DPP-45 to 46 Physical Answer

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 45 to 46 Class : XIII Course : DPP No.1 Total Marks : 35 Max. Time : 35 min. Single choice Objective (no negative marking) Q.1 to Q.4 (3 marks 3 min.) [12, 12] Comprehension ('–1' negative marking) Q.5 to Q.9 (3 marks 3 min.) [15, 15] Subjective Questisons (no negative marking) Q.10 to Q.11 (4 marks 4 min.) [8, 8] 1. The ionic product of water at 45 ºC is 4  1014. What is pH of pure water at this temperature. [Take : log 2 = 0.3] 45ºCij ty dk vk;fud xq.kuQy 4  1014 gSA bl rki ij 'kq) ty dk pHD;k gS \ [Take : log 2 = 0.3] (1*) 6.7 (2) 7 (3) 7.3 (4) 13.4 2. For which temperature the pKw of pure water can be greater than 14. fdl rki ij 'kq) tyds pKw dk eku 14ls T;knk gks ldrk gS \ (1*) 20 ºC (2) 30 ºC (3) 40 ºC (4) 50 ºC 3. For pure water at 10 ºC and 60 ºC , the correct statement is 10ºCrFkk60ºCij 'kq) ty ds fy, lgh dFku gSA (1) pOH10ºC = pOH60ºC (2*) pOH10ºC > pOH60ºC (3) pOH60ºC > pOH10ºC (4)Can'tsay ugha dgk tk ldrk gSA 4. For pure water at 25 ºC and 50 ºC the correct statement is 25ºCrFkk50ºCij 'kq) ty ds fy, lgh dFku gS \ (1) pH25ºC = pH50ºC (2*) pH25ºC > pH50ºC (3) pH50ºC > pH25ºC (4) Can't say ugha dgk tk ldrk gSA Passage : (Q.5 to Q.9) Relative strengths of conjugate acid base pairs : HCIO4  CIO4 –  very weak bases HCI   Strong acids CI–  Negligible tendency Stronger acid  2 4  100% dissociated in aqueous solution HSO4 –  to be protonated in Weaker base HNO3  NO3 –  aqueous solution H O+ H O 3   2 HSO4  H3PO4   SO2   H PO  HNO2  HF   2 4 NO  F  CH3COOH  H2CO3  CH3CO  Weak bases H2S NH  Weak acids  Exist in solution as  a mixture of HA HCO   HS   Moderate tendency to be protonated HCN  HCO   H2O  A ,and H3O NH3 CN CO2 OH  in aqueous solution       NH3    Very Weak acids NH  2  Strong bases 100% Weaker acid OH  Negligibletendency O2  protonated in Strong base H2  vuqPNsn % (Q.5lsQ.9) to dissociate.   H  aqueous solution la;qXehvEy{k j;qXeksadhvkisf{kdlkeF;Z% HCIO4   izcyvEy CIO4 – CI–   cgqr nqc{Zkkyj izcy vEy HCI  tyh; foy; u esa HSO –  tyh; foy;u esizaksVdk`sr nqcZyre {kkj H2SO4  100% fo; ksftr 4 NO3 –  gksduhs izo`kfÙux.;  HNO3  H O+ H O 3 HSO  H3PO4  HNO2  HF   2 SO2   H2PO  NO  F  CH3COOH CH CO  H2CO3  nqcZvEy y  foy; u esHaA] 3 HCO 2  nqcZ{ykkj tyh; foy; u H2S  A  rFkHk O ds 3  esizaksVdk`srughksduhs NH  3 HS  4 HCN  feJ.k ds:  i esamifLFkr gsrkgs NH3  lkekU; iz`ofÙk   HCO   H2O  CN  CO2  OH  NH3   cgqr nqrceZyvEy NH  2  OH  fo; ksftrgksduhs O2  izcy {kkj tyh; foy; u nqcZyrevEy  2  