Tiêu đề Environmental Chemistry Engg Question Bank Solution (HSC 26).pdf ✅

cwi‡ek imvqb  Engineering Question Bank Solution (HSC 26) ..................................................................................... 1 WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. (a) KZ ZvcgvÎvq CO2 Gi eM©g~j Mo eM©‡eM 20°C ZvcgvÎvq Cl2 Gi eM©g~j Mo eM©‡e‡Mi mgvb n‡e? [BUET 23-24] (b) GKwU M ̈v‡mi e ̈vcb nvi NH3 Gi 2 2 ̧Y n‡j M ̈vmwU Kx? mgvavb: (a) Crms(CO2 ) = Crms(Cl2 )  3RT1 MCO2 = 3RT2 MCl2  T1 MCO2 = T2 MCl2  T1 44 = 273 + 20 71  T1 = 181.58 K = – 91.42C (b) awi, AÁvZ M ̈vmwU x rNH3 rx = Mx MNH3      1 2 2 2 = Mx 17 [ rNH3 rx = 1 2 2 ]  Mx = 2.125 g/mol  M ̈vmwU n‡jv H2| 2. CO2 Gi e ̈vc‡b mgq jv‡M 65 sec, CO wgwkÖZ CO2 Gi e ̈vc‡b mgq jv‡M 58 sec| wgkÖ‡Y CO Gi kZKiv cwigv‡Yi nvi KZ? [BUET 22-23] mgvavb: t2 t1 = M2 M1  58 65 = M2 44  M2 = 35.03 awi, CO Av‡Q x%  CO2 Av‡Q (100 – x)%  44 × 100 – x 100 + 28 × x 100 = 35.03  x = 56.063%  CO Gi kZKiv nvi 56.063%| 3. 950 mol M ̈vm enb Kivi Rb ̈ 200 L AvqZb wewkó wmwjÛvi •Zwi‡Z 250 MPa kw3 wewkó ÷xj e ̈envi Kiv n‡e| MigKv‡j ZvcgvÎv 50C ch©šÍ e„w×i we‡ePbvq D3 ÷xjwU e ̈envi Kiv hv‡e wK bv Zv wbY©q Ki| [BUET 20-21] mgvavb: Avgiv Rvwb, PV = nRT  P = nRT V = 950 × 8.314 × (50 + 273) 200 × 10–3  P = 12.75  106 Pa = 12.75 MPa < 250 MPa  ÷xjwU e ̈envi Kiv hv‡e| 4. A ̈vjwfIwj bvgK †QvU _‡j †_‡K Aw·‡Rb i‡3 e ̈vwcZ nq| A ̈vjwfIwji Mo e ̈vmva© 5.0 × 10–3 cm Ges Gi wfZ‡ii evZv‡m Aw·‡R‡bi cwigvY 14 mol%, hw` A ̈vjwfIwji wfZ‡ii Pvc 1 atm I kix‡ii ZvcgvÎv 98.6F nq, Z‡e GKwU A ̈vjwfIwji wfZi Aw·‡Rb AYyi msL ̈v wbY©q Ki| [BUET 19-20] mgvavb: Avgiv Rvwb, F – 32 9 = K – 273 5  98.6 – 32 9 = K – 273 5 K = 310   GLb, PV = nRT P  4 3 r 3 = nRT  4 3    (5  10–3  10–2 ) 3 = n  8.314  310  n = 2.058  10–11 mol Aw·‡R‡bi cwigvY 14 mol% 100 mol evZv‡m Aw·‡Rb Av‡Q 14 mol  2.058  10–11 mol evZv‡m Aw·‡Rb Av‡Q = 14  2.058  10–11 100 mol = 2.88  10–12 mol  Aw·‡Rb AYyi msL ̈v = (2.88  10–12  6.023  1023) wU   =   wU 5. wb‡Pi †hŠM ̧‡jv †_‡K cÖvBgvwi evqy `~lK kbv3 Ki: NO2, CO, SO2, H2S, O3 [BUET 19-20] mgvavb: cÖvBgvwi evqy `~lK: CO, SO2, H2S|
