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Class : XIII PART TEST- 1 INSTRUCTIONS 1. The question paper contain pages and 3-parts. Part-A contains 6 objective question , Part-B contains 2 questions of "Match the Column" type and Part-C contains 4 subjective type questions. All questions are compulsory. Please ensure that the Question Paper you have received contains all the QUESTIONS and Pages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator. PART-A (i) Q.1 to Q.6 have One or More than one is / are correct alternative(s) and carry 4 marks each. There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. PART-B (iii) Q.1 to Q.2 are "Match the Column" type which may have one or more than one matching options and carry 8 marks for each question. 2 marks will be awarded for each correct match within a question. There is NO NEGATIVE marking. Marks will be awarded onlyif all the correct alternatives are selected. PART-C (iv) Q.1 to Q.4 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded only if all the correct bubbles are filled in your OMR sheet. 2. Indicate the correct answer for each question by filling appropriate bubble(s) in your answer sheet. 3. Use only HB pencil for darkening the bubble(s). 4. Use of Calculator, Log Table, Slide Rule and Mobile is not allowed. 5. The answer(s) of the questions must be marked by shading the circles against the question by dark HB pencil only. PART-B For example if Correct match for (A) is P, Q; for (B) is P, R; for (C) is P and for (D) is S then the correct method for filling the bubble is P Q R S (A) (B) (C) (D) PART-C Ensure that all columns (4 before decimal and 2 after decimal) are filled. Answer having blank column will be treated as incorrect. Insert leading zero(s) if required after rounding the result to 2 decimal places. e.g. 86 should be filled as 0086.00 Class - XIII Chemistry Part Test - 1 PART A Q.1 Calculate total maximum mass (kg) which can be lifted by 10 identical balloon (each having volume 82.1 lit. and mass of balloon & gas = 3 kg) at a height 83.14 m at Mars where g = 5 m/s2 & atmosphere contains only Ar (At. wt.40). At Mars temperature is 10 K and density of atmosphere at ground level is 2 k gm/lit. [Given : e–0.1 = 0.9] (Assume d = d eMgh RT to be applicable). 0.821 H 0 (A*) 2000 × 0.81–30 (B) 1970 (C) 2000 × 0.9 –30 (D) 2000 × 0.81 – 3 – Mgh [Sol: (A) d = d0 2 d = 0.821 e RT – 40103583.14 e 8.31410 2 = 0.821 e–0.2 = 2 (0.9)2 0.821 Now at height h = 83.14 m V × d = m1 a4ss4of4 b4al4loo4n4s +2 m4as4s 4of 4ga4s 4fil4le3d + mass of load 82.110 (0.9)2 0.821 = (3 × 10) + m m = (2000 × 0.81) – 30 Kg ] Q.2 How many mg of quick lime is required to remove hardness of 1 kg of hard water having 366 ppm of HCO3– and contains Ca2+ as the only cation (A) 72

- 1. Class : XIII (Sterling) Time : 3 hour Max. Marks : 60 INSTRUCTIONS 1. The question paper contain pages and 3-parts. Part-A contains 6 objective question , Part-B contains 2 questions of "Match theColumn" type and Part-C contains 4 subjective type questions.All questions are compulsory. Please ensure that the Question Paper you have received contains all the QUESTIONS and Pages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator. PART-A (i) Q.1 to Q.6 have One or More than one is / are correct alternative(s) and carry 4 marks each. There is NEGATIVE marking. 1 mark will be deducted for eachwrong answer. PART-B (iii) Q.1 to Q.2 are"Match the Column" typewhich mayhave oneor more than onematching options and carry8 marks foreach question. 2 marks will be awarded for each correct match within a question. ThereisNONEGATIVEmarking.Markswillbeawardedonlyifallthecorrectalternativesareselected. PART-C (iv) Q.1 to Q.4 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded onlyifall the correct bubbles are filled in yourOMR sheet. 2. Indicate the correctanswer for each question byfillingappropriate bubble(s) inyour answer sheet. 3. Use onlyHB pencil fordarkeningthe bubble(s). 4. Use ofCalculator, LogTable, SlideRule and Mobile is not allowed. 5. Theanswer(s)ofthequestionsmustbemarked byshadingthecirclesagainst thequestionbydarkHBpencil only. PART TEST-1 PART-B For example if Correct match for (A) is P, Q; for (B) is P, R; for (C) is P and for (D) is S then the correct method for filling thebubbleis P Q R S (A) (B) (C) (D) PART-C Ensure that all columns (4 before decimal and 2 after decimal) are filled. Answer having blank column willbe treated as incorrect. Insert leading zero(s) if required after rounding the result to 2 decimalplaces. e.g. 86 should befilled as 0086.00 . . . . . . . . . . PART-A For example if only 'B' choice is correct then, the correct method for fillingthebubbleis A B C D Forexampleifonly'B& D' choices are correct then, the correct method forfillingthebubblesis A B C D The answer of the question in any other manner (such as putting , cross , or partial shading etc.) will be treated as wrong.
