Tiêu đề 4. (1st Paper) Che. Written Practice Sheet_With Solve 12-11-2023.pdf ✅

ivmvqwbK cwieZ©b  Academic Practice Sheet 1 PZz_© Aa ̈vq ivmvqwbK cwieZ©b Chemical Changes ACS Chemistry Department Gi g‡bvbxZ wjwLZ cÖkœmg~n ivmvqwbK wewμqv I wMÖb †Kwgw÷a 1| UvBUvwbqvg `ywU wfbœ wfbœ c×wZ Øviv AvKwiK †_‡K wb®‹vkb Kiv hvq| (i) AwaKZi mwμq avZzi e ̈envi : TiO2+ 2Mg  Ti + 2MgO (ii) AvKwi‡Ki Zwor we‡kølY : TiO2  Ti + O2 Kvw•ÿZ Drcv‡` wewμqK cigvYyi me©vwaK Dcw ̄’wZi aviYv e ̈envi K‡i Dc‡ii †Kvb c×wZwU wMÖbvi wbY©q Ki| [Ti = 47.88 Ges Mg = 24.3] [BUET 18-19] mgvavb: (i) GUg BKbwg = 47.88 47.88 + 2  (24.34 + 16)  100% = 37. 2665% (ii) GUg BKbwg = 47.88 47.88 + 32  100 = 59.93%  (ii) bs c×wZ AwaKZi wMÖbvi| 2| wMÖb †Kwgw÷a Kx? GUg BKbwg Kv‡K e‡j? mgvavb: imvq‡bi †h kvLvq Kg cwi‡ek `~lY K‡i Ggb me cÖwμqv I Drcv`b c×wZ wb‡q M‡elYv Kiv nq Zv‡K wMÖb †Kwgw÷a e‡j| †Kv‡bv ivmvqwbK wewμqvi †ÿ‡Î Drcbœ Kvw•ÿZ Drcv‡`i fi I Drcbœ mKj Drcv‡`i f‡ii Abycv‡Zi 100 ̧wYZK msL ̈v gvb‡K GUg BKbwg e‡j| 3| B-d ̈v±i Kv‡K e‡j? CH2 = CH – CH2Cl + H2O  CH2 = CH – CH2OH + HCl wewμqvq CH2 = CH – CH2OH Drcv` Ges HCl eR© ̈| wewμqvwUi ‘E’ d ̈v±i KZ? mgvavb: Drcv`b cÖwμqvq †gvU Drcv‡`i f‡ii Zzjbvq Kx cwigvY eR© ̈ Drcbœ nq Zvi AbycvZ‡K B-d ̈v±i e‡j| E = †gvU eR© ̈ (kg) †gvU Drcv` (kg) = 36.5 58 = 0.629 = 0.63 (Ans.) 4| Bw_wjb A·vBW (C2H4O) Drcv`‡bi `ywU mgxKiY wb¤œiƒc : (i) C2H4 + Cl2 + Ca(OH)2  C2H4O + CaCl2 + H2O (ii) C2H4 + 1 2 O2 Ag250C –––––––– C2H4O G `ywU c×wZi g‡a ̈ †KvbwU wMÖbvi c×wZ n‡e, Zv e ̈vL ̈v Ki| mgvavb: GUg B‡Kvbwg (% AE) = Kvw•ÿZ Drcv‡`i †gvU †gvj  †gvjvi fi  100 †gvj msL ̈vmn †gvU Drcv‡`i ms‡KZ f‡ii mgwó Kvw•ÿZ Drcv`  C2H4O (i) C2H4 + Cl2 + Ca(OH)2  C2H4O + CaCl2 + H2O % AE = 44  100 44 + 111 + 18 = 25.43% (ii) C2H4 + 1 2 O2 Ag250C –––––––– C2H4O; % AE = 100%  (ii) bs c×wZ wMÖbvi| wewμqvi nvi 5| A ̈v‡gvwbqv Aw·‡R‡bi mv‡_ wb‡¤œi mgxKiY Abyhvqx wewμqv K‡i| 4NH3(g) + 5O2(g) = 4NO(g) + 6H2O(g) †Kv‡bv gyn~‡Z© A ̈v‡gvwbqv 0.24 mol L–1 s –1 nv‡i wewμqv Ki‡j wewμqvwUi nvi mgxKiYwU