Tiêu đề 1. P2C1. Che. For FRB-2024_With Solve_Jakir_15.4.24.pdf ✅

cwi‡ek imvqb  Final Revision Batch 1 cwi‡ek imvqb Environmental Chemistry cÖ_g Aa ̈vq Topicwise Board Analysis eûwbe©vPwb cÖkœ UwcK 2017 2018 2019 2021 2022 2023 †gvU Av`k© M ̈v‡mi m~Îmg~n 7 1 6 11 11 16 52 M ̈v‡mi AvYweK MwZZË¡ 3 Ñ 4 6 5 4 22 MÖvnv‡gi e ̈vcb m~Î, Wvë‡bi AvswkK Pvc m~Î 4 Ñ 2 5 6 7 25 Av`k© M ̈vm I ev ̄Íe M ̈vm 2 Ñ 5 1 1 5 14 GwmW-ÿviK gZev`, cvwbi weï×Zvi gvb`Û 6 1 10 13 13 14 57 m„Rbkxj cÖkœ UwcK 2017 2018 2019 2021 2022 2023 †gvU Av`k© M ̈v‡mi m~Îmg~n Ñ 1 3 3 4 9 20 M ̈v‡mi AvYweK MwZZË¡ 2 Ñ 4 6 1 5 18 MÖvnv‡gi e ̈vcb m~Î, Wvë‡bi AvswkK Pvc m~Î 3 1 2 3 8 14 31 Av`k© M ̈vm I ev ̄Íe M ̈vm Ñ Ñ 2 Ñ 1 5 8 GwmW-ÿviK gZev`, cvwbi weï×Zvi gvb`Û 1 1 1 6 2 3 14 *we.`a.: 2020 mv‡j GBPGmwm cixÿv AbywôZ nqwb| ACS Chemistry Department Gi g‡bvbxZ enywbev©Pwb cÖkœmg~n Av`k© M ̈v‡mi m~Îmg~n 1. SATP †Z 2 †gvj O2 M ̈v‡mi AvqZb KZ? [iv. †ev. 23] 22.789 L 24.789 L 45.578 L 49.578 L DËi: 49.578 L e ̈vL ̈v: SATP †Z 1 †gvj O2 M ̈v‡mi AvqZb = 24.789 L  SATP †Z 2 †gvj M ̈v‡mi AvqZb = (2  24.789) L = 49.578 L 2. 8 g He M ̈v‡mi Rb ̈ Av`k© M ̈vm mgxKiY †KvbwU? [Kz. †ev. 23] PV = nRT PV = RT 2 PV = 2RT PV = RT DËi: PV = 2RT e ̈vL ̈v: n = W M = 8 4 = 2 mol Av`k© M ̈vm mgxKiY, PV = nRT PV = 2RT 3. †Kvb †jLwPÎwU AvB‡mv_vg© mg_©b K‡i? [Kz. †ev. 23]  1 P V  T3 T2 T1 T V  P3 P2 P1 T P  V3 V2 V1 T PV  DËi:  1 P V  T3 T2 T1 e ̈vL ̈v: ÔKÕ Ack‡bi MÖv‡d cÖwZwU †iLvi Rb ̈ ZvcgvÎv wbw`©ó| ÔLÕ, ÔMÕ, ÔNÕ Ack‡bi MÖv‡d X-A‡ÿ ZvcgvÎv (T) _vKvq †iLv ̧‡jv AvB‡mv_vg© †iLv bq|
2  Chemistry 2 nd Paper Chapter-1 4. w ̄’i AvqZ‡b wbw`©ó f‡ii Av`k© M ̈v‡mi P T ebvg T †jLwPÎ n‡e- [h. †ev. 23] T P T T P T T P T T P T DËi: T P T e ̈vL ̈v: †M-jymv‡Ki m~Î n‡Z, P  T ev P T = aaæeK| †jLwP‡Î P T Gi wecix‡Z T ̄ vcb Ki‡j Avbyf~wgK mij‡iLv ev X A‡ÿi mgvšÍivj mij‡iLv cvIqv hvq| 5. w ̄’i Pv‡c 0C ZvcgvÎvq O2 M ̈v‡mi AvqZb 3.5 L n‡j 20C ZvcgvÎvq M ̈vmwUi AvqZb n‡e- [h. †ev. 