2 ....................................................................................................................................  Chemistry 2nd Paper Chapter-1 6. GKwU †jŠn wmwjÛv‡i 250 kPa Pv‡c Ges 300 K ZvcgvÎvq wnwjqvg M ̈vm fwZ© Av‡Q| wmwjÛviwU 1 × 103 kPa Pvc mn ̈ Ki‡Z cv‡i Ges Mjbv1⁄4 1800 K| wmwjÛviwU wK M‡j hv‡e bvwK we‡ùvwiZ n‡e? [BUET 18-19] mgvavb: Avgiv Rvwb, P1 T1 = P2 T2  250 × 103 300 = P2 1800  P2 = 1.5 × 106 Pa = 1.5 × 103 kPa GLv‡b, P1 = 250 × 103 Pa T1 = 300 K T2 = 1800 K P2 = ?  wmwjÛviwU Mjbv‡1⁄4 †cu.Qv‡bvi Av‡MB we‡õvwiZ n‡e| 7. evqygÛ‡ji cÖvPz‡h©i wfwˇZ cuvPwU wMÖbnvDm M ̈v‡mi bvg wjL| [BUET 18-19] mgvavb: wMÖb nvDR M ̈vm evqy‡Z kZKiv cwigvY Kve©b WvB A·vBW 49% wg‡_b 18% †K¬v‡iv‡d¬v‡iv Kve©b †h.Mmg~n 14% I‡Rvb 8% bvBUavm A·vBW 6% 8. eRacv‡Zi d‡j msNwUZ wewμqvmg~n wjL Ges †`LvI wKfv‡e Bnv DwTM¢R I cÖvwYR †cÖvwUb •Zwi‡Z mvnvh ̈ K‡i| [BUET 18-19] mgvavb: eRacv‡Z: N2 + O2 3000C we`y ̈r ÿiY 2NO RviY: 2NO + O2 (50 – 100C 2NO2 cvwb‡Z †kvlY: 2NO2 + 1 2 O2 + H2O  2HNO3 gvwU‡Z cÖkgb: CaCO3 + 2HNO3  Ca(NO3)2 + CO2 + H2O wkK‡oi mvnv‡h ̈ DwTM¢‡` MÖnY DwTM¢3⁄4 †cÖvwUb Lv` ̈ wn‡m‡e cÖvYx MÖnY K‡i Ca(NO3)2 cÖvwYR †cÖvwUb 9. GKwU 1.0 L AvqZ‡bi ﮋ evZv‡mi bgybvq 25C ZvcgvÎvq I 786 mm (Hg) Pv‡c 0.925 g N2 Ges ARvbv cwigvY Aw·‡Rb, AvM©b I Kve©b WvB A·vBW mn Ab ̈vb ̈ M ̈vm Av‡Q| GB ZvcgvÎvq evZv‡mi bgybvq N2 Gi AvswkK Pvc KZ? [BUET 17-18] mgvavb: Pvc, P = 786 mm (Hg) = 786 760 = 1.034 atm AvqZb, V = 1 L ZvcgvÎv, T = (25 + 273) K = 298 K PV = ntotalRT  ntotal = PV RT = 1.034  1 0.0821  298 = 0.04226 mol nN2 = 0.925 28 = 0.033 mol  N2 Gi AvswkK Pvc = nN2 ntotal  P = 0.033 0.04226  786 = 613.77 mm (Hg) 10. wbw`©ó AvqZ‡bi weï× Aw·‡Rb M ̈vm GKwU †QvU wQ`a w`‡q wbtmwiZ n‡Z 80 second mgq jv‡M| GKB Ae ̄’vq mgvb AvqZ‡bi 20% ARvbv M ̈vm wgwkÖZ Aw·‡Rb wbtmi‡Yi Rb ̈ 