Tiêu đề 3. Che. Varsity Practice Sheet (2nd Paper) With Solve.pdf ✅

cwigvYMZ imvqb  Varsity Practice Sheet ........................................................................................................................ 1 MCQ weMZ mv‡j DU-G Avmv cÖkœvejx 1. 2.5 MÖvg CaCO3 †_‡K NTP †Z Kx cwigvY CO2 Drcbœ n‡e? [DU 23-24] 22.4 L 0.56 L 5.6 L 11.2 L DËi: 0.56 L e ̈vL ̈v: CaCO3  CaO + CO2 100 g 22.4 L 100 g CaCO3 †_‡K CO2 cvIqv hvq = 22.4 L  2.5 g CaCO3 †_‡K CO2 cvIqv hvq = 22.4 × 2.5 100 = 0.56 L weKí: wewμqvwUi stoichiometry Abyhvqx, nCaCO3 1 = nCO2 1  WCaCO3 MCaCO3 = VCO2 22.4  VCO2 = 2.5 100 × 22.4 = 0.56 L 2. 100 mL NaOH Gi `ae‡Y 0.5 g NaOH Av‡Q| GB `ae‡Yi NbgvÎv ppm GK‡K KZ n‡e? [DU 23-24] 50000 50 500 5000 DËi: 5000 e ̈vL ̈v: NaOH Gi NbgvÎv = W(g) V(mL) × 106 = 0.5 100 × 106 = 5000 ppm 3. (NH4)3[Fe(CN)6] †hŠ‡M Avqi‡bi RviY Ae ̄’v KZ? [DU 23-24] + 5 + 4 + 3 + 2 DËi: + 3 e ̈vL ̈v: (NH4)3[Fe(CN)6] †hŠ‡M Fe Gi RviY msL ̈v x n‡j (+ 1)  3 + x + (– 1) × 6 = 0  3 + x – 6 = 0  x = + 3 4. 100 mL 0.2 M Na2CO3 Gi Rjxq `aeY‡K cÖkwgZ Ki‡Z KZ AvqZ‡bi 0.4 M HCl cÖ‡qvRb n‡e? [DU 22-23] 50 mL 25.0 mL 100.0 mL 10.0 mL DËi: 100.0 mL e ̈vL ̈v: (e1V1S1)acid = (e2V2S2)base  1 × V1 × 0.4 = 2 × 100 × 0.2  V1 = 100 mL HCl Gi Zzj ̈ msL ̈v, e1 = 1 Na2CO3 Gi Zyj ̈ msL ̈v, e2 = 2 5. wb‡Pi †KvbwU AmvgÄm ̈KiY wewμqv? [DU 21-22] 2Na + Cl2  2NaCl AgNO3 + NaCl  NaNO3 + AgCl Cl2 + H2O  HCl + HOCl FeCl3 + SnCl2  SnCl4 + FeCl2 DËi: Cl2 + H2O  HCl + HOCl e ̈vL ̈v: †h wewμqvq GKwU wewμqK c`v‡_©i GKBmv‡_ RviY I weRviY N‡U Zv‡K AmvgÄm ̈Zv wewμqv e‡j| C 0 l2 + H2O  HCl –1 + HOCl +1 GLv‡b Cl Gi GKB mv‡_ RviY Ges weRviY n‡q‡Q| ZvB GB wewμqvwU AmvgÄm ̈KiY wewμqv| 6. 0.98 g H2SO4 e ̈envi K‡i 1.0 L Rjxq `aeY •Zwi Kiv n‡j `aeYwUi NbgvÎv KZ? [DU 19-20] 0.1 M 0.2 M 0.01 M 0.001 M DËi: 0.01 M e ̈vL ̈v: S = W MV = 0.98 98 × 1 = 0.01 M 7. BaMnF4 Ges Li2MgFeF6 †hŠM؇q Mn I Fe Gi RviY msL ̈v h_vμ‡gÑ [DU 19-20] + 5, + 3 + 5, + 2 + 4, + 3 + 2, + 2 DËi: + 2, + 2 e ̈vL ̈v: BaMnF4 †hŠ‡M Mn Gi RviY msL ̈v x n‡j, + 2 + x + (– 1 × 4) = 0  x = + 2 Li2MgFeF6 †hŠ‡M Fe Gi RviY msL ̈v x n‡j, (+ 1 × 2) + 2 + x + (– 1 × 6) = 0  x = + 2
