DPP-51 to 52-(Physical)-Faculty Copy

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 51 to 52 Class : XIII Course : DPP No.1 Total Marks : 38 Max. Time : 38 min. Single choice Objective ('–1' negative marking) Q.1 to Q.10 (3 marks 3 min.) [30, 30] Subjective Questions ('–1' negative marking) Q.11 to Q.12 (4 marks 4 min.) [8, 8] 1. (4) 2. (2) 3. (2) 4. (1) 5. (2) 6. (2) 7. (2) 8. (3) 9. (1) 10. (3) 1. The solubility of A X is y mol dm 3 . Its solubility product is : A X dhfoys;rk yeksydm3 .gSAbldk foys;rkxq.kuQy gSA (1) 6 y4 (2) 64 y4 (3) 36 y5 (4*) 4 y3 2. The solubility of sparingly soluble electrolyte MmAa in water is given by the expression : ty esavYifoys;fo|qrvi?kV~;dhfoys;rkfdlO;atdds}kjknhtkrhgSA  (1) s =  Ksp m  a   (2*) s =  Ksp 1/ m  a   (3) s =  Ksp m  a   (4) s =  Ksp 1/ m  a   mm aa   mm aa   ma am   ma am          3. Three sparingly soluble salts M2X, MX and MX3 have the solubility product are in the ratio of 4: 1 : 27. Their solubilities will be in the order rhu vYifoys; yo.k M2X,MXrFkkMX3dsfoys;rk xq.kuQydk vuqikr4:1:27gSAbudh foys;rkdk ØegksxkA (1) MX3 > MX > M2 X (2*) MX3 > M2X > MX (3) MX > MX3 > M2X (4) MX > M2X > MX3 4. A particular saturated solution of silver chromate, Ag CrO , has [Ag+]= 5×10–5 and [CrO 2–] = 4.4×10–4 M. What is value of Ksp for Ag2CrO4 ? flYoj ØksesV] Ag CrO ds fuf'prlar`Ir foy;uesa [Ag+]=5×10–5 rFkk [CrO 2–]=4.4×10–4 MgSaA Ag CrO dsfy, 2 4 Ksp dkekuD;kgSa\ 4 2 4 (1*) 1.1 × 10–12 . (2) 1.5 × 10–12 (3) 2 × 10–6 (4) 1 × 1012. 5. If the solubility product of silver oxalate is 5 × 10–10, what will be the weight of Ag C O in 2.5 litres of a saturated solution ? (Ag = 108, C = 12, O = 16). 2 2 4 ;fnflYojvkDlysVdkfoys;rkxq.kuQy5×10–10 gSA,dlar`Irfoy;uds2.5yhVjesaAg C O dkHkkjD;kgksxk 2 2 4 (Ag = 108, C = 12, O = 16). (1) 0.50 gm (2*) 0.38 gm (3) 0.30 gm (4) 0.45 gm. 6. A student wants to prepare a saturated solution of Ag+ ion . He has got three samples AgCl (K = 10 10) , AgBr (Ksp = 1.6 × 10 13) and Ag CrO (Ksp = 3.2 × 10 11) . Which of the above compound will be used by him using minimum weight to prepare 1 lit. of saturated solution. (1) AgCl (2*) AgBr (3) Ag2 CrO4 (4) all the above . ,d fo|kFkhZ Ag+ vk;u dk ,d lar`Ir foy;u rS;kj djuk pkgrk gSA mlus rhu uewus AgCl (K = 1010), AgBr(K =1.6×1013)rFkk Ag CrO (K =3.2×1011). dks ç;qDr fd;kA 1yhVj lar`Ir foy;u dks rS;kj djus sp 2 4 sp dsfy,U;wureHk jdkmi;ksxdjrsgq,smlsfdl;kSfxddksç;qDrdjukpkfg, (1) AgCl (2*) AgBr (3) Ag2 CrO4 (4)mijksDrlHkhdks 7. If the solubility of Ag SO in 10–2 M Na SO solution be 2 × 10–8 M then K of Ag SO will be: 10–2 MNa SO foy;u esa Ag SO dh foys;rk 2×10–8 MgS rc Ag SO dk K gksxkA 2 4 2 4 2 4 sp (1) 32 × 10–24 (2*) 16 × 10–18 (3) 32 × 10–18 (4) 16 × 10–24 Sol. Ag SO 2Ag+ + SO –2 2s' (s' + 10–2)  0–2 K = (2s')2 (10–2) = (2 × 2 × 10–8)2 (10–2) = 16 × 10–18 . 