO4.6H2O have 5 ions each. & FeSO4.(NH4)2SO4.6H2O & KCl. MgCl2.6H2O have 5 ions each. Select correct statement : (A*) Higher the value of k H (Henry’s law constant) at a given pressure, the lower is the solubility of the gas in the liquid (B) Solubility of a gas in a liquid decreases with increase in temperature and pressure (C*) to minimise the painful effects accompanying the decompression of deep sea divers, O 2 diluted with less soluble He gas is used as breathing gas (D*) The solubility of a gas in a liquid is governed by Henry’s law

lgh dFku dk pquko dhft,A (A*) fn;s x;s nkc ij k H (gSujh fu;e fu;rkad) dk mPp eku gS] mruk gh nzo esa xSl dh mPp foys;rk n'kkZrk gSA (B) ,d nzo esa xSl dh ?kqyu'khyrk] rki o nkc esa o`f) ds lkFk de gksrh gSA (C*) xgjk;h esa tkus okys xksrk[kksjksa ds fy, dksf'kdkvksa esa nokc ds dkj.k mRiUu nnZ dks de djus ds fy, 'olu xSl ds :i esa de foy;'khy He xSl ds lkFk ruqd`r O2 xSl dk mi;ksx fd;k tkrk gSA (D*) nzo esa xSl dh ?kqyu'khyrk] gSujh fu;e }kjk le>k;h tk ldrh gSA Sol.

Solublity of a gas in a liquid increases with increase in pressure and decrease in temperature.

gy .

nzo esa xSl dh ?kqyu'khyrk] nkc esa o`f) rFkk rki esa deh ds lkFk c<+rh gSA

11.

The value of Hºr for the given reaction cannot be calculated using which of the following informations : CH4(g) + 2O2(g)  CO2(g) + 2H2O() (A*) Bond enthalpies of C=O, O=O, O–H and C–H bonds. (B) Hºƒ(CO2(g)), Hºƒ(H2O()) & Hºƒ(CH4(g)) (C*) Bond enthalpies of C–H, C=O & O–H bonds and Hºvap.(H2O()) (D*) Hºatomisation(C, graphite, s), Hºatomisation(CH4(g)),Hºƒ(H2O() and Bond enthalpies of O=O & C=O bonds. fuEu esa ls fdu lwpukvksa dk mi;ksx djus ij uhps nh xbZ vfHkfØ;k ds fy, Hºr dk eku Kkr ugha fd;k tk

ldrk gS%

(A) C=O, O=O, O–H ,oe~ C–H ca/kksa

dh ca/k ,UFkSYih rFkk Hºƒ(CH4(g)) (C) C–H, C=O o O–H ca/kksa dh ca/k ,UFkSYih ,oe~ Hºvap.(H2O()) (D) H°ijek.ohdj.k (C,xzsQkbV]s), H°ijek.ohdj.k (CH4(g)),Hºƒ(H2O() rFkk O=O o C=O ca/kksa dh ca/k ,UFkSYih (B) Hºƒ(CO2(g)), Hºƒ(H2O())

Sol.

(A) All calculations done by bond enthalpy data involves species in gaseous state only. But in question, H2O () is there. (C) O = O bond energy is not given. (D) H – H bond energy is not given.

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g y-

(A) ca/k

,UFkSYih ds vkadM+ksa }kjk dh x;h lHkh x.kuk;sa dsoy xSlh; Lih'kht ds fy, gksrh gSaA ijUrq ç'u esa H2O () fn;k

x;k gSA (C) O = O ca/k

ÅtkZ ugha nh x;h gSA (D) H – H ca/k ÅtkZ ugha nh x;h gSA Comperhension # The pressure of two pure liquid A and B which form an ideal solutions are 400 mm Hg and 800 mm Hg respectively at temperature T. A liquid containing 3 : 1 molar composition pressure can be varied. The solutions is slowly vaporized at temperature T by decreasing the applied pressure starting with a pressure of 760 mm Hg. A pressure gauge (in mm) Hg is connected which given the reading of pressure applied. rki T ij A rFkk B nks 'kq) nzo ds nkc Øe'k% 400 mm rFkk 800 mm Hg gSa, tks fd vkn'kZ foy;u cukrs gSA ,d nzo]

tks fd 3 : 1 eksyj laxBu j[krk gS] dk nkc ifjofrZr gks ldrk gSA foy;u rki T ij /khjs &/khjs ok"ihd`r gksrk gS tc yxk;s x;s nkc esa deh dh tkrh gS] ;g yxk;k x;k nkc 760 mm Hg ls izkjEHk gksrk gSA yxk;s x;s nkc ds ekiu ds fy;s ,d nkcekih (mm esa) yxk;k x;k gSA 12.

