DPP-DPP 57 to 58 With Answer

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 57 to 58 Class : XIII Course : DPP No.1 Total Marks : 30 Max. Time : 30 min. Single choice Objective (no negative marking) Q.1 to Q.3 (3 Marks, 3 Min.) [9, 9] Comprehension ('–1' negative marking) Q.4 to Q.6 (3 marks 3 min.) [9, 9] Subjective Questions (no negative marking) Q.7 to Q.9 (4 marks 4 min.) [12, 12] 1. A current is passed through 2 voltameters connected in series. The first voltameter contains XSO4 (aq.) and second has Y2SO4 the relative atomic masses of X and Y are in the ratio of 2 : 1. The ratio of the mass of X liberated to the mass of Y liberated is Js.khesala;ksftr2oksYVehsVjesals/k jkizokfgrdhtkrhgSAizFkeoksYVehVj]XSO4(tyh;);qDrgSrFk f}rh;Y2SO4 ;qDrgSaAXrFk Ydkvkisf{kdijek.kqnzO;eku2:1vuqikresagSaAeqDrgq,XrFk eqDrgq,YdsnzO;ekuksadkvuqikr gksxkgSA (1*) 1 : 1 (2) 1 : 2 (3) 2 : 1 (4)NoneoftheabovemijksDresalsdksbZugha Sol. XSO4 (aq.) Y2SO4 (aq.) Mol.mass X Mol.massY 2 = 1 mol. mass X eq. wt. of X = 2 eq. wt. of Y = mass of Y mol. mass 1 eq.wt.of Y Y mol.mass Y 2 mass of X = eq.wt.of X = mol.mass X × 1 = 1:1. 2. For the electrolytic production of NaClO4 from NaClO3 as per the equation, NaClO3 + H O  NaClO + H2, how many faradays of electricity will be required to produce 0.5 mole of NaClO4, assuming 60% efficiency ? NaClO3 ls NaClO4 dk fo|qrvi?kVuh; mRiknu vfHkfØ;k NaClO3 +H2ONaClO4 +H2,ds vuqlkj gksrk gSA 60%n{krkekursgq,]NaClO4ds0.5eksycukusdsfy,fdrusQSjkMsfo|qr/k jkdhvko';drkgksxhA& (1) 0.835 F (2*) 1.67 F (3) 3.34 F (4) 1.6 F Sol. NaCI O3 + H O  NaCIO + H2 At cathode : 2H+ + 2e–  H For 1 mole of NaCIO4 and H2 we require 2 F of charge For 0.5 mole of NaCIO4 we require 1 F of charge ∴ Electricity required = 1×100/60 = 1.67F 3. To observe the effect of concentration on the conductivity, electrolytes of different nature are taken in two vessel ‘A’ and ‘B’. ‘A’ contains weak electrolyte e.g., NH4OH and ‘B’ contains strong electrolyte e.g., NaCl. In both container concentration of respective electrolyte is increased and conductivity observed : (1) in ‘A’ conductivity increases, in ‘B’ conductivity decreases (2) in ‘A’ conductivity decreases while in ‘B’ conductivity increases (3*) in both ‘A’ and ‘B’ conductivity increases (4) in both ‘A’ and ‘B’ conductivity decreases nks ik=k ‘A’ rFkk ‘B’ esa fofHkUu izd`fr ds fo|qr vi?kV~; fy;s tkrs gSa ftudh pkydrk ij lkUnzrk dk izHkko izsf{kr fd;k tkrk gSA ‘A’ ,d nqcZy