PW Mole Concept PDF

PRARAMBH EXERCISE-1 (TOPICWISE) LAW OF CHEMICAL COMBINATIONS 8. 4.4 g of an oxide of nitrogen gives 2.24 L of nitrogen and 60 g of another oxide of nitrog...

PRARAMBH EXERCISE-1 (TOPICWISE) LAW OF CHEMICAL COMBINATIONS 8. 4.4 g of an oxide of nitrogen gives 2.24 L of nitrogen and 60 g of another oxide of nitrogen gives 22.4 L of nitrogen 1. In Haber's process, the volume at S.T.P of ammonia relative at S.T.P. The data illustrates: to the total volume of reactants at STP is: (1) Law of conservation of mass (1) One fourth (2) One half (3) Same (4) Three fourth (2) Law of constant proportions (3) Law of multiple proportions 2. 6 g of carbon combines with 32 g of sulphur to form CS2, 12 g of C also combine with 32 g oxygen to form CO2. 10 g (4) Law of reciprocal proportions of sulphur combines with 10 g of oxygen to form Sulphur 9. “The total mass of reactants is always equal to the total mass of dioxide. Which law is illustrated by this? products in a chemical reaction”. This statement is konwn as: (1) Law of multiple proportions (1) Law of conservation of mass (2) Law of constant composition (2) Law of definite proportions (3) Law of reciprocal proportions (3) Law of equivalent weights (4) Gay Lussac’s law (4) Law of combining masses 3. Which of the following data illustrates the law of conservation 10. The law of multiple proportions is illustrated by the two of mass? compounds: (1) 56 g of C reacts with 32 g of Oxygen to produce 44 g (1) Sodium chloride and sodium bromide of CO2 (2) Ordinary water and heavy water (2) 1.70 g of AgNO3 reacts with 100 ml of 0.1M HCl to (3) Caustic soda and caustic potash produce 1.435 g of AgCl and 0.63 g of HNO3 (4) Sulphur dioxide and sulphur trioxide (3) 12 g of C is heated in vacuum and on cooling, there is 11. If law of conservation of mass was to hold true, then 20.8 g no change in mass of BaCl2 on reaction with 9.8 g of H2SO4 will produce 7.3 g (4) 36 g of S reacts with 16 g of O2 to produce 48 g of SO2 of HCl and BaSO4 equal to: 4. One part of an element A combines with two parts of another (1) 11.65 g (2) 23.3 g element B, 6 parts of element C combines with 4 parts (3) 25.5 g (4) 30.6 g of (B) If A and C combine together the ratio of their weights, 12. One of the following combinations which illustrates the law will be governed by: of reciprocal proportions is: (1) law of definite proportion (1) N2O3, N2O4, N2O5 (2) law of multiple proportion (2) NaCl, NaBr, NaI (3) law of reciprocal proportion (3) CS2, CO2, SO2 (4) law of conservation of mass (4) PH3, P2O3, P2O5 5. The law of conservation of mass holds good for all of the following except. ATOMIC AND MOLECULAR MASSES (1) All chemical reactions 13. Insulin contains 3.4% sulphur by mass. What will be the (2) Nuclear reaction minimum molecular weight of insulin? (3) Endothermic reactions (1) 94.117 u (2) 1884 u (4) Exothermic reactions (3) 941 u (4) 976 u 6. The % of copper and oxygen in samples of CuO obtained 14. Boron has two isotopes