wjL Ges H2O(g) Drcv`‡bi nvi wbY©q Ki| [BUET 17-18] mgvavb: 4 NH3(g) + 5O+2(g) = 4 NO(g) + 6 H2O(g) wewμqvi nvi = –1 4 [NH3] t = –1 5 [O2] t = 1 4 [NO] t = 1 6 [H2O] t –[NH3] t = 0.24 GLb, –1 4 [NH3] t = 1 6 [H2O] t  [H2O] t = 6 4     –[NH3] t = 6 4  0.24 = 0.36 mol L–1 s –1 (Ans.) 6| KI Gi `ae‡Y †Kvb wbw`©ó NbgvÎvi H2O2 `aeY †hvM Ki‡j 10 †m‡K‡Û 10–5molL–1 Av‡qvwWb gy3 nq| Av‡qvwWb Drcbœ nIqvi G wewμqvwUi Mo MwZ‡eM wbY©q Ki| [KU 15-16] mgvavb: 2KI + H2SO4 + H2O2  K2SO4 + 2H2O + I2 wewμqvwUi Mo MwZ‡eM = [I2] t = 10–5 molL–1 10 sec = 10–6 molL–1 sec–1 (Ans.) 7| wewμqvi nvi Kx? NbgvÎv e„wׇZ wewμqv nvi e„w× cvq †Kb? mgvavb: GKK mg‡q wewμq‡Ki NbgvÎv hZUzKz n«vm cvq ev Drcv‡`i NbgvÎv hZUzKz e„w× cvq Zv‡K wewμqvi nvi e‡j| aiv hvK, †Kvb wewμqv A + B ⇋ C + D NbgvÎv e„w× Ki‡j AYymg~‡ni g‡a ̈ msNl© e„w× cvq hvi d‡j wewμqvi nvi e„w× cvq| Avi ZvQvov NbgvÎvi mv‡_ wewμqvi nvi wb‡¤œv3fv‡e wbf©ikxj| Kc = – A t = – B t = + C t = + D t 8| wewμqvi nvi I nvi aaæe‡Ki g‡a ̈ wZbwU cv_©K ̈ wjL| mgvavb: wewμqvi nvi I nvi aaæe‡Ki g‡a ̈ wZbwU cv_©K ̈ n‡jvÑ wewμqvi nvi wewμqvi nvi aaæeK 1. cÖwZ GKK mg‡q †Kv‡bv ivmvqwbK wewμqvi wewμqK ev Drcv‡`i NbgvÎvi cwieZ©b‡K wewμqvi nvi ejv nq| 1. wewμqvi nvi I wewμq‡Ki NbgvÎvi AbycvZ‡K wewμqvi nvi aaæeK ejv nq| 2. wewμqvi nvi wewμq‡Ki NbgvÎvi Dci wbf©ikxj| 2. wewμqvi nvi aaæeK ZvcgvÎvi Dci wbf©ikxj|
2  Chemistry 1st Paper Chapter-4 3. wewμqvi nv‡ii GKK mol L –1 s –1 | 3. wewμqvi nvi aaæeK wewfbœ wewμqvi †ÿ‡Î wewfbœ| GwU cixÿv wbf©i gvÎ| 9| wb‡Pi wewμqvi Rb ̈ wewμqvi nv‡ii ivwkgvjv †jLÑ 5Br– (aq) + BrO– 3 (aq) + 6H+ (aq)  3Br2(aq) + 3H2O(l) mgvavb: G‡ÿ‡Î g‡b ivL‡Z n‡e †h wewμqK ̧‡jvi mnM w`‡q fvM w`‡jB wewμqvi cÖK...Z nvi cv‡ev Avgiv| A_©vr wewμqvi nvi = – 1 5 [Br – ] t = – [BrO3 – ] t = – 1 6 [H+ ] t = 1 3 [Br2] t = 1 3 [H2O] t 10| CCl4 Gi Dcw ̄’wZ‡Z 318K ZvcgvÎvq N2O5 Gi we‡qvRb Kiv n‡jv| wewμqvi ïiæ‡Z N2O5 Gi NbgvÎv 2.33 M Ges 184 min ci N2O5 Gi NbgvÎv cvIqv †Mj 2.08 M wewμqvwU wb¤œiƒcÑ 2N2O5(g)  4NO2(g) + O2(g) wewμqvwUi Mo nvi