fo|qr vi?kV~; tSls NH4OH rFkk ‘B’ izcy fo|qr vi?kV~; tSls NaCl j[krk gS nksuks ik=kksa esa fo|qr vi?kV~; dh lkUnzrk esa o`f) dh tkrh gS rFkk pkydrk izsf{kr dh tkrh gS] rc % (1) ‘A’ esa pkydrk ckrkgSfd;fn N2O5 dhlkUnzrknqxquhdjnhtkrhgSrc N2O5dsfo?kVudhnj]NO2dscu s dhsnj]rFk O2dscu sdhnj]lHkhnqxquhgkstkrhgSA 4. If rate constants for decomposition of N2O5, formation of NO2 and formation of O2 are K1, K2 and K3 respectively, then : N2O5 dsfo?kVudsfy,njfu;rkad]NO2dscu sdsfy,njfu;rkad]rFkkO2dscu sdsfy,njfu;rkadØe'k%K1, K2rFkk K3gSrc & (1) K1 = K2 = K3 (2*) 2K1 = K2 = 4K3 (3) K1 = 2K2 = K3 (4) K1 = K2 = 2K3 5. If rate of formation of O2 is 16 gm/hr, then rate of decomposition of N2O5 and rate of formation of NO2 respectively are : ;fnO2 dsfuekZ.kdhnj 16gm/hr, gSrc N2O5 dsfo?kVudhnjrFkkNO2 dscuusdhnj Øe'k%gSa % (1) cannot be calculated with out knowing rate constants njfu;rkadKkrfd;sfcukx.kukughadhtkldrhgSA (2*) 108 gm/hr, 92 gm/hr (3) 32 gm/hr, 64 gm/hr (4) 54 gm/hr, 46 gm/hr 6. The container of 2 litre contains 4 mole of N2O5. On heating to 100°C, N2O5 undergoes complete dissociation to NO and O . Select the correct answers if rate constant for decomposition of N O is 6.3 x 10–4 sec–1. (𝑙n2 2 2 2 5 = 0.693) 1. The mole ratio before and after dissociation (of total gaseous moles) is 4 : 2. 2. Half life of N2O5 is 1100 sec and it is independent of temperature. 3. Time required to complete 87.5% of reaction is 55 min. 4. If volume of container is doubled, the rate of decomposition becomes half of the initial rate : ,d2yhVjdkik=kN2O5ds4eksyj[krkgS\100°CrkiijxeZdjusij]N2O5 ,NO2 rFkkO2esaiw.kZr%fo?kfVrgks tkrk gSA ;fn N O ds fo?kVu dk nj fu;rkad 6.3x 10–4 sec–1 gSA rc lgh mÙkj dk p;u dhft;s (𝑙n2=0.693) 2 5 1. fo?kVudsiwoZrFk i'pkr~eksyvuqikr¼dqyxSlh;eksyksadk½4:2gSA 2. N2O5dhv)Zvk;q1100sec.gSrFkk;grki lsLora=kgSA 3.87.5%vfHkfØ;kdksiw.kZgksusdsfy, 55feuVdkle;vko';dgSA 4.;fnik=kdkvk;runqxqukdjfn;ktk,]rcfo?kVudhnj]izkjfEHkdnjdhvk/khgkstkrhgSA (1) 1, 3, 4 (2) 1, 2, 3, 4 (3*) 3, 4 (4) 2, 3, 4 Sol. 4 to 6 1 1 d [NO2 ] d [O2 ] (1) Rate of reaction = – 2 d(N2O5] = 4 dt = dt0 d [N2O5 ] d [NO2 ] and d [O2 ] = K [N O ] dt 3 2 5  K1  K 2 = K 2 4 3 d [O2 ] 16 (2) = 16 g/hr = mol hr–1 dt d [NO2 ] 32 4 d [O2 ] 16 dt = dt = 4 x 32 = 2 mol hr–1 = 2 x 46 = 92 g hr–1  d[N2O5 ] dt = 2 d [O2 ] dt 16 = 2 x 32 = 1 mol hr–1 1 x 108 = 108 g hr–1 (3) N O  2NO + 1 O 2 5 2 2 2 Initial mole    moles after diss 0 8 2 4  Mole ratio = 10 = 2.5 t = 