B 10 and B 11 whose relative by different methods were found to be same. This proves abundances are 20% and 80% respectively. Atomic weight of Boron is: the law of: (1) 10 (2) 11 (3) 10.5 (4) 10.8 (1) Constant Proportion (2) Reciprocal Proportion (3) Multiple Proportion (4) Conservation of mass. 15. Avogadro’s number is the number of molecules present in: (1) 1 g of molecule (2) 1 atom of molecule 7. Two elements X and Y combine in gaseous state to form XY in the ratio 1 : 35.5 by mass. The mass of Y that will be (3) gram molecular mass (4) 1 litre of molecule required to react with 2 g of X is: 16. One amu is equal to: (1) 7.1 g (2) 3.55 g (1) 1.66 × 10–8 g (2) 1.66 × 10–4 g (3) 71 g (4) 35.5 g (3) 1.66 × 10–16 g (4) 1.66 × 10–24 g Some Basic Concepts of Chemistry 21 Itachi Uchiha [ @DRDIC ]  17. The number of molecules present in one milli litre of a gas 28. A person adds 1.71 gram of sugar (C12H22O11) in order to at STP is known as: sweeten his tea. The number of carbon atoms added are: (1) Avogadro number (2) Boltzman number (mol mass of sugar = 342) (3) Loschmidt number (4) Universal gas constant (1) 3.6 × 1022 (2) 7.2 × 1021 (3) 0.05 (4) 6.6 × 1022 MOLE CONCEPT AND MOLAR MASSES 29. The number of atoms present in 0.1 mole of P4 (at. mass 18. 1 g-atom of nitrogen represents: = 31) are: (1) 6.02 × 1023 N2 molecules (1) 2.4 × 1024 atoms (2) 22.4 L of N2 at S.T.P (2) Same as in 0.05 mol of S8 (3) 11.2 L of N2 at S.T.P (3) 6.02 × 1022 atoms (4) 28 g of nitrogen (4) Same as in 3.1g of phosphorus 19. Which is correct for 10 g of CaCO3? 30. Which one contains maximum number of molecules? (1) It contains 1 g atom of carbon (1) 2.5 g molecule of N2 (2) 4 g atom of nitrogen (2) It contains 0.3 g atoms of oxygen (3) 3.01 × 1024 atoms of H2 (4) 82 g of dinitrogen (3) It contains 12 g of calcium 31. Out of 1.0 g dioxygen, 1.0 g (atomic) oxygen and 1.0 g ozone, (4) It refers to 0.1 g equivalent of CaCO3 the maximum number of oxygen atoms are contained in: 20. The number of oxygen atoms present in 14.6 g of magnesium (1) 1.0 g of atomic oxygen bicarbonate is: (2) 1.0 g of ozone (1) 6 NA (2) 0.6 NA (3) NA (4) N A (3) 1.0 g of oxygen gas 2 (4) All contain same number of atoms 21. Which of the following has the highest mass? 32. The maximum volume at S.T.P. is occupied by: (1) 20 g of sulphur (1) 12.8 g of SO2 (2) 4 mol of carbon dioxide (2) 6.02 × 1022 molecules of CH4 (3) 12 × 1024 atoms of hydrogen (3) 0.5 mol of NO2 (4) 11.2 L of helium at N.T.P. (4) 1g molecule of CO2 22. If isotopic distribution of C 12 and C 14 is 98% and 33. If NA is Avogadro’s number, then the number of oxygen 2% respectively, then the number of C14 atoms in 12 g of atoms in one g-equivalent of oxygen is [O2 + 4e → 2O2–]: carbon is: (1) NA (2) NA/2 (3) NA/4 (4) 2NA (1) 1.2 × 1022 (2) 3.01 × 1022 34. If 224 ml. of a triatomic gas has a mass of 1g at 273 K and (3) 5.88 × 1023 (4) 6.02 × 1023 1 atm pressure, then the mass of one atom is: 23. 