wbY©q Ki| GB mg‡q NO2 •Zwii nvi KZ wQj? mgvavb: Mo nvi = 1 2       – [N2O5] t = – 1 2     (2.08 – 2.33) molL–1 186 min = 6.79  10–4mol L–1min–1 wewμqvi Mo nvi = 1 4      [NO2] t 1 4 [NO2] t = 6.79  10–4mol L–1min–1  [NO2] t = 2.72  10–3mol L –1min–1 (Ans.) wewμqvi μg 11| M ̈vmxq Ae ̄’vq mvB‡K¬v‡cÖv‡cb †_‡K mvB‡K¬v‡cÖvwcb •Zwi GKwU cÖ_g μg wewμqv hvi nvi aaæeK 500C G 6.7  10–4 s –1 | (a) mvB‡K¬v‡cÖv‡cb Gi cÖv_wgK NbgvÎv hw` 0.25 M nq Z‡e 8.8 min ci Gi NbgvÎv KZ n‡e? (b) mvB‡K¬v‡cÖv‡cb Gi NbgvÎv 0.25 M †_‡K n«vm †c‡q 0.15 M n‡Z KZ mgq jvM‡e? [BUET 16-17] mgvavb: (a) C = C0e –Kt = 0.25  e –6.710–48.860 M = 0.1755 M (b) C = C0e –Kt  ln C C0 = – Kt  t = ln C C0 –K = ln 0.15 0.25 –6.7  10–4 s = 762.43 s (Ans.) 12| cÖvPxb mf ̈Zvi GKwU Kv‡Vi cyZz‡ji †ZRw®ŒqZvi gvb 12 cpm g–1 cvIqv †Mj| cÖv_wgK Ae ̄’vq cyZzjwUi †ZRw®ŒqZvi gvb wQj 20 cpm g–1 | cÖvPxb mf ̈Zvi eqm wbY©q Ki| (†ZRw®Œq Kve©‡bi Aa©vqy 5600 eQi) [CUET 09-10, BUET 10-11] mgvavb: T1 2 = 0.693 k = 5600 yr  k = 0.693 5600 yr–1 = 1.2375  10–4 yr–1 Avgiv Rvwb, kt = ln C0 Ct  t = 1 k ln C0 Ct = 1 1.2375  10–4 ln 20 12 yr = 4127.88 yr (Ans.) 13| 500C ZvcgvÎvq NO2 Gi we‡qvRb wewμqvi †eM aaæeK 2.4  10–3 dm3mol–1 sec–1 KZ cÖv_wgK NbgvÎvq wewμqvwUi Aa©vqyKvj 15 minutes n‡e? [BUET 08-09] mgvavb: GwU GKwU wØZxq μg wewμqv| AZGe, t1 2 = 1 ka  a = 1 t1 2  k = 1 900  2.4  10–3 = 0.4629 mole dm–3 (Ans.) 14| GKwU 1g μg wewμqvi nvi aaæeK 1.05  10–2 sec–1 | hw` wewμq‡Ki cÖviw¤¢K NbgvÎv 0.8 mol/dm3 nq, Zvn‡j 3 wgwbU c‡i nvi KZ? mgvavb: 3 wgwbU c‡i wewμqvi nvi = K Ct , Avgiv Rvwb, K = 2.303 t log C0 Ct  1.05  10–2 = 2.303 180 log 0.8 Ct  log     0.8 Ct = 0.8206  0.8 Ct = 100.8206  Ct = 0.8 6.617  Ct = 0.12089 mol L–1  3 wgwbU c‡i wewμqvwUi nvi = K  Ct = 1.05  10–2  0.1209 = 1.692  10–3mol L–1 s –1 (Ans.) 15| †iwWqv‡gi Aa©vqyKvj 1590 eQi| †iwWqv‡gi GKwU bgybv cÖviw¤¢K NbgvÎvi 10% G n«vm †c‡Z KZ mgq jvM‡e? mgvavb: t1/2 = 0.693 / k  k = 0.693/1590 = 4.36 10–4 year–1 GLb, kt 2.303 = log [A]0 [A]t