- 2. Class - XIII Chemistry Part Test - 1 PART A Q.1 Calculatetotalmaximummass(kg)whichcanbeliftedby10identicalballoon(eachhavingvolume82.1 lit. and mass of balloon & gas = 3 kg) at a height 83.14 m at Mars where g = 5 m/s2 & atmosphere containsonlyAr(At. wt.40).AtMars temperature is 10K and densityofatmosphereat groundlevel is 821 . 0 2 k gm/lit. [Given : e–0.1 = 0.9] (Assume dH = d0 RT Mgh e to be applicable). (A*) 2000 × 0.81–30 (B) 1970 (C) 2000 × 0.9 –30 (D) 2000 × 0.81 – 3 [Sol: (A) d = d0 RT Mgh e d = 821 . 0 2 10 314 . 8 14 . 83 5 10 40 3 e = 821 . 0 2 e–0.2 = 2 ) 9 . 0 ( 821 . 0 2 Now at height h = 83.14 m V × d = load of mass filled gas of mass balloons of mass 2 ) 9 . 0 ( 821 . 0 10 1 . 82 = (3 × 10) + m m = (2000 × 0.81) – 30 Kg ] Q.2 How manymg of quick lime is required to remove hardness of 1 kg of hard water having 366 ppm of HCO3 – and contains Ca2+ as the only cation (A) 72 mg (B) 84 mg (C*) 168 mg (D) 170 mg [Sol: In 1 Kg water – 3 HCO w = 366 × 10–3 gm – 3 HCO n = 6 × 10–3 CaO + Ca(HCO3 –)2 2CaCO3 + H2O moles of CaO required = 2 n – 3 HCO = 3 × 10–3 wt. of CaO required = 3 × 10–3 × 56 = 168 × 10–3 gm = 168 mg [C] ] Q.3 A 10 litre box contains O3 and O2 at equilibrium at 2000K. Kp = 4.17 × 1014 for 2O3 3O2. Assume that 3 2 O O P P and if total pressureis 7.33 atm, then partial pressure of O3 will be (A) 9.71 × 10–5 atm (B*) 9.71 × 10–7 atm (C) 9.71 × 10–6 atm (D) 9.71 × 10–2 atm [Sol: 2O3 3O2 Kp = 2 3 3 2 O O P P Here 3 2 O O P P so the total pressure 2 O P
- 3. 4.17 × 1014 = 2 3 O 3 P 33 . 7 3 O P = 9.71 × 10–7 atm ] Q.4 KOH Br2 If the reactant is (d) (dextrorotatory) then in final product (A)inversionofconfigurationtakes place (B)racemisationtakes place (C*) retentionofconfigurationtakes place (D) none of the above [Sol: r.d.s.is1,2shift,shiftingtakes placewith retentionofconfiguration ] Q.5 Select the correct statement(s) If S = f (E, V, n1, n2) &A= f (T, V, n1, n2) (A*) S is equal to f (E, V, n1, n2) (B) S is equal to f (E, V, n1, n2) (C*) Ais equal to f (T, V, n1, n2) (D) Ais equal to f (E, V, n1, n2) [Sol: S = f (E, V, n1, n2) & E, V, n1, n2 all are extensive vairable S = f (E, V, n1, n2) A= f (T, V, n1, n2) hereTisintensivevariable A= f (T, V, n1, n2) [A, C] ] Q.6 Compound Bhas a neutralisationequivalent112. G is adichloro alkane. (A*) A = , B = , C = , E = ; F = & G = ClCH2CH2CH2CH2Cl (B) A = , B = , C = , E = & G = ClCH2CH2CH2CH2CH2Cl (C) A = , B = , C = , E = , F = & G = ClCH2CH2CH2CH2CH2Cl
- 4. (D) A = , B = , C = ; E = ; F = & G = ClCH2CH2CH2CH2CH2Cl [Sol: PART B Q.1 Matchthefollowing ColumnI ColumnII (A) Canaryyellowprecipitate with (NH4)2MoO4 (P) NO3 – (B) Brownringtest (Q) NO2 – (C)Acidradical whichevolves gas with conc. HCl (R) AsO4 3– (D)Acidradical whichgives gas withdil. H2SO4 (S) PO4 3– [Ans: (A) R, S; (B) P, Q; (C) P; (D) Q] [Sol: (A) (NH4)2MoO4 + AsO4 3– + H+ (NH4)3Mo12AsO40 Canaryyellow ppt. (NH4)2MoO4 + PO4 3– + H+ (NH4)3M12PO40 Canaryyellow ppt. (B) NO3 – + Fe2+ + H2SO4 Fe3+ + NO + SO4 2– + H2O Fe2+ + NO [Fe(NO)]2+ Brownring