The reading of pressure Gauge at which only liquid phase exists.

tc dsoy nzo voLFkk mifLFkr gS] rc nkcekih dk ikB~;kad gS \ (A) 499 13.

(B) 399

(C) 299

(D*) None

(C) 700

(D) None

The reading of pressure Gauge at bubble point is

cqycqyk fcUnq ij nkc ekih dk ikB~;kad gSA (A*) 500 14.

(B) 600

The reading of pressure Gauge at which only vapour phase exists is

tc dsoy ok"i voLFkk mifLFkr gS] rc nkcekih dk ikB~;kad fuEu gSA Sol.

(A) 501 (B) 457.14 (C*) 425 (Q.12 to Q.14) XA = 0.75 XB = 0.05 above 500 mm Hg only liquid phase exists. XA = 0.75 XB = 0.05 500 mm Hg ds Åij dsoy nzo voLFkk mifLFkr Pbubble point = XA P°A + XB P°B Pcqycqfyr fcUnq = XA P°A + XB P°B = 0.75 + 400 + 0.25 × 800 = 500 mm. yA = 0.75, yB = 0.5 At dew point (vksl

(D) 525

gksrh gSA

fcUnq ij)

1 y y 1 0.75 0.25 1.5  0.25  A  B PT PAº PBº  PT = 400 + 800 = 800 

PT =

800 = 457.14 mm Hg 1.75

Below dew point only vapour phase exists. (vksl fcUnq ds uhssps dsoy ok"i voLFkk mifLFkr gksrh 15. (A) (B) (C) (D) (A) (B) (C)

Column-I n-hexane + n-heptane. Acetone + chloroform Chloro-benzene and bromo-benzene

(p) (q) (r)

Ethanol + water.

(s)

LrEHk  n-gSDlsu + n-gSIVsu ,lhVksu + DyksjksQkWeZ Dyksjks&csathu rFkk czksekscsathu

(p) (q) (r)

gSA ) Column-II Can be separated by fractional distillation. Maximum boiling azeotrope. Cannot be separated by fractional distillation completely. Minimum boiling azeotrope. LrEHk 

çHkkth vklou }kjk i`Fkd~ fd;k tk ldrk gSaA vf/kdre fLFkj DokFkha rkiekuA iw.kZ :i ls çHkkth vklou }kjk i`Fkd ugha fd;k tk ldrk gSaA

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Ans. Sol.

(D) ,sFksukWy + ty (s) U;wure fLFkjDokFkh rkiekuA [A – p] ; [B – q,r] ; [C – p] ; [D – r,s]. Hexane & Heptane solution do not form azeotrope, but have different boiling points, so can be separated by fractional distillation. Acetione & chloroform form maximum boiling azeotrope ethanol & water form minimum boiling azeotrope. Azeotropes cannot be separated completely by fractional distillation chlorobenzene & bromobenzene do not form azeotrope but have different boiling points. so can be separated by fractional distillation.

ChemINFO-1.3

Solution & Colligative Properties

Daily Self-Study Dosage for mastering Chemistry

Non ideal solution A non-ideal solution is defined as the one which does not obey Raoult’s law. These are divided into two types. 1. Non-ideal positive deviation solution 2. Non-ideal negative deviation solution. Positive deviation solution

Negative deviation solution

(i)

PT  X APA0  XBPB0

(i)

PT  X APA0  XBPB0

(ii)

The inter molecular interaction is weaker as

(ii)

The inter molecular interaction is more than the interaction found in any of the pure components

interaction found in any one of the pure components.

(iii) (iii)

Ex.

There is decrease in volume on mixing and

There is increase in volume on mixing and

evolution of heat takes place on mixing

absorption of heat.

Vmix  ve

Hmix   ive

Smix   ive

Gmix   ive

H2O + HCOOH,

H2O + CH3COOH

H2O + HNO3,

CHCl3 + CH3OCH3

Vmix  positive

Hmix  positive

Smix   ive

Gmix   ive

H2O + CH3OH,

C2H5OH + Hexane

Ex.

C2H5OH + Cyclohexane, CHCl3 + CCl4 Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

Identify the mixture that shows negative deviation solutions (A) Benzene + (CH3)2CO (B*) (CH3)2CO+ C6H5NH2 (C) CHCl3 + C6H6 (D) (CH3)2CO + CS2

17.

At 550C, ethanol has a vapour pressure of 168 mm, and the vapour pressure of methyl cyclohexane is 280 mm. A solution of the two in which the mole fraction of ethanol is 0.68, has a total vapour pressure of 376 mm. This solution is formed from its components with (A) The evolution of heat (B*) The absorption of heat. (C) Neither absorption nor evolution of heat (D) Nothing can be predicted on the basis of given information 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.- 6

18.