8. The solubility of CaF2 NaF solution. in water at 1518ºC is 2 × 10–4 mole/litre. Calculate K of CaF2 and its solubility in 0.1M 1518ºCij ty esa CaF dhfoys;rk 2×10–4 eksy@yhVj gSA CaF dsK dh x.kuk djks rFkk 0.1MNaFfoy;u esa 2 bldhfoys;rkcrkvksA 2 sp (1) 3.5 × 108 mole/litre (2) 3.0 × 109 mole/litre. (3*) 3.3 × 10–9 mole/litre (4) 4.0 × 107 mole/litre Ans. K = 3.2 × 10–11, [Ca2+] = 3.3 × 10–9 mole/litre. 9. Calculate F— in a solution saturated with respect of both MgF and SrF2. Ksp (MgF ) = 9.5 x 10-9, K (SrF ) = 4 x 10-9. MgF rFk SrF .nksuksdslkis{klar`Irfoy;uesaF— dhx.kukdjksA 2 2 Ksp (MgF ) = 9.5 x 10-9, K (SrF ) = 4 x 10-9. (1*) 3 × 10–3 M. (2) 4 × 10–2 M. (3) 3.5 × 10–3 M (4) 1 × 10–3 M. Sol. [F—] = 3 × 10-3M. 10. A solution is saturated with respect to SrCO & SrF . The [CO 2] was found to be 1.2 x 103 M. The concentration of F in the solution would be : K (SrCO ) = 10–9, K (SrF ) = 3 × 10–11. ,dfoy;uSrCO rFkkSrF .dslkis{klar`IrgSA[CO 2]1.2x103 M.ikbZx;hfoy;uesaFdhlkUnzrkD;kgksxhA Ksp 3 (SrCO ) = 10–9, K 2 3 (SrF ) = 3 × 10–11 (1) 3 x 103 M (2) 2 x 102 M (3*) 6 x 102 M (4) 6 x 107 M 11. Calculate the solubility of AgCl (s) in (a) pure water (b) 0.1 M NaCl (c) 0.01 M CaCl2 at 25º C . Ksp (AgCl) = 2.56  10 10. Comment on the influence of [ Cl ] on the solubility of AgCl. 25ºCij AgCI(s)dsfoys;rk dh x.kuk djks (a) 'kq)tyrFkk (b) 0.1M NaCl esa (c) 0.01MCaCl2 esa Ksp (AgCl)= 2.56 1010. AgCl. ds foys;rk ij [Cl] ds izHkko dks crkb;s A Ans. (a) 1.6  10 5 mol/lit. (b) 2.56  10 9 mol/lit. (c) 1.28 × 10–8 mol/lit. 12. Find the solubility of CaF2 in 0.5 M solution of CaCl2 and water. How many times in solubility in the second case greater than in the first ? Ksp (CaF ) = 3.2 × 10–11. CaCI2ds0.5Mfoy;uesarFk tyesaCaF2dhfoys;rkKkrdjksAf}rh;ifjfLFkfresafoys;rkizFkeifjfLFkfrlsfdrus xqukizkIrgksrhgSA Ans. 4 × 10–6 M, 2 × 10–4 M, 50 times. Subject : Physical/Inorg.Chemistry Date : DPP No. 52 Class : XIII Course : DPP No.2 Total Marks : 38 Max. Time : 38 min. Single choice Objective ('–1' negative marking) Q.1 to Q.11 (3 marks 3 min.) [33, 33] Subjective Questions ('–1' negative marking) Q.12 (4 marks 5 min.) [4, 5] ANSWER KEY DPP No. 52 1. (2) 2. (1) 3. (2) 4. (4) 5. (1) 6. (4) 7. (2) 8. (2) 9. (4) 10. (1) 11. (1) 12. (a) Will dissolve, 32% saturation (b) will not dissolve, 100% saturation. 