23] 3.25 L 3.76 L 7.0 L 8.0 L DËi: 3.76 L e ̈vL ̈v: V1 T1 = V2 T2  V2 = V1  T2 T1 = 3.5  293 273 L = 3.76 L T1 = 0C = 273 K T2 = (273 + 20) K = 293 K 6. SATP †Z bvBUavm A·vBW M ̈v‡mi NbZ¡ gL–1 GK‡K KZ? [P. †ev. 23] 1.77 1.96 1.85 1.21 DËi: 1.77 e ̈vL ̈v: Avgiv Rvwb, d = PM RT = 0.987  44 0.082  298 = 1.77 g L–1 7. †M-jymv‡Ki Pv‡ci m~Î wb‡Pi †KvbwU? [e. †ev. 23] P  T (V, n w ̄ i) V  n (P,T w ̄ i) V  T (P, n w ̄ i) V  1 P (n, T w ̄ i) DËi: P  T (V, n w ̄ i) 8. †M-jymv‡Ki Pvcxq m~‡Îi mgxKiY n‡jv- [w`. †ev. 23] V  P P  T V  1 P P  1 T DËi: P  T 9. 1.80  10–3 g Møy‡KvR AYy‡Z KZwU Aw·‡Rb cigvYy Av‡Q? [wm. †ev. 23] 6.02  1018 3.61  1018 3.61  1019 6.02  1019 DËi: 3.61  1019 e ̈vL ̈v: Møy‡KvR (C6H12O6) Gi AvYweK fi, M = 12  6 + 1  12  n = W M = N NA  1.80  10–3 180 = N 6.023  1023  1  10–5 = N 6.023  1023  N = 6.023  1018 O cigvYyi msL ̈v = 6  6.023  1018 = 3.61  1019 wU 10. S.I. GK‡K R –Gi gvb †KvbwU? [wm. †ev. 23] 0.082 L atm K–1mol–1 8.314 J K–1 mol–1 8.314  107 erg K –1mol–1 1.987 cal. K–1 mol– 1 DËi: 8.314 J K–1 mol–1 e ̈vL ̈v: R-Gi gvb GKK wjUvi evqyPvc GKK 0.0821 L atm K–1 mol–1 CGS GKK 8.314  107 erg K–1 mol–1 SI GKK 8.314 J K–1 mol–1 K ̈vjwi GKK 1.987 cal mol–1 K –1 11. w ̄’i ZvcgvÎvq wbw`©ó f‡ii †Kv‡bv M ̈v‡mi AvqZb I Pv‡ci m¤úK©hy3 †iLv †Kvb cÖK...wZi? [w`. †ev. 23] cive„Ë g~jwe›`yMvgx mij‡iLv Awae„Ë Y-Aÿ †Q`Kvix mij‡iLv DËi: Awae„Ë e ̈vL ̈v: w ̄ i ZvcgvÎvq wbw`©ó f‡ii †Kvb M ̈v‡mi AvqZb H M ̈v‡mi Dci cÖhy3 Pv‡ci e ̈ ̄ÍvbycvwZK; V  1 P | Pvc (P) AvqZb (V)
cwi‡ek imvqb  Final Revision Batch 3 12. STP †Z 2 mol CaCO3 I HCl Gi wewμqvq Drcbœ CO2 M ̈v‡mi AvqZb KZ wjUvi? [w`. †ev. 23] 11.2 22.4 34.8 44.8 DËi: 44.8 e ̈vL ̈v: CaCO3 + 2HCl  CaCl2 + CO2 + H2O 1 mol 22.4 L STP †Z, 1 mol CaCO3 n‡Z CO2 Drcbœ nq = 22.4 L 2 mol CaCO3 n‡Z CO2 Drcbœ nq = (22.4  2) L = 44.8 L 13. e‡q‡ji m~‡Îi mgxKi‡Yi †jLwPÎ †Kvb ai‡bi? [g. †ev. 23] AvB‡mv_vg© AvB‡mvevi AvB‡mv‡Kvi AvB‡mv‡gvj DËi: AvB‡mv_vg© e ̈vL ̈v: P1V1 = P2V2 = ....... PnVn mKj †ÿ‡Î ZvcgvÎv aaæe| 14. 