Total vapour pressure of mixture of 1 mole of A PA0  150 mm and 2 mole of B PB0  240 mm is 200 mm. In this case : (A) There is a positive deviation from Raoult’s law. (B*) There is a negative deviation from Raoult’s law (C) There is no deviation from Raoult’s law. (D) Molecular masses of A & B are also required for calculating deviation.

19.

Identify the mixture showing positive deviation of non-ideal solution. (A*) CH3COCH3 + CS2 (B) Hexane + Heptane (C) Chlorobenzene + Bromo benzene (D) Acetone + Chloroform

ChemINFO-1.3

Solution & Colligative Properties

Daily Self-Study Dosage for mastering Chemistry

Non ideal solution

,d vukn'kZ foy;u dks og gS tks jkÅYV fu;e dk ikyu ugha djrk gSA bUgs nks çdkj ls foHkkftr djrs gSa 1. vukn'kZ

/kukRed fopyu foy;u

2. vukn'kZ

_.kkRed fopyu foy;u _.kkRed fopyu foy;u

/kukRed fopyu foy;u (i)

PT  X APA0  XBPB0

(ii)

fdlh ,d 'kq+) vo;o esa vUrjfØ;k dh vis{kk vUrj vkf.od vUrjfØ;k nqcZy gksrh gS

(iii)

;gk¡ feJ.k ij vk;ru esa o`f+) gksrh gS o Å"ek dk vo'kks"k.k gksrk gSA

Ex.

(i)

PT  X APA0  XBPB0

(ii)

fdlh ,d 'kq+) vo;o esa vUrjfØ;k dh vis{kk vUrj vkf.od vUrjfØ;k vf/kd gksrh gS

(iii)

fefJr djus ij ;gk¡ feJ.k ij vk;ru esa deh gksrh gS o Å"ek dks fu"dklu gksrk gSA

V



H



V

 ve

H

  ive

S

  ive

G

  ive

S

  ive

G

  ive

H2O + CH3OH,

C2H5OH + gSDlsu

C2H5OH + lkbDyksgSDlsu, CHCl3 + CCl4

Ex.

H2O + HCOOH,

H2O + CH3COOH

H2O + HNO3,

CHCl3 + CH3OCH3

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

og feJ.k igpkfu;s tks _.kkRed fopyu dk foy;u n'kkZrk gS % (A) csUthu + (CH3)2CO (B*) (CH3)2CO+ C6H5NH2 (C) CHCl3 + C6H6

17.

(D) (CH3)2CO + CS2

550C, ij

,FksukWy ds fy, ok"i&nkc 168 mm gS o esfFky lkbDyksgSDlsu ds fy, ok"i&nkc 280 mm gSA bu nks foy;uksa esa ftlesa ,FksukWy dk eksy&fHkUUk 0.68 gS o dqy ok"i nkc 376 mm gSA ;g foy;u fuEu ds lkFk blds vo;oksa ls curk gSA (A) Å"ek ds fu"dklu (B*) Å"ek ds vo'kks"k.k (C) u rks Å"ek ds vo'kks"k.k vkSj uk gha Å"ek ds fu"dklu lsA (D) nh xbZ tkudkjh ds vk/kkj ij buesa ls fdlh dks crk;k ugha tk ldrk gSA 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.- 7

18.

1 eksy A PA0  150 mm

o 2 eksy B PB0  240 mm ds feJ.k dk dqy ok"i nkc 200 mm gSaA bl fLFkfr esa

(A) jkÅYV

fu;e ls /kukRed fopyu gksrk gSA fu;e ls _.kkRed fopyu gksrk gSA (C) jkÅYV fu;e ls dksbZ fopyu ugha gksrk gSA (D) fopyu dks tk¡pus ds fy, A o B ds v.kqHkkjksa dh Hkh vko';drk gksrh gSA (B*) jkÅYV

19.

ml feJ.k dks igpkfu, tks vukn'kZ foy;u ds /kukRed fopyu dks n'kkZrk gSA (A*) CH3COCH3 + CS2 (B) gSDlsu + gsIVsu (C) DyksjkscsUthu + czksekscsUthu (D) ,lhVksu + DyksjksQkWeZ

DPP No. # 06 (JEE-ADVANCED) Total Marks : 75

Max. Time : 46 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.5 Multiple choice objective ('–1' negative marking) Q.5 to Q.10 Integer type Questions ('–1' negative marking) Q.11 to Q.14 Match the Following (no negative marking) Q.15 ChemINFO : 4 Questions ('–1' negative marking) Q.16 to Q.19

1.

