Tiêu đề 3. Chemistry FRB-26.03.24_With Solve.pdf.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


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