100C ZvcgvÎvq 2.05 atm Pv‡c CO2 M ̈v‡mi NbZ¡ KZ? [g. †ev. 23] 1.50 g L–1 1.76 g L–1 2.34 g L–1 2.95 g L–1 DËi: 2.95 g L–1 e ̈vL ̈v: Avgiv Rvwb, d = PM RT = 2.05  44 0.082  373 = 2.95 g L–1 MCO2 = 44 g R = 0.082 Latm.K–1 mol–1 T = (273 + 100) K = 373 K myZivs M ̈v‡mi NbZ¡, d = 2.95 g L–1 15. †Kv‡bv M ̈v‡mi ZvcgvÎv I Pvc wØ ̧Y Kiv n‡j AvqZ‡bi Kx cwieZ©b n‡e? [Xv. †ev. 22] wØ ̧Y n‡e †Kvb cwieZ©b n‡e bv Pvi ̧Y A‡a©K n‡e DËi: †Kvb cwieZ©b n‡e bv e ̈vL ̈v: Avgiv Rvwb, P1V1 T1 = P2V2 T2  P1V1 T1 = 2P1  V2 2T1  P1V1 T1 = P1V2 T1  V1 = V2 A_©vr, AvqZ‡bi †Kv‡bv cwieZ©b n‡e bv| 16. w ̄’i ZvcgvÎvq P ebvg 1 V †jLwPÎ njÑ [iv. †ev. 22]  1 V P   1 V P  1 V P   1 V P  DËi:  1 V P  e ̈vL ̈v: e‡q‡ji m~Îvbymv‡i, V  1 P |  P  1 V  P = 1 V C [C = aaæeK]    y x m †jLwPÎ g~jwe›`yMvgx mij‡iLv|  1 P V   1 V P  17. 18C ZvcgvÎvq 0.8 atm Pv‡c M ̈v‡mi NbZ¡ 2.25 g/L , Gi AvYweK fi KZ? [Kz. †ev. 22] 67.11 g/mol 36.24 g/mol 24.36 g/mol 23.63 g/mol DËi: 67.11 g/mol e ̈vL ̈v: Avgiv Rvwb, d = PM RT  M = dRT P  M = 2.25  0.0821    M = 67.1 g/mol  AvYweK fi 67.1 g/mol 18. 1.5 atm Pv‡c 25C ZvcgvÎvq GKwU M ̈v‡mi AvqZb 0.5 L n‡j D3 ZvcgvÎvq wØ ̧Y Pv‡c M ̈vmwUi AvqZb KZ n‡e? [h. †ev. 22] 0.45 L 0.35 L 0.25 L 0.15 L DËi: 0.25 L e ̈vL ̈v: Avgiv Rvwb, P1V1 = P2V2  V2 = P1V1 P2 = 1.5  0.5 2  1.5 L = 0.25 L
4  Chemistry 2 nd Paper Chapter-1 19. mwÜ ZvcgvÎvi wb‡P c`v‡_©i Ae ̄’v †KvbwU? [e. †ev. 22] ev®ú Zij Zij ùwUK cøvRgv DËi: Zij e ̈vL ̈v: mwÜ ZvcgvÎvi Ic‡i †Kv‡bv M ̈v‡mi Ici h‡Zv Pvc cÖ‡qvM Kiv †nvK bv †Kb, GK Zi‡j iƒcvšÍi Kiv hvq bv| ZvB M ̈vm ZijxKi‡Y Gi ZvcgvÎv mwÜ ZvcgvÎvi wb‡P Avb‡Z nq| †hgb: CO2 Gi mwÜ ZvcgvÎv = 31.1C | 20. SATP I STP †Z ZvcgvÎvi cv_©K ̈ KZ C? [e. †ev. 22] 273 25 0 –273 DËi: 25 e ̈vL ̈v: STP †Z, ZvcgvÎv = 0C ev 273 K Pvc = 1 atm ev 101.325 kPa †gvjvi AvqZb = 22.414 Lmol–1 SATP †Z, ZvcgvÎv = 25C ev 298 K Pvc = 100 kPa †gvjvi AvqZb = 24.789 Lmol–1  ZvcgvÎvi cv_©K ̈ = (25 – 0) C = 25C 21. †KvbwU cigk~b ̈ ZvcgvÎv? [w`. †ev. 22] 0C 25C 273 K – 273C DËi: – 273C e ̈vL ̈v: – 273.15C ev, – 273C ev, 0 K †K cigk~b ̈ ZvcgvÎv e‡j| 22. 