(3 (4 (4 (8 (4

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

[15, [20, [16, [08, [16,

10] 10] 12] 06] 08]

For the given process on an ideal gas, which of the following statements is true :

,d vkn'kZ xSl ds fn, x, izØe ds fy;s fuEu esa ls dkSulk dFku lR; gS %

(A) Work done by the gas during complete cycle is 6P0V0. (B) Work done on the gas during complete cycle is 6P0V0. (C) Work done on the gas during complete cycle is (D*) Work done by the gas during complete cycle is

5P0 V0 . 2 5P0 V0 . 2

(A) ,d

iw.kZ pØ ds nkSjku xSl }kjk 6P0V0 dk;Z fd;k tkrk gSA (B) ,d iw.kZ pØ ds nkSjku xSl ij 6P0V0 dk;Z fd;k tkrk gSA (C) ,d (D*) ,d 2.

iw.kZ pØ ds nkSjku xSl ij

5P0 V0 2

iw.kZ pØ ds nkSjku xSl }kjk

dk;Z fd;k tkrk gSA

5P0 V0 2

dk;Z fd;k tkrk gSA

3 R) is allowed to expand adiabatically and reversibly from initial volume of 8L 2 at 300 K to a volume of V2 at 250 K. The value of V2 is : (Given (4.8)1/2 = 2.2) A monoatomic ideal gas (CV =

,d ,dyijek.kfod vkn'kZ xSl (CV = 250 K ij vfUre (A*) 10.5 L

3 R) dks :)ks"e rFkk mRØe.kh; :i ls 300 K rki ij izkjfEHkd vk;ru 8L ls] 2

vk;ru V2 rd izlkfjr gksus fn;k tkrk gSA rc V2 dk eku gksxk % (fn;k x;k gS % (4.8)1/2 = 2.2) (B) 23 L

(C) 8.5 L

(D) 50.5 L

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PAGE NO.- 8

Sol.

TV–1 = constant (fu;rkad) =  

3.

5 3



–1=

2 3

300 × (8)2/3 = 250 × (V2)2/3  V2 = (4.8)3/2  4.8 × 2.2 = 10.5 L

(V2)2/3 = 4.8

Three solutions are prepared by adding 'w' gm of 'A' into 1kg of water, 'w' gm of 'B' into another 1 kg of water and 'w' gm of 'C' in another 1 kg of water (A, B, C are non electrolytic). Dry air is passed from these solutions in sequence (A  B  C). The loss in weight of solution A was found to be 2 gm while solution B gained 0.5 gm and solution C lost 1 gm. Then the relation between molar masses of A, B and C is : rhu foy;u 1kg ty esa 'A' ds 'w' xzke] vU; 1kg ty esa 'B' ds 'w' xzke] rFkk vU; 1kg ty esa 'C' ds 'w' xzke dks

feykdj cuk;s tkrs gaSA (A, B, C fo|qr vu~vi?kV~; gS)A bu foy;uksa ls 'kq"d ok;q fuEu Øe (A  B  C) esa izokfgr dh tkrh gSA foy;u A ds nzO;eku esa deh 2 xzke ik;h x;h tcfd B dk 0.5 xzke c<+rk gS rFkk foy;u C ds nz O;eku es a 1 xz ke dh deh gksr h gSa A A, B rFkk C ds eks yj nzO ;ekuks a ds e/; lEca /k gks xkA (A) M A : M B : M C = 4 : 3 : 5 Sol.

(B) MA : MB : MC =

1 1 1 : : 5 4 3

(C*) MC > MA > M B (D) M B > M A > M C The loss in weight should be proportional to vapour pressure above that solution : So,

PSA  2gm PSB  1.5gm PSC  2.5 gm

So, maximum vapour pressure is above C solution hence, it is having minimum lowering and hence minimum mole fraction (hence minimum number of moles of solute) So max. molar mass of substance. Sol.

ok"i nkc] Hkkj esa gkfu ds lekuqi krh gksxk PSA  2gm PSB  1.5gm PSC  2.5 gm

pwafd C foy;u dk ok"inkc vf/kdre gS] vr% blesa U;wure deh vk;h gksxh blfy, bldk eksy izHkkt U;wure gksxk rFkk blfy, bldk v.kqHkkj vf/kdre gksxkA 4.

A solution of 0.2 mole KI ( = 100%) in 1000 g water freezes at T 1 0C. Now to this solution 0.1 mole Hg2 is added and the resulting solution freezes at T 2 0C. Which of the following is correct : (A) T1 = T2 (B) T1 > T2 (C*) T1 < T2 (D) Cannot be predicted.

esa 0.2 eksy K ( = 100%) dk foy;u T1 0C ij terk gSaA vc bl foy;u esa 0.1 eksy Hg2 Mkyk tkrk gS rFkk ifj.kkeh foy;u T2 0C ij terk gSA rks fuEu esa ls dkSulk lR; gSA (A) T1 = T2 (B) T1 > T2 (C*) T1 < T2 (D) dqN ugha dgk tk ldrkA 1000 g ty

Sol.

