Title পরিমানগত রসায়ন With Solve.pdf ✅

cwigvYMZ imvqb  Final Revision Batch 1 cwigvYMZ imvqb Quantitative Chemistry Z...Zxq Aa ̈vq Topicwise Board Analysis eûwbe©vPwb cÖkœ UwcK 2017 2018 2019 2021 2022 2023 †gvU ivmvqwbK MYbv, †gvj msL ̈v 2 2 2 5 5 Ñ 16 mgxKiY wfwËK MYbv Ñ Ñ 2 Ñ 1 4 7 NbgvÎv 12 1 14 20 19 14 90 A¤øwgwZ, ÿviwgwZ, UvB‡Uakb, wb‡`©kK Ñ 1 5 7 5 5 23 RviY-weRviY, RviY-weRviY Aa©wewμqv, mgZvKiY 7 Ñ 7 14 14 16 58 m„Rbkxj cÖkœ UwcK 2017 2018 2019 2021 2022 2023 †gvU ivmvqwbK MYbv, †gvj msL ̈v Ñ Ñ Ñ Ñ 1 Ñ 1 mgxKiY wfwËK MYbv Ñ Ñ 1 1 1 1 4 NbgvÎv 12 1 12 13 15 14 67 A¤øwgwZ, ÿviwgwZ, UvB‡Uakb, wb‡`©kK 3 1 4 5 3 3 19 RviY-weRviY, RviY-weRviY Aa©wewμqv, mgZvKiY 10 2 12 20 14 17 75 *we.`a.: 2020 mv‡j GBPGmwm cixÿv AbywôZ nqwb| ACS Chemistry Department Gi g‡bvbxZ eûwbe©vPwb cÖkœmg~n ivmvqwbK MYbv, †gvj msL ̈v 1. 10% w w Na2CO3 Gi Rjxq `ae‡Y cvwbi †gvj fMœvsk KZ? [e. †ev. 23] 0.0185 0.98 0.9815 0.9833 DËi: 0.9815 e ̈vL ̈v: 10% w w Na2CO3 Gi Rjxq `ae‡Y 10 g Na2CO3 I 90 g cvwb we` ̈gvb|  Na2CO3 Gi †gvj msL ̈v, n1 = 10 106 = 0.0943 mol H2O Gi †gvj msL ̈v, n2 = 90 18 = 5 mol  cvwbi †gvj fMœvsk, Xn = n2 n1 + n2 = 5 5 + 0.0943 = 0.9815 2. cÖgvY Ae ̄’vq 10 cm3 NH3 M ̈v‡mi fi KZ? [iv. †ev. 22] 5.583 × 10–3 g 6.589 × 10–3 g 7.589 × 10–2 g 7.589 × 10–3 g DËi: 7.589 × 10–3 g e ̈vL ̈v: STP †Z, 22.4 × 103 cm3 NH3 M ̈v‡mi fi = 17 g  10 cm3 NH3 M ̈v‡mi fi = 17 × 10 22.4 × 103 g = 7.589 × 10–3 g 3. GKwU Aw·‡Rb cigvYyi fi KZ? [Kz. †ev. 22] 2.66 × 10–23 g 3.76 × 10–23 g 1.33 × 10–22 g 1.88 × 10–22 g DËi: 2.66 × 10–23 g e ̈vL ̈v: 6.023 × 1023 wU Aw·‡Rb cigvYyi fi = 16 g  1wU Aw·‡Rb cigvYyi fi = 16 6.023 × 1023 g = 2.66 × 10–23 g
2  HSC Chemistry 2nd Paper Chapter-3 4. 2H2O2(aq)  2H2O(l) + O2(g) GB wewμqvi gva ̈‡g 16 g O2 •Zwi‡Z KZ MÖvg H2O2 jvM‡e? [P. †ev. 22] 64 34 17 8.5 DËi: 34 e ̈vL ̈v: 2H2O2(aq)  2H2O(l) + O2(g) {2 × (1 × 2 + 16 × 2)} g (16 × 2) g = 68 g = 32 g 32 g O2 •Zwi‡Z H2O2 jv‡M = 68 g  16 g O2 •Zwi‡Z H2O2 jv‡M = 68 × 16 32 g = 34 g 5. STP †Z 3.2 g GKwU M ̈vm 2.24 wjUvi AvqZb `Lj Ki‡j M ̈vmwU n‡Z