5.6 L of a gas at S.T.P. weights equal to 8 g. The vapour (1) 8.30 × 10–23 g (2) 6.24 × 10–23 density of gas is: –23 (3) 2.08 × 10 g (4) 5.54 × 10–23 g (1) 32 (2) 16 (3) 8 (4) 40 35. The rest mass of an electron is 9.11 × 10–31 kg. Molar mass 24. One atom of an element weighs 1.8 × 10–22 g, its atomic of the electron is: mass is: (1) 1.5 × 10–31 kg mol–1 (2) 9.11 × 10–31 kg mol–1 (1) 29.9 g (2) 18 g (3) 5.5 × 10–7 kg mol–1 (4) 6.02 × 1023 kg mol–1 (3) 108.36 g (4) 154 g 36. A sample of ammonium phosphate, (NH4)3 PO4, contains 25. If H2SO4 ionises as H2SO4 + 2H2O →2H3O+ + SO42–. Then 3.18 moles of hydrogen atoms. The number of moles of total number of ions produced by 0.1 mol H2SO4 will be: oxygen atoms in the sample is: (1) 9.03 × 1021 (2) 3.01 × 1022 22 (1) 0.265 (2) 0.795 (3) 1.06 (4) 3.18 (3) 6.02 × 10 (4) 1.8 × 1023 37. What is the total number of atoms present in 25.0 mg of 26. Which of the following will not have a mass of 10 g? camphor, C10H16O? (1) 0.1 mol CaCO3. (2) 1.51 × 1023 Ca2+ ions (1) 9.89 × 1019 (2) 6.02 × 1020 (3) 0.16 mol of CO32- ions (4) 7.525 × 1022 Br atom (3) 9.89 × 1020 (4) 2.67 × 1021 27. x L of N2 at S.T.P. contains 3 × 1022 molecules. The number 38. Which of the following samples contains the largest number of molecules in x/2 L of ozone at S.T.P. will be: of atoms? (1) 3 × 1022 (2) 1.5 × 1022 (1) 1 g of CO2 (2) 1 g of C8H18 (3) 1.5 × 1021 (4) 1.5 × 1011 (3) 1 g of C2H6 (g) (4) 1 g of LiF (s) 22 P NEET Dropper Module-1 CHEMISTRY W Itachi Uchiha [ @DRDIC ]  39. 4.0 g of caustic soda (NaOH) (mol mass 40) contains same PERCENTAGE COMPOSITION AND number of sodium ions as are present in: EMPIRICAL & MOLECULAR FORMULA (1) 10.6 g of Na2CO3 (mol. mass 106) 50. The percentage of C, H and N in an organic compound (2) 58.5 g of NaCl (Formula mass 58.5) are 40%, 13.3% and 46.7% respectively then empirical (3) 100 ml of 0.5 M Na2SO4 (Formula mass 142) formula is: (4) 1 mol of NaNO3 (mol. mass 85) (1) C3H13N3 (2) CH2N 40. Total number of atoms present in 64 gm of SO2 is: (3) CH4N (4) CH6N (1) 2 × 6.02 × 1023 (2) 6.02 × 1023 (3) 4 × 6.02 × 1023 (4) 3 × 6.02 × 1023 51. B1 g of an element gives B2 g of its chloride, the equivalent mass of the element is: 41. The total number of protons, electrons and neutrons in 12 gm of 6C12 is: (1) B1 (2) B2 ´35.5 ´35.5 (1) 1.084 × 1025 (2) 6.022 × 1023 B2 - B1 B2 - B1 (3) 6.022 × 10 22 (4) 18 (3) B2 - B1 ´35.5 (4) B2 - B1 ´35.5 42. Number of Ca+2 and Cl– ions in 111 g of anhydrous CaCl2 B1 B2 respectively are: 52. 60 g of a compound on analysis gave 24 g C, 4 g H and (1) NA, 2NA (2) 2NA, NA 32 g O. The empirical formula of the compound is: (3) NA, NA (4) None of these (1) C2H4O2 (2) C2H2O2 43. The maximum volume at N.T.P. is occupied by: (3) CH2O2 (4) CH2O (1) 12.8 gm of SO2 53. A compound made of two elements A and B are found to (2) 6.02 × 1022 molecules of CH4 contain 25% A (at mass 12.5) and 75% B (at mass 37.5). (3) 0.5 mol of NO2 The simplest formula of the compound is: (4) 1 gm-molecule of CO2 (1) AB (2) AB2 (3) AB3 (4) A3B 44. Number of moles of water in 488 g of BaCl2.2H2O are - 54. 