Mole Concept Exercise Mock Test

MOCK TEST SECTION - I : STRAIGHT OBJECTIVE TYPE 1. The charge on 1 gram ions of Al3+ is : (NA = Avogadro number, e = charge on one electron) (A) 2. 1 N e coulomb 27 A (D) 3 × NAe coulomb (B) 1.24 × 10–21 g (C) 5.025 × 10–23 g (D) 16.023 × 10–23 g (B) x = 2 (C) x = 4 (D) none (B) CaBr2 (C) CaBr (D) Ca2Br A compound possess 8% sulphur by mass. The least molecular mass is : (A) 200 6. 1 × NAe coulomb 9 Calculate the molecular formula of compound which contains 20% Ca and 80% Br (by wt.) if molecular weight of compound is 200. (Atomic wt. Ca = 40, Br = 80) (A) Ca1/2Br 5. (C) 16 g of an ideal gas SOx occupies 5.6 L. at STP. The value of x is (A) x = 3 4. 1 × NAe coulomb 3 The weight of a molecule of the compound C60H22 is : (A) 1.09 × 10–21 g 3. (B) (B) 400 (C) 155 (D) 355 Equal weight of 'X' (At. wt. = 36) and 'Y' (At. wt. = 24) are reacted to form the compound X2Y3. Then : (A) X is the limiting reagent (B) Y is the limiting reagent (C) No reactant is left over and mass of X2Y3 formed is double the mass of ‘X’ taken (D) none of these 7. The mass of 70% H2SO4 required for neutralisation of 1 mol of NaOH. (A) 49 gm 8. (C) 70 gm (D) 34.3 gm What weights of P4O6 and P4O10 will be produced by the combustion of 31g of P4 in 32g of oxygen leaving no P4 and O2. (A) 2.75g, 219.5g 9. (B) 98 gm (B) 27.5g, 35.5g (C) 55g, 71g (D) 17.5g, 190.5g NX is produced by the following step of reactions M + X2  M X2 ; 3MX2 + X2  M3X8 ; M3 X8 + N2CO3  NX + CO2 + M3O4 How much M (metal) is consumed to produce 206 gm of NX. (Take at wt of M = 56, N=23, X = 80) (A) 42 gm 10. (C) 14 gm 3 (D) 7 gm 4 In FeCr2O4, the oxidation numbers of Fe and Cr are : (A) + 2 and + 3 11. (B) 56 gm (B) 0 and + 2 (C) + 2 and + 6 (D) + 3 and + 6 (C) 2 (D) 3 The average oxidation state of Fe in Fe3O4 is : (A) 2 and 3 (B) 8/3 MOLE CONCEPT 2. A solution of FeCl3 is (A) 13. M its molarity for Cl– ion will be : 30 M 90 (B) M 30 (C) M 10 (D) M 5 The molarity of Cl¯ in an aqueous solution which was (w/V) 2% NaCl, 4% CaCI2 and 6% NH4Cl will be (A) 0.342 (B) 0.721 (C) 1.12 (D) 2.18 SECTION - II : MULTIPLE CORRECT ANSWER TYPE 14. 15. Which is/are correct statements about 1.7 gm of NH3 (A) It contain 0.3 mol H – atom (B) it contain 2.408  1023 atoms (C) Mass % of hydrogen is 17.65% (D) It contains 0.3 mol N-atom Which of the following are examples of disproportionation reaction : (A) HgO  Hg + O2 (B) KClO3  KCl + O2 (C) KClO3  KClO4 + KCl (D) Cl2 + OH–  ClO– + Cl– + H2O 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 and Statement-2 is correct explanation for Statement-1. (B) Statement-1 is true, Statement-2 is true and Statement-2 is not correct explanation for Statement-1. (C) Statement-1 is true, Statement-2 is false. (D) Statement-1 is false, Statement-2 is true. (E) Both Statements are false. 16. Statement-1 : The average mass of one Mg atom is 24.305 amu, which is not the actual mass of one Mg atom. Statement-2 : Three isotopes, 24Mg, 25Mg and 26Mg, of Mg are found in nature. 17. Statement-1 : A molecule of butane, C4 H10 has a mass of 58.12 amu. Statement-2 : One mole of butane contains 6.022 × 1023 molecules and has a mass of 58.12 g. 18. Statement-1 : Both 12 g. of carbon and 27 g. of aluminium will have 6.02 × 1023 atoms. Statement-2 : Gram atomic mass of an element contains Avogadro’s number of atoms. 19. Statement-1 : For calculating the molality or the mole fraction of solute, if the molarity is known, it is necessary to know the density of the solution. Statement-2 : Molality, molarity and the mole fraction of solute can be calculated from the weight percentage and the density of the solution 20. Statement-1 : A one molal solution prepared at 20°C will retain the same molality at 100°C, provided there is no loss of solute or solvent on heating. Statement-2 : Molality is independent of temperature. SECTION - IV : COMPREHENSION TYPE Read the following comprehensions carefully and answer the questions. Comprehension # 1 A chemist decided to determine the molecular formula of an unknown compound. He collects following informations (I) Compounds