2 ....................................................................................................................................  Chemistry 2nd Paper Chapter-3 8. wb‡Pi †KvbwU‡Z †ewk cigvYy Av‡Q? [DU 18-19] 1.10 g of hydrogen atoms 14.7 g of chromium atoms 2.0 g of helium atoms 7.0 g of nitrogen atoms DËi: 1.10 g of hydrogen atoms e ̈vL ̈v: n = N NA  n  N A_©vr, hvi †gvjmsL ̈v me©vwaK Zvi cigvYy msL ̈vI me‡P‡q †ewk| nH = 1.10 1 = 1.10 mol NH = nH  NA = (1.10  6.023  1023) = 6.63  1023wU cigvYy 9. cÖgvY ZvcgvÎv I Pv‡c (STP) †Kvb M ̈v‡mi 1.0 MÖvg me‡P‡q †ewk AvqZb `Lj K‡i? [DU 18-19] N2 H2 O2 Ar DËi: H2 e ̈vL ̈v: n = W M = VSTP 22.4  VSTP  1 M (W w ̄’i) A_©vr, w ̄’i f‡i †h M ̈v‡mi AvYweK fi hZ Kg Zvi AvqZb ZZ †ewk nq| AvYweK f‡ii μg: H2 < N2 < O2 < Ar  H2 me‡P‡q †ewk AvqZb `Lj Ki‡e| weKí: VN2 = 1 28 × 22.4 = 0.8 L VH2 = 1 2 × 22.4 = 11.2 L VO2 = 1 32 × 22.4 = 0.7 L VAr = 1 40 × 22.4 = 0.56 L 10. 2.2 g C3H8 c~Y© `nb K‡i CO2 I H2O •Zwi Ki‡Z KZ †gvj O2 cÖ‡qvRb? [DU 18-19] 0.05 0.15 0.25 0.50 DËi: 0.25 e ̈vL ̈v: C3H8 + 5O2  3CO2 + 4H2O 44 g 5 mol 44 g C3H8 †_‡K O2 Drcbœ nq = 5 mol  2.2 g C3H8 †_‡K O2 Drcbœ nq = 5 × 2.2 44 = 0.25 mol 11. 0.125 M HCl Gwm‡Wi 500 wg.wj. `aeY‡K 0.1 M jNy `ae‡Y cwiYZ Ki‡Z KZUzKz cvwb †hvM Ki‡Z n‡e? [DU 18-19] 100 mL 150 mL 125 mL 75 mL DËi: 125 mL e ̈vL ̈v: V1S1 = V2S2  V2 = 500 × 0.125 0.1 = 625 mL  cvwb †hvM Ki‡Z n‡e = (625 – 500) = 125 mL 12. 2KNO3  2KNO2 + O2 wewμqvwU‡Z RvwiZ I weRvwiZ †gŠj h_vμ‡g Kx Kx? [DU 18-19] nitrogen and oxygen oxygen and nitrogen potassium and oxygen nitrogen and potassium DËi: oxygen and nitrogen e ̈vL ̈v: +1 +5 –6 weRviY +1 +3 –4 0 2KNO3  2KNO2 + O2 RviY GLv‡b, Aw·‡R‡bi RviY msL ̈v e„w× cvq Ges bvB‡Uav‡R‡bi RviY msL ̈v n«vm cvq| ZvB Aw·‡Rb RvwiZ nq Ges bvB‡Uav‡Rb weRvwiZ nq| 13. wb¤œwjwLZ RviY-weRviY wewμqvi Drcv`mg~n Kx? K2Cr2O7 + H2SO4 + FeSO4  Products (Drcv`mg~n) [DU 18-19] K2SO4, Fe2(SO4)3, H2O K2SO4, Cr2(SO4)3, H2O Cr2(SO4)3, Fe2(SO4)3, H2O