cv‡i- [P. †ev. 22] Cl2 CO2 N2 O2 DËi: O2 e ̈vL ̈v: 2.24 L M ̈v‡mi fi = 3.2 g  22.4 L M ̈v‡mi fi = 3.2 × 22.4 2.24 g = 32 g O2 Gi AvYweK fi = 32 myZivs, M ̈vmwU O2 6. cÖgvY Ae ̄’vq 10.0 L CH4 M ̈v‡m AYyi msL ̈v KZ? [e. †ev. 22] 0.2689 × 1023 2.689 × 1023 26.89 × 1023 0.02689 × 1023 DËi: 2.689 × 1023 e ̈vL ̈v: Avgiv Rvwb, VSTP 22.4 = N NA  N = VSTP 22.4 × NA  N = 10 22.4 × 6.023 × 1023  N = 2.689 × 1023 7. Aa©‡gvj CO2 M ̈v‡m Aw·‡Rb cigvYyi msL ̈v KZ? [wm. †ev. 22] 1wU 2wU 3.01 × 1023 wU 6.023 × 1023 wU DËi: 6.023 × 1023 wU e ̈vL ̈v: Avgiv Rvwb, 1 mol CO2 G Aw·‡Rb cigvYyi msL ̈v = 2 × 6.023 × 1023  1 2 mol CO2 G Aw·‡Rb cigvYyi msL ̈v = 2 × 6.023 × 1023 × 1 2 wU = 6.023 × 1023 wU 8. cÖgvY Ae ̄’vq 9.0 g cvwb‡Z KqwU nvB‡Wav‡Rb cigvYy _v‡K? [iv. †ev. 21] 6.023 × 1023 3.0115 × 1023 6.023 × 1021 12.046 × 1023 DËi: 6.023 × 1023 e ̈vL ̈v: W M = N NA  N = 9 × 6.023 × 1023 18 = 3.0115 × 1023 wU  H cigvYyi msL ̈v = (2 × 3.0115 × 1023) wU = 6.023 × 1023 wU 9. 1.8 kg cvwb‡Z KZ ̧‡jv nvB‡Wav‡Rb cigvYy we` ̈gvb? [h. †ev. 21] 6.023 × 1023 wU 12.046 × 1023 wU 6.023 × 1025 wU 12.046 × 1025 wU DËi: 12.046 × 1025 wU e ̈vL ̈v: GLb, 18 g cvwb‡Z H-cigvYy Av‡Q = 2 × 6.023 × 1023 wU  1.8 × 103 g cvwb‡Z H-cigvYy Av‡Q = 2 × 6.023 × 1023 × 1.8 × 103 18 wU = 12.046 × 1025 wU 10. STP †Z wb‡Pi †Kvb M ̈v‡mi GK wg.wj. Gi fi Kg? [h. †ev. 21] N2 O2 CO2 NO2 DËi: N2 e ̈vL ̈v: Avgiv Rvwb, STP †Z 1 mol M ̈v‡mi AvqZb, V = 22.4 L = 22.4 × 103 mL * 22.4 × 103 mL N2 M ̈v‡mi fi = 28 g  1 mL N2 M ̈v‡mi fi = 28 22.4 × 103 g = 1.25 × 10–3 g *1 mL O2 M ̈v‡mi fi = 32 22.4 × 103 g = 1.42 × 10–3 g * 1 mL CO2 M ̈v‡mi fi = 44 22.4 × 103 g = 1.96 × 10–3 g * 1 mL NO2 M ̈v‡mi fi = 46 22.4 × 103 g = 2.05 × 10–3 g
cwigvYMZ imvqb  Final Revision Batch 3 11. 32 g O2 Gi A_© n‡jv- [P. †ev. 21] i. 1 mol O2 ii. cÖgvY Ae ̄’vq 24.8 L AvqZb iii. A ̈v‡fv‡M‡Wav msL ̈vi mgvb AYy wb‡Pi †KvbwU mwVK? i i, ii i, iii i, ii, iii DËi: i, iii e ̈vL ̈v: STP †Z, 1 mol O2  32 g O2  22.4 L O2  6.023 × 1023 wU O2 AYy SATP †Z 1 mol Aw·‡Rb M ̈v‡mi AvqZb 24.8 L| 12. 