1. The standard oxidation potentials, E°, for the half-reaction are as v)ZvfHkfØ;kdsfy,ekudvkWDlhdj.kfoHkoE°gSA Zn = Zn2+ + 2e– ; E° = + 0.76 V Fe = Fe2+ + 2e– ; E° = + 0.41 V the E° Fe+2 + Zn Zn2+ + Fe gSaA cell is - E° lsy gksxkA (1) –0.35 V (2*) + 0.35 V (3) +1.17 V (4) – 1.17 V Sol. Zn + Fe2+  Zn2+ + Fe. E0 = E0 2 – E0 2 = – 0.41 – (– 0.76) = 0.35 V. cell Fe / Fe Zn / Zn 2. From the following E° values of half cells - v)Z lsy ds fy, E°dkeku fuEu gS& (i) A + e–  A¯ ; E° = – 0.24 V (iii) C¯ + 2e–  C3– ; E° = –1.25 V (ii) B¯ + e–  B2– ; (iv) D + 2e–  D2– ; E° = + 1.25 V E° = + 0.68 V What combination of two half cells would result in a cell with the largest potential dkSulsnksv)Zlsyksdkla;kstu]vf/kdrefoHkookyslsydkfuekZ.kdjsxk\ (1*) (ii) and (iii) (2) (ii) and (iv) (3) (i) and (iii) (4) (i) and (iv) (1*) (ii) rFkk (iii) (2)(ii) rFkk (iv) (3) (i) rFkk (iii) (4)(i) rFkk (iv) 3. The Ni/Ni2+ and F–/F electrode potentials are listed as +0.25 V and –2.87 V respectively (with respect to the standard hydrogen electrode). The cell potential when these are coupled under standard conditions is (1) 2.62 V and dependent on the reference electrode chosen. (2*) 3.12 V and independent of the reference electrode chosen. (3) 3.12 V and dependent on the reference electrode chosen. (4) 2.62 V and independent of the reference electrode chosen. Ni/Ni2+ rFk F–/F dsfy,bysDVªkWMfoHko(ekudgkbMªkstubysDVªkWMdslUnHkZesa)Øe'k%+0.25VrFk –2.87Vfn;kx;k gStc;gekudifjfLFkfr;ksaesala;ksftrfd,tkrsgSrkslsyfoHkogksxk\ (1)2.62VrFk pqusx,ekudbysDVªkWMijfuHkZjdjrkgSA (2*)3.12VrFk pqusx,ekudbysDVªkWMlsLora=kgksrkgSA (3)3.12VrFk pqusx,ekudbysDVªkWMijfuHkZjdjrkgSA (4)2.62VrFk pqusx,ekudbysDVªkWMlsLora=kgksrkgSA Sol. Electrode potential values depend on reference electrode chosen but not cell potential. Cr2O2 4. For the cell prepared from electrode A and B, electrode A : Fe3 7 Cr3 0 red = + 1.33 V and electrode B : , E0 Fe = 0.77 V, which of the following statement is not correct ? (1) The electrons will flow from B to A (in the outer circuit) when connections are made. (2) The standard emf of the cell will be 0.56 V. (3) A will be positive electrode. (4*) None of the above. bysDVkWM A o B ls cus ,d lSy ds fy,] bysDVªkWM+ A : Cr2O2 Cr3 , Eºred = + 1.33 V o Fe3 b y s D V ª k W M + B: Eºred=0.77VgSAfuEuesalsdkSulkdFkulghughgS& (1) tcla;ksftrfd;ktkrkgS]rksbysDVªkWuBlsA dhvksj ¼ckà;ifjiFkesa½izokfgrgks axsA (2) lsy dk ekud emf 0.56Vgksaxk A (3)A/kukRedbysDVªkWMgksxkA (4*)mijksDresalsdksbZughaA 5. The standard reduction potentials at 25°C for the following half reactions are given against each - 25ºCijfuEuv)ZvfHkfØ;kvksadsfy,ekudvip;ufoHkoizR;sddslkeusfn;kx;kgSA Zn2+(aq) + 2e¯ Zn(s), –0.762 V Cr3+(aq) + 3e¯ Cr(s), –0.740 V 2H+ + 2e¯ H (g), 0.00 V Fe3+ + e¯ Fe2+, 0.77 V Which is the strongest reducing agent - dkSulkizcyrevipk;dvfHkdeZdgSA (1*) Zn (2) Cr (3) H (g) (4) Fe3+(aq) Fe2 , Ans. Higher oxidation potential are reducing tendency. 6. Hydrogen gas will not reduce - (1) heated cupric oxide (2) heated feric oxide (3) heated stannic oxide (4*) heated aluminium oxide [ E 4 /Sn 2  0.15V ; E 2 /Cu   0.167V ; E 3 / Fe 2  0.771V ; E 3 /Al  1.67V ] gkbMªkstuxSlfdlsvipf;rughadjsxh& (1)xeZD;wfizdvkWDlkbMdks (2)xeZQsfjdvkWDlkbMdks (3)xeZLVsfudvkWDlkbMdks (4*)xeZ,Y;qfefu;evkWDlkbMdks [ E 4 /Sn 2  0.15V ; E 2 /Cu   0.167V ; E 3 / Fe 2  0.771V ; E 3 /Al  1.67V ] Sol. Eº  = 0 V. 2 Hydrogen gas will reduce to those metals which have reduction potential greater than H2 gas. 7. Four colourless salt solutions are placed in separate test tubes and a strip of copper is dipped in each. Which solution finally turns blue ? (use data from electrochemical series) pkjjaxghuyo.kfoy;uksdksi`Fkdµi`Fkdij[kuyhesaj[k x;kgSrFk izR;sdesadkWijdhiryhiV~VhdkMqcks;htkrh gSdkSulkfoy;uvUr %uhykgkstkrkgS(fo|qrjklk;fudJs.khlsvk¡dM+ksdksiz;qDrdhft,) (1) Pb(NO3)2 (2*) AgNO3 (3) Zn(NO3)2 (4) Cd(NO3)2 Sol. Cu will reduce to those metal which have might reduction potential than Cu. So, Ag+ higher potential. 8. Red hot carbon will remove oxygen from the oxide XO and YO but not from ZO. Y will remove oxygen from XO. Use this evidence to deduce the order of activity of the three metals X, Y, and Z putting the most active first - ykyrIrdkcZu]vkWDlkbMXOrFk YOlsvkWDlhtudksi`FkddjrkgSysfduZOlsugha]Y,XOlsvkWDlhtudksi`Fkd djrkgSblrF;dksiz;qDrdjrhu/k rqvksaX,Y,rFk ZdhfØ;k'khyrkdsØedksfu/k Zfjrdhft,rFk vf/kdfØ;k'khy dks igys fyf[k, \ (1) XYZ (2*) ZYX (3) YXZ (4) ZXY 9. Two Ist order reactions have half-lives in the ratio 3 : 2. Then the ratio of time intervals t : t , will be? Where 2 t1 is the time period for 25% completion of the first reaction and t2 is time required for 75% completion of the second