Redox Reaction and Equivalent Concept Mock Test

REDOX REACTION AND EQUIVALENT CONCEPT MOCK TEST SECTION - I : STRAIGHT OBJECTIVE TYPE 1. The number of moles of KMnO4 that will be needed to react completely with one mole of ferrous oxalate in acid solution is (A) 3/5 (B) 2/5 (C) 4/5 (D) 1 2. The number of moles of KMnO4 that will be needed to react with one mole of sulphite ions in acidic solution is (A) 2/5 (B) 3/5 (C) 4/5 (D) 1 3. The normality of 0.3 M phosphorus acid (H3PO3) is (A) 0.1 (B) 0.9 (C) 0.3 (D) 0.6 4. An aqueous solution of 6.3 gm of oxalic acid dihydrate is made upto 250 ml. The volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution is (A) 40 ml (B) 20 ml (C) 10 ml (D) 4 ml 5. Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is (A) 3 (B) 4 (C) 5 (D) 6 SECTION - II : MULTIPLE CORRECT ANSWER TYPE 6. Consider the redox reaction 2S2O32– + 2  S4O62– + 2 – : (A) S2O32– gets reduced to S4O62– (B) S2O32– gets oxidised to S4O62– (C) 2 gets reduced to – (D) 2 gets oxidised to – 7. There are two sample of HCl having molarity 1N and 0.25 N. Find volume of these sample taken in order to prepare 0.75 N HCl solution. (Assume no water is used) : (A) 20 mL, 10 mL (B) 100 mL, 50 mL (C) 40 mL, 20 mL (D) 50 mL, 25 mL 8. To a 25 ml H2O2 solution excess acidified solution of KI was added. The iodine liberated 20 ml of 0.3 N sodium thiosulphate solution. Use these data to choose the correct statements from the following : (A) The weight of H2O2 present in 25 ml solution is 0.102 g (B) The molarity of H2O2 solution is 0.12 M (C) The weight of H2O2 present in 1 L of the solution is 0.816 g (D) The volume strength of H2O2 is 1.344 L SECTION - III : ASSERTION AND REASON TYPE Each question has 5 choices (A), (B), (C), (D) and (E) out of which only one is correct. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. (E) Statement-1 and Statement-2 both are False. 9. Statement-1 : In the redox reaction 8 H+ (aq) + 4 NO3– + 6 Cl– + Sn(s)  SnCl62– + 4 NO2 + 4 H2O, the reducing agent is Sn(s). Statement-2 : In balancing half reaction, S2O32–  S(s), the number of electrons added on the left is 4. 10. Statement-1 : Among Br –, O22– , H– and NO3–, the ions that cannot act as oxidising agents are Br – and H– Statement-2 : Br – and H– cannot be reduced. SECTION - IV : COMPREHENSION TYPE Read the following comprehensions carefully and answer the questions. Comprehension # 1 Chromium exists as FeCr2O4 in the nature and it contains Fe0.95O & other impurity. To obtain pure chromium from FeCr2O4, the ore is fused with KOH and oxygen is passed through the mixture when K2CrO4 and Fe2O3 are produced. FeCr2O4 + KOH + O2  K2CrO4 + Fe2O3 Fe0.95O + O2  Fe2O3 2 g of ore required 270 mL of O2 at 273 K and 1 atm for complete oxidation of ore. K2CrO4 is precipitated as BaCrO4 when Barium salt is added . To remaining solution 10 mL of 1 M K4Fe(CN)6 is added when Fe3+ ions react with it to form KFe[Fe(CN)6], often called ‘Prussian Blue’. To determine excess of K4Fe(CN)6 in solution 7mL of 0.2 N of Fe2+ is added when all the K4Fe(CN)6 is precipitated as K2Fe[Fe(CN)6]. 11. 12. 13. Weight of BaCrO4 precipitated (A) 1.64 (B) 6.29 (C) 0.82 (D) 3.29 % by mass of Fe0.95O in the ore (A) 9.6% (B) 10.1% (C) 8.55% (D) 20.2% n factor for Fe0.95O (A) 0.9 14. (B) 0.85 (C) Weight of impurities present in the ore (A) 0.421 (B) 0.123 2 0.95 (C) 0.341 (D) 1.8 (D) 0.206 SECTION - V : MATRIX - MATCH TYPE 15. Column  (A) Sn+2 3.5 mole (B) H2C2O4 8.4 mole (C) S2O3–2 7.2 mole (D) Fe+2 9.2 mole Column  + + + + MnO4– (acidic) 1.2 mole MnO4– (acidic) 3.6 mole I2 3.6 mole Cr2O7–2 (acidic) 1.6 mole (p) Amount of oxidant available decides the number of electrons transfer (q) Amount of reductant available decides the number of electrons transfer (r) Number of electrons involved per mole of oxidant > Number of electrons involved per mole of reductant (s) Number of electrons involved per mole of oxidant < Number of electrons involved per mole of reductant. SECTION - VI : SUBJECTIVE TYPE 16. H2O2 is reduced rapidly by Sn2+, the products being Sn4+ & water. H2O2 decomposes slowly at room temperature to yield O2 & water. Calculate the volume of O2 produced at 273 & 1.00 atm when 200g of 10.0 % by mass H2O2 in water is treated with 88.2 ml of 1 M Sn2+ & then the mixture is allowed to stand until no further reaction occurs. 17. 80 gm of a sample of Anhydrous CuSO4 was dissolved in water and made to 250ml. 25 ml of this solution after taking usual precautions was treated with a little excess of KI solution. A white ppt. of Cu2I2 and iodine was evolved. The iodine so evolved required 40 ml of 1 M of hypo solution. What is the approximate purity of CuSO4 solution.[Mol. wt. of CuSO4 = 160] MOCK TEST 1. A 13. B 2. A 14. C 3. D 4. A 5. D 6. B,C 7. A,B,C,D 8. A,B,D 15. A (p,r), B (q,r), C (p,q,r), D (q,r) 16. 5.6 L 9. C 10. B 17. 80% 11. D 12. B