ivmvqwbK cwieZ©b  Academic Practice Sheet 3 4.36  10–4 2.303 t = log 100 10  t = 5282.11 year 5282.11 eQi c‡i cÖviw¤¢K NbgvÎvi 10% †iwWqvg Aewkó _v‡K| mwμqY kw3 16| mwμqY kw3 Kx? A‡Uv cÖfveK Kv‡K e‡j? mgvavb: †Kv‡bv ivmvqwbK wewμqv m¤úbœ n‡Z b~ ̈bZg †h cwigvY kw3 cÖ‡qvRb Zv‡K mwμqY kw3 e‡j| †Kv‡bv ivmvqwbK wewμqvq wewμqvRvZ †h‡Kv‡bv GKwU Dcv`vb cÖfveK wn‡m‡e wewμqvi †eM e„w× K‡i, G ai‡bi cÖfveK‡K A‡Uv cÖfveK e‡j| 17| †Kv‡bv wewμqvi †ÿ‡Î 338 K ZvcgvÎvq †eM aaæe‡Ki gvb 328 K ZvcgvÎvq †eM aaæe‡Ki gv‡bi wØ ̧Y n‡j wewμqvwUi mwμqY kw3i gvb MYbv Ki| mgvavb: Avi‡nwbqvm mgxKiY Abyhvqx, GLv‡b, log K2 K1 = Ea 2.303 R     T2 – T1 T1T2  log     2K1 K1 = Ea  103 2.303  8.314     338 – 328 328  338  Ea = 63.19 kJ mol–1 (Ans.) 18| A + B ⇌ C + D 500 K I 530 K ZvcgvÎvq wewμqvwUi †eM aaæe‡Ki gvb h_vμ‡g 1.95  10–4 L mol–1 s –1 I 1.75  10–3 L mol–1 s –1 n‡j wewμqvwUi mwμqY kw3i gvb wbY©q Ki| mgvavb: Avgiv cvB, log     K2 K1 = Ea 2.303 R     T2 – T1 T1T2  log K2 K1 = Ea 2.303 R     T2 – T1 T1T2  log 1.75  10–3 1.95  10–4 = Ea  103 2.303  8.314     530 – 500 500  530  Ea = 161.18 kJ mol–1 (Ans.) 19| wewμqv nv‡ii Dci ZvcgvÎvi cÖfve K = Ae –Ea/RT mgxKiY Øviv cÖKvk Kiv hvq| GLv‡b K n‡jv wewμqv nvi aaæeK I Ea nj mwμqY kw3| K Gi gvb 0C ZvcgvÎvq 0.237  10–4L mol–1 s –1 I 25C ZvcgvÎvq 2.64  10–4L mol–1 s –1 n‡j mwμqY kw3 wbY©q Ki| [BUTex 19-20] mgvavb: K = Ae –Ea/RT  K1 = Ae –Ea/RT2  ln K1 = ln A – Ea RT1 ... ... ... (i)  K2 = Ae –E a /RT2  ln K2 = ln A – Ea RT2 ... ... ... (ii) (i) bs mgxKiY n‡Z (ii) bs mgxKiY we‡qvM K‡i cvB,  ln K1 K2 = Ea R     1 T2 – 1 T1  ln 0.237  10–4 2.64  10–4 = Ea 8.314     1 298 – 1 273  Ea = 65215.58 J mol–1 = 65.21558 kJ mol–1 (Ans.) cafveK I Gi e ̈envi 20| cÖfve‡Ki 5wU •ewkó ̈ wjL| mgvavb: 1| cÖfveK ïaygvÎ wewμqvi †eM e„w× ev n«vm K‡i| wewμqv ïiæ A_ev mgvwß wba©vi‡Y Gi †Kvb f~wgKv †bB| 2| wbw`©ó wewμqvq wbw`©ó ivmvqwbK c`v_© cÖfveK wn‡m‡e KvR K‡i| A_©vr cÖfve‡Ki Kvh©KvwiZv mywbw`©ó| 3| wewμqv †k‡l cÖfve‡Ki fi I ivmvqwbK MVb AcwiewZ©Z _v‡K| 4| mvgvb ̈ cwigvY cÖfve‡Ki Dcw ̄’wZB wewμqvi MwZ‡K cÖfvweZ Kivi Rb ̈ h‡_ó| 5| cÖfveK mswkøó wewμqvi mwμqY kw3 n«vm ev e„w× K‡i wewμqvi Rb ̈ weKí c_ m„wó K‡i| ivmvqwbK mvg ̈ve ̄’v 21| N2(g) + 3H2(g) ⇌ 2NH3(g) ; H = – 92.38 kJ. wb‡¤œ cÖ`Ë cÖkœ ̧‡jvi DËi `vI: (a) mg‡qi mv‡_ N2 I NH3 Gi cwigv‡Yi cwieZ©b wP‡Î †`LvI| Df‡qi mv‡c‡ÿ m¤§yL wewμqvi nvi †jL| (b) wewμqvwUi mvg ̈ve ̄’vi Dci Zvc I Pv‡ci cÖfve wK n‡e? (c) mvg ̈aaæeK (K) Gi Dci cÖfve‡Ki †Kvb cÖfve i‡q‡Q Kx? [DU 19-20] mgvavb: (a) N2 NH3 time Amount © wewμqvi nvi = – [N2] t = 1 2 [NH3] t (b) wewμqvwU Zv‡cvrcv`x| mvg ̈v ̄’vq ZvcgvÎv evov‡j wewμqv cðvrw`‡K Mgb Ki‡e Ges ZvcgvÎv Kgv‡j wewμqv m¤§yLw`‡K Mgb Ki‡e| Avevi, mvg ̈ve ̄’vq Pvc e„w× Kiv n‡j wewμqv m¤§yLw`‡K Mgb Ki‡e Ges Pvc n«vm Kiv n‡j cðvrgyLx n‡e| (c) mvg ̈aaæeK (K) Gi Dci cÖfve‡Ki †Kvb cÖfve †bB| 22| GKwU dv‡b©‡m 35% CO2 – 65% CO M ̈vm wgkÖY e ̈envi K‡i 700C G ÷xj‡K Zvc w`‡j Gi Dci gwiPv co‡e wKbv Zv wbY©q Ki| [†`Iqv Av‡Q; Fe(s) + CO2(g) = FeO(s) + CO(g); K = 1.43] [BUET 20-21] mgvavb: Qp = PCO PCO2 = 0.65 0.35 = 1.86 > K. myZivs, gwiPv co‡e bv|
4  Chemistry 1st Paper Chapter-4 23| wb‡¤œv3 ivmvqwbK wewμqvmg~‡n mvg ̈ve ̄’vi Pvc n«vm K‡i M ̈vmxq AvqZb e„w× Ki‡j wewμqvi wgkÖ‡Y Drcv‡`i †gvj msL ̈vi Kx cwieZ©b n‡e e ̈vL ̈v Ki| (i) CaO(s) + CO2(g) ⇋ CaCO3(s) (ii) 3Fe(s) + 4H2O(g) ⇋ Fe3O4(s) + 4H2(g) [BUET 20-21] mgvavb: (i) Drcv` Kg‡e, n = –1 (ii) AcwiewZ©Z _vK‡e, n = 0 24| A2(g) + B2(g) ⇋2AB(g) wewμqvwU 2L AvqZ‡bi cv‡Î m¤úbœ Kiv nq| wewμqvi ïiæ‡Z A2 Gi †gvjmsL ̈v 2, B2 Gi †gvjmsL ̈v 2 Ges mvg ̈ve ̄’vq Drcv‡`i †gvjmsL ̈v 3.12| Kp I Kc Gi gvb wbY©q Ki| [BUTex 21-22] mgvavb: A2(g) + B2(g) ⇌ 2AB(g) cÖv_wgK Ae ̄’vq: 2 mol 2 mol 0 mol mvg ̈ve ̄’vq †gvj:(2 – x) mol (2 – x) mol 2x mol GLv‡b, 2x = 3.12 mol, x = 1.56 mol  [A2] = 2 – 1.56 2 = 0.22 M [B2] = 2 – 1.56 2 = 0.22 M; [AB] = 3.12 2 = 1.56 M  Kc = [AB]2 [A2]  [B2] = 1.562 0.22  0.22 = 50.28  n = 0,  Kp = Kc = 50.28 (Ans.) 25| GKwU DfgyLx ivmvqwbK wewμqvi n = 1 2 | KZ †Kjwfb ZvcgvÎvq Kp I Kc Gi gvb h_vμ‡g 40.5 Ges 5.5 n‡e? [R = 0.082 