DPP-59 to 60 With Answer Physical Chemistry

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 59 to 60 Class : XIII Course : DPP No.1 Total Marks : 27 Max. Time : 27 min. Single choice Objective (no negative marking) Q.1 to Q.5 (3 Marks, 3 Min.) [15, 15] Multiple choice objective (no negative marking) Q.6 to Q.7 (4 marks 4 min.) [8, 8] Subjective Questions (no negative marking) Q.8 (4 marks 4 min.) [4, 4] ANSWER KEY DPP No.-59 1. 4 2. 3 3. 3 4. 4 5. 4 6. 2,4 7. 2,3 8. 5.3. 1. Assuming the salts to be completely ionized in solution, which of the following has highest osmotic pressure. (1) 1% CsCl w/w (2) 1% RbCl w/w (3) 1% KCl w/w (4*) 1% NaCl w/w ekukfdyo.kfoy;uesaiw.kZr;kvk;furgkstkrsgSrcfuEuesafdldkijklj.knkcmPpregSA (1) 1%CsClHk j@Hkkj (2) 1%RbClHkkj@Hk j (3) 1%KClHkkj@Hkkj (4*) 1%NaClHk j@Hkkj 2. What weight of glucose dissolved in 100 grams of water will produce the same lowering of vapour pressure as one gram of urea dissolved in 50 grams of water, at the same temperature? lekurkiijtyds50xzkeesa],dxzke;wfj;kfoy;u}kjkiznf'kZrok"inkcesadehdslekudehdksmRiUudjus dsfy;s100xzketyesaXywdksldkfdrukHk jfoys;fd;ktkukpkfg,\ (1) 3 gms (2) 5 gms (3*) 6 gms (4) 4 gms Sol. Since same lowering in vapour pressure. X solute of urea = Xsolute of glucose. 1 60 w 180 1 60 w 180 1  50 60 18 = w 180  100 .  18 50 = 18 100 18 (For dilute solution).  w = 6 gm. of glucose. 3. 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 1kgty esa 'A' ds 'w' xzke] vU;1kgty esa 'B' ds 'w' xzke] rFkk vU; 1kgty esa 'C' ds 'w' xzkefeykdj cuk;stkrsgaSA(A,B,Cfo|qrvu~vi?kV~;gSa)Abufoy;uksals'kq"dok;qfuEuØe(A  B  C)esaizokfgr dhtkrhgSfoy;uA dsnzO;ekuesadeh2 xzkeik;hx;htcfdB dk0.5 xzkec<+rkgSrFkkfoy;uCdsnzO;eku esa1 xzkedhdehgksrhgSaAA,B rFk C dseksyjnzO;ekuksadse/;lEca/kgksxkA (1) M : M : M = 4 : 3 : 5 (2) M : M : M = 1 : 1 : 1 A B C A B C 4 3 5 (3*) MC > MA > MB (4) MB > MA > MC Sol. The loss in weight should be proportional to vapour pressure above that solution : So, PS  2gm ; P  1.5gm B P  2.5 gm C 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"inkc]Hk jesagkfudslekuqikrhgksxk PS  2gm ; PS  1.5gm ; P  2.5 gm C pwafdCfoy;udkok"inkcvf/kdregS]vr%blesaU;wuredehvk;hgksxhblfy,bldkeksyizHk tU;wuregksxk rFk blfy,bldkv.kqHk jvf/kdregksxkA 4. 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 ; K = 1.86 K moIality–1, The degree of dissociation of electrolyte is 500gtyesa,df}ydoS|qrvi?kV~; ds20g(vkf.od Hkkj=100)dks?kksyktkrkgSaAfoy;udkfgekad–0.74°C; K =1.86Keksy rk–1,oS|qrvi?kV~;dsfo;kstudhek=k fuEugS& (1) 50% (2) 75% (3) 100% (4*) Zero Sol. T = iK m f f 0.74 = i × 1.36 × 0.4 ⇒ i = 0.9945  1 i = 1 +   1 ⇒   0 5. 