1 atm = KZ c ̈vm‡Kj? [g. †ev. 22] 1.01325  102 Pa 1.01325  10–2 Pa 1.01325  105 Pa 1.01325  10–5 Pa DËi: 1.01325  105 Pa e ̈vL ̈v: 1 atm = 76.0 cm (Hg) = 760 mm (Hg) = 101.325 kPa = 1.01325  105 Pa 23. Av`k© M ̈v‡mi •ewkó ̈m~PK gvb`Ð n‡jv- [Xv. †ev. 22] i. PV = nRT ii.     dU dV T = 0 iii. STP †Z †gvjvi AvqZb 22.414 L wb‡Pi †KvbwU mwVK? i I ii ii I iii i I iii i, ii I iii DËi: i, ii I iii e ̈vL ̈v: (i) Av`k© M ̈vm mKj ZvcgvÎv I Pv‡c PV = nRT mgxKiY †g‡b P‡j| (ii) w ̄ i ZvcgvÎvq Av`k© M ̈v‡mi Af ̈šÍixY kw3 AvqZ‡bi Dci wbf©ikxj bq| A_©vr,     U V T = 0| (iii) Av`k© ZvcgvÎv I Pv‡c (STP-†Z) Av`k© M ̈v‡mi †gvjvi AvqZb 22.414 L| 24. STP -†Z †Kvb M ̈v‡mi 1 g me‡P‡q †ewk AvqZb `Lj Ki‡e? [Xv. †ev. 21] H2 N2 O2 Ar DËi: H2 e ̈vL ̈v: STP †Z, mKj M ̈v‡mi †gvjvi AvqZb = 22.4 L STP †Z, 1 mol H2 ev 2 g H2 Gi AvqZb = 22.4 L  1 g H2 Gi AvqZb = 22.4 2 L = 11.2 L Abyiƒcfv‡e, 1 g N2 Gi AvqZb = 22.4 28 L = 0.8 L 1 g O2 Gi AvqZb = 22.4 32 L = 0.7 L 1 g Ar Gi AvqZb = 22.4 40 L = 0.56 L myZivs, 1g H2 me‡P‡q †ewk AvqZb `Lj K‡i| 25.  V T T Y P †jLwPÎwU †Kvb m~·K mg_©b K‡i- [iv. †ev. 21] Pvj©‡mi m~Î Pvj©‡mi m~Î A ̈v‡fv‡M‡Wavi m~Î †M-jymv‡Ki m~Î DËi: Pvj©‡mi m~Î e ̈vL ̈v: Pvj©‡mi m~Îvbymv‡i, w ̄ i Pv‡c wbw`©ó f‡ii †Kv‡bv M ̈v‡mi AvqZb Gi cig ZvcgvÎv ev †Kjwfb ZvcgvÎvi mgvbycvwZK nq| V  T [GLv‡b n I P w ̄ i Av‡Q] V = KT  V T = K 26. 100C ZvcgvÎvq I 1.0526 atm Pv‡c CO2 M ̈v‡mi NbZ¡ (g\L) KZ? [iv. †ev. 21] 0.00291 0.0149 1.5212 1.49  1022 DËi: 1.5212 e ̈vL ̈v: d = PM RT = 1.0526  44 0.0821  373 = 1.512 g\L 27. 22 g CO2 Gi Rb ̈ Av`k© M ̈vm mgxKiY †KvbwU? [e. †ev. 21] 2PV = RT PV = 2RT PV = 22RT PV = RT DËi: 2PV = RT e ̈vL ̈v: CO2 Gi AvYweK fi, M = 12 + 16  2 = 44 †gvj msL ̈v, n = W M = 22 44 = 1 2 mol PV = nRT  PV = 1 2 RT  2PV = RT