0.693 1/2 K 0.693 = 6.3104 1 = 1100 sec but it depends upon temperature as K also depends upon tempear 100 t87.5% = 6.3 104 𝑙n 100  87.5 = 3300 sec = 55 min Rate = K[N2O5] ; Thus r1 = K [N2O5] If V is doubled the concentration becomes half 1  r = K [N O ] 2 2 2 5 r1 2  r2 = 1 7. Find the thickness of the electro deposited silver if the surface area over which deposition occurred was 100 cm2 and a current of 0.2 A flowed for 1hr with the cathodic efficiency of 80%. Density of Ag = 10 g/cc (Ag = 108). ;fn100cm2dsi`"Bh;lrg{ks=kijfu{ksi.kgksrkgSrFk 80% dSFk sMhdn{krkdslkFk0.2Adhfo|qr/k jk1?k.Vsds fy,izokfgrdhtkrhgS]oS|qr:ilsfu{ksfirflYoj dheksVkbZKkrdhft,AAgdk?kuRo=10g/cc(Ag=108). 108  0.2  0.8  3600 Sol. th × Area × density = 96500 ; 108  0.2  0.8  3600 th = 96500 100  10 cm = 6.4 × 10–4 cm 8. A 300 mL solution of NaCl was electrolysed for 6.00 min. If the pH of the final solution was 12 calculate the average current used. NaCl dk300mLdk foy;u6.00feuV ds fy, oS|qrvi?kfVr fd;k x;kA ;fn vfUre foy;u dhpH12Fkh] rc ç;qDr dhx;hvkSlr/k jkdhx.kukdjksA Ans. 0.8 A Sol. Anode : 2CI–  Cl2 (g) + 2e– Eº = – 1.36 V Cathods : 2H2O + 2e–  H2(g) + 2OH– Eº = – 0.83 V pH = 12 ⇒ [OH–] = 0.01 M Moles of OH– = 0.3 × 0.01 = 0.3 × 10–2 moles   360 0.3 × 0.01 = 96500  = 0.8 A 9. An acidic solution of Cu2+ salt containing 0.635 g of Cu2+ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of the solution kept at 100 mL and the current at 1.93 amp. Calculate the volume of gases evolved at NTP during the entire electrolysis. Cu2+ yo.k dk ,d vEyh; foy;u] 0.635gCu2+ ;qDr gS] bldk rc rd fo|qr vi?kVu fd;k tkrk gS tc rd fd lHkh dkWijfu{ksfirugkstk,A100mLfoy;udsvk;rudslkFk1.93ampdh/k jkizokfgrdhtkrhgSrFk fo|qrvi?kVu dksvkSjlkrfeuV rdyxkrkj fd;ktkrkgSrks NTPijfu"dkflrxSldk vk;ruKkrdhft,A Ans. 253.12 ml. Sol. Anode : 2H2O  O + 4H+ + 4e–. Eº = – 1.13 V. Cathode : Cu2+ + 2e–  Cu Eº = 0.34 V. 2H+ + 2e–  H Eº = 0 V. 2H O + 2e–  H + 2OH–. Eº = – 0.83 V. 1 H2O  2 O2 + 2H + 2e . + – Cu2+ + 2e–  Cu. H O + Cu2+ 1  1 2 O2 + 2H+. 2 moles of Cu deposited = moles of O2 at NTP. Moles of O2 1 at NTP = 2 0.635 × 63.5 = 5 × 10–3. After complete electrolysis of Cu. Charge passed = 1.93 × 7 × 60 = 504 coulomb. 