400 mg of capsule contains 100 mg of ferrous fumarate. The (Ba = 137): percentage of Fe present in the capsule is approximately: (1) 2 moles (2) 4 moles (3) 3 moles (4) 5 moles (formula of ferrous fumarate is (CHCOO)2 Fe). 45. 4.4 g of CO2 and 2.24 litre of H2 at STP are mixed in a (1) 8.2% (2) 25% container. The total number of molecules present in the (3) 16% (4) Unpredictable container will be: 55. Simplest formula of compound containing 50% of element (1) 6.022 × 1023 (2) 1.2044 × 1023 X (at mass 10) and 50% of element Y (at mass 20) is: (3) 2 moles (4) 6.023 × 1024 (1) XY (2) X2Y (3) XY2 (4) X2Y3 46. The total number of electrons in 1.6 g of CH4 to that in 1.8 g 56. A compound having the empirical formula (C3H4O) has a of H2O is: molecular mass of 170 ± 5. The molecular formula of it’s (1) Same (2) One half compound is: (3) One fourth (4) Double (1) C3H4O (2) C6H8O2 47. One mole of nitrogen gas has volume equal to: (3) C6H12O3 (4) C9H12O3 (1) 1 litre of nitrogen at S.T.P. 57. Two oxides of a metal contains 50% and 40% metal (M) (2) 32 litre of nitrogen at S.T.P. respectively. If the formula of first oxide is MO2, the formula (3) 22.4 litre of nitrogen at S.T.P. of second oxide will be: (4) 11.2 litre of nitrogen at S.T.P. (1) MO2 (2) MO3 (3) M2O (4) M2O5 48. 23g of sodium will react with ethyl alcohol to give: 58. The vapour density of gas A is four times that of B. If molecular mass of B is M, then molecular mass of A is: (1) 1 mole of H2 (2) 1/2 mole of H2 (3) 1 mole of O (4) 1 mole of NaOH (1) M (2) 4 M (3) M (4) 2 M 4 49. If we assume 1/24 th part of mass of carbon instead of 1/12 th 59. A metal nitride M3N2 contains 28% of nitrogen. The atomic part of it as 1 amu., mass of 1 mole of a substance will: mass of metal M is: (1) Remain unchanged (2) get doubled (1) 24 (2) 54 (3) 9 (4) 87.62 (3) Get halved (4) can’t be predicted Some Basic Concepts of Chemistry 23 Itachi Uchiha [ @DRDIC ]  60. A container of volume V, contains 0.28 g of N2 gas. If same 72. The molarity of pure water is: volume of an unknown gas under similar conditions of (1) 100 M (2) 55.6M (3) 50 M (4) 18 M temperature and pressure weights 0.44 g, the molecular mass 73. The mass of 70% H2SO4 by mass is required for neutralisation of gas is: of 1 mole of NaOH is: (1) 22 (2) 44 (3) 66 (4) 88 (1) 65 (2) 98 (3) 70 (4) 54 61. A gaseous hydrocarbon on complete combustion gives 74. If potassium chlorate is 80% pure then 48 g of oxygen would 3.38 g of CO2 and 0.690 g of H2O and no other products. be produced from: The empirical formula of hydrocarbon is: (1) 153.12 g of KClO3 (2) 120 g of KClO3 (1) CH (2) CH2 (3) 20 g of KClO3 (4) 90 g of KClO3 (3) CH3 (4) The data is not complete 75. Density of a solution containing x% by mass of H2SO4 is y. 62. The percentage of Carbon in CO2 is: The normality is: (1) 27.27% (2) 29.27% (3) 30.27% (4) 26.97% xy × 10 xy × 10 (1) (2) ×2 63. The haemoglobin from red blood corpuscles of most 