Title Chemistry- Subject final-A Solution.pdf ✅

1 Varsity Subject Final Chemistry [Set-A (Solution)] c~Y©gvb: 140 †b‡MwUf gvK©: 0.25 mgq: 1 NÈv 30 wgwbU 1. 228 90 Th  212 83 Bi GB †ZRw ̄Œq wewμqvq wbM©Z  I  KYvi msL ̈v KZ?  [ 228 90 Th  212 83 Bi What is the number of  and  particles released in this radioactive reaction?] 1wU, 4wU 4wU, 1wU 7wU, 4wU none DËi: 4wU, 1wU e ̈vL ̈v: 228 90 Th  212 83 Bi + m 4 2He2+ + n 0 –1 e    m =  KYv msL ̈v n =  KYv msL ̈v  228 = 212 + 4m  m = 4   KYv 4wU Avevi, 90 = 83 + 2m – n  90 = 83 + 8 – n  n = 1   KYv 1wU       Gi ms‡KZ  4 2He+2  Gi ms‡KZ o –1e 2. wkLv cixÿvq Na wK eY© †`q? [What color does Na give in the flame test?] †mvbvwj njy` njy`vf meyR bxj jvj‡P †e ̧wb DËi: †mvbvwj njy` e ̈vL ̈v: †kÖwY avZzmg~n wkLvi m„ó eY© MÖæc- IA Li D3⁄4¡j jvj, wμgmb ev m~h©v‡ ̄Íi eY© Na D3⁄4¡j †mvbvjx njy` K †e ̧wb Rb jvj‡P †e ̧wb Cs bxj MÖæc-IIA Ca B‡Ui jvj e©Y (A ̄'vqx) Sr D3⁄4¡j jvj, wμgmb Ba njy`vf meyR 3. mnKvix †Kvqv›Uvg msL ̈v 4 n‡j, †PŠ¤^Kxq †Kvqv›Uvg msL ̈v m Gi gvb KZwU n‡e? [If the azimuthal quantum number is 4, what will be the value of the magnetic quantum number m?] 6 9 7 8 DËi: 9 e ̈vL ̈v: mnKvix †Kvqv›Uvg msL ̈v l Ges †Pؤ^Kxq †Kvqv›Uvg msL ̈v m n‡j, m Gi gvb (2l + 1)wU  l = 4 n‡j m Gi gvb = (2 × 4 + 1) = 9wU 4. RH wiWevM© aaæeK n‡j, H– eY©vwji c ̈v‡ðb wmwi‡Ri 2q jvB‡bi Zi1⁄2‰`N© ̈ KZ n‡e? [If RH is the Rydberg constant, what is the wavelength of the 2nd line of the Paschen series of the H spectrum?] 5 36RH 16 3 RH 36RH 5 225 16RH DËi: 225 16RH e ̈vL ̈v: c ̈v‡ðb wmwi‡Ri 2q jvB‡bi Rb ̈, n2 = n1 + 2 = 5  1  = RH     1 n1 2 – 1 n2 2  1  = RH     1 9 – 1 25   = 225 16RH 5. H cigvYyi B‡jKUab n kw3 ̄Íi n‡Z f‚wg ͇̄i wd‡i Avm‡j 3wU eY©vwj †iLv cvIqv †M‡j, n = ? [3 spectral lines are obtained when the electron of H atom returns from the n energy level to the ground energy level, n = ?] 6 3 4 2 DËi: 3 e ̈vL ̈v: eY©vwj †iLvi msL ̈v = (n2 – n1) (n2 – n1 + 1) 2  3 = (n – 1) (n – 1 + 1) 2  n 2 – n – 6 = 0  (n – 3) (n + 2) = 0  n – 3 = 0 n = 3 6. 0.001M AB `ae‡Y CB3 Gi †gvjvi `ave ̈Zv KZ? [Ksp(CB3) = 5.4  10–21] [What is the molar solubility of CB3 in a 0.001M AB solution?[Ksp(CB3) = 5.4 × 10−21]] 5.4  10–20 6.73  10–18 5.4  10–12 5.5  10–18 DËi: 5.4  10–12 e ̈vL ̈v: CB3 ⇌ C 3+ + 3B– S 3S GLb, wgkÕ‡Yi ci, [B– ] = (3S + 0.001) M [Ksp(CB3) = 5.4  10–21] S (3S + 0.001)3 = 5.4  10–21 S  (0.001) 3 = 5.4  10–21 S = 5.4  10–12 7. 25C ZvcgvÎvq GKwU xM AgCl `ae‡Y MgCl2 Gi `ave ̈Zv 9  10–9 | x-Gi gvb wbY©q Ki| Ksp (MgCl2) = 8.1  10–10 .
