Title Chemistry- Subject final-Home Practice-Solution.pdf ✅

1 Varsity Subject Final Chemistry [Home Practice (Solution)] c~Y©gvb: 140 †b‡MwUf gvK©: 0.25 mgq: 1 NÈv 30 wgwbU 1. a aM I b cO †gŠj `yBwUi g‡a ̈ b–a = 5 n‡j, O †gŠjwU †Kvb †kÖwYi? [If b–a = 5 between elements a aM and b cO, then element O is of which class?] 4 18 14 16 DËi: 14 e ̈vL ̈v: a aM = 1 1 H  a = 1 b – a = 5 b = 6 b cM = 6C C = 1s2 2s2 2P2 MÖæc = 4 + 10 = 14 2. Mg cigvYyi cvigvYweK eY©vwji evgvi wmwi‡Ri Z...Zxq jvB‡bi Zi1⁄2‰`N© ̈ KZ? [What is the wavelength of the third line of the bamer series of atomic spectrum of Mg atom?] 33 784 RH 25 756RH 784 3993RH 784 33 RH DËi: 25 756RH e ̈vL ̈v: eav‡KU wmwi‡Ri 3q jvBb, n2 = n1 + 3 = 2 + 3 = 5  1  = RH     1 n 2 1 – 1 n 2 2  Z 2  1  = RH     1 2 2 – 1 5 2  122   = 25 756RH 3. H cigvYyi 3q kw3 ͇̄i e – Av‡Q| H + Avqb ˆZwi‡Z wK cwigvY kw3 †`qv jvM‡e? [H atom has e– in 3rd energy level. What amount of energy will be required to produce H+ ions?] 1.51eV – 0.85 eV – 13.6 eV 3.4 eV DËi: 1.51eV e ̈vL ̈v: H + Avqb ˆZwi Ki‡Z cÖ‡qvRbxq kw3. E = E() – E(3) = 0 – (– 1.51) eV = 1.51 eV 4. S = – 1 2 , 0, 1 2 n‡j n = 5 Gi Rb ̈ †gvU B‡jKUab msL ̈v KZ? [What is the total number of electrons for n = 5, if S = – 1 2 , 0, 1 2 ?] 72 wU 108 wU 75 wU 102 wU DËi: 75 wU e ̈vL ̈v: hLb, S =  1 2 ZLb, cÖavb †Kvqv›Ug msL ̈v n n‡j kw3 ͇̄i †gvU e – msL ̈v 2n2  S = 0,  1 2 n‡j, e – msL ̈v = 3n2 B‡jKUab msL ̈v = 3  5 2 = 75 wU 5. Hg2I4 + 2NH3  NH2[Hg2I3] + NH4I Aat‡ÿcwUi eY© Kx? [What is the color of the precipitate formed in the reaction Hg2I4 + 2NH3  NH2[Hg2I3] + NH4I?] bxj nvjKv njy` meyR jvj‡P ev`vwg DËi: jvj‡P ev`vwg e ̈vL ̈v: cixÿY: GKwU cixÿvb‡j 1 – 2 mL g~j `aeY wb‡q Zv‡Z 2 – 4 †duvUv †bmjvi `aeY †hvM Kiv nq| ch©‡eÿY: jvj‡P ev`vwg Aat‡ÿc c‡o wewμqv: 2NH+ 4 + 2KOH  2NH3 + 2K+ + 2H2O 2K2[HgI4]  4KI + 2HgI2 2HgI2 ⇌ Hg2I4 Hg2I4 + 2NH3  NH2[Hg2I3] + NH4I A ̈vwg‡bv gviwKDwiK Av‡qvWvBW (jvj‡P ev`vwg) wm×všÍ: NH+ 4 g~jK Dcw ̄’Z| 6. 