izo`fÙk.;ux   H  es1a00% izksVk`srughkdsgr izcy {kkj 5. Account for the acidic properties of nitrous acid in terms of (i) Arrhenius theory and (ii) Bronsted Lowry theory ukbVªlvEydsvEyh;xq.kdksfuEudslUnHkZesale>kb;s& (i)vkjgsfu;lfl)kUrrFk (i) czkWUlVsMyksjhfl)kUr Sol. (i) HNO2 (aq) H (aq) + NO (aq) + – Since, HNO2 ionises to give H+ hence, it is a Arrhenius acid. (ii) HNO (l) + H O(aq) H O+ (aq) + NO –(aq) 2 2 3 2 acid 1 base 2 acid 2 base 1 HNO donates a proton hence, it is an acid and changes to NO – (conjugate base). H O(l) accepts the proton 2 2 2 hence, it is a base and change to H O+ (conjugate acid). Thus HNO is a Bronsted Lowry acid. 6. Write a balanced equation for the dissociation of each of the following Bronsted Lowry acids in water. (1) H SO (2) H O+ (3) HSO – 2 4 3 4 Also write conjugate base of the acid tyesafuEuczkWUlVsMyksjhvEyksadsfo;kstudsfy,lUrqfyrjklk;fudvfHkfØ;kfyf[k;s& (1) H SO (2) H O+ (3) HSO – 2 4 3 4 vEyksadsla;qXeh{kjHkhfyf[k;sA Sol. (1) H SO + H O H O+ + HSO – 2 4 2 3 4 acid1 base2 acid2 base1 (conjugatete) 7. Which of the following reactions proceeds to the right and which proceeds to the left if you mix equal concentrations of reactants and products ? fuEuesalsdkSulhvfHkfØ;k,anka;hvksjrFk dkSulhvfHkfØ;k,acka;hvksjesaizxfrdjrhgS];fnvkivfHkdkjdomRikn nksuksadhlekulkUnzrk,¡feykrsgSA (1) HF(aq) + NO –(aq) HNO (aq) + F– (aq) (2) NH + (aq) + CO 2– (aq) HCO –(aq) + NH (aq) 3 3 Ans. (1) to the left (2) to the right 4 3 3 3 Reaction proceed to words the side favoring formation of weak acid and weak base Ans. (1)cka;hvksj (2)nka;hvksj Hint: Reaction proceed to words the side favoring formation of weak acid and weak base 8. Which of the following species behave as a strong acids or as strong base in aqueous solutions ? fuEuesadkSulhLih'khttyh;foy;uesaizcyvEyvFkokizcy{k jdslekuO;ogkjdjrhgSa\ (a) HNO (b) HNO (c) NH + (d) Cl– (e) H– (f) O2– (g) H SO 2 3 4 2 4 9. Consider following reactions : fuEuvfHkfØ;kvksaijfopkjdhft,% (1) H CO (aq) + HSO– (aq) H SO (aq) + HCO – (aq) 2 3 4 2 4 3 (2) HF (aq) + Cl¯ (aq) HCl (aq) + F¯ (aq) (3) HF (aq) + NH (aq) NH + + F¯ (aq) (4) HSO – (aq) + CN¯ (aq) HCN (aq) + SO2– (aq) 4 4 Reactions proceeding to the right are : nka;hvksjizxfrdjusokyhvfHkfØ;k,¡gSA (1) a, b (2*) c, d (3) a, c (4) b, d Sol. (1) H CO (aq) + HSO– (aq) H SO (aq) + HCO – (aq) 2 3 4 2 4 3 W.A. W.B. S.A. S.B. Reaction will go backward. (2) HF (aq) + Cl¯ (aq) HCl (aq) + F¯ (aq) W.A. W.B. S.A. S.B. Reaction will go backward (3) HF (aq) + NH (aq) NH + (aq) + F¯ (aq) 3 4 