2 [At 25°C, in an x M AgCl solution, the solubility of MgCl2 is 9 × 10−9 . Determine the value of x. Ksp(MgCl2) = 8.1 × 10−10.] 0.3 M 0.01 M 0.2 M 0.02 M DËi: 0.3 M e ̈vL ̈v: MgCl2 ⇌ Mg2+ + 2Cl– Ksp(MgCl2) 8.1  10–10 = S (2S + x)2 8.1  10–10 = 9  10–9 x 2 x 2 = 0.09 x = 0.3 M 8. wb‡Pi †Kvb wewμqv Na+ kbv3Ki‡Y e ̈envi nq? [Which of the following reactions is used to identify Na+?] Na+ + Cl –  NaCl 2Na+ + k2H2 Sb2O7  Na2H2Sb2O7 Na+ + H2 Sb2O7  NaH2 Sb2O7 Na+ + H2 SO4  Na2 (SO4) DËi: 2Na+ + k2H2 Sb2O7  Na2H2Sb2O7 9. 1.2 Av‡cwÿK ̧iæ‡Z¡i 80 ml `ae‡Y †Kv‡bv `a‡ei 25 g `aexf‚Z Av‡Q| H ZvcgvÎvq `a‡ei `ave ̈Zv KZ? [In a 80 ml solution with a specific gravity of 1.2, 25 g of solute is dissolved. What is the solubility of the solute at that temperature?] 25 20 15 35.2 DËi: 35.2 e ̈vL ̈v: `ae‡Yi fi, M = 1.2  80 = 96 g `ave ̈Zv, S = m M – m  100 = 25  100 96 – 25 = 35.2 10. 1 2 N2(g) + 3 2 N2(g) ⇌ NH3(g) + 47 kJ/mol wewμqvq cÖfve‡Ki Dcw ̄’wZ‡Z †Kvb Ae ̄’vq NH3 Gi Drcv`b me©wb¤œn‡e? [In the reaction 1 2 N2(g) + 3 2 N2(g) ⇌ NH3(g) + 47 kJ/mol, under which conditions will the production of NH3 be lowest in the presence of catalyst?] D”PZvcgvÎv I D”PPvc wb¤œ ZvcgvÎv I wb¤œPvc wb¤œ ZvcgvÎv I AwaKPvc D”PZvcgvÎv I wb¤œPvc DËi: D”PZvcgvÎv I wb¤œPvc e ̈vL ̈v: wewμqvwU Zv‡cvrcv`x, GKB mv‡_ wewμq‡Ki †gvj msL ̈v > Drcv‡`i †gvj msLv|  ZvcgvÎv Kgv‡j Ges Pvce„w× Ki‡j mvg ̈e ̄'v Wvbw`‡K hv‡e m‡ev©”P Drcv` cvIqv hv‡e| ZvcgvÎv evo‡j Ges Pvc Kg‡j mvg ̈e ̄'v evgw`‡K m‡i hv‡e Ges m‡ev©wb¤œDrcv` cvIqv hv‡e| 11. †Kvb mvg ̈wewμqvi Rb ̈ Kp > Kc? [For which equillibrium reaction Kp > Kc?] N2(g) + 3H2(g) ⇌ 2NH3(g) H2(g) + I2(g) ⇌ 21+ I(g) 2SO2(g) + O2(g) ⇌ 2SO3(g) COCl2 (g) ⇌ CO(g) + Cl2(g) DËi: H2(g) + I2(g) ⇌ 21+ I(g) e ̈vL ̈v: n = (2 – 2) = 0; Kp = Kc(RT)n n = 1 n‡j, Kp = Kc(RT)  Kp = Kc 12. DfgyLx ivmvqwbK wewμqvi mvg ̈aaæeK Kp I Kc Gi gvb wb‡Pi †KvbwU Øviv cÖfvweZ nq? [The values of the equilibrium constants Kp and Kc for a reversible chemical reaction are influenced by which of the following?] wewμq‡Ki cÕv_wgK NbgvÎv Pvc ZvcgvÎv cÖfveK DËi: ZvcgvÎv e ̈vL ̈v: ivmvqwbK wewμqvi mvg ̈aaæeK Kp I Kc Gi gvb ïaygvÎ ZvcgvÎvi Dci wbf©ikxj| 13. 