85 second mgq jv‡M| ARvbv M ̈vmwUi AvYweK fi wbY©q Ki| [BUET 17-18] mgvavb: MÖvnv‡gi e ̈vcb m~Îvbymv‡i, M2 M1 = t2 t1  M2 M1 = t2 2 t1 2  M2 =     85 80 2 × 32  M2 = 36.125 g/mol g‡b Kwi, AÁvZ M ̈v‡mi AvYweK fi = x    AÁvZ M ̈v‡mi AvYweK fi  × 20 100 +    O  2 Gi AvYweK fi  80 100 = 36.125  (x × 0.2) + (32 × 0.8) = 36.125  x = 52.625 g/mol 11. Ca2+ Avqb‡K CaCO3 wnmv‡e m¤ú~Y©iƒ‡c Aatwÿß Ki‡Z 200 mL AvqZ‡bi GKwU cvwbi bgybvq 0.025 M NbgvÎvi 5.0 mL Na2CO3 `aeY cÖ‡qvRb nj| cvwbi bgybvwUi LiZv ppm-G wnmve Ki| [BUET 17-18] mgvavb: Ca2+ + Na2CO3  CaCO3 + 2Na+ 1 mol CaCO3  1 mol Na2CO3  0.025 × 5 1000 = w 100  w = 0.0125 g = 12.5 mg 1000 mL Gi Rb ̈ LiZv = (12.5 × 5) = 62.5 ppm 12. wb‡Pi kãms‡ÿc ̧‡jvi c~Y©iƒc wjLÑ (a) ETP (b) MRI (c) PPM (d) TDS (e) BOD [BUET 16-17] mgvavb: (a) ETP: Effluent Treatment Plant (b) MRI: Magnetic Resonance Imaging (c) PPM: Parts Per Million (d) TDS: Total Dissolved Solids (e) BOD: Biochemical Oxygen Demand 13. c~Y©iƒc wjL: BOD, COD, TDS, ETP [BUET 15-16] mgvavb: BOD: Biochemical Oxygen Demand COD: Chemical Oxygen Demand TDS: Total Dissolved Solids ETP: Effluent Treatment Plant 14. CFC-Gi Dcw ̄’wZ‡Z O3 Gi fvOb cÖwμqv wjL| [BUET 15-16] mgvavb: i. H – F | C | H – Cl AwZ‡e ̧wb iwk¥  396 kJ/mol H – F | C | H  + Cl  ii. O3 + Cl  O2 + ClO  iii. O2 + Cl   O  + ClO  iv. ClO + O Cl  + O2 v. O3 + ClO   O2 + ClO  2
cwi‡ek imvqb  Engineering Question Bank Solution (HSC 26) .................................................................................... 3 15. GKwU Aw·‡Rb wmwjÛvi 250 atm Pvc mn ̈ Ki‡Z cv‡i| wmwjÛviwU 125 atm Pvc Ges 27C ZvcgvÎvq Aw·‡Rb Øviv c~Y© Kiv n‡jv| KZ ZvcgvÎvq wmwjÛviwU we‡ùvwiZ n‡e? [BUET 13-14; CUET 04-05] mgvavb: P1 T1 = P2 T2  T2 = P2 × T1 P1 = 250 × 300 125 = 600 K = 327C P1 = 125 atm T1 = 300 K P2 = 250 atm 16. wbw`©ó ZvcgvÎvq I 1 atm Pv‡c a~wjKYv wgwkÖZ Aw·‡Rb M ̈v‡mi AvqZb 100 mL| ZvcgvÎv AcwiewZ©Z †i‡L Pvc 75% e„w× Kiv n‡j a~wjKYvmn Aw·‡R‡bi AvqZb n«vm †c‡q 60 mL nq| a~wjKYvi AvqZb KZ? [BUET 13-14] mgvavb: awi, a~wjKYvi AvqZb x mL P1V1 = P2V2 1 × (100 – x) = 1.75  (60 – x)  100 – x = 105 – 1.75 x  0.75 x = 5  x = 6.67 mL P1 = 1 atm V1 = (100 – x) mL P2 = (1 + 1 Gi 75%) atm = (1 + 0.75) atm = 1.75 atm V2 = (60 – x) mL 17. 