2.56g of sulfur in 100g of CS has depression in freezing point of 0.010C.K = 0.10molal1. Hence, the atomicity 2 f of sulfur in CS2 is 100g,CS esa2.56glYQj0.010Cdkfgekadesavoueuj[krkgSAK =0.10eksyy1Avr%CS esalYQjdhijek.kqdrk 2 f 2 fuEugS& (1) 2 (2) 4 (3) 6 (4*) 8 Sol. T = K m f f 10–2 = 0.1 × m ⇒ m = 0.1 m 2.56  1000 molality = M100 256 ⇒ M = 256 atomicity = 32 = 8. 6. In which of the following pairs of solutions will the values of the vant Hoff factor be the same? (1) 0.05 M K4 [Fe(CN)6] and 0.10 M FeSO4 (2*) 0.10 M K4[Fe(CN)6] and 0.05 M FeSO4 (NH4)2SO4. 6H2O (3) 0.20 M NaCl and 0.10 M BaCl2 (4*) 0.05 M FeSO4 (NH4)2SO4 . 6H2O and 0.02 M KCl . MgCl2 . 6H2O fuEuesalsdkSulsfoy;u;qXeksadsfy,okUVgkWQdkjd(factor)dkekulekugksxkA (1) 0.05 M K4 [Fe(CN)6] rFkk 0.10 M FeSO4 (2*) 0.10 M K4[Fe(CN)6] rFkk 0.05 M FeSO4 (NH4)2SO4. 6H2O (3) 0.20 M NaCl rFkk0.10 M BaCl2 (4*) 0.05 M FeSO4 (NH4)2SO4 . 6H2O rFkk 0.02 M KCl . MgCl2 . 6H2O. 7. Which is the correct relation between osmotic pressure of 0.1 M NaCl solution & 0.1 M Na2SO4 solution : (1) the osmotic pressure of Na2SO4 is less than NaCl solution (2*) the osmotic pressure of Na2SO4 is more than NaCl solution (3*) the osmotic pressure of Na2SO4 is 1.5 times that of NaCl solution (4) the osmotic pressure of NaCl is 1.5 times that of Na2SO4 solution 0.1MNaClfoy;u rFkk 0.1M Na2SO4 foy;u ds ijklj.k nkc ds chp lgh lEcU/k dkSulk gS& (1)Na2SO4 dk ijklj.knkc] NaClfoy;udh vis{kkde gksrkgSA (2*)Na2SO4 dkijklj.knkc]NaClfoy;udhvis{kkvf/kdgksrkgSA (3*)Na2SO4 dkijklj.knkc]NaClfoy;udhvis{kk1.5xqukgksrkgSA (4)NaCldkijklj.knkc]Na2SO4foy;udhvis{kk1.5xqukgksrkgSA 8. At 12°C the osmotic pressure of a urea solution is 500 mm. The solution is diluted and the temperature is raised to 27°C, when the osmotic pressure is found to be 100 mm. determine the extent of dilution. 12°Crkiij,d;wfj;kdsfoy;udkijklj.knkc500mmgSAtcfoy;udksruqfd;ktkrkgSrFk rkiekudks27°C, rdc<+k;ktkrkgSAtcfoy;udkijklj.knkc100mmik;ktkrkgSrksfoy;udhruqrkdhijkl(extentofdilution) dkfu/kZj.kdhft;s\ Sol. Suppose V1 titres of the solution contains n moles of the solute at 12°C which was diluted to V2 litres at 27°C. Thus we have 500 760 = n V1 x 0.082 x 285 ...(i) 100 and 760 = n V2 x 0.082 x 300 V2 Dividing (1) by (2), we get V1 = 5.3. DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 60 Class : XIII Course : DPP No.2 Total Marks : 29 Max. Time : 29 min. Single choice Objective (no negative marking) Q.1 to Q.3 (3 Marks, 3 Min.) [9, 9] Multiple choice objective (no negative marking) Q.4 to Q.5 (4 marks 4 min.) [8, 8] Subjective Questions (no negative marking) Q.6 to Q.7 (4 marks 4 min.) [8, 8] Match the Following (no negative marking) Q.8 (4 marks 4 min.) [4, 4] 1. 