S.A. S.B. W.A. W.B. Reaction will go forward (4) HSO – (aq) + CN¯ (aq) HCN (aq) + SO2– (aq) 4 4 S.A. S.B. W.A. W.B. Reaction will go forward Reaction will favour weak acid and eak base. Integer Answer Type This section contains 2 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9. iw.k±dmÙkjizdkj bl[k.Mesa2iz'ugSaAizR;sdiz'udkmÙkj0ls9 rdiw.k ±dgSA 10. The equilibrium N2(g) + 3H2(g) 2NH3(g) is established in a closed container by initially taking only NH3. Find out the number of moles of H2 present in 102 g of the equilibrium mixture if the molar mass of the equilibrium mixture is observed to be 51 g/mole. 4 ,dcanik=kesaizkjEHkesadsoy]NH3dksysdj]lkE;N2(g)+3H2(g) 2NH3(g)dksLFk firfd;kx;kAlkE;feJ.k ds102gesamifLFkrH dseksyksadhla[;kKkrdhft,;fnfeJ.kdkeksyjnzO;eku 51 g/moleizsf{krfd;kx;kA Ans. 3 Sol. M obs = 2 4 MTh 1 (n  1)  51 = 4 17 1 (2  1)   = 1 3 Mass of eq. mixture = Initial mass = 102 g lkE;feJ.kdknzO;eku=izkjfEHkdnzO;eku=102g  izkjEHkesafy;sx;s nNH3 = 102 =6moles 17 N2(g) + 3H2(g) 2NH3(g) t = 0 6 moles eq. 6(/2) 6(3/2) 6(1 – )  3  1  nH2 present = 6    = 9 = 9 × 2  = 3 moles 11. For the following equilibrium PCl5(g) PCl3(g) + Cl2(g), vapour density at equilibrium is found to be 100. Initially 1 mole of PCl5 is taken in 12.82 lt. flask at 27ºC. Calculate equilibrium pressure (in atm) of the system. fuEulkE; PCl5(g) PCl3(g)+Cl2(g)dsfy,]lkE;ij ok"i?kuRo 100ik;k x;kAizkjEHk esa]tc 27ºCrkiij 12.82yhVjdsik=kesaPCl5ds1eksyfy,tkrsgSArcra=kdslkE;nkcdhx.kuk(atmesa)dhft,A Ans. 2 Sol. Vapour density of PCl5 (D) = 208.5 2 = 104.25 So,  = D  d d(n  1) 104.25  100 = 100 (2  1) = 4.25 × 10–2 Total moles at equilibrium = 1 +  = 1 + 0.0425 = 1.0425  Equilibrium pressure = nRT 1.0425  0.082  300 = = 2 atm V gy- PCl (D) dk ok"i ?kuRo = 208.5 = 104.25 12.82 5 blfy,,  = D  d d(n  1) 2 104.25  100 = 100 (2  1) = 4.25 × 10–2 lkE; ij dqy eksy = 1 +  = 1 + 0.0425 = 1.0425  lkE; nkc = nRT = 1.0425  0.082  300 = 2 atm. V 12.82 DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 46 Class : XIII Course : DPP No.2 Total Marks : 50 Max. Time : 50 min. Single choice Objective (no negative marking) Q.1 to Q.6 (3 marks 3 min.) [18, 18] Subjective Questisons (no negative marking) Q.7 to Q.14 (4 marks 4 min.) [32, 32] 1. At – 50ºC autoprotolysis of NH gives [NH+ ] = 1 × 10–15 M hence, autoprotolysis constant of NH is: 3 4 3 –50ºCrkiijNH Lor%izksVksuhd`rgksdj[NH+ ]=1×10–15 MnsrkgSAblizdkjveksfu;kdkLor%izksVksuhdj.kfLFkjkad gSA (1) 3 4 (2*) 1 × 10 –30 (3) 1 × 10–15 (4) 2 × 10–15 2. The self ionization constant for