1.93  7  60 Moles of e– = 96500 = 8.4 × 10–3. Moles of H2 Moles of O2 at NTP = 4.2 × 10–3 . at NTP = 2.1 × 10–3. Total moles of gases at NTP = 11.3 × 10–3 moles. Volume of gases evolved at NTP = 253.12 ml. DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 58 Class : XIII Course : DPP No.2 Total Marks : 34 Max. Time : 34 min. Single choice Objective (no negative marking) Q.1 to Q.6 (3 Marks, 3 Min.) [18, 18] Subjective Questions (no negative marking) Q.7 to Q.10 (4 marks 4 min.) [16, 16] 1. For a saturated solution of AgCl at 25°C,  = 3.4×10–6 S cm–1 and that of H O (𝑙) used is 2.02 × 10–6 S cm1  for AgCl is 138 S cm2 mol–1 then the solubility of AgCl in moles per liter will be - 25°CijAgClds ,d lar`Ir foy;u ds fy, =3.4×10–6 Scm–1 rFkk H O(𝑙)ds 2.02×10–6 Scm–1 iz;qDr fd;k tkrk gS] AgCl dsfy,m 138Scm mol gSrksAgCldheksy@yhVjesafoys;rkgksxh& 2 –1 (1*) 10–5 (2) 10–10 (3) 10–14 (4) 10–16 Sol. Kelectrolyte = Ksolution – Ksolvent = 3.4 × 10–6 – 2.02 × 10–6 = 1.38 × 10–6 Scm–1 Solubility = Kelectrolyt e 1000 º = m 1.38  10–6  1000 138 = 10–5 M. 2. Given that (in S cm2 eq–1) at T = 298 K : eq for Ba(OH2), BaCl2 & NH4Cl are 228.8, 120.3 & 129.8 respectively. Specific conductance for 0.2 N NH OH solution is 4.766 × 10–4 S cm–1, then value of pH of the given solution of NH4OH will be nearly 298K ij fn;k x;k gS fdBa(OH) ,BaCl rFkkNH Clds fy, Øe'k%228.8, 120.3rFkk129.8(Scm2 eq–1 esa) gSaA 2 2 4 eq 0.2NNH OHfoy;u ds fy, fof'k"V pkydrk4.766×10–4 Scm–1gSANH OHds fn;s x;s foy;u dkpHyxHkxfdruk 4 4 gksxk\ (1) 9.2 (2*) 11.3 (3) 12.1 (4) 7.9 Sol.  Ba(OH) = º Ba2+ + ºeq, OH– 2 ºeq BaCI = ºeq Ba2+ + ºeq,CI– ºeq NH CI = ºeq NH+ + ºeq,CI– ºeq, NH OH = ºeq,NH + + ºeq,OH– 4 4 I + III + II ºeq,NH OH = (228.8 + 129.8) – 120.3 = 238.33 cm2eq–1 eq, NH OH = 4.766  10–14  1000 0.2 = 2.383   = eq,NH4OH º eq,NH4OH = 10–2 NH OH NH + + OH– 4 4 c(1–) c c [OH–] = 0.2 × 10–12 = 2 × 10–3 POH = 3– log2 ⇒ PH = 14 –(3–log2) = 11.3 3. In which case van’t Hoff factor is maximum ? (1) KCl, 50% ionised (2) K2SO4, 40% ionised (3) SnCl4, 20% ionised (4*) FeCl3, 30% ionised fdlifjfLFkfresaokUVgkWQdkjdvf/kdregS\ (1) 50% vk;fur KCl (2) 40% vk;fur K2SO4 (3) 20% vk;fur SnCl4 (4*) 30% vk;fur FeCl3 4. Solution having osmotic pressure nearer to that of an equimolar solution of K4[Fe(CN)6] is: og foy;u ftldk ijklj.k nkc K4[Fe(CN)6]ds leeksyjfoy;u ds yxHkx cjkcj gksxk %& (1) Na2SO4 (2) BaCl2 (3*) Al2(SO)3 (4)C12 H22O11 5. Osmotic pressure [atm] of a 0.1 M solution of K4[Fe(CN)6], which undergoes 50% dissociation, will be atm at 270C. (1*) 7.38 (2) 3.69 (3) 405.9 (4) none of these 27°Cij 50%fo;kstu ds vUrxZr K4[Fe(CN)6],ds 0.1Mfoy;u dk ijklj.k nkc [atm]esa gksxkA (1*) 7.38 (2) 3.69 (3) 405.9 (4)buesalsdksbZugha 