Sol. 5.

2KI + Hg2  K2[Hg4].1. 0.1 mole of which of the following when added to 1 L water, the aqueous solution obtained will have the lowest freeing point? 1L ty esa fuEu ls fdldk 0.1 eksy feykus ij izkIr foy;u dk fgekad U;wure gksxkA (A) C6H5NH3Cl (B) K3[Fe(CN)6] (C*) K4[Fe(CN)6] (D) Al(NO3)3. Greater the number of ions greater will be lowering in freezing point. Which of the following compound has bond angles of 120° only, either in cationic or anionic species ? fuEu esa ls fdl ;kSfxd esa ;k rks /kuk;fud ;k _.kk;fud Lih'kht+ esa] cU/k dks.k 120° gS \ (A*) N2O5 (s) (B) PCl5 (s) (C) I2Cl6 () (D) XeF6(s)

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PAGE NO.- 9

Sol.

N2O5(s) —

PCl5(s) — XeF6(s) — I2Cl6(l) —

6.

Sol.

NO+2 sp PCl4+ sp3 XeF+5 sp3d2 ICl2+ sp3

+

+ + +

NO3–

(trigonal planar, so bond angle 120°) (f=kdks.kh; leryh;] vr% cU/k dks.k 120°)

sp2 PCl6– sp3 d2 F– ICl4– sp3d2

Consider the following statements with regard to magnesium sulphate : 1. It is sparingly soluble in water. 2. Its solubility in water is less than calcium sulphate. 3. When heated to 900 °C, it gives magnesium oxide. Which of the above statements is/are correct? (A) 1 and 3 (B) 2 only (C*) 3 only (D) 2 and 3 MgSO4 is more soluble than CaSO 4 in water and MgSO4 gives MgO on heating to 900ºC.

fuEu dFkuksa dks eSXuhf'k;e lYQsV ds lanHkZ esa voyksdu dhft,& 1. ;g ty esa vYi foys; gksrk gSA 2. bldh ty esa foys;rk dSfY'k;e lYQsV ls de gksrh gSA 3. tc bls 900°C ij xeZ djrs gS rks ;g eSfXuf'k;e vkWDlkbM nsrk gSA fuEu esa ls dkSulk dFku lgh gS@gaS& (A) 1 rFkk 3 (B) dsoy 2 (C*) dsoy 3 Sol.

7.

(D) 2 rFkk 3 MgSO4 is more soluble than CaSO 4 in water and MgSO4 gives MgO on heating to 900ºC. MgSO4 ty esa CaSO4 ls vfèkd foys; gksrk gS o MgSO4 900°C ij xeZ djus ij MgO nsrk gSA An ideal gas at initial pressure Pi and volume Vi undergoes reversible expansion to the same volume V f either isothermally or adiabatically. Which of the following statement(s) is (are) correct ? Where the symbols have their usual meaning. (A*) |Pf (adiabatic) < Pf (isothermal)| (B*) |W(adiabatic)| < |W(isothermal)| (C*) |Tf (adiabatic) < Tf (isothermal)| (D*) |q(adiabatic)| < |q(isothermal)|

izkjfEHkd nkc Pi vk;ru Vi ij ,d vkn'kZ xSl ;k rks lerkih; ;k :)ks"eh; ij leku vk;ru Vf ds lkFk mRØe.kh; :i ls izlkfjr gksrs gSA fuEufyf[kr dFkuksa esa ls dkSu dFku lgh gS@gSa \ tgk¡ izrhd Loa; dk lkekU; vFkZ j[krs gSA (A*) |Pf (:)ks"eh;) < Pf (lerkih;)| (B*) |W(:)ks"eh;)| < |W(lerkih;)| (C*) |Tf (:)ks"eh;) < Tf (lerkih;)| (D*) |q(:)ks"eh;)| < |q(lerkih;)|

Sol.

bls vkjs[k }kjk Li"V fd;k tkrk gSA |Pf (:)ks"eh;) < Pf (lerkih;)| |W(:)ks"eh;)| < |W(lerkih;)| |Tf (:)ks"eh;) < Tf (lerkih;)| |q(:)ks"eh;)| < |q(lerkih;)|

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PAGE NO.- 10

8.