1wU bvB‡Uav‡Rb AYyi fi KZ? [mw¤§wjZ. †ev. 18] 2.32 × 10–26 kg 2.32 × 10–23 kg 4.65 × 10–26 kg 4.65 × 10–23 kg DËi: 4.65 × 10–26 kg e ̈vL ̈v: m M = x NA  m = x × M NA = 1 × 28 6.023 × 1023 = 4.65 × 10–23 g = 4.65 × 10–26 kg mgxKiY wfwËK MYbv 13. 50 g CaCO3 Gi Zvcxq we‡qvR‡b Drcbœ CO2 Gi fi KZ MÖvg? [Kz. †ev. 23] 11 22 44 88 DËi: 22 e ̈vL ̈v: CaCO3  CaO + CO2 100 g 44 g 100 g CaCO3 Gi we‡qvR‡b Drcbœ CO2 Gi fi = 44 g  50 g CaCO3 Gi we‡qvR‡b Drcbœ CO2 Gi fi = 44 × 50 100 g = 22 g 14. A ̈v‡gvwbqvg †K¬vivBW I K ̈vjwmqvg A·vBW wewμqv K‡i STP †Z 44.8 L NH3 M ̈vm cÖ ̄‘Z Ki‡Z e ̈eüZ K ̈vjwmqvg A·vB‡Wi cwigvY KZ? [Xv. †ev. 23] 56 g 28 g 14 g 7 g DËi: 56 g e ̈vL ̈v: wewμqvwU wb¤œiƒc: 2NH4Cl + CaO  CaCl2 + 2NH3 + H2O 56 g (2 × 22.4) L = 44.8 L  STP †Z 44.8 L NH3 cÖ ̄‘wZ‡Z CaO cÖ‡qvRb = 56 g 15. 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 GLb, 1 mol CaCO3 †_‡K CO2 Drcbœ nq = 22.4 L  2 mol CaCO3 †_‡K CO2 Drcbœ nq = (2 × 22.4)L = 44.8 L 16. A·vwjK Gwm‡Wi ÿviKZ¡ KZ? [Kz. †ev. 21] 1 2 4 3 DËi: 2 e ̈vL ̈v: A‡¤øi cÖkgb wewμqvi gva ̈‡g wbi‡cÿ `aeY I cvwb •Zwi Ki‡Z hZmsL ̈K g‡bv‡cÖvwUK ÿviK cÖ‡qvRb Zv‡K A‡¤øi ÿviKZ¡ e‡j| H2C2O4 + 2NaOH  Na2C2O4 + 2H2O GK †gvj H2C2O4 cÖkg‡bi Rb ̈ `yB †gvj NaOH (g‡bv‡cÖvwUK ÿviK) cÖ‡qvRb| ZvB H2C2O4 Gi ÿviKZ¡ 2| 17. KZ MÖvg KClO3 †K DËß Ki‡j cÖgvY Ae ̄’vq 20 L Aw·‡Rb cvIqv hv‡e? [h. †ev. 19] 36.49 g 54.73 g 61.01 g 72.98 g DËi: 72.98 g e ̈vL ̈v: 2 KClO3   2KCl + 3O2 2 × 122.5 3 × 22.41 3 × 22.4 L O2 †c‡Z KClO3 cÖ‡qvRb 2 × 122.5 g  20 L O2 †c‡Z KClO3 cÖ‡qvRb = 2 × 122.5 × 20 3 × 22.4 g = 72.92 g  72.98 g 18. 95% (w/w) weï× Pzbvcv_‡ii 120 g wb‡q HCl Gwm‡W `aexf~Z Ki‡j STP-†Z KZ wjUvi CO2 M ̈vm cvIqv hv‡e? [w`. †ev. 19] 29.75 28.26 26.89 25.55 DËi: 25.55 e ̈vL ̈v: 95% weï× Pzbvcv_‡i, 100 g cwigv‡Y Pzbvcv_i Av‡Q 95 g  120 g cwigv‡Y Pzbvcv_i Av‡Q = 95 × 120 100 g = 114 g CaCO3 (Pzbvcv_i) + 2HCl  2NaCl + CO2 + H2O 100 g 22.4 L 100 g Pzbvcv_i †_‡K CO2 Drcbœ nq 22.4 L  114 g Pzbvcv_i †_‡K CO2 Drcbœ nq = 22.4 × 114 100 L = 25.54 L