27C ZvcgvÎvq GKwU e ̄‘mn †Kv‡bv M ̈v‡mi AvqZb 100 cm3 | ZvcgvÎv 54C G ewa©Z Kiv n‡j Ges Gi Pvc wØ ̧Y Kiv n‡j KwVb e ̄‘mn AvqZb 59.3 cm3 nq| KwVb e ̄‘wUi AvqZb wbY©q Ki| [BUET 13-14] mgvavb: g‡b Kwi, KwVb e ̄‘i AvqZb = x cm3  V1 = (100 – x) cm3 V2 = (59.3 – x) cm3 Avgiv Rvwb, P1V1 T1 = P2V2 T2  P(100 – x) 300 = 2P × (59.3 – x) 327  32700 – 35580 = 327x – 600x  x = 10.55 cm3 18. HNO3 I H3PO4 Gi g‡a ̈ †KvbwU AwaK kw3kvjx? †Zvgvi Dˇii h_v_©Zv e ̈vL ̈v Ki| [BUET 13-14] mgvavb: HNO3 Ges H3PO4 Gi g‡a ̈ HNO3 AwaK kw3kvjx| KviY HN +5 O3 I H3P +5 O4 A‡·vGwm‡Wi †K›`axq cigvYyi RviY gvb mgvb nIqvq GwmW ̧‡jvi g‡a ̈ †hwUi †K›`axq cigvYyi AvKvi †QvU †mwU †ewk Zxea| †h‡nZz cigvYyi AvKv‡ii †ÿ‡Î N < P ZvB, HNO3 > H3PO4 19. MÖvnv‡gi e ̈vcb m~ÎwU wjL| [BUET 06-07] mgvavb: MÖvnv‡gi M ̈vm e ̈vcb m~Î: wbw`©ó ZvcgvÎvq I Pv‡c M ̈v‡mi e ̈vcb nvi H M ̈v‡mi †gvjvi fi (M) Gi eM©g~‡ji e ̈ ̄ÍvbycvwZK  e ̈vcb nvi, r  1 M 20. 32 g Aw·‡Rb M ̈v‡mi Rb ̈ f ̈vbWvi Iqvjm-Gi mgxKiYwU wjL| D3 mgxKiY a I b Gi A_© wjL| [BUET 06-07] mgvavb: 32 g Aw·‡Rb = 1 mol Aw·‡Rb  n = 1     P + a V 2 (V – b) = RT [†hLv‡b, a I b h_vμ‡g Pvc I AvqZb ms‡kvab aaæeK|] 21. wb‡Pi A¤ø ̧wj‡K Zv‡`i kw3i μge„w× Abymv‡i mvRvI| H2SO4, H2SeO4, H2TeO4 [BUET 04-05] mgvavb: H2SO4 > H2SeO4 > H2TeO4 22. k~b ̈ ̄’vb c~iY Ki: eab‡÷W-jvDwi gZev` Abymv‡i NH3, H2O BZ ̈vw` ÿviK wnmv‡e wPwýZ, A_P ––––– gZev` Abymv‡i G ̧‡jv ÿviK bq| [BUET 04-05] mgvavb: Avi‡nwbqvm 23. †mjwmqvm † ̄‹‡j KZ ZvcgvÎvq Cl2 M ̈v‡mi eM©g~j Mo eM©‡eM NTP-†Z SO2 M ̈v‡mi eM©g~j Mo eM© †e‡Mi mgvb n‡e? [BUET 03-04] mgvavb: T1 MSO2 = T2 MCl2  T2 = T1 MSO2 × MCl2 = 273 64 × 71 = 302.86 K = 29.86C 24. wngvj‡qi cv`‡`‡k evZv‡mi Pvc 102.0 kPa Ges ZvcgvÎv 15C| Gfv‡i‡÷i k„‡1⁄2 