6. Two aqueous solutions, one of the NaCl in water (1) and the other of C8H15O2Na in water (2) are isotonic. If wA and wB are weight fractions of NaCl and C8H15O2Na in solution A and B respectively, then (assuming that both the salts dissociate completely) : (1) wA > wB (2) wA = wB (3*) wA < wB (4) none of these nks tyh;foy;u] ftuesals,d(1) tyesaNaClrFkknwljk(2) tyesaC8H15O2NagS]leijkljhgSA;fnfoy;u(1) rFk (2)esaNaClrFk C8H15O2Na dseksyfHkUuØe'k%wArFk wBgksa]¼;gekudjfdnksuksayo.kiw.kZr;kfo;ksftrgks tkrsgSa½rc& (1) wA > wB (2) wA = wB (3*) wA < wB (4)mijksDresalsdksbZugh 7. A conductivity cell has a resistance of 250  when filled with 0.02 M KCl at 250C and one of 105  when filled with 6 × 10–5 M NH4OH solution, the specific conductivity of 0.02 M KCl is 0.00277 S cm–1 and the NH + and OH– are 73.4 and 198 respectively. Calculate the cell constant and the degree of dissociation of NH4OH in the 6 × 10–5 M solution. ,dpkydrklsydkizfrjks/k250gS]bls25ºCij0.02MKCldslkFkHkjktkrkgSrFk ,dvU;pkydrklsyftldk izfrjks/k105gS]tcblesa6×10–5MNH4OHfoy;uHkjktkrkgSrks0.02MKCldhfof'k"Vpkydrkdkeku0.00277 Scm–1 rFkkNH +o OH–Øe'k%73.4rFk 198gSrks6×10–5 Mfoy;uesaNH OHdsfo;kstudhek=kkrFkklsy fu;rkaddhx.kukdhft,A Ans. 0.69 cm–1, 0.425. 8. Calculate the osmotic pressure of a decimolar solution of cane sugar at 27°C. 27°CrkiijxUusdh'kdZjkdsMslheksyjfoy;udsijklj.knkcdhx.kukdhft;s\ n Sol. = molar concentration = 0.1 M  R = 0.082 litre atm K–1 mol–1 , T = 300 K V  we have P = n RT  = 0.1 x 0.082 x 300 = 2.46 atm V 9. An aqueous solution contains 18 g of glucose (mol wt. = 180) per 0.5 L. Assuming the solution to be ideal, calculate osmotic pressure at 27°C. ,dtyh;foy;u izfr0.5Lesa18gXywdksl(v.kqHk j=180) j[krkgSAvkn'kZfoy;uekursgq;s27°Crkiijijklj.k nkcdhx.kukdhft;s\ 18 Sol. Mole of glucose = 180 R = 0.082 , T = 300 K n 1 = 10 , V = 0.5 liter 0.1 we have , P = V RT = 0.5 x 0.082 x 300 = 4.92 atm. 10. A solution of 1.73 g of ‘A’ in 100 cc of water is found to be isotonic with a 3.42% (weight/volume) solution sucrose (C12H22O11) Calculate molecular weight of A (C12H22O11 = 342) ty ds 100ccesa Ads 1.73gdk ,d foy;u] 3.42%(Hkkj@vk;ru)lwØksl (C12H22O11)ds foy;u ds lkFk leijkljh gSAA(C12H22O11 =342) dsv.kqHk jdhx.kukdhft;s Sol. We know that the isotonic solution have the same molar concentration (i.e moles.litre) Let the molecular weight of A be M moles of A = 1.73/M 3.42 1000  molar conc. of sucrose = 342 x 100 = 0.1 17.3 M = 0.1 ; M = 173.