1g of arsenic dissolved in 86 g of benzene brings down the freezing point to 5.31 °C from 5.50 °C. If Kf of C benzene is 4.9 m , the atomicity of the molecule is : (As – 75) 1gvklsZfuddks86gcsUthuesafoys;djusijfgekadfcUnq5.50°Cls5.31°CrdfxjtkrkgSA;fncsUthudkKf , C 4.9 m gSrc v.kqdh ijek.kqdrkgS (As–75) (1) 8 (2) 2 (3) 3 (4*) 4 Sol. T = K m f f 0.19 = 4.9 × 1 M 86 . 1000 1000 0.19 = 4.9 × M 86 .  M = 4.9  1000 86  0.19 300 = 300. Atomicity = 75 = 4. 2. Liquids A and B form an ideal solution. At a certain temperature the total vapour pressure of a mixture of A and B is 400 mm. The mole fraction of A in the liquid mixture in equilibrium with the vapour phase is 0.4. If the vapour pressure ratio (P 0/P 0) for the pure liquids at this temperature is 1/6, what is the partial pressure of A A B in the vapour phase? ArFkk Bdk nzo ,d vkn'kZ foy;u cukrk gSA ,d fuf'pr rki ij] feJ.k ArFkk Bdk dqy ok"i nkc 400mm. gSAok"ivoLFk dslkE;esanzofeJ.kesaAdkeksyizHk t0.4gSA;fnblhrkiij'kq)nzodsfy,ok"inkcvuqikr (P 0/P 0)1/6gSAok"ivoLFkkesaAdkvkaf'kdnkcfuEugSA A B (1) 50mm (2) 60mm (3) 70mm (4*) 40mm 3. Mixture of volatile components A and B has total vapour pressure (in Torr) p = 254 – 119 xA where xA is mole fraction of A in mixture. Hence p0 and p0 are (in Torr) ok"i'khy?kVdArFkkBdsfeJ.kdkdqyok"inkc(Vksjesa)p=254–119xAgStgk¡xAfeJ.kesaAdkeksyizHk tgS blfy, p0 rFk p0 (Vksjesa)gSA (1) 254, 119 (2) 119, 254 (3*) 135, 254 (4) 119, 373 Sol. P = X P 0 + X P 0 = (P 0 – P 0)X + P 0 A A B B A B A B So P 0 = 254 P 0 – P 0 = –119 P 0 = 135 A B A 4. Which of the following are true for ideal solutions : (1*) Vmix = 0 (2*) Hmix = 0 (3) Smix = 0 (4) Gmix = 0 (5*) Raoult's law is obeyed for entire concentration range and temperatures. vkn'kZ foy;u ds fy, fuEu esa ls dkSulk lR; gSA (1*) V mix = 0 (2*) H = 0 (3) S mix = 0 (4) G mix = 0 (5*)lEiw.kZlkUnzrkijklrFk rkidsfy,jkÅYVfu;edkikyufd;ktkrkgSA 5. Consider the following solution in water (I) 1 M Sucorse (II) 1 M KCl (III) 1M Benzoic acid (IV) 1 M (NH4)3PO4 Which of the following is/are true : (1) All solutions are isotonic (2) III is hypertonic of II, IV (3*) I, II and III are hypotonic of IV (4*) IV is hypertonic of I, II and III fuEutyh;foy;uksaijfopkjdhft;s% (I) 1MlqØksl (II) 1 M KCl (III)1McsUtksbdvEy (IV) 1 M (NH4)3PO4 fuEuesalsdkSulk@dkSulsdFkulR;gS: (1) lHkhfoy;uleijklj.kh;gSA (2)II,IVdhrqyukesaIIIvfrijklj.kh;gSA (3*)I,IIrFkkIIIfoy;u,IVdhrqyukesafuEuijklj.kh;gSA (4*)IVfoy;u,I,IIrFkkIIIdhrqyuk esavfrijklj.kh; gSA Sol. Osmotic pressure is in order (ijklj.k nkc dk Øe) I < III < II < IV 6. 