DPP-59 to 60 English Physical Chemistry

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 59 to 60 Class : XIII Course : DPP No.59 DPP No.1 Total Marks : 27 Single choice Objective (no negative marking) Q.1 to Q.5 (3 Marks, 3 Min.) Max. Time : 27 min. [15, 15] Multiple choice objective (no negative marking) Q.6 to Q.7 (4 marks 4 min.) [8, 8] Subjective Questions (no negative marking) Q.8 (4 marks 4 min.) [4, 4] 1. Assuming the salts to be completely ionized in solution, which of the following has highest osmotic pressure. (1) 1% CsCl w/w (2) 1% RbCl w/w (3) 1% KCl w/w (4) 1% NaCl w/w 2. What weight of glucose dissolved in 100 grams of water will produce the same lowering of vapour pressure as one gram of urea dissolved in 50 grams of water, at the same temperature? (1) 3 gms (2) 5 gms (3) 6 gms (4) 4 gms 3. Three solutions are prepared by adding 'w' gm of 'A' into 1kg of water, 'w' gm of 'B' into another 1 kg of water and 'w' gm of 'C' in another 1 kg of water (A, B, C are non electrolytic). Dry air is passed from these solutions in sequence (A  B  C). The loss in weight of solution A was found to be 2 gm while solution B gained 0.5 gm and solution C lost 1 gm. Then the relation between molar masses of A, B and C is : (1) M : M : M = 4 : 3 : 5 (2) M : M : M = 1 : 1 : 1 A B C A B C 4 3 5 (3) MC > MA > MB (4) MB > MA > MC 4. 20g of a binary electrolyte (molecular weight = 100) are dissolved in 500 g of water. The freezing point of the solution is – 0.74°C ; K = 1.86 K moIality–1, The degree of dissociation of electrolyte is : (1) 50% (2) 75% (3) 100% (4) Zero 5. 2.56g of sulfur in 100g of CS has depression in freezing point of 0.010C.K = 0.10molal1. Hence, the atomicity 2 f of sulfur in CS2 is : (1) 2 (2) 4 (3) 6 (4) 8 6. In which of the following pairs of solutions will the values of the vant Hoff factor be the same? (1) 0.05 M K4 [Fe(CN)6] and 0.10 M FeSO4 (2) 0.10 M K4[Fe(CN)6] and 0.05 M FeSO4 (NH4)2SO4. 6H2O (3) 0.20 M NaCl and 0.10 M BaCl2 (4) 0.05 M FeSO4 (NH4)2SO4 . 6H2O and 0.02 M KCl . MgCl2 . 6H2O 7. Which is the correct relation between osmotic pressure of 0.1 M NaCl solution & 0.1 M Na2SO4 solution : (1) the osmotic pressure of Na2SO4 is less than NaCl solution (2) the osmotic pressure of Na2SO4 is more than NaCl solution (3) the osmotic pressure of Na2SO4 is 1.5 times that of NaCl solution (4) the osmotic pressure of NaCl is 1.5 times that of Na2SO4 solution 8. At 12°C the osmotic pressure of a urea solution is 500 mm. The solution is diluted and the temperature is raised to 27°C, when the osmotic pressure is found to be 100 mm. determine the extent of dilution. DPP No.60 DPP No.2 Total Marks : 29 Max. Time : 29 min. Single choice Objective (no negative marking) Q.1 to Q.3 (3 Marks, 3 Min.) [9, 9] Multiple choice objective (no negative marking) Q.4 to Q.5 (4 marks 4 min.) [8, 8] Subjective Questions (no negative marking) Q.6 to Q.7 (4 marks 4 min.) [8, 8] Match the Following (no negative marking) Q.8 (4 marks 4 min.) [4, 4] 1. 