50 mL 0.01 M A3B2 Ges 50 mL 0.02 M XY `aeYØq wgwkÖZ Ki‡j, `ae‡Y Aa:‡ÿc co‡e wKbv? †hLv‡b Ksp(AY2) = 2.3  10–11 [If 50 mL of 0.01 M A3B2 solution is mixed with 50 mL of 0.02 M XY solution, will a precipitate form in the solution? Where Ksp(AY2) = 2.3 × 10−11.] Aa:‡ÿc co‡e bv Aa:‡ÿc co‡e (i) I (ii) DfqB mwVK †Kv‡bvwU bq DËi: Aa:‡ÿc co‡e e ̈vL ̈v: A3B2 ⇌ 3A2+ + 2B–3 XY2 ⇌ X 2+ + 2Y– 0.02 0.02 0.04  AY2 ⇌ A 2+ + 2Y–  Kip = [A+2][Y– ] 2 =     S1V1 V1 + V2      S2V2 V1 + V2 2 =     0.03  50 100      0.04  50 100 2
2 = 6  10–6 > Ksp  Aa:‡ÿc co‡e| 7. cvwb‡Z Mg(OH)2 Gi †gvjvi `ave ̈Zv 2  10–3 M. pH = 12 wewkó evdv‡i Mg(OH)2 Gi `ave ̈Zv KZ? [If the molar solubility of Mg(OH)2 in water is 2 × 10−3 M.What is the solubility of Mg(OH)2 in a buffer solution with a pH = 12?] 2.64  10–4 M 6.2  10–4 M 32  10–5 M 3.2  10–4 M DËi: 3.2  10–4 M e ̈vL ̈v: pH = 12, pOH = 14 – 12 = 2 [OH– ] = 10–2 [evdvi `ae‡Y] Mg(OH)2 ⇌ Mg2+ + 2OH– S 2S Ksp = S  (2S)2 = 4S3 = 4  (2  10–3 ) 3 = 4  8  10–9 = 32  10–9  [Mg2+] = 32  10–9 10–4 = 3.2  10–4 M 8. Fe2+ Gi mv‡_ wewμqvq K3[Fe(CN)6] (aq) †Kvb e‡Y©i Aat‡ÿc †`q? [What color precipitate is formed when Fe2+ reacts with K3 Fe(CN)6?] Mvp bxj ev`vwg jvj nvjKv bxj DËi: Mvp bxj 9. wb‡Pi †KvbwU †hvM Ki‡j AgCN Gi `ave ̈Zv e„w× cv‡e? [Which of the following, when added, will increase the solubility of AgCN?] H2S KI KCN KCl DËi: KCN 10. 1 L AvqZ‡bi GKwU Ave× cv‡Î 92 g N2O4(g) wb‡q DËß K‡i we‡qvRb Kiv n‡jv- N2O4(g) ⇌ 2NO2(g), mvg ̈ve ̄’vq 50% N2O4(g) we‡qvwRZ n‡j, mvg ̈aaæeK Kc Gi gvb- [92 g N2O4(g) was heated in a 1 L closed vessel. The reaction N2O4(g) ⇌ 2NO2(g), took place. If 50% of N2O4(g) dissociated at equilibrium, what is the value of the equilibrium constant ?] 0.01 2 0.2 0.4 DËi: 2 e ̈vL ̈v: N2O4 Gi MÖvg GK‡K AvYweK fi = 14 × 2 + 16 × 4 = 92 gmol–1 92 g N2O4 Gi g‡a ̈ Dnvi †gvj msL ̈v = 92 92 = 1 mol N2O4(g) ⇌ 2NO2(g) we‡qvR‡bi c~‡e©: 1 mol 0 mvg ̈ve ̄’vq:     1 – 1 × 50 100 2 × 1 × 50 100 0.5 mol 1 mol cv‡Îi AvqZb 1 L nIqvq mvg ̈ve ̄’vq N2O4 Gi NbgvÎv (N2O4) (N2O4) = 0.5 mol L–1 Kc = [NO2] 2 [N2O4] = (1) 2 (0. 5) = 2 11. 