600 K ZvcgvÎvq PCl5(g) ⇌ PCl3(g) + Cl2 mvg ̈wewμqvi Kc Gi gvb 0.08| 2 L AvqZ‡bi GKwU d¬v‡ ̄‹ KZ †gvj PCl5 †hvM Ki‡j PCl5 Gi we‡qvR‡b Drcbœ Cl2 Gi NbgvÎv 0.4 mol/L n‡e? [At a temperature of 600 K, the equilibrium constant Kc for the reaction PCl5(g) ⇌ PCl3(g) + Cl2 is 0.08. How many moles of PCl5 should be added to a 2L flask so that the concentration of Cl2 produced due to the decomposition of PCl5 is 0.4 mol/L?] 1.25 1.50 4.8 2.5 DËi: 4.8 e ̈vL ̈v: g‡b Kwi, x mol PCl5 †bqv n‡qwQj| mvg ̈wgkÕ‡Yi Cl2 Gi †gvj msL ̈v = 0.4 × 2 = 0.8 mvg ̈we‡qvR‡bi mgxKiYwU: PCl5(g) ⇌ PCl3(g) + Cl2(g) cÕv_wgK Ae ̄'vq NbgvÎv: x 2 0 0 mvg ̈ve ̄'vq NbgvÎv:     x 2 – 0.8 2 0.4 0.4 =     x 2 – 0.4  mvg ̈aaæeK, Kc = [PCl3] × [Cl2] [PCl5] = 0.4 × 0.4     x 2 – 0.4 = 0.08  0.16  x – 0.8 = 0.08  x – 0.8 = 4  x = 4.8 14. †Kv‡bv wbw`©ó ZvcgvÎvq HI Gi 60% we‡qvwRZ n‡q H2 I I2 Drcbœ K‡i| wewμqvwUi mvg ̈aaæe‡Ki gvb KZ? [In a certain temperature, HI decomposes 60% to produce H2 and I2. What is the value of the equilibrium constant for this reaction?] 1 4 9 16 4 †Kv‡bvwUB bq DËi: 9 16
3 e ̈vL ̈v: HI Gi cÖv_wgK NbgvÎv 1 mol/L 2HI ⇌ H2 + I2 mvg ̈ve ̄'vq: 1 – 0.6 0.3 0.3 0.4 mol 0.3 mol 0.3 mol  Kc = [H2] [I2] [HI]2 = 0.32 0.42 = 9 16 15. 0.04 M NH4OH (Kb = 1.6  10–5 ) Gi Rjxq `ae‡Yi pOH KZ? [What is the pOH of a 0.04 M NH4OH (Kb = 1.6 10–5 ) aqueous solution?] 10 + 2 log2 10 + log 4 10 – log 4 3.1 DËi: 3.1 e ̈vL ̈v: [OH] = Kb . C = 1.6  10–5  0.04 = 8  10–4 M  pOH = 4 – log8 = 3.1 16. wb‡Pi †Kvb μgwU mwVK? [Which of the following is the correct order?] C2H5COOH > CH3COOH > HCOOH CH3COOH > HCOOH > C2H5COOH HCOOH > CH3COOH > C2H5COOH HCOOH > C2H5COOH > CH3COOH DËi: HCOOH > CH3COOH > C2H5COOH e ̈vL ̈v: ˆRe Gwm‡Wi †ÿ‡Î A ̈vjKvBj MÕæ‡ci Kve©b msL ̈v evovi mv‡_ mv‡_ ZxeaZv n«vm cv‡e| 17. GKwU g„`y GK ÿviKxq Gwm‡Wi 0.01 M NbgvÎvi Rjxq `ae‡Y pH Gi gvb 5| G GwmW `ae‡Y GwmWwUi we‡qvRb gvÎv- [A weak monoprotic acid has a 0.01 M aqueous solution with a pH of 5. What is the degree of dissociation of the acid in the solution?] 