ZvcgvÎv – 23C| Gfv‡i‡÷i k„‡1⁄2 evZv‡mi Pvc I cÖgvY Pv‡ci Zdvr wnmve Ki| [BUET 02-03] mgvavb: P1 T1 = P2 T2  P2 = P1 T1 × T2 = 102 × 250 288 = 88.54 kPa P1 = 102 kPa T1 = 288 K T2 = 250 K cÖgvY Pv‡ci mv‡_ cv_©K ̈ = (101.325 – 88.54) kPa = 12.78 kPa 25. wb‡Pi M ̈vm ̧‡jvi †KvbwU GwmW e„wói Rb ̈ `vqx? Gi cÖavb Drm ̧‡jv Kx? CO2, NO2, H2S, CO [BUET 02-03] mgvavb: H2S, NO2; (cÖavb Drm wkíRvZ d¬z M ̈vm) 26. g‡b Ki, m~‡h©i †K‡›`a †h M ̈vm ̧‡jv Av‡Q Zv‡`i Mo AvYweK fi, NbZ¡ Ges Pvc h_vμ‡g 5.6, 1.05 kg/m3 Ges 1.1 × 1010 N/m2 | m~‡h©i †K‡›`ai ZvcgvÎv MYbv Ki| [BUET 01-02] mgvavb: PV = nRT  P = w M RT V = w V RT M = dRT M  T = PM dR = 1.1 × 1010 × 5.6 × 10–3 1.05 × 8.314 = 7.05 × 106 K AvYweK fi, M = 5.6 NbZ¡, d = 1.05 kg/m3 Pvc, P = 1.1  1010 N/m2 27. wb‡¤œi AYy/Avqb ̧‡jv‡K ÿviK kw3i ewa©Z μgvbymv‡i mvRvI| F – , H2O, NH3, OH– , CO2– 3 [BUET 01-02] mgvavb: ÿvi wnmv‡e kw3 : H2O < F– < NH3 < CO2– 3 < OH–
4 ....................................................................................................................................  Chemistry 2nd Paper Chapter-1 28. 27C ZvcgvÎvq I GK A ̈vU‡gvwùqvi Pv‡c 30 wjUvi 0.5 N H2SO4, 327 g Zn ̧uovi mv‡_ wewμqvq KZUzKz H2 M ̈vm Drcbœ n‡e? [Zn Gi AvYweK fi = 65.4] [BUET 97-98] mgvavb: Zn + H2SO4  ZnSO4 + H2 nZn 1 = wZn MZn = 327 65.4 = 5 mol nH2 SO4 1 = SH2 SO4 × VH2 SO4 = NH2 SO4 eH2 SO4 × VH2 SO4 = 0.5 2 × 30 = 7.5 mol  nZn 1 < nH2 SO4 1 ;  Zn wjwgwUs wewμqv| STP †Z, 327 g Zn n‡Z H2 Drcvw`Z nq = 22.4 65.4 × 327 = 112 L Avgiv Rvwb, P1V1 T1 = P2V2 T2  1 × 112 273 = 1 × V2 (27 + 273)  V2 = 123.07692 L 29. GKwU bgybv Kve©b WvB A·vBW 23C ZvcgvÎvq, 748 mm (Hg) Pv‡c 212 cm3 AvqZb `Lj K‡i| Kve©b WvB A·vB‡Wi fi I NbZ¡ wbY©q Ki| [BUET 95-96] mgvavb: P1V1 T1 = P2V2 T2  V2 = P1V1 T1 × T2 P2 = 748 × 212 (23 + 273) × 273 760 = 192.439 cc  CO2 Gi fi = 44 × 192.439 × 10–3 22.4 = 0.378 g  NbZ¡ = 0.378 212 = 1.783 × 10–3 g/cm3 weMZ mv‡j RUET-G Avmv cÖkœvejx 1. NTP †Z O2 M ̈v‡mi NbZ¡ 1.429 MÖvg