60C ZvcgvÎvq WvBbvB‡Uav‡Rb †UUavA•vB‡Wi 80% we‡qvRb N‡U| ZLb †gvU Pvc 1 atm n‡j, we‡qvR‡bi †ÿ‡Î mvg ̈aaæeK Kp Gi gvb- [At 60 C, dinitrogen tetroxide undergoes 80% dissociation. If the total pressure is 1 atm, what is the value of the equilibrium constant Kp?] 1 3 atm 2 3 atm 64 9 atm 2 atm DËi: 64 9 atm e ̈vL ̈v: N2O4 ⇌ 2NO2 1– .8 1.6 †gvj msL ̈v 1.6 + .2 = 1.8 mol  PNO2 = 1.6 1.8 × 1 PN2O4 = 0.2 1.8 × 1 Kp = P 2 NO2 PN2O4 =     1.6 1.8 2 × 12 0.2 1.8 = 64 9 12. KZ ZvcgvÎvq I Pv‡c CO2 GKwU Super Critical fluid wn‡m‡e KvR K‡i? [At what temperature and pressure does CO2 act as a Supercritical fluid?] – 267.65C, 1.26 atm – 290C, 12.8 atm 31.1C, 72.9 atm CO2 KL‡bv Super critical fluid wnmv‡e KvR K‡i bv DËi: 31.1C, 72.9 atm 13. 1 mole cUvwkqvg †K¬v‡iU‡K DËß Ki‡j KZ †gvj O2 M ̈vm cvIqv hv‡e? [When 1 mole potassium chlorate is heated, how many moles of O2 gas are obtained?] 1 mole 3 2 mole 2 mole 3 mole DËi: 3 2 mole e ̈vL ̈v: 2kClO3(s)  2kCl(s) + 3O2(g) 14. GKwU M ̈vmxq wewμqvq nv‡ii mgxKiY K[A][B]| hw` cv‡Îi AvqZb 1 3 ̧Y Kiv nq, Z‡e bZzb nvi cÖ_g nv‡ii KZ ̧Y n‡e? [For a gaseous reaction, the rate equation is K[A][B] If the volume of the vessel is 1 3 , by what factor will the new rate be compared to the initial rate?] 1 10 1 8 3 9 DËi: 9 e ̈vL ̈v: R1 = K[A][B]
3 R2 = K[A][B] [A] = 3[A] [B] = 3[B]  R2 = 9K[A][B] = 9R1 15. weï× cvwbi NbgvÎv KZ? [What is the molarity of pure water?] 58.5 M 59.25 21.50 55.5M DËi: 55.5M e ̈vL ̈v: H2O  1L  1000 ml = 1000 gm 18 gm = 55.5 mole d = 1 gm ml d = M v  1 = M 1000  M = 1000 gm 16. H–COOH `ae‡Yi P H gvb 3 Gi `ae‡Y H– COOH Gi cwigvY 1  10–2 molL–1 n‡j, H–COOH Gi †ÿ‡Î Ka Gi gvb wb‡Pi †KvbwU? [If the amount of H-COOH is 1 × 10-2 molL-1 in a solution of PH value 3 of H-COOH solution, which of the following is the value of Ka for H-COOH?] 1.70  10–5 10–4 2.56  10–4 8.42  10–4 DËi: 10–4 e ̈vL ̈v: Ka = [HCOO– ] [H+ ] [HCOOH] P H = – log [H+ ]  [H+ ] = antilog [–P H ] = antilog (– 3) = 10–3 M [H+ ] = [HCOO– ] = 10–3M Ka = (10–3 ) 2 1  10–2 Ka = 10–4 17. wb‡Pi †KvbwU ÿvixq evdvi `aeY? [Which of the following is an alkaline buffer solution?] HCN + CN – H3PO4 + NaH2 PO4 C5H5N + C5H5NH+ Na2HPO4 + Na3 PO4 DËi: C5H5N + C5H5NH+ e ̈vL ̈v: ÿvixq evdvi [`ye©j ÿviK + AbyeÜx GwmW] 18. 