L atm mol–1 K –1 ] [CKRUET 21-22] mgvavb: Kp Kc = (RT)n =  40.5 5.5 = (0.082  T) 1 2  T = 661.26K (Ans.) 26| A2 + B2 ⇌ 2AB wewμqvq mvg ̈ aaæeK Kp n‡j, AB ⇌ 1 2 A2 + 1 2 B2 wewμqvi mvg ̈ aaæeK KZ? [KUET 19-20] mgvavb: Kp = [AB]2 [A2]  [B2] AB ⇌ 1 2 A2 + 1 2 B2 wewμqvi mvg ̈aaæeK, K = [A2] 1 2  [B2] 1 2 [AB] =     [A2]  [B2] [AB]2 1 2 =     1 Kp 1 2 = 1 Kp (Ans.) 27| jv-kv‡Zwjqvi bxwZ †jL| fiwμqvi m~Î Kv‡K e‡j? mgvavb: †Kv‡bv DfgyLx wewμqv mvg ̈ve ̄’vq _vKvKv‡j hw` H Ae ̄’vi GKwU wbqvgK, †hgb ZvcgvÎv, Pvc A_ev NbgvÎv cwieZ©b Kiv nq, Z‡e mv‡g ̈i Ae ̄’vb Wv‡b ev ev‡g Ggbfv‡e cwiewZ©Z n‡e, hv‡Z wbqvgK cwieZ©‡bi djvdj cÖkwgZ nq| wbw`©ó ZvcgvÎvq, wbw`©ó mg‡q †h †Kv‡bv wewμqvi nvi H mg‡q Dcw ̄’Z wewμqK ̧‡jvi mwμq f‡ii (A_©vr †gvjvi NbgvÎv ev AvswkK Pv‡ci) mgvbycvwZK nq| 28| mvg ̈aaæeK Kc Gi gvb k~b ̈ nq bv †Kb? mgvavb: GKwU mij wewμqv we‡ePbv Kwi, A(g) ⇌ B(g); Kc = [B] [A]; Kp = PB PA mvg ̈e ̄’vq A Gi cwigvY 0 n‡j [A] = 0; PA = 0 ZLb, Kc = ; Kp =  n‡e| Avevi, B Gi cwigvY 0 n‡j (B) = 0; PB = 0 ZLb, Kc = 0; Kp = 0 wKš‘ Avgiv Rvwb, mvg ̈aaæeK †KejgvÎv DfgyLx wewμqvi †ÿ‡Î cÖ‡hvR ̈| Avi DfgyLx wewμqvq wewμqK ev Drcv‡`i NbgvÎv KLbI 0 n‡Z cv‡i bv| ZvB mvg ̈aaæeK KLbI 0 ev  n‡Z cv‡i bv| 29| DfgyLx wewμqv KL‡bv †kl nq bv †Kb? ev ivmvqwbK mvg ̈ve ̄’vi MwZkxjZv e ̈vL ̈v Ki| mgvavb: DfgyLx wewμqvq m¤§yLgyLx I cðvrgyLx wewμqvi nvi hLb mgvb nq, ZLb wewμqvwU mvg ̈ve ̄’v AR©b K‡i| GB mvg ̈ve ̄’vq wewμqv‡Z †h cwigvY wewμqK †f‡1⁄2 Drcv‡` cwiYZ nq, wVK GKB cwigvY Drcv` †f‡1⁄2 wewμq‡K cwiYZ nq| A_©vr wewμqK Ges Drcv‡`i cwigvY wVK _vK‡jI GLv‡b ivmvqwbK c`v‡_©i fv1⁄2v-Mov AbeiZ Pj‡Z _v‡K| ZvB DfgyLx wewμqv †kl nq bv| NbgvÎv mvg ̈ve ̄’v Drcv‡`i NbgvÎv mgq wewμq‡Ki NbgvÎv DfgyLx GKwU wewμqvi D`vniY : CaCO3 ⇌  CaO + CO2 (e× cvÎ) 30| DuPz RvqMvq eid ax‡i M‡j †Kb? mgvavb: f~c„‡ô evqygЇji Pvc †ewk| f~c„ô n‡Z hZB Dc‡i hvIqv hvq evqygЇji Pvc ZZB n«vm cvq| DuPz cvnv‡oi Dci Pvc, we‡kl K‡i Gfv‡i‡÷i P~ovq Pvc f~-c„‡ôi Zzjbvq A‡bK Kg| KwVb eid M‡j Zij cvwb‡Z cwiYZ nq| G‡ÿ‡Î cÖwZ 110 mL