4  HSC Chemistry 2nd Paper Chapter-3 NbgvÎv 19. 0.5 mol L–1 H2SO4 `ae‡Y H + Gi NbgvÎv KZ wcwcGg? [Xv. †ev. 23] 10000 1000 100 10 DËi: 1000 e ̈vL ̈v: H2SO4 H2O  2H+ + SO2– 4 0.5 M (2 × 0.5) M  [H+ ] = 1 mol L–1 = (1 × 1 × 1000) mg L–1 [⸪ ppm = S × M × 103 ] = 1000 ppm 20. 500 mL 0.05 M Na2CO3 `ae‡Y KZ MÖvg Na2CO3 _v‡K? [iv. †ev. 23] 2.65 5.30 6.30 10.60 DËi: 2.65 e ̈vL ̈v: W = SMV 1000 = 0.05 × 106 × 500 1000 = 2.65 g 21. 5% Na2CO3 `ae‡Yi NbgvÎv KZ †gvjvi? [Kz. †ev. 23] 0.98 0.89 0.74 0.47 DËi: 0.47 e ̈vL ̈v: S = 10 x M = 10 × 5 106 = 0.47 M 22. 0.15 M NaOH `ae‡Yi ppm NbgvÎv KZ? [h. †ev. 23] 4000 5000 7000 6000 DËi: 6000 e ̈vL ̈v: NaOH Gi NbgvÎv = 0.15 mol L–1 = (0.15 × 40) g L–1 = (0.15 × 40 × 103 ) mg L–1 = 6000 ppm 23. wb‡Pi †KvbwUi Rb ̈ W W cÖ‡hvR ̈? [h. †ev. 23] †gvjvwiwU †gvjvwjwU bigvwjwU digvwjwU DËi: †gvjvwjwU e ̈vL ̈v: cÖwZ †KwR `ave‡K `aexf~Z `a‡ei †gvj msL ̈v‡K H `ae‡Yi †gvjvwjwU e‡j| †gvjvwjwU = `a‡ei †gvjmsL ̈v (W) `ave‡Ki fi (W) 24. †Kvb †hŠMwU †m‡KÛvwi ÷ ̈vÛvW© c`v_©? [P. †ev. 23; h. †ev. 21; P. †ev. 21] Na2CO3 K2Cr2O7 Na2C2O4 KMnO4 DËi: KMnO4 tricks: C I Cr hy3 †hŠMmg~n mvaviYZ cÖvBgvwi ÷ ̈vÛvW© c`v_©| 25. †KvbwU cÖvBgvwi ÷ ̈vÛvW© c`v_©? [e. †ev. 23] KMnO4 NaOH K2Cr2O7 HCl DËi: K2Cr2O7 26. 100 mL 0.01 M K2Cr2O7 `ae‡Yi ppm NbgvÎv KZ? [e. †ev. 23] 2.94 29.4 294 2940 DËi: 2940 e ̈vL ̈v: `ae‡Yi NbgvÎv = 0.01 mol L–1 = (0.1 × 294) g L–1 = (2.94 × 103 ) mg L–1 = 2940 ppm GLv‡b, K2Cr2O7 Gi AvYweK fi = (39 × 2 + 52 × 2 + 16 × 7) = 294 g mol–1 27. `ae‡Yi †Kvb GKKwU ZvcgvÎvi Dci wbf©ikxj bq? [wm. †ev. 23; iv. †ev. 22; Ky. †ev. 21] †gvjvwjwU †gvjvwiwU bigvwjwU wcwcGg DËi: †gvjvwjwU e ̈vL ̈v: †gvjvwjwU = `a‡ei †gvj msL ̈v `ave‡Ki fi (kg) `a‡ei †gvj msL ̈v ev `ave‡Ki fi †Kv‡bvwUB ZvcgvÎvi Dci wbf©ikxj bq| ZvB, †gvjvwjwUI ZvcgvÎvi Dci wbf©ikxj bq| 28. 50 mL `ae‡Y 4.9 g H2SO4 `aexf~Z Av‡Q| `aeYwUi NbgvÎv- [w`. †ev. 23] i. 1 M ii. 9800 ppm iii. 9.8 × 104 g/mL wb‡Pi †KvbwU mwVK? i, ii i, iii ii, iii i, ii, iii DËi: i, iii e ̈vL ̈v: `ae‡Yi NbgvÎv, S = 1000W MV = 1000 × 4.9 98 × 50 = 1 M ppm = SM × 103 = 1 × 98 × 103 = 98000 ppm = 98000 mg L–1 = 98000 g mL–1 = 9.8 × 104 g mL–1