For free expansion of an ideal gas (expansion against vacuum) adiabatically, which of the following will have zero value :

,d vkn'kZ xSl ds :)ks"e :i ls eqDr izlkj ds fy, ¼fuokZr ds fo:) izlkj½] fuEu esa ls fdu jkf'k;k¡ dk eku 'kwU; gksxk% Sol.

Sol.

(A*) W (B*) q (C*) U Adiabatic process. So, q = 0. Expansion against vacuum. So, Pext = 0. Therefore, W = 0. So, from Ist law, U = 0. So, T = 0. So, H = 0. Ssys > 0, since expansion of gas occurs. :)ks"e izØe%] vr%] q = 0.

(D*) H

fuokZr ds fo:) izlkj] blfy, Pext = 0. vr%] W = 0. blfy,] izFke fu;e ls, U = 0. blfy,, T = 0. blfy,, H = 0. Ssys > 0, pwafd xSl dk izlkj gksrk gSA 9.

Which is the correct relation between osmotic pressure of 0.1 M NaCl solution & 0.1 M Na2SO4 solution : 0.1 M NaCl foy;u rFkk 0.1 M Na2SO4 foy;u ds ijklj.k nkc ds chp lgh lEcU/k dkSulk gS& (A) the osmotic pressure of Na2SO4 is less than NaCl solution (B*) the osmotic pressure of Na2SO4 is more than NaCl solution (C*) the osmotic pressure of Na2SO4 is 1.5 times that of NaCl solution (D) the osmotic pressure of NaCl is 1.5 times that of Na2SO4 solution (A) Na2SO4 dk ijklj.k nkc] NaCl foy;u dh vis{kk de gksrk gSA (B*) Na2SO4 dk ijklj.k nkc] NaCl foy;u dh vis{kk vf/kd gksrk gSA (C*) Na2SO4 dk ijklj.k nkc] NaCl foy;u dh vis{kk 1.5 xquk gksrk gSA (D) NaCl dk ijklj.k nkc] Na2SO4 foy;u dh vis{kk 1.5 xquk gksrk gSA

10.

For the depression of freezing point experiment involving non-volatile solute. The correct statement(s) is/are: (A*) Vapour pressure of pure solvent is more than that of solution (B) Vapour pressure of pure solvent is less than that of solution (C) Only solute molecules solidify at the freezing point (D*) Only solvent molecules solidify at the freezing point fgekad fcUnq ds vou;u iz;ksx ds fy, , fuEu esa ls dkSulk@dkSuls dFku lgh gS /gSa : (A*) 'kq ) foyk;d dk ok"inkc ,foy;u fd rq yuk esa vf/kd gksr k gS A (B) 'kq ) foyk;d dk ok"inkc ,foy;u fd rq yuk esa de gksr k gSA (C) fgeka d fcUnq ij dso y foy; ds v.kq] Bks l esa ifjofrZ r gks rs gSA (D*) fgekad fcUnq ij ds oy foyk;d ds v.kq Bks l esa ifjofrZ r gks rs gSA From Roult’s Law.

Sol.

jkmYV fu;e ls 11.

Ans. Sol.

12.

20g of a binary electrolyte (molecular weight = 100) are dissolved in 500 g of water. The freezing point of the solution is – 0.74°C ; Kf = 1.86 K moIality–1, Calculate percentage degree of dissociation of electrolyte. 500g ty esa ,d f}yd oS|qr vi?kV~; ds 20g (vkf.od Hkkj = 100) dks ?kksyk tkrk gSaA foy;u dk fgekad – 0.74°C ; Kf = 1.86 K eksyjrk–1 gSA oS|qr vi?kV~; dk izfr'kr fo;kstu Kkr djksA 0 Tf = 2Kf m 0.74 = 2 × 1.36 × 0.4  2 = 0.9945  1 i = 1 +   1   0 Consider the following trends in properties of hydrogen halides : 1. Acidic strength : HF < HCl < HBr < HI 2. Thermal stability : HF > HCl > HBr > HI 3. Dipole moment : HF < HCl < HBr < HI 4. Reducing power : HF > HCl > HBr > HI Find number of correct trends.

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PAGE NO.- 11

gkbMªkstu gSykbM ds xq.kèkeksZ esa fuEu izo`fÙk dk voyksdu dhft,& 1. vEyh; lkeF;Z : HF < HCl < HBr < HI 2. rkih; LFkkf;Ro : HF > HCl > HBr > HI 3. f}èkzqo vkèkw.kZ : HF < HCl < HBr < HI 4. vipk;d {kerk : HF > HCl > HBr > HI lgh izo`fr dh la[;k Kkr djks A Ans.

2

13.