K2SO4, Cr2(SO4)3, Fe2(SO4)3, H2O DËi: K2SO4, Cr2(SO4)3, Fe2(SO4)3, H2O e ̈vL ̈v: K2Cr2O7 + 7H2SO4 + 6FeSO4  3Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + 7H2O 14. cÖgvY Ae ̄’vq 10.0 L wg‡_b M ̈v‡m AYyi msL ̈v KZ? [DU 17-18] 2.689 × 1023 26.89 × 1023 0.2689 × 1023 26.89 × 1025 DËi: 2.689 × 1023 e ̈vL ̈v: n = N NA = VSTP 22.4  N = VSTP × NA 22.4 = 10 × 6.023 × 1023 22.4 = 2.689 × 1023 wU 15. wb‡Pi †KvbwU RviY-weRviY wewμqv bq? [DU 17-18, 05-06] 2Na + Cl2  2NaCl SnCl2 + FeCl3  SnCl4 + FeCl2 Cu + HNO3  Cu(NO3)2 + NO + H2O NaCl + AgNO3  NaNO3 + AgCl DËi: NaCl + AgNO3  NaNO3 + AgCl e ̈vL ̈v: NaCl + AgNO3  NaNO3 + AgCl D3 wewμqvq †Kv‡bv †gŠ‡ji RviY msL ̈vi †Kv‡bv cwieZ©b N‡U wb, ZvB GwU RviY-weRviY wewμqv bq|
cwigvYMZ imvqb  Varsity Practice Sheet ....................................................................................................................... 3 16. MnO– 4 Avqb‡K ethanedioate ion w`‡q weRvwiZ Ki‡j Mn Gi RviY gvb n‡jvÑ [DU 17-18] 7 + 4 + 2 + 3 + DËi: 2 + e ̈vL ̈v: 2MnO – 4 + 16H+ + 5C2O 2– 4  2Mn2+ + 8H2O + 10CO2 17. 5 L 0.1 M `aeY •Zwi Ki‡Z Kx cwigvY Na2CO3 cÖ‡qvRb? [DU 16-17] 106 g 53 g 10.6 g 5.3 g DËi: 53 g e ̈vL ̈v: W = SMV = (0.1 × 106 × 5) = 53 g 18. 1 wjUvi †Wwm‡gvjvi Na2CO3 `ae‡Y KZ MÖvg Na2CO3 _vK‡e? [DU 15-16] 5.3 g 10.6 g 16.6 g 53.6 g DËi: 10.6 g e ̈vL ̈v: W = SMV = (0.1 × 106 × 1) = 10.6 g 19. weï× cvwbi NbgvÎv (†gvj/wjUvi) n‡jvÑ [DU 15–16; CU 14-15] 35.5 1.0 55.5 18.0 DËi: 55.5 e ̈vL ̈v: S = 1000W MV = 1000 × 1 18 × 1 = 55.5 mol L–1 [1 g cvwbi AvqZb 1 cm3 ev 1 mL] 20. SnCl2 + 2FeCl3  SnCl4 + 2FeCl2 wewμqvq †KvbwU mZ ̈? [DU 15-16; RU-C 21-22] Sn RvwiZ n‡q‡Q Fe RvwiZ n‡q‡Q Cl RvwiZ n‡q‡Q Cl weRvwiZ n‡q‡Q DËi: Sn RvwiZ n‡q‡Q e ̈vL ̈v: weRviY +2 –2 SnCl2 + FeCl3 SnCl4 + 2FeCl2 RviY +3 –3 +4 –4 +2 –2 GLv‡b, Sn2+ RvwiZ n‡q Sn4+ G cwiYZ nq Ges Fe3+ weRvwiZ n‡q Fe2+ G cwiYZ nq| 21. wb‡¤œi mgZvK...Z wewμqvi mnM ̧‡jvi gvb n‡jvÑ aNH3 + bO2  cNO + dH2O [DU 14-15] a = 2, b = 3, c = 2 and d = 3 a = 4, b = 7, c = 4 and d = 4 a = 4, b = 5, c = 4 and d = 6 a = 6, b = 7, c = 6 