1g of arsenic dissolved in 86 g of benzene brings down the freezing point to 5.31 °C from 5.50 °C. If Kf of C benzene is 4.9 m , the atomicity of the molecule is : (As – 75) (1) 8 (2) 2 (3) 3 (4) 4 2. Liquids A and B form an ideal solution. At a certain temperature the total vapour pressure of a mixture of A and B is 400 mm. The mole fraction of A in the liquid mixture in equilibrium with the vapour phase is 0.4. If the vapour pressure ratio (P 0/P 0) for the pure liquids at this temperature is 1/6, what is the partial pressure of A A B in the vapour phase? (1) 50mm (2) 60mm (3) 70mm (4) 40mm 3. Mixture of volatile components A and B has total vapour pressure (in Torr) p = 254 – 119 xA where xA is mole fraction of A in mixture. Hence p0 and p0 are (in Torr) (1) 254, 119 (2) 119, 254 (3) 135, 254 (4) 119, 373 4. Which of the following are true for ideal solutions : (1) V = 0 (2) H = 0 (3) S = 0 (4) G = 0 (5) Raoult's law is obeyed for entire concentration range and temperatures. 5. Consider the following solution in water (I) 1 M Sucorse (II) 1 M KCl (III) 1M Benzoic acid (IV) 1 M (NH4)3PO4 Which of the following is/are true : (1) All solutions are isotonic (2) III is hypertonic of II, IV (3) I, II and III are hypotonic of IV (4) IV is hypertonic of I, II and III 6. A membrane permeable only to water separates a 0.01 M solution of sucrose from a 0.001 M one. On which solution must pressure be applied to bring the system into equilibrium? Find this pressure if the T = 300 K. 7. The osmotic pressure of blood is 7 atm at 30°C. What is the molarity of the isotonic saline solution if the ‘i’ factor for sodium chloride is taken to be 1.9? 3 2 3 4 4 6 ANSWER KEY DPP No. 57 1. (1) 2. (2) 3. (3) 4. (2) 5. (2) 6. (3) 7. 6.4 × 10–4 cm 8. 0.8 A 9. 253.12 ml DPP No. 58 1. (1) 2. (2) 3. (4) 4. (3) 5. (1) 6. (3) 7. 0.69 cm–1, 0.425. 8. 2.46 atm 9. 4.92 atm. 10. 173 Informaiton About Commercial Cell Primary cell : Redox reaction occurs only once. These cell cannot be reacharged. Dry Cell Mercury Cell Anode Zn container Zn-Hg amalgam as anode. Cathode Carbon rod (graphite) Paste of Hg and carbon. Electrolyte Paste of NH4Cl and ZnCl2 Paste of KOH and ZnO. Reaction at anode Zn  Zn2+ + 2e– Zn (Hg) + 2OH–  ZnO +H2O + 2e– Reaction at cathode MnO2 + NH + + e–  MnO(OH) + NH 4 3 HgO + H2O + 2e–  Hg(l) + 2OH– Cell potential remain constant during its life. Secondary Batteries : This type of cell can be reacharged by passing current through it in the opposite direction so that it can be used again. Lead storage battery Ni—Cd cell Anode Lead Cadmium Cathode Grid of lead packed with lead dioxide (PbO2) Metal grid containing NiO2 Electrolyte 38% solution of H2SO4 KOH solution Reaction at anode 2– – Pb + SO4  PbSO4 + 2e Cd(s) + 2OH–  Cd(OH)2(s) + 2e– Reaction at cathode 2– + – PbO2 + SO4 + 4H + 2e  PbSO4 + 2H2O NiO2 + 2H2O + 2e–  Ni(OH)2(s) + 2OH– Fuel Cell : Fuel Celll Anode H2 is passed through porous carbon electrode containing suitable catalyst (generally finely divided Pt and Pd). Cathode O2 is passed through porous carbon electrode containing suitable catalyst. Electrolyte KOH or NaOH solution. Reaction at anode 2H2 + 4OH–  4H2O + 4e– Reaction at cathode O2 + 2H2O + 4e–  4OH– Over all reaction 2H2(g) + O2(g)  2H2O (l) Corrosion : The process of slowly eating away of the metal due to attack of atmoshperic gases on the surface of the metal resulting into the formation of compound such as oxide, sulphides, carbonates, sulphates etc. is called corrosion. Oxidation : Fe (s)  Fe2+ (aq) + 2e– Reduction : O (g) + 4H+ (aq) + 4e–  2H O (l) Atmospheric oxidation : 2Fe2+ (aq) + 2H O (l) + 1 (g)  Fe O 2 2 3 (s) + 4H+ (aq) Prevention of corrosion :  Converting the surface with paint or by some chemical (bisphenol).  Cover the surface by other metal (Sn, Zn).  On electrochemical method is to provide a sacrificial electrode of another metal (Like, Mg, Zn etc.) Which corrodes itself but saves the object.

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