DPP-5 to 6 Mole1 + Mole2-Answer

DAILY PRACTICE PROBLEMS ( DPP) Subject : Physical/Inorg. Chemistry Date : DPP No. 5 Class : XIII Course : DPP No.1 Max. Time : 25 Total Time : 25 min. Subjective Questions ('–1' negative marking) Q.1 to Q.4 (4 marks 4 min.) [16, 16] Comprehension ('–1' negative marking) Q.4 to Q.6 (3 marks 3 min.) [9, 9] 1. From the following reaction sequence Cl2 + 2KOH  KCl + KClO + H2O 3KClO  2KCl + KClO3 4KClO3  3KClO4 + KCl Calculate the mass of chlorine needed to produce 138.5 g of KClO4. Ans. 284 g 2. Zinc and hydrochloric acid react according to the reaction. Zn(s) + 2HCl(aq.)  ZnCl (aq.) + H (g) 2 2 If 0.30 mole of Zn are added to hydrochloric acid containing 0.52 mole HCl, how many moles of H2 are produced ? Ans. 0.26 3. A mixture of 1.0 mole of Al and 3.0 mole of Cl2 are allowed to react as : 2Al (s) + 3Cl2 (g)  2AlCl3 (s) (a) Which is limiting reagent ? (b) How many moles of AlCl3 are formed (c) Moles of excess reagent left unreacted is Ans. (a) Al, (b) 1.0 (c) 1.5 Passage : Read the following passage carefully and answer the questions (4 to 6). Iodine is an important substance needed by the body of a human being. We consume it in the form of salt which has very-very small % content of 2. Iodine has various industrial application also. The following process has been used to obtain iodine from oil-field brines in California. Nal + AgNO3  Agl + NaNO3 AgI + Fe  Fel2 + Ag FeI2 + Cl2  FeCl3 + 2 [Atomic mass Ag–108, –127, Fe–56, N-14, Cl–35.5] 4. If 381 kg of iodine is produced per hour then mass of AgNO3 required per hour will be - (1) 170 kg (2) 340 kg (3) 255 kg (4*) 510 kg 381 103 Sol. Moles of 2 produced = 254 for this much moles of 2 3 103 = 2 moles of AgNO3 required = 3 × 2 × 103 2  mass of AgNO3 required = 3 × 170 × 103 = 510 kg 5. Above reaction is carried out by taking 150 kg of Na and 85 kg of AgNO3 then number of moles of iodine formed is - (1) 0.5 (2) 500 (3*) 250 (4) 0.25 150 Sol. moles of Na = 150 × 103 = 103 85 moles of AgNO3 = 170 × 103 = 5 × 102 clearly AgNO3 is limiting reagent  moles of  formed = moles of AgNO3 2 2 5 102 = 2 = 250 6. If 324 g of Ag is recovered in pure form then minimum amount of Na required will be - (1*) 450 g (2) 150 g (3) 300 g (4) 600 g 324 Sol. Moles of Ag recovered = 108 = 3 Hence moles of Na required to produce this Ag = 3  mass of Na = 3 × 150 = 450 g DAILY PRACTICE PROBLEMS ( DPP) Subject : Physical/Inorg. Chemistry Date : DPP No. 6 Class : XIII Course : DPP No.2 Max. Marks : 28 Total Time : 28 min. Single choice Objective ('–1' negative marking) Q.1 to Q.5 (3 marks 3 min.) [15, 15] Short Subjective Questions ('–1' negative marking) Q.6 to Q.8 (3 marks 3 min.) [9, 9] Match the Following (no negative marking) Q.9 (4 marks 4 min.) [4, 4] 1. 