and d = 9 DËi: a = 4, b = 5, c = 4 and d = 6 e ̈vL ̈v: 4NH3 + 5O2  4NO + 6H2O 22. GKwU †ivMxi i‡3i Møy‡Kv‡Ri cwigvY 10 mmol L–1 | mg/dL GK‡K Gi gvb KZ? [DU 13-14] 180 18.0 1.80 None DËi: 180 e ̈vL ̈v: 10 mmol/L = 10  10–3  180  1000 10 = 180 mg/dL Shortcut: Møy‡Kv‡Ri cwigvY‡K mg/dL †K mmol/L G wb‡Z n‡j 18 w`‡q fvM Ki‡Z nq| Avevi, mmol/L †K mg/dL G wb‡Z n‡j 18 w`‡q ̧Y Ki‡Z n‡e| 23. GKwU 1.0 M †mvwWqvg nvB‡Wav·vB‡Wi Rjxq `ae‡Yi 100 mL †K m¤ú~Y©iƒ‡c wbi‡cÿ Ki‡Z KZ AvqZ‡bi 0.5 M A·vwjK GwmW cÖ‡qvRb co‡e? [DU 13-14] 50 mL 100 mL 200 mL 400 mL DËi: 100 mL e ̈vL ̈v: (e1V1S1)acid = (e2V2S2)base  2 × V1 × 0.5 = 1 × 100 × 1  V1 = 100 mL A·vwjK GwmW (HOOC – COOH) Gi Zzj ̈ msL ̈v, e1 = 2 24. S2O 2– 3 Ges S4O 2– 6 mvjdv‡ii RviY msL ̈v njÑ [DU 13-14] – 2 and – 2.5 + 2 and + 2.5 + 4 and + 6 + 2 and – 2 DËi: + 2 and + 2.5 e ̈vL ̈v: (i) S2O 2– 3 Avq‡b S Gi RviY msL ̈v, 2x + (– 2  3) = – 2  x = + 2 (ii) S4O 2– 6 Avq‡b S Gi RviY msL ̈v, 4x + (– 2  6) = – 2  x = + 2.5 25. 10 g Aw·‡R‡b AYyi msL ̈v KZ? [DU 12-13, 07-08; Agri. Guccho 20-21, 19-20] 3.76 × 1023 6.02 × 1022 9.63 × 1023 1.88 × 1023 DËi: 1.88 × 1023 e ̈vL ̈v: n = W M = N NA  N = W M × NA = 10 32 × 6.023 × 1023 = 1.88 × 1023wU 26. 10.0 wg.wj. NaOH `aeY‡K 0.12 M NbgvÎvi 15.0 wg.wj. A·vwjK GwmW Øviv UavB‡Uakb Ki‡j cÖkgb we›`y cvIqv hvq| NaOH Gi NbgvÎv KZ? [DU 12-13] 0.25 M 0.36 M 0.32 M 0.40 M DËi: 0.36 M e ̈vL ̈v: (e1V1S1)base = (e2V2S2)acid  1 × 10 × S1 = 2 × 15 × 0.12  S1 = 0.36 M
4 ....................................................................................................................................  Chemistry 2nd Paper Chapter-3 27. 0.5 M H2SO4 Gwm‡Wi 20.5 mL Øviv 20 mL Kw÷K †mvWv `aeY cÖkwgZ nq| H ÿvi `ae‡Yi †gvjvwiwU KZ?[DU 11-12] 1.025 mol L –1 0.1025 mol L–1 0.5125 mol L–1 0.025 mol L–1 DËi: 1.025 mol L –1 e ̈vL ̈v: (e1V1S1)base = (e2V2S2)acid  1 × 20 × S1 = 2 × 20.5 × 0.5  S1 = 1.025 mol L–1 28. 18 g Møy‡KvR AYy‡Z KZwU Kve©b cigvYy _vK‡e? [DU 10-11; SUST 18-19] 6.0 × 1023 6.0 × 1022 3.6 × 1023 3.6 × 1024 DËi: 3.6 × 1023 e ̈vL ̈v: 1wU Møy‡KvR AYy‡Z 6wU C cigvYy we` ̈gvb| MC6H12O6 = 180 mol–1  N = W M × NA × 6 = 18 180 × 6.023 × 1023 × 6 = 3.6 × 1023wU 29. 