contains 2 : 1 'H' to 'O' atoms(number of atoms). (II) Compounds has 40% C by mass (III) Approximate molecular mass of the compound is 178 g (IV) Compound contains C, H and O only. 21. What is the % by mass of oxygen in the compound (A) 53.33% 22. (C) 33.33% (D) None of these (C) C2H2O (D) CH3O2 What is the empirical formula of the compound (A) CH3O 23. (B) 88.88% (B) CH2O Which of the following could be molecular formula of compound (A) C6H6O6 (B) C6H12O6 (C) C6H14O12 (D) C6H14O6 Comprehension # 2 According to the Avogadro’s law, equal number of moles of gases occupy the same volume at identical condition of temperature and pressure. Even if we have a mixture of non-reacting gases then Avogadro’s law is still obeyed by assuming mixture as a new gas. Now let us assume air to consist of 80% by volume of Nitrogen (N2) and 20% by volume of oxygen (O2). If air is taken at STP then its 1 mol would occupy 22.4 L. 1 mol of air would contain 0.8 mol of N2 and 0.2 mol of O2 hence the mole fractions of N2 and O2 are given by XN2  0.8 , X O2  0.2 24. Volume occupied by air at NTP containing exactly 11.2 gm of Nitrogen : (A) 22.4 L 25. (C) 11.2 L (D) 2.24 L If air is treated as a solution of O2 and N2 then % W/W of oxygen is : (A) 26. (B) 8.96 L 10 9 (B) 200 9 (C) 700 9 (D) 350 9 (B) 9 g/L 7 (C) 2 g/L 7 (D) can’t be determined Density of air at NTP is : (A) 1 g/L SECTION - V : MATRIX - MATCH TYPE 27. Column - I (A) A gaseous organic compound containing C = 52.17%, H = 13.04% & O = 34.78% Column - II (p) One mole of compound contains 4NA atoms of Hydrogen. (by weight) having molar mass 46 g/mol. (B) 0.3 g of an organic compound containing C, H and O on combustion yields 0.44 g of CO2 and 0.18 g of H2O, with two O atoms per molecule. (q) The empirical formula of the compound is same as its molecule formula. (C) A hydrocarbon containing C = 42.857% and H = 57.143% (by mole) containing 3C atoms per molecule. (r) Combustion products of one mole of compound contains larger number of moles of CO2 than that of H2O. (D) A hydrocarbon containing 10.5 g carbon per gram of hydrogen having vapour density 46. (s) CO2 gas produced by the combustion of 0.25 mole of compound occupies a volume of 11.2 L at NTP. MOLE CONCEPT 28. Column I Column II (A) Zn(s) + 2HCl(aq)  ZnCl2(s) + H2(g) (p) 50% of excess reagent left above reaction is carried out by taking 2 moles each of Zn and HCl (B) AgNO3(aq) + HCl(aq)  AgCl(s) + HNO3(g) (q) 22.4 L of gas at STP is liberated above reaction is carried out by taking 170 g AgNO3 and 18.25 g HCl (Ag = 108) (C) CaCO3(s)  CaO(s) + CO2(g) (r) 1 moles of solid (product) obtained. 100 g CaCO3 is decomposed (D) 2KClO3(s)  2KCl(s) + 3O2(g) (s) HCl is the limiting reagent 2/3 moles of KClO3 decomposed SECTION - VI : SUBJECTIVE TYPE 29. (i) K4Fe(CN)6 + 3H2SO4  2K2SO4 + FeSO4 + 6HCN (ii) 6HCN + 12H2O  6HCOOH + 6NH3 (iii) (a) 6NH3 + 3H2SO4  3(NH4)2 SO4 H2SO 4 (b) 6HCOOH    6CO + 6H2O Above steps of reactions occur in a container starting with one mole of K4[Fe(CN)6], 5 mole of H2SO4 and enough water. Find out the limiting reagent in step (i) and calculate maximum moles of CO gas and (NH4)2 SO4 that can be produced. 30. 1 1 mol of B2 (A + B2  AB), 100 Kcal heat is liberated and when 1 mol of A 2 2 reacted with 2 mol of B2 (A + 2B2  AB4) , 200 Kcal heat is liberated. When 1 mol of A is completely reacted with When 1 mol of A reacts with excess of B2 to form AB as well as AB4, 140 Kcal heat is liberated Calculate the mol of B2 used. [Write your answer as (No. of moles of B2 used  1000)] 31. 1 g of dry green algae absorbs 4.7 x 10–3 mole of CO2 per hour by photosynthesis. If the fixed carbon atoms were all stored after photosynthesis as starch (C6H10 O5)n how long would it take for the algae to double their own weight assuming photosynthesis takes place at a constant rate ? MOCK TEST 1. 8. 14. 21. D B A, B, C A 2. 9. 15. 22. B A C, D B 3. 10. 16. 23. B A A B 27. A  (q, s), B  (p, s), C  (p, q, r), D  (q, r) 28. A  (p, q, r, s), B  (p, s), C  (q, r), D  (q) 4. 11. 17. 24. B B A C 5. I2. 18. 25. B C A B 6. 13. 19. 26. C D B B 7. C 20. A