- 5. NO2 – + CH3COOH HNO2 + CH3COO– HNO2 NO + HNO3 + H2O Fe2+ + SO4 2– + NO [Fe(NO)]2+ Brown ring (C) NO3 – + H+ NO2 + H2O (D) NO2 – + H+ HNO2 NO + HNO3 + H2O ] Q.2 Matchthefollowing List I List II (A) CuCl (P)Freeradical mechanism (B) CH3–CH=CH2 (Q) Non Classical Carbocation Cl2/ Cl | CH CH CH 2 2 (C) CH3–CH=CH2 Cu / Zn / I CH 2 2 (R) Carbenoid (D) CH3–CH=CH2 OH / NaBH ) ii ( O H / ) OAc ( Hg ) i ( 4 2 2 (S)Through carbocation [Ans: (A) P; (B) P; (C) R; (D) Q] [Sol: (A) PhCl CuCl N Ph 2 Mechanism 2 2 o 2 CuCl N Ph CuCl Cl N Ph CuCl PhCl CuCl Ph 2 o (B) Cl2 2Clo (C)
- 6. (D) PART C Q.1 1.0gofamixturecontainingSb2O3 andSb2O5 andsomeinert impurities required 100ml0.01 Niodine fortitration.Theresultingsolutionis thenacidifiedandexcessofKIwasadded.Theliberated I2 required 100 ml 0.02 M Na2S2O3·5H2O for complete reaction. Calculate the % of Sb2O5 in the mixture. The reactions are [at. wt. Sb = 122] Sb2O3 + 2I2 + 2H2O Sb2O5 + 4H+ + 4I– Sb2O5 + 4H+ + 4I– Sb2O3 + 2I2 + 2H2O [Sol: Meq of I2 used = 100 × 0.01 = 1 meq Let meq of Sb2O3 and Sb2O5 are a and b, on addition of I2 to mixture Sb+3 converts into Sb+5 Meq. of Sb+5 formed = meq of I2 = 1 meq = meq of Sb2O3 a = 1 meq After reaction,.mixturecontains all theSb in+5oxidation state. Meq of Sb+5 = meq of liberated I2 = meq of hypo solution used a + b = 100 × 0.02 × 1 = 2 meq b = 1 meq = 4 324 . 0 4 324 1000 1 g = 0.081 g Sb2O5 = 8.1 % ] Q.2 Aballooncontaining1moleairat 1 atminitiallyis filled furtherwithairtill pressure increasesto 3 atm. The initial diameter of the balloonis 1 m and the pressure at eachstate is proportion to diameter of the balloon.Ifballoonwillburst whenpressureincreasesto7 atm.Calculatethenumberofmolesofairthat must be addedafterinitial condition toburst theballoon. [Sol: P diameter d = Kp Initially 1 = K × 1 K = atm m 1 For Pf = 7 atm
- 7. d = 7m V = 3 2 7 3 4 m3 7 × 3 2 7 3 4 = nf RT T ..................(3) Dividing (3) by (1) i f n n = 7 × 7 × 7 × 7 nf = 2401 moles added = 2400 ] Q.3 Thekeyreactioninthemanufactureofsyntheticcryolightforaluminiumelectrolysisis HF(g) + Al(OH)3 (s) + NaOH (aq) Na3AlF6(aq) + H2O(l) Assuminga96%yieldofdried,crystallized product, whatmass (inkg)ofcryolitecanbeobtainedfrom the reaction of 351 kg ofAl(OH)3, 1.10 m3 of 50.0% by mass aqueous NaOH (d = 1.50 g/mL), and 225 m3 of gaseous HF at 312.08 kPa and 87oC? (assume that the ideal gas law holds) [Given: Al = 27, O = 16, H = 1, Na = 23, F = 19, R = 8.3 JK–1 mol–1] [Sol: 6HF + Al(OH)3 + 3NaOH Na3AlF6 + 6 H2O(l) moles ofAl(OH)3 = 78 10 351 3 = 4500 moles of NaOH = 40 5 . 0 5 . 1 10 1 . 1 6 = 20625 moles of HF = RT PV = 360 3 . 8 225 312080 = 23500 HereHFislimitingreagent so moles of Na3AlF6 formed = 100 96 6 23500 = 3760 wt. of Na3AlF6 formed = 1000 210 3760 = 789.6 kg ] Q.4 The vapour pressure of water at 300 K is 25 torr. If the standard state pressure is defined as 1 bar (750 torr) estimate the G° [inkJ/mol] for the process. H2O (l) H2O (g) at 300 K (Neglect variation of H and S with pressure for liquid) Use [R = 8.314 JK–1 mol–1 ; X og X n l l = 2.3, log 3 = 0.48] [Ans. 8.49 ] [Sol: G0 = –2.3RT log Kp = – 2.3 × 8.314 × 300 log O H2 P (bar) = – 2.3 × 8.314 × 300 log 750 25 = – 2.3 × 8.314 × 300 log 30 1 = 8490J/mol = 8.49 kJ/mol ]