2.00 g NaOH 50.00 mL `ae‡Y `aexf~Z _vK‡j H NaOH `ae‡Yi †gvjvwiwU KZ? [DU 10-11] 0.10 M 0.50 M 1.00 M 2.00 M DËi: 1.00 M e ̈vL ̈v: S = 1000W MV = 1000 × 2 40 × 50 = 1.00 M 30. cUvwkqvg cvig ̈v1⁄2v‡bU-†mvwWqvg A·v‡jU UvB‡Uak‡b wb‡¤œi †KvbwU Zzwg e ̈envi Ki‡e? [DU 10-11; CU 14-15; Agri. Guccho 20-21] methyl orange starch diphenylamine no indicator DËi: no indicator e ̈vL ̈v: cUvwkqvg cvig ̈v1⁄2v‡bU w`‡q UvB‡Uak‡b †Kv‡bv wb‡`©k‡Ki cÖ‡qvRb nq bv KviY KMnO4 Gi eY© AZ ̈šÍ Zxea Ges 100 mL cvwb‡Z 0.1 mL 0.02 M KMnO4 †hvM Ki‡j Gi my ̄úó nvjKv wc1⁄2j eY© †`Lv hvq| A_©vr cUvwkqvg cvig ̈v1⁄2v‡bU ̄^-wb‡`©kK wn‡m‡e KvR K‡i| 31. 7.1 g †K¬vwi‡bi g‡a ̈ KZ †gvj Cl2 i‡q‡Q? [DU 09-10] 0.1 mol 1.0 mol 0.2 mol 0.4 mol DËi: 0.1 mol e ̈vL ̈v: n = W M = 7.1 71 = 0.1 mol 32. 300 mL 0.25 M `aeY •Zwi Ki‡Z Kx cwigvY Na2CO3 cÖ‡qvRb? [DU 08-09; BSMRSTU 19-20] 8.0 g 7.95 g 5.3 g 10.6 g DËi: 7.95 g e ̈vL ̈v: W = SMV 1000 = 0.25 × 106 × 300 1000 = 7.95 g 33. 20 mL 0.1 M Fe2+ `aeY UvB‡Uak‡bi Rb ̈ 0.1 M NbgvÎvi KZ AvqZ‡bi KMnO4 cÖ‡qvRb n‡e? [DU 08-09] 4 mL 5 mL 10 mL 20 mL DËi: 4 mL e ̈vL ̈v: (e1V1S1)KMnO4 = (e2V2S2)Fe2+  5 × V1 × 0.1 = 1 × 20 × 0.1  V1 = 4 mL 34. 250.0 cm3 0.5 M Na2CO3 `aeY m¤ú~Y©fv‡e cÖkwgZ Ki‡Z Kx cwigvY 0.25 M HCl `ae‡Yi cÖ‡qvRb nq? [DU 07-08] 250 cm3 125 cm3 1000 cm3 500 cm3 DËi: 1000 cm3 e ̈vL ̈v: (e1V1S1)acid = (e2V2S2)base  1 × V1 × 0.25 = 2 × 250 × 0.5  V1 = 1000 cm3 35. wb‡Pi †Kvb wewμqvwU‡K RviY-weRviY wewμqv wn‡m‡e MY ̈ Kiv †h‡Z cv‡i? [DU 07-08; JU 22-23; RU 22-23] Cu2+ + 4NH3  [Cu(NH3)4] 2+ Cl2 + 2OH–  Cl – + ClO – + H2O NH3 + H+  NH+ 4 Ca2+ + 2F–  CaF2 DËi: Cl2 + 2OH–  Cl – + ClO – + H2O e ̈vL ̈v: Cl2 + 2OH –  Cl – + ClO – + H2O weRviY RviY 0 –1 +1 G wewμqvq RviY msL ̈vi n«vm-e„w× N‡U| ZvB GwU GKwU RviY- weRviY wewμqv| 36. KwóK †mvWv `ae‡Yi cÖwZ wjUv‡i 5 g NaOH _vK‡j `aeYwUi †gvjvwiwU KZ? [DU 06-07] 1.25 M 12.5 M 0.125 M 1.52 M DËi: 0.125 M e ̈vL ̈v: S = W MV = 5 40 × 1 = 0.125 M