120 g of solution containing 40% by mass of NaCl are mixed with 200 g of a solution containing 15% by mass NaCl. Determine the mass percent of sodium chloride in the final solution. (1*) 24.4% (2) 78% (3) 48.8% (4) 19.68% Sol. mass of NaCl is st solution = 120 × 0.4 = 48 g mass of NaCl in nd solution = 200 × 0.15 = 30 g Total mass of NaCl = 30 + 48 = 78 g Total mass of solution = 120 + 200 = 320 g mass % of NaCl = 78  100 320 = 24.375 % 2. What is the molality of the above solution. (1) 4.4 m (2*) 5.5 m (3) 24.4 m (4) none Sol. mass of solvent = 320 – 78 = 242 . 78 molality = 58.5 242 × 1000 = 5.5 m 3. What is the mole fraction of the solute. (1) 0.18 (2) 0.75 (3*) 0.09 (4) 0.25 78 Sol. Mole fraction of solute = 58.5 = 0.09 78 58.5  242 18 4. What is the molarity of solution if density of solution in 1.6 g/ml (1) 5.5 M (2*) 6.6 M (3) 2.59 M (4) None 78 Sol. Molarity = 58.5 320 1.6 × 1000 = 6.66 M 5. Percentage (weight / vol) of NaCl present in the solution. (1) 24.4 % (2) 40% (3*) 39% (4) 3.9% Sol. % w/v = 78 × 100 = 39% 320 1.6 6. 10 ml of sulphuric acid solution (sp. gr. = 1.84) contains 98% by weight of pure acid. Calculate the volume of 2.5 M NaOH solution required to just neutralize the acid. Ans. 147.2 ml Sol. For neutralisation m moles of H2SO4 = 2 × m moles of NaOH 98 98 × 1.84 × 10 × 10 = 2 × 2.5 × V V = 147.2 ml 7. Mole fraction of I2 in C6H6 is 0.2. Calculate molality of I2 in C6H6. Ans. 3.205 Sol. m = XC6H5 XI2 1000  Mol. wt. of C6H6 0.2  1000 = 0.8  78 = 3.205 8. Calculate individual and average Oxidation number (if required) of the marked element and also draw the structure of the following compounds or molecules. (1) Na2 S2 O3 (2) Na2 S4O6 (3) H2S O5 (4) H2 S2O8 (5) H2S2O7 (6) S8 NOTE : This question will not be discusse in class Ans. (1) +2 (6, –2) (2) +5/2(5, 5, 0, 0) (3) +6 (4) +6 (+6, +6) (5) +6 (+6, +6) (6) 0 Sol. (1) Na2S2O3 (Sodium thiosulphate). 2 × 1 + 2x + 3 × (– 2) = 0. (+2) + 2x + (–6) = 0. 2x = (+6) + (–2). 2x = + 4. x = + 2. (average oxidation number). (2) Na2S4O6 (Sodium tetra-thionate) 2 × (+1) + 4x + 6 × (– 2) = 0. 4x = (+12) + (–2). 4x = +10. x = + 10 4 5 = + 2 . (average oxidation number). (3) H2SO5 (Caro's acid) 2 × (+1) + x + 2 × (–1) + 2 × (–1) + 3 × (– 2) = 0. (+2) + x + (–2) + (–6) = 0. x = +6. (average oxidation number). (4) H2S2O8 (Marshall's acid). 2 × (+1) + 2x + 2 × (–1) + 6 × (–2) = 0. (+2) + 2x + (–2) + (–12) = 0. 2x = +12. x = +6. (average oxidation number). (5) H2S2O7 (Pyro sulphuric acid). 2 × (+1) + 2x + 7 × (–2) = 0. (+2) + 2x = + 14. 2x = (+14) + (–2). 2x = +12. x = + 6.(average oxidation number). (6) S8 (Crown sulphur). Oxidation no. of element in homogeneous molecule will be zero. 3 7 50  3  150  1 3 Sol. (A) [Cl–] = 200 600 = 200 = 3 M (B) molality = 0.1 0.9 18 1000 = 6.17 m (C) Molality = 10  1000 60  90 = 1.85 m. 10.95 (D) Molarity of HCl = 36.5 × 1000 = 3 M 100

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