cÖwZ wjUvi| O2 M ̈vm AYyi eM©g~j Mo eM©‡eM SI GK‡K wnmve Ki| [RUET 18-19] mgvavb: Avgiv Rvwb, C = 3P d  C = 3 × 101325 1.429 = 461.21 ms–1 GLv‡b, P = 101325 Pa d = 1.429 g/dm3 = 1.429 kg/m3 2. DB1⁄4jvi Av‡qv‡WvwgwZK c×wZ‡Z cvwb‡Z `aexf~Z Aw·‡Rb wbY©‡q NwUZ ivmvqwbK wewμqvmg~n D‡jøL Ki| [RUET 18-19] mgvavb: MnSO4 + 2KOH  Mn(OH)2 + K2SO4 2Mn(OH)2 + O2  2MnO(OH)2 MnO(OH)2 + 2H2SO4  3H2O + Mn(SO4)2 Mn(SO4)2 + 2KI  K2SO4 + MnSO4 + I2 2Na2S2O3 + I2  Na2S4O6 + 2NaI 3. (a) wU Gj wf (m‡e©v”P wbivc` gvÎv) Kx? (b) Av‡m©wb‡Ki TLV KZ? (c) wb¤œwjwLZ wMÖb nvDm M ̈vm mg~‡ni evqy‡Z kZKiv cwigvY I Zv‡`i Zzjbvg~jK †Møvevj Iqvwg©s ÿgZv D‡jøL K‡iv| (i) CO2 (i) N2O (iii) CFC [RUET 18-19] mgvavb: (a) wU Gj wf (m‡ev©”P wbivc` gvÎv): cwi‡e‡ki †Kv‡bv `~l‡Ki †h wbw`©ó gvÎv AwZμg Ki‡j cwi‡ek I RxeRMr ÿwZMÖ ̄’ nq, H wbw`©ó gvÎv‡K H `~l‡Ki Threshold Limit Value (TLV) ejv nq| (b) Av‡m©wb‡Ki TLV (Threshold Limit Value) 0.05 mgL–1 (c) wMab nvDm M ̈vm evqy‡Z % cwigvY Zzjbvg~jK wMÖb nvDm cÖfve (i) CO2 49% 1 ̧Y (ii) N2O 6% 270 ̧Y (iii) CFC 14% 10, 000 ̧Y 4. 1 evqygÐjxq Pv‡c Ges 27C ZvcgvÎvq 1 cm3 AvqZ‡b KZ ̧wj wnwjqvg AYy _v‡K Zv wbY©q Ki| [RUET 15-16, 08-09, 07-08] mgvavb PV = nRT n = PV RT = 1  10–3 0.082  300 mol = 4.065  10–5 mol  AYyi msL ̈v = †gvj msL ̈v  A ̈v‡fvMv‡Wav msL ̈v = (4.065  10–5  6.02  1023) wU = 2.4471  1019 wU 5. wewfbœ GK‡K R Gi gvb wbY©q Ki| [RUET 15-16, 06-07] mgvavb: i. M.K.S. System: R = PV nT = 101325 × 22.4 × 10–3 1 × 273.15 J mol–1 K –1 = 8.31 J mol–1 K –1 ii. L – atm system: R = PV nT = 1 × 22.4 1 × 273 L atm mol–1 K –1 = 0.0821 L atm mol–1 K –1 iii. C.G.S system: R = PV nT = 76 × 13.6 × 981 × 22400 1 × 273 erg mol–1 K –1 = 8.32 × 107 erg mol–1 K –1 6. 1.5 L Br2 M ̈vm‡K 27C ZvcgvÎv I 700 mm (Hg) Pvc †_‡K cÖgvY Ae ̄’vq †bIqv n‡j, cÖgvY Ae ̄’vq KZ ̧‡jv AYy Av‡Q Zv wbY©q Ki| [RUET 12-13] mgvavb: P0V0 T0 = P1V1 T1  V0 = 700  1.5 300  273 760  V0 = 1.257 L  1.257 L G AYyi msL ̈v = 6.022  1023 22.4  1.257 = 3.38  1022 wU