1.0% 0.1% 10% 20% DËi: 0.1% e ̈vL ̈v: GwmW `ae‡Y pH = 5| myZivs `ae‡Y H + Avq‡bi NbgvÎv, [H+ ] = 1.0  10–5 g-ion.L–1 Avgiv Rvwb, `ye©j GK ÿviKxq Gwm‡Wi †ÿ‡Î- [H+ ] =   C   = [H+ ] C = 1.0  10–5 0.01 = 10–3  kZKiv we‡qvR‡bi cwigvY 10–3  100 = 0.1  0.1% we‡qvwRZ nq| 18. †Kvb †ÿ‡Î pH = pKa? [In which case is pH = pKa?] 100 mL 0.1 M CH3COOH + 100 mL 0.1 M NaOH 100 mL 0.1 M CH3COOH + 100 mL 0.1 M NH4OH 100 mL 0.1 M CH3COOH + 30 mL 0.1 M NaOH none DËi: none e ̈vL ̈v: ÔK× Dw3i †ÿ‡Î, 100 mL 0.1 M CH3COOH `aeY 100 mL 0.1 M NaOH †K c~Y© cÕkwgZ Ki‡e| G‡ÿ‡Î AwZwi3 Avi †Kv‡bv GwmW AweK...Z _vK‡e bv| d‡j evdvi `aeYB m„wó Ki‡e bv| d‡j pH = pKa nIqvi cÕkœB Av‡m bv| ÔL× bs Dw3i †ÿ‡Î, 100 mL 0.1 M CH3COOH `aeY 100 mL 0.1 M NH4OH `aeY‡K c~Y© cÕkwgZ Ki‡e| G‡ÿ‡ÎI AwZwi3 Avi †Kv‡bv GwmW ev ÿvi AweK...Z _vK‡e bv| d‡j evdvi `aeYB m„wó Ki‡e bv| d‡j pH = pKa nIqvi cÕkœB Av‡m bv| ÔM× bs Dw3i †ÿ‡Î 30 mL 0.1 M NaOH `aeY 100 mL 0.1 M – COOH `aeY‡K cÕkwgZ Kivi ci AviI AwZwi3 70 mL 0.1 M CH3– COOH †_‡K hv‡e| CH3COOH(aq) + NaOH(aq)  CH3–COONa(aq) + H2O(l) Drcbœ CH3–COONa Gi cwigvY 30 mL Ges gvÎv 0.1 M nq| evdvi `ae‡Yi †gvU AvqZb (100 + 30) = 130 mL pH = pKa + log [CH3COONa] [CH3COOH] = pKa + log 30  0.1 130 70  0.1 130  pH = pKa 19. 0 C ZvcgvÎvq weï× cvwbi H3O + Avq‡bi †gvjvi NbgvÎv, 25C ZvcgvÎvq weï× cvwbi NbgvÎvi mv‡_ m¤úK©Ñ [The molar concentration of H3O + ions in pure water at 0°C is related to the molar concentration of pure water at 4°C as:] Kw (25C) < Kw (0C) Kw (25C) > Kw (0C) [H3O + ] = Kw [H3O + ] Kw 2 DËi: Kw (25C) > Kw (0C) e ̈vL ̈v: T KW ZvB Kw (25C) > Kw (0C) weï× cvwbi [H3O + ] = Kw 20. 27C ZvcgvÎvq GKwU 250 mL †KvìwWasK‡mi †evZ‡j 0.25 gm CO2 M ̈vm Dcw ̄’Z _vK‡j, †evZv‡ji wfZ‡i M ̈v‡mi Pvc KZ? [If 0.25 gm CO2 gas is present in a 250 mL colddrinks bottle at 27°C, what is the gas pressure inside the bottle?] 0.09 atm 0.57 atm 0.15 atm 0.056 atm DËi: 0.57 atm e ̈vL ̈v: PV = nRT  P = W M RT V  P = 0.25 44  1 12  300 0.25  P = 0.57 atm (Ans.) 21. NH3 wK‡mi b ̈vq AvPiY K‡i? [What does NH3 behave like?] jyBm A¤ø jyBm ÿviK †cÕvUbxq GwmW †Kv‡bvwUB bq DËi: jyBm ÿviK e ̈vL ̈v: H – .. N | H – H; †K›`axq cigvYy N GK‡Rvov B‡jKUab MÕn‡Y mÿg| 22. wbw`©ó ZvcgvÎvq †Kvb M ̈v‡mi AYymg~‡ni e ̈vcb nvi me‡P‡q †ewk?