A membrane permeable only to water separates a 0.01 M solution of sucrose from a 0.001 M one. On which solution must pressure be applied to bring the system into equilibrium? Find this pressure if the T= 300 K. Report your answer as (pressure × 100) dsoy ty ds fy, ikjxE; ,d f>Yyh] ,d 0.001 M lqØksl ds tyh; foy;u dks 0.01 M foy;u ls i`Fkd djrh gS] buesa ls fdl foy;u ij nkc yxkuk pkfg, fd] fudk; lkE; esa vk tk;sA og nkc Kkr dhft;s] ;fn T = 298 K gksA vius mÙkj dks (nkc × 100) ds :i esa O;Dr djksA 22 Pressure will be applied to higher conc. side.  = P1 – P2 = (C1 – C2) × RT = (10–2 – 10–3) × 0.082 × 300 = 0.2214 atm. 0.22 × 100 = 22

Ans. Sol.

14. Ans. Sol.

Calculate the osmotic pressure of a decinormal solution of cane sugar at 27°C. 27°C rki ij xUus dh 'kdZjk ds MslhukWjey foy;u ds ijklj.k nkc dh x.kuk dhft;s \ 2.46 atm

n = molar concentration = 0.1 M V  we have P =

15.

Ans.

n RT V

Match the Column Column-I (A) 0.1 M Al2(SO4)3 (B) 0.1 M AlPO4 (C) 0.1 M urea. (D) 0.1 M MgCl2 LrEHk -I (A) 0.1 M Al2(SO4)3 (B) 0.1 M AlPO4 (C) 0.1 M urea. (D) 0.1 M MgCl2 [A – p,q,s] ; [B – q] ; [C – r] ; [D – q]



R = 0.082 litre atm K–1 mol–1 , T = 300 K



= 0.1 x 0.082 x 300 = 2.46 atm

(p) (q) (r) (s)

Column-II Solution with highest boiling point. Van't Hoff factor is greater than 1. Solution with lowest osmotic pressure. Solution with lowest freezing point. LrEHk -II

(p) (q) (r) (s)

mPpre DoFkukad dk foy;uA okWUV gkWQ dkjd 1 ls vf/kdA U;wure ijklj.k nkc dk foy;uA U;wure fgekad dk foy;uA

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PAGE NO.- 12

ChemINFO-1.4

Solution & Colligative Properties Immiscible Liquid

Daily Self-Study Dosage for mastering Chemistry

Immiscible Liquid When a volatile liquid is added to another volatile liquid, which are completely immiscible with each other, each liquid will behave independent of the other and will exert its own vapour pressure. The more dense liquid (A) will form lower layer and less dense liquid (B) is passed into liquid (A). The total vapour pressure of the system will be the sum of vapour pressure of both the volatile liquids.



Ptotal = PA + PB = PA0 + PB0 (as XA & XB = 1)

The boiling point of a liquid system is the temperature at which its vapour pressure becomes equal to the external atmospheric pressure. The total vapour pressure of a mixture containing immiscible volatile liquids is greater than either of the pure constituents, thus the mixture will boil at a temperature, which is lower than the boiling point of either of the pure constituents. This forms the basis of steam distillation, in which one of the volatile liquids is water. This composition of the vapour above the liquid mixture can be calculated using Dalton’s law of partial pressures. PA = PA0 = PTotal XA’

.......(i)

PB = PB0 = PTotal XB’

.......(ii)

XA' and XB' represents the mole fraction of A and B in the vapour phase.

PTotalXB ' PB  PTotalXA ' PAº

WB PB  MB  W A PA  MA

The masses of the constituents in the vapour (distillate) will be proportional to their respective molar mass of vapour pressure. Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

A mixture of immiscible liquids of an organic liquid and water boils at 90 0C and 732 torr pressure. The distillate contains 73% by weight of the organic liquid. Calcualte the molar mass of the organic liquid (vapour pressure of water at 900C is 526 torr.) (A) 168.2 g mol–1

17.

(B) 228.8 g mol–1

(C*) 122.2 g mol–1

(D) 62.7 g mol–1

When a mixture of water and chlorobenzene (mutually immiscible) is distilled at an external pressure of 740.2 mm Hg, the mixture is found to boil at 90.30C, at which temperature the vapor pressure of pure water is 350.1 mmHg. Calculate the per cent by weight of chlorobenzene in the disillate. (Molecular weight is 112). (A*) 0.71 (71%)

18.

(B) 0.48 (48%)

(C) 0.25 (25%)

(D) 0.86 (86%)

Two liquids X and Y are perfectly immisible. If X and Y have molecular masses in ratio 1 : 2, the total vapour pressure of a mixture of X and Y prepared in weight ratio 2 : 3 should be : Px0  400 torr , Py0  200 torr

(A*) 600 torr 19.