pure formic acid , K = [ HCOOH + ] [HCOO ] has been estimated as 10 6 at room temperature .The density of formic acid is 1.15 g/cm3. The percentage of formic acid converted to formate ion are : dejsdsrkiij'kq)QkfeZdvEydsfy,Lor%vk;u fLFkjkad10–6 vafdrfd;kx;kgSAQkfeZdvEydk?kuRo1.15 g/cm3 gSArksQkfeZdvEydkfdrukçfr'krQksesZVvk;uesaifjofrZrgksrkgS\K=[HCOOH +] [HCOO] (1) 0.002 % (2*) 0.004 % (3) 0.006 % (4) 0.008 %  3. What is the Kb of a weak base that can produce one OH per molecule if its 0.04 M solution is 2.5% ionized. ;fn0.04Mfoy;u2.5%vk;furgksrkgSrksçfrv.kqls,dOHmRiUugksrkgSnqcZy{k jdk K D;kgksxk\ (1) 7  108 (2) 1.6  106 (3*) 2.5  105 (4) 2  1011 4. [Cl¯ ] in a mixture of 200mL of 0.01 M HCl and 100 ml of 0.01 M BaCl2 is : 0.01M HCl ds 200mLrFkk 0.01M BaCl2ds 100mLfeJ.k esa [Cl¯ ] gS& (1) 0.01 M (2*) 0.0133 M (3) 0.03 M (4) 0.02 M 5. 10–2 mole of NaOH was added to 10 litre of water. The pH will change by 10yhVjtyesaNaOHds10–2 eksyfeyk;stkrsgSrcpHifjorZugksxk& (1*) 4 (2) 3 (3) 11 (4) 7 Sol. Initially pH = 7 finally [NaOH] = 10–3 so pOH = 3 pH = 11 so (pH) = 4 6. Blue litmus turns red in the following mixture of acid and base : fuEuvEyrFk {k jfeJ.kesalsfdlesauhykfyVelykyesaifjofrZrgkstkrkgS& (1) 100 mL of 1 × 10–2 M H SO + 100 mL of 1 × 10–2 M Ca (OH) (2) 100 mL of 1 × 10–2 M HCl + 100 mL of 1 × 10–2 M Ba (OH) (3*) 100 mL of 1 × 10–2 M H SO + 10 mL of 1 × 10–2 M NaOH (4) 100 mL of 1 × 10–2 M HCl + 100 mL of 1 × 10–2 M NaOH 7. K for HCN is 5 x 10–10, calculate K for CN–. HCN ds fy, K 5 x 10–10, CN– ds fy, K dh x.kuk dhft,A a b Ans. K [CN–] = 2 × 10–5 8. K for trimethylamine is 6.4 x 10–5. Calculate K for trimethyl ammonium ion (CH ) NH+. VªkbZfeFk by,ehudsfy,K 6.4x10–5 gSAVªkbZesfFkyveksfu;evk;u(CH ) NH+ dsfy,K dhx.kukdhft,A Ans. Ka b = 1.56 × 10–10 3 3 a 9. For the following equilibrium : NH + H O NH + + OH– equilibrium constant is 5.55 x 10–10. Calculate equilibrium constant for the equilibrium, NH + + H O NH OH + H+ 4 2 4 fuEulkE;dsfy, NH +H O NH + +OH– lkE;koLFkk fLFkjkad 5.55x 10–10gSA 3 2 4 NH ++ H O NH OH+H+ lkE; ds fy, lkE; fLFkjkad dh x.kuk dhft,A 4 2 4 Ans. 1.8 × 10–5 . 10. If equilibrium constant of CH COO– + H O CH COOH + OH– is 5.55 x 10–10, calculate equilibrium constant of CH COOH + H O CH COO– + H O+. ;fn CH COO– +H O CH COOH+OH– ds fy, lkE; fLFkjkad 5.55x 10–10 gS rks 3 2 3 CH COOH+H O CH COO– +H O+ lkE; fLFkjkad dh x.kuk dhft,A 3 2 3 3 Ans. 1.8 × 10–5 11. CO in aqueous solution shows following ionic equilibrium : 2H O + CO HCO – + H O+ If hydronium ion (H