4 [Which gas molecules have the highest diffusion rate at a certain temperature?] NH3 O2 N  N CO2 DËi: NH3 e ̈vL ̈v: N  N  N2 M ̈vm e ̈vcb nvi, r  1 M NH3  17 O2  32 N2  28 CO2  44  NH3 Gi e ̈vcb nvi me‡P‡q †ewk| 23. 0.2 mol ev ̄Íe M ̈v‡mi †ÿ‡Î f ̈vÛviIqvjm mgxKiY Kxiƒc n‡e? [What is the Vanderwaals equation for 0.2 mol of ideal gas?]     5P + a 5V 2 (V – 0.2b) = RT     P + 0.04a V 2 (V – 0.2b) = RT PV = 4RT     P + 0.2a V 2 (V – 0.2b) = 0.2RT DËi:     5P + a 5V2 (V – 0.2b) = RT e ̈vL ̈v:     P + n 2 a V 2 (V – nb) = nRT f ̈vÛviIqvjm mgxKiY| 24. wb‡Pi †KvbwU mwVK? [Which of the following is correct?]  T P  O V1 V2 V3 V3 > V2 > V1 V2 > V1 > V3 V1 = V2 = V3 V1 > V2 > V3 DËi: V3 > V2 > V1 e ̈vL ̈v:  T P  O V1 V2 V3 (T  fixed) n‡j P  1 V  P; V evo‡j Kg‡e  V3 > V2 > V1 25. C cv‡Î NH3 M ̈vm †h Pvc cÖ‡qvM Ki‡e Zvi gvb KZ? [What is the value of pressure that NH3 gas will exert in container C?] 3.4 g NH3 4.4 g CO2 15 atm  2 L + A-cvÎ B-cvÎ C-cvÎ 14.28 atm 15 atm 4.7 atm 10 atm DËi: 10 atm e ̈vL ̈v: C-cv‡Î NH3 I CO2 M ̈vm GK‡Î _vK‡e|  NH3 AvswkK Pvc cÕ`vb Ki‡e| PNH3 = 3.4 17 3.4 17 + 4.4 44  15 = 0.2  15 0.2 + 0.1  PNH3 = 10 atm 26. pH Gi gvb –log(2  108 ) n‡j †dbjd_ ̈vwjb wb‡`©kK †Kvb eY© aviY K‡i? [What color does phenolphthalein indicator have when the pH value is –log(2  108 )?] †e ̧wb bxj eY©nxb njy` DËi: †e ̧wb e ̈vL ̈v: †dbjd_ ̈vwjb A¤øxq gva ̈‡g eY©nxb, ÿvixq gva ̈‡g jvj‡P †e ̧wb| Gi eY© cwieZ©‡bi pH cwimi 8.2 – 10| 27. H2 I O2 Gi wewμqvq cvwb Drcbœ n‡j, 2 g O2 n‡Z KZ MÖvg cvwb cvIqv hv‡e? [If H2 and O2 react to produce water, how many grams of water can be obtained from 2 g of O2?] 0.225 g 2.25 g 20.5 g 22.5 g DËi: 2.25 g e ̈vL ̈v: 2H2(g) + O2(g)  2H2O(l) wewμqv n‡Z, nO2 = nH2O 2  2 32 = W 18  2  W = 36  2 32 = 2.25 g 28. †h cwigvY Kve©b WvB A•vBW 5 g CaCO3 †K DËß Ki‡j cvIqv hvq Zv m¤ú~Y©iƒ‡c Kw÷K †mvWvi mv‡_ wewμqv Kiv‡bv n‡jv| wewμqv m¤ú~Y© Ki‡Z wK cwigvY Kw÷K †mvWv jvM‡e? [The amount of carbon dioxide obtained by heating 5g of CaCO3 is completely reacted with caustic soda. What amount of caustic soda will be required to complete the reaction?] 1.8 g 4 g 6.2 g 5 g DËi: 4 g e ̈vL ̈v: CaCO3  CaO + CO2 100 g 44 g 2NaOH + CO2  Na2CO3 + H2O wewμqvØq n‡Z, nCaCO3 = nNaOH 2