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. dsoytyds fy,ikjxE; ,df>Yyh],d 0.001MlqØksl dstyh; foy;udks0.01Mfoy;uls i`FkddjrhgS]buesa lsfdl foy;uij nkcyxkuk pkfg,fd] fudk;lkE; esavk tk;s];fnT=300K rks ognkcKkrdhft;sA . Ans. 0.22 atm. Sol. Pressure will be applied to higher conc. side.  = P – P2 = (C1 – C ) × RT = (10–2 – 10–3) × 0.082 × 300 = 0.22 atm. 7. The osmotic pressure of blood is 7 atm at 30°C. What is the molarity of the isotonic saline solution if the ‘i’ factor for sodium chloride is taken to be 1.9? 30°CijjDrdkijklj.knkc7atmgksrkgSAleijkljh{k jh;(saline) foy;udheksyjrkD;kgS\;fnblesalksfM;e DyksjkbMdsfy,‘i’xq.k ad1.9fy;ktkrkgSA Ans. 0.148 M. Sol. For isotonic solution. =  7 = 1.9 × M × 0.082 × 303 = 0.148 M. 3 2 3 4 4 6 Informaiton About Commercial Cell Primary cell : Redox reaction occurs only once. These cell cannot be reacharged. Dry Cell Mercury Cell Anode Zn container Zn-Hg amalgam as anode. Cathode Carbon rod (graphite) Paste of Hg and carbon. Electrolyte Paste of NH4Cl and ZnCl2 Paste of KOH and ZnO. Reaction at anode Zn  Zn2+ + 2e– Zn (Hg) + 2OH–  ZnO +H2O + 2e– Reaction at cathode MnO2 + NH4+ + e–  MnO(OH) + NH3 HgO + H2O + 2e–  Hg(l) + 2OH– Cell potential remain constant during its life. Secondary Batteries : This type of cell can be reacharged by passing current through it in the opposite direction so that it can be used again. Lead storage battery Ni—Cd cell Anode Lead Cadmium Cathode Grid of lead packed with lead dioxide (PbO2) Metal grid containing NiO2 Electrolyte 38% solution of H2SO4 KOH solution Reaction at anode 2– – Pb + SO4  PbSO4 + 2e Cd(s) + 2OH–  Cd(OH)2(s) + 2e– Reaction at cathode 2– + – PbO2 + SO4 + 4H + 2e  PbSO4 + 2H2O NiO2 + 2H2O + 2e–  Ni(OH)2(s) + 2OH– Fuel Cell : Fuel Celll Anode H2 is passed through porous carbon electrode containing suitable catalyst (generally finely divided Pt and Pd). Cathode O2 is passed through porous carbon electrode containing suitable catalyst. Electrolyte KOH or NaOH solution. Reaction at anode 2H2 + 4OH–  4H2O + 4e– Reaction at cathode O2 + 2H2O + 4e–  4OH– Over all reaction 2H2(g) + O2(g)  2H2O (l) Corrosion : The process of slowly eating away of the metal due to attack of atmoshperic gases on the surface of the metal resulting into the formation of compound such as oxide, sulphides, carbonates, sulphates etc. is called corrosion. Oxidation : Fe (s)  Fe2+ (aq) + 2e– Reduction : O2 (g) + 4H+ (aq) + 4e–  2H2O (l) Atmospheric oxidation : 2Fe2+ (aq) + 2H O (l) + 1 (g)  Fe O 2 2 3 (s) + 4H+ (aq) Prevention of corrosion :  Converting the surface with paint or by some chemical (bisphenol).  