reaction. [log 2 = 0.3, log 3 = 0.477] nksizFkedksfVdhvfHkfØ;kvksadhv)Z&vk;q3:2dsvuqikresagSArcle;kUrjkyt1:t2 dkvuqikrD;kgksxk\tgk¡t izFkevfHkfØ;kds25%iw.kZgksusesayxkle;rFk t2nwljhvfHkfØ;kds75%iw.kZgksusaesayxkle;gS[log2=0.3,log 3 = 0.477] (1) 0.199 : 1 (2) 0.420 : 1 (3) 0.273 : 1 (4*) 0.311 : 1 Sol. Time taken for 25% completion C = C e–1k t  3 = e–k t t 0 1 1 4 1 1 1  4  as ln(3/4) = – k b so t = ln   . 1 1 1 k1  3  ln 2 time taken for 75% completion t2 = 2. k 2 so required ratio t1 t2 ln(4 / 3)k2 = k1.2ln2 = 3 × 2 (ln 4  ln3) ln 4 = 0.311 : 1 10. Identify the correct order of wavelength of light absorbed for the following complex ions. [Co(H O) ]3+ ; Co(CN) ]3– ; [Co(I) ]3– ; [Co(en) ]3+ 2 6 6 6 3 I II III IV (1*) III > I > IV > II (2) II > IV > I > III (3) III > I > II > IV (4) I > III > IV > II fuEuladqyvk;uksadsfy,vo'k sf"krizdk'kdhrjax}S/;ZdslghØedhigpkudhft;s& [Co(H O) ]3+ ; Co(CN) ]3– ; [Co(I) ]3– ; [Co(en) ]3+ 2 6 6 6 3 I II III IV (1*) III > I > IV > II (2) II > IV > I > III (3) III > I > II > IV (4) I > III > IV > II hc Sol.  corresponds to the wave length of the visible light and  =  The order of ability to produce d-orbital splitting of various ligands ; CO > CN– > NO – > en > NH > H O > OH– > F– > Cl– > Br– > I– 2 3 2 strong field weak field ligands ligands (larger ) (smaller ) 11. Which one of the following complexes exhibit chirality ? (1*) [Cr(ox) ]3 (2) cis - [PtCl (en)] 3 2 (3) cis - [RhCl (NH ) ]– (4) mer - [Co(NO ) (NH ) ] 2 3 2 2 3 3 3 fuEuesalsdkSulkladqyfdjsyrkn'kZrkgS\ (1*) [Cr(ox) ]3 (2) lei{k- [PtCl (en) 3 2 (3) lei{k- [RhCl (NH ) ]– (4) ej- [Co(NO ) (NH ) ] 2 3 2 2 3 3 3 Sol. (1) has three symmetrical bidentate lignds, which lack plane of symmetry. (2) and (3) has square planar geometry and has plane of symmetry. (4) has one of the plane of symmetry. So (2), (3) and (4) will not show chirality. 12. If you place the amounts given below in pure water, will all of the salt dissolve before equilibrium can be established, or will some salt remain undissolved ? (a) 4.96 mg of MgF2 in 125 ml of pure water, Ksp = 3.2 x 10-8 (b) 3.9 mg of CaF2 in 100 ml of pure water, Ksp = 4 x 10-12 Also find the percentage saturation in each case. ;fnuhpsnhx;hek=k dks'kq)tyesafeyk;ktkrkgSrksD;klkE;koLFk vkuslsigyslHkhyo.kfoys;gksxsa;kdqN yo.kvfoys;jgsaxs? (a) 125 ml 'kq) ty esa MgF2 ds 4.96 mg, Ksp (b) 100 ml 'kq} ty esa CaF2 ds 3.9mg, Ksp izR;sdifjfLFkfresalar`Irkdhizfr'krKkrdjksA = 3.2 x 10-8 = 4 x 10-12