0.02 M NH4OH `ae‡Y P H = ? [†hLv‡b Kb = 2  10–6 ] [P H in 0.02 M NH4OH solution =? [where Kb = 2  10–6 ]] 14 + 3 + log10 2 11 + log10 2 13 + log10 2 10 + log10 2 DËi: 10 + log10 2 e ̈vL ̈v:  = Kb c = 2  10–5 0.02 = 10–2  100  = 1% < 10% [O– H] = c = 0.01  0.02 = 2  10–4 P OH = – log 10 [OH– ] = 4 – log10 2 P OH = 14 –4 + log10 2 = 10 + log10 2 19. 3.0 PH Gi evdvi `aeY ˆZwi Ki‡Z 50 ml 0.5 M diwgK Gwm‡Wi `aeY KZ mL 1M †mvwWqvg di‡gU `aeY †hvM Ki‡Z n‡e? [awi, HCOOH Gi pka = 2] [How many mL of 1M sodium formate solution must be added to 50 mL of 0.5 M formic acid solution to make a buffer solution of PH 3.0? [Assume, pka of HCOOH is 2]] 200 ml 500 ml 400 ml 250 ml DËi: 250 ml e ̈vL ̈v: P H = pka + log [salt] [Acid]  3 = 2 + log ns na  1 = log     1  x 0.5  50  10 = x 0.5  50  x = 250 ml 20. wb‡Pi †KvbwU mwVK? [Which of the following is correct?] V  T3 PV  T2 T1 T1 > T2 > T3 T3 < T2 < T1 T3 > T2 > T1 both K I L DËi: both K I L e ̈vL ̈v: "V  w ̄’i n‡j" kZ©g‡Z, V  T3 PV  T2 T1 V  fixed PV  T AZGe, PV evov‡j; T evo‡e PV Kg‡j; T Kg‡e  T1 > T2 > T3 Avevi, T3 < T2 < T1 GKB 21. H2 M ̈v‡mi NbZ¡ 25 ZvcgvÎvq 2.1 gm dm–3 ; Pvc w ̄’i _vK‡j 596K ZvcgvÎvq M ̈vmwUi NbZ¡ KZ n‡e? [The density of H2 gas at 25° is 2.1 gm dm–3 ; What will be the density of the gas at 596K if the pressure is constant?]
4 0.525 gm dm–3 0.25 gm dm–3 1.25 gm dm–3 1.05 gm dm–3 DËi: 1.05 gm dm–3 e ̈vL ̈v: p  Constant T1d1 = T2d2  d2 = T1d1 T2 = 298  2.1 596  d2 = 1.05 gm dm–3 (Ans.) 22. GKwU 12.5 L AvqZ‡bi cv‡Î CxHy ms‡KZ wewkó M ̈vm i‡q‡Q| 27C ZvcgvÎvq I 2 atm Pv‡c M ̈vm AYywUi Kve©b I nvBWa‡Rb cigvYyi msL ̈v h_vμ‡g 1.8  1024 wU I 3.61  1024 wU, M ̈vmwUi cÖK...Z ms‡KZÑ [A container of volume 12.5 L contains CxHy, the characteristic gas. At 27°C and 2 atm pressure, the number of carbon and hydrogen atoms in a gas molecule is 1.8  1024 and 3.61  1024 respectively, the actual sign of the gas is-] C2H6 C2H4 C3H6 C3H8 DËi: C3H6 e ̈vL ̈v: GLv‡b, n = PV RT = 2  12.5 1 12  300 = 1 mol kZ©g‡Z, x  1 mol  6.023  