Cover the surface by other metal (Sn, Zn).  On electrochemical method is to provide a sacrificial electrode of another metal (Like, Mg, Zn etc.) Which corrodes itself but saves the object. O;olkf;dlSydsckjsesatkudkjh çkFkfedlSy:mikip;vfHkfØ;k,¡dsoy,dckjgksrhgSA;glSyfujkosf'krughagksldrkgSA Dry Ce ll ¼' kq"d l Say ½ Mercury Cell ¼edZj hl Say ½ , uksM Zn i k=k , uksM i j Zn-Hg vey xe dSFkksM d kcZu NM+¼xzsQkbV½ Hg o d kcZu dk y si oS| qr vi ?kV~; NH4Cl o ZnCl2 d k y si KOH o ZnO d k y si , uksM i j v fHkfØ; k Zn  Zn2+ + 2e– Zn (Hg) + 2OH–  ZnO +H2O + 2e– dSFkksM i j v fHkfØ; k MnO2 + NH4+ + e–  MnO(OH) + NH3 HgO + H2O + 2e–  Hg(l) + 2OH– ¼bl d sl e; v Ur j ky d snkSj ku l sy foHko fu; r j gr k gS½ f}rh;dcSVjh:lSydsblçdkjdkfujkos'kufoifjrfn'k esa/k jkçokfgrdjfd;ktkldrkgSaAblfy,blsiqu%ç;qDr fd;ktkldrkgSA l hl kl apk; d cSVj h Ni—Cd l Sy , uksM l hl k d SMfe; e dSFkksM y sM MkbZvkWDl kbM (PbO2) dsl kFk l adqfy r y sM dk fxzM ¼t ky h½ /kkr qfxzM ; qDr NiO2 oS| qr vi ?kV~; H2SO4 dk38% foy ; u KOH foy ; u , uksM i j vfHkfØ; k Pb + SO42–  PbSO4 + 2e– Cd(s) + 2OH–  Cd(OH)2(s) + 2e– dSFkksM i j vfHkfØ; k PbO2 + SO42– + 4H+ + 2e–  PbSO4 + 2H2O NiO2 + 2H2O + 2e–  Ni(OH)2(s) + 2OH– b±/kulSy% b±/ku l Sy , uksM H2 dksmi ; qDr mRçsj d ; qDr ¼l kekU; r %vPNhr j g pwf.kZr Pt r Fkk Pd) l aj /kzhdkcZu by sDVªkWM esal sçokfgr fd; k t kr k gSA dSFkksM O2 dksmi ; qDr mRçsj d ; qDr l aj /kzhdkcZu by sDVªkWM esal sçokfgr fd; k t kr k gSA oS| qr vi ?kV~; KOH vFkokNaOH foy ; u , uksM i j vfHkfØ; k 2H2 + 4OH–  4H2O + 4e– dSFkksM i j vfHkfØ; k O2 + 2H2O + 4e–  4OH– l Ei w. kZvfHkfØ; k 2H2(g) + O2(g)  2H2O (l) la{k j.k:/k rqdsi`"Bijok;qe.Myh;xSlksadsçHk odsdkj.k/k rqdk/khjs&/khjslekIrgksusdkçØeftldsQyLo:i vkWDlkbMksa]lYQkbMksa]dkcksZusVksa]lYQsVksadkfuekZ.kgksrkgS]la{kj.kdgykrkgSA vkWDlhdj.k: Fe (s)  Fe2+ (aq) + 2e– vip;u : O (g) + 4H+ (aq) + 4e–  2H O (l) ok;qe.Myh;vkWDlhdj.k% 2Fe2+ (aq) + 2H O (l) + 1 (g)  Fe O 2 2 3 (s) + 4H+ (aq) la{kkj.k dk laj{k.k(Preventionofcorrosion) : isUVvFkokdqNjlk;u¼fcl~fQukWy½dslkFklrgdkscnyukA vU;/kkrq(Sn,Zn)}kjklrgdks<+dukA oS|qrjklk;fudfof/kijvU;/k rq¼tSlsMg,ZnbR;kfn)ds,dç;qDr(sacrificial) bysDVªkWMdksçnkudjuktksLoa; dksla{kfjrdjrkgksaysfduvU;dkscpkrkgkasA