O+) concentration, is 2 x 10–6 M, what is hydroxide ion (OH–) concentration ? tyh; foy;u esa CO fuEu vk;fud lkE; n'kkZrk gS] 2H O+CO HCO – +H O+ 2 2 2 3 3 ;fngkbMªksfu;evk;u(H O+) dhlkUnzrk2x10–6 M,rksgkbMªkWDlkbMvk;u(OH–)dhlkUnzrkD;kgS\ Ans. [OH–] = 5 × 10–9 M 12. Several acids are listed below with their respective equilibrium constants. HF(aq) + H O(𝑙) H O+ (aq) + F (aq) K = 7.2 x 10 4 2 3 a HS (aq) + H O(𝑙) H O+ (aq) + S2  (aq) K = 1.3 x 10 11 2 3 a CH COOH(aq) + H O(𝑙) H O+(aq) + CH COO (aq) K = 1.8 x10 5 3 2 3 3 a (i) Which is the strongest acid ? Which is the weakest ? (ii) What is the conjugate base of the acid HF ? (iii) Which acid has the weakest conjugate base ? (iv) Which acid has the strongest conjugate base ? cgqrlkjsvEyksadkslkE;fLFkjkaddsvk/k jijuhpslwphc)fd;kx;kgSA HF(aq) + H O(𝑙) H O+ (aq) + F (aq) K = 7.2 x 10 4 2 3 a HS (aq) + H O(𝑙) H O+ (aq) + S2  (aq) K = 1.3 x 10 11 2 3 a CH COOH(aq) + H O(𝑙) H O+(aq) + CH COO (aq) K = 1.8 x10 5 3 2 3 3 a (i)çcyrevEydkSulkgS \nqcZyredkSulkgS \ (ii)HFvEydkla;qXeh{kkjD;kgS\ (ii)dkSulkvEynqcZyrel;qaXeh{k jj[krkgS\ (iv)dkSulkvEyçcyrel;aqXeh{k jj[krkgS\ Ans. (i) strongest acid,HF; weakest acid HS—. (ii) fluoride F—. (iii) The strongest acid (HF) has the weakest conjugate base. (iv) The weakest acid (HS—) has the stongest conjugate base. Integer Answer Type This section contains 2 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9. iw.k±dmÙkjizdkj bl[k.Mesa2iz'ugSaAizR;sdiz'udkmÙkj0ls9 rdiw.k ±dgSA 13. In the following reaction, 3A + B 2C + D, initial moles of B is four times of A. At equilibrium moles of A and C are equal. Find % dissociation of B. fuEuvfHkfØ;k3A+B 2C+Desa]BdsizkfjEHkdeksy]Adhrqyukesa4xqukgSrFk lkE;ijArFk Cdseksy leku gSA rc Bds fo;kstu dk %Kkrdhft,A Ans. 5 Sol. 3A + B 2C + D Initial moles a 4a 0 0 izkjfEHkdeksy a 4a 0 0 At equilibrium a – 3x 4a – x 2x x lkE;ij a – 3x 4a – x 2x x At equilibrium moles of A = moles of C lkE; ij Ads eksy =molesof C a – 3x = 2x a or x = 5 so, % dissociation of B = a / 5 4a × 100 = 5% blfy, B dk % fo;kstu = a / 5 4a × 100 = 5% 14. CH3–CO–CH3(g) CH3–CH3(g) + CO(g) 1 Initial pressure of CH3COCH3 is 12 atm. When equilibrium is set up mole fraction of CO(g) is 3 . Find Kp. CH3–CO–CH3(g) CH3–CH3(g) + CO(g) 1 CH3COCH3dkizkjfEHkdnkc12atmgSAtclkE;LFk firgksrkgSrcCO(g)dkeksyizHkkt 3 gSrcKpdkekuKkr dhft,A Ans. 6 Sol. CH3COCH3(g) CH3–CH3(g) + CO(g) 12 0 0 12 – x x x So PTotal = 12 + x Mole fraction of CO = x 12  x COdkeksyizHk t= x 12  x 1 or 3 x = 12  x x  x 6  6 so x = 6 Now, KP = 12  x = 6 = 6 atm.