1023 = 1.8  1024  x = 3  y  1 mol  6.023  1023 = 3.61  1024  y = 6  wb‡Y©q ms‡KZ, CxHy  C3H6 23. A ̈v‡gvwbqv M ̈vm 2 atm Pv‡c I dmwdb M ̈vm 1 atm Pv‡c e ̈vwcZ n‡Z ïiæ K‡i, †KvbwUi e ̈vcb nvi me©vwaK? [Ammonia gas begins to diffuse at 2 atm pressure and phosphine gas begins to diffuse at 1 atm pressure, which has the highest rate of expansion?] dmwd‡bi Dfq GK‡Î e ̈vwcZ n‡e A ̈v‡gvwbqvi mwVK DfqwU †bB DËi: A ̈v‡gvwbqvi e ̈vL ̈v: e ̈vcb nvi, r  P Ges r  1 M r1  NH3 r2  PH3(dmwdb) r1 r2 = P1 P2  M2 M1  r1 r2 = 2 atm 1 atm  34 17  r1 r2 = 2  2  r1 = 2 2 r2  r1 > r2 A_©vr, A ̈v‡gvwbqvi e ̈vcb nvi †ewk| 24. [Cu(NH3)4] 2+ Avqb MV‡b NH3 Kx wn‡m‡e KvR K‡i? [What role does NH3 play in forming [Cu(NH3)4] 2+ ion?] jyBm GwmW jyBm ÿviK †cÖvUbxq ÿviK RwUj Avqb DËi: jyBm ÿviK e ̈vL ̈v: 4       H – H | N: | H Cu2+| NH3 GK‡Rvov B‡jKUab Cu2+ Avqb‡K `vb K‡i ZvB jyBm ÿviK| 25. cvwb‡Z Ca2+ Gi cwigvY KZ n‡j, H cvwb‡Z ̄’vqx LiZv ˆZwi n‡e? [What amount of Ca2+ in water will cause permanent ordity in that water?] 100 ppm < 10 ppm > 150 ppm < 100 ppm DËi: > 150 ppm e ̈vL ̈v: cvwb‡Z Ca2+ I Mg2+ Avq‡bi cwigvY 150 ppm (mg L–1 ) Gi †ewk n‡j (> 150 ppm) cvwb‡Z ̄’vqx LiZv ˆZwi nq| 26. dzUe‡ji e ̈vmva©2 cm n‡j, dzUe‡ji †fZi Pvc KZ? [If the diameter of the football is 2 cm, what is the pressure inside the football?] 27C P = 15 kPa V = 180 cm3 dzUej 80 kPa 80 Pa 10 Pa 15 kPa DËi: 80 kPa e ̈vL ̈v: GLv‡b, V2 = 4 3 r 3 = 4 3  3.14  2 3 = 33.5 cm3 kZ©g‡Z, P1V1 = P2V2  P2 = P1V1 V2 = 15  180 33.5  80 kPa 27. 100 mL 0.1 M HCl Ges 100 mL 0.1 N H2SO4 Gi wgkÖY †K cÖkwgZ Ki‡Z 0.2 N Na2CO3 Gi KZ AvqZb jvM‡e? [What volume of 0.2 N Na2CO3 is required to neutralize a mixture of 100 mL 0.1 M HCI and 100 mL 0.1 N H2SO4?] 220 mL 100 mL 300 mL 200 mL DËi: 100 mL e ̈vL ̈v: (eVS)H2SO4 + (eVS)HCl = (eVS)NaOH  0.1  100 + 0.1  100 = 0.2  V V = 100 mL 28. 17 g PH3 M ̈v‡m †gvU cigvYyi msL ̈v KZ? [What is the total number of atoms in 17 g of PH3 gas?] 1.204  1024 3.01  1023 12.04  1024 1.8  1023 DËi: 1.204  1024 e ̈vL ̈v: n = W M = 17 34 = 0.5 mol †gvU cigvYy= 4  NA  0.5 = 1.204  1024