(B) 700 torr

(C) 850 torr

(D) 900 torr

An unknown compound is immiscible with water. It is steam distilled at 98.00C and P = 737 torr, P0(H2O) = 707 torr at 98.00C. This distillate was 75% by weight water. Calculate the molecular weight of the unknown. (A*) M = 141.4

(B) M = 92.1

(C) M = 126.2

(D) M = 186.8

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ChemINFO-1.4

Solution & Colligative Properties vfeJ.kh; nzo

Daily Self-Study Dosage for mastering Chemistry

vfeJ.kh; nzo tc ,d ok"i'khy nzo dks vU; ok"i'khy nzo esa feyk;k tkrk gS] tks ,d&nwljs ds lkFk iw.kZ :i ls vfeJ.kh; gS] izR;sd nzo vU; nzo ds lkFk Lora=k :i ls O;ogkj djrk gS o blds Lo;a dk Okk"i&nkc yxk;sxkA vf/kd ?kuRo okyk nzo (A) fupyh ijr cuk,xk o de ?kuRo okyk nzo (B) dks nzo (A) esa ls izokfgr djsaxsA rU=k ds dqy ok"inkc] nksuksa ok"i'khy nzoksa ds ok"inkc dk ;ksx gksxkA 

Pdqy = PA + PB = PA0 + PB0 (pwafd XA & XB = 1)

,d nzo dk DoFkukad og rkieku gS ftl ij bldk ok"i&nkc blds cká nkc ds cjkcj gksxkA vfeJ.kh; ok"i'khy nzo ;qDr ,d feJ.k dk dqy ok"i&nkc] buesa ls izR;sd 'kq) vo;o ls vf/kd gksrk gSA bl izdkj feJ.k ,d rkieku ij mcyrk gS tks fd buesa ls fdlh Hkh 'kq) vo;o ds DoFkukad ls de gksrk gSA ;g Hkkih; vklou dk vk/kkj cukrk gS] ftlesa ,d ok"i'khy nzo ty gSA nzo feJ.k ds Åij ok"i ds bl laxBu dks fuEu dks iz;qDr dj ifjdfyr fd;k tkrk gSA vkaf'kd nkcksa dk MkWYVu fu;e PA = PA0 = PTotal XA’

.......(i)

PB = PB0 = PTotal XB’

.......(ii)

XA' o XB' ;gka

ok"i izkoLFkk esa A o B ds eksy&fHkUu dks iznf'kZr djrs gSA

PTotalXB ' PB  PTotalX A ' PAº

WB PB  MB  WA PA  MA

ok"i ¼vklou½ esa vo;oksa dk nzO;eku] ok"inkc ds Øe'k% eksy nzO;eku ds lekuqikrh gksaxsA Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

900C

o 732 torr nkc ij ,d dkcZfud nzo o ty ds vfeJ.kh; nzo ds ,d feJ.k dks mckyk tkrk gSA vklou dkcZfud nzo ds Hkkj dk 73% j[krk gSA dkcZfud nzo dk eksyj&nzO;eku ifjdfyr dhft, (900C ij ty dk ok"inkc 526 torr gS) (A) 168.2 g mol–1

17.

(C*) 122.2 g mol–1

(D) 62.7 g mol–1

tc 740.2 mm Hg, ds ,d cká nkc ij ty o DyksjkscUs thu ds ,d feJ.k dks vklfor fd;k tkrk gS] rks feJ.k dks 90.30C ij mckyk x;k] ftl rkieku ij 'kq) ty dk ok"inkc 350.1 mmHg gSA vklou eas DyksjkcsUthu ds Hkkj dk izfr'kr ifjdfyr dhft,A( v.kqHkkj 112). (A*) 0.71 (71%)

18.

(B) 228.8 g mol–1

(B) 0.48 (48%)

(C) 0.25 (25%)

(D) 0.86 (86%)

nks nzo X o Y iw.kZ :i ls vfeJ.kh; gaSA ;fn X o Y ds fy, v.kqHkkj 1 : 2 vuqikr esa gS] rks Hkkjkuqikr 2 : 3 esa cuk;s x;s X o Y ds ,d feJ.k dk dqy ok"i nkc fuEu gksuk pkfg, : Px0  400 torr , Py0  200 torr

(A*) 600 torr 19.

(B) 700 torr

(C) 850 torr

(D) 900 torr

,d vKkr ;kSfxd ty ds lkFk vfeJ.kh; gSA bldk Hkkih; vklou 98.00C ij gS o P = 737 torr, P0(H2O) = 707 torr gSA ;g vklou ty ds Hkkj dk 75% gSA vKkr dk v.kqHkkj ifjdfyr dhft,A (A*) M = 141.4

(B) M = 92.1

(C) M = 126.2

(D) M = 186.8

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