98 98 y mammals contain approximately 0.33% of iron by mass. xy × 10 x × 10 The molecular mass of haemoglobin is 67200. The number (3) ×2 (4) 98 98 y of iron atoms in each molecule of haemoglobin is: 76. Number of gram equivalents of solute in 100 ml of 5 N HCl (1) 3 (2) 4 (3) 2 (4) 6 solution is: 64. On analysis a certain compound was found to contain iodine (1) 50 (2) 500 (3) 5 (4) 0.5 and oxygen in the ratio of 254 g of iodine (at mass 127) and 77. If 1.26 grams of oxalic acid is dissolved in 250 ml of solution 80 g oxygen (at mass 16). What is the formula of compound? then its normality is: (1) IO (2) I2O (3) I5O3 (4) I2O5 (1) 0.05 (2) 0.04 (3) 0.02 (4) 0.08 65. 0.5 mol of potassium ferrocyanide contains carbon equal to: 78. 100 ml of ethylalcohol is made upto a litre with distilled (Formula of potassium ferrocyanide is K4[Fe(CN)6]. water. If the density of C2H5OH is 0.46 gm/ml. Then its (1) 1.5 mol (2) 36 g (3) 18 g (4) 3.6 g molality is: (1) 0.55 m (2) 1.11m (3) 2.22 m (4) 3.33m STOICHIOMETRY & CONCENTRATION TERMS 79. When 100 ml of M/10 H2SO4 is mixed with 500 ml of M/10 66. ‘X’ litres of carbon monoxide is present at STP. It is NaOH then nature of resulting solution and normality of completely oxidized to CO2. The volume of CO2 formed is excess of reactant left is: 11.207 litres at STP. What is the value of ‘X’ in litres? (1) Acidic, N/5 (2) Basic, N/5 (1) 22.414 (2) 11.207 (3) 5.6035 (4) 44.828 (3) Basic, N/20 (4) Acidic, N/10 67. The volume of phosgene formed at STP when 11.2 lit of 80. If 20 g of CaCO3 is treated with 100 mL 20% HCl solution. chlorine reacts with carbon monoxide is: The amount of CO2 produced is: (1) 11.2 lit (2) 22.4 lit (3) 5.6 lit (4) 44.8 lit (1) 22.4l g (2) 8.8 g (3) 2.2 g (4) 8 l 68. What mass of CaCl2 in grams would be enough to produce 81. The mass of CaCO3 required to react with 25 mL of 0.75 molar 14.35 gm of AgCl? HCl is: CaCl2 + 2AgNO3 → Ca (NO3)2 + 2 AgCl (1) 0.94 g (2) 0.68 g (3) 0.76 g (4) 0.52 g (1) 5.55 g (2) 8.29 g (3) 16.59 g (4) 10 g 82. 2 moles of H2S and 11.2 L of SO2 at S.T.P. reacts to form x moles of sulphur. The value of x is: 69. Benzene burns in oxygen according to the equation 2C6H6(I) (1) 1.5 (2) 3.5 (3) 7.8 (4) 12.7 +15O2(g) → 12CO2(g) + 6H2O(l). How many litres of oxygen are required at STP for the complete combustion of 39 g of 83. The volume of CO 2 obtained at STP by the complete liquid benzene? decomposition of 9.85 g Na2CO3 is: (1) 74L (2) 84L (3) 22.4L (4) 11.2L (1) 2.24L (2) Zero (3) 0.85L (4) 0.56L 70. H2O2 is sold as a solution of approximately 5.0 g H2O2 84. The specific gravity of 98% H2SO4 is 1.8 g/cc. 50 ml of this per 100 mL of the solution. The molarity of this solution is solution is mixed with 1750 ml of pure water. Molarity of approximately: resulting solution is: (1) 0.15 M (2) 1.5 M (3) 3.0 M (4) 3.4 M (1) 0.2 M (2) 0.5 M (3) 0.1 M (4) 1 M 71. The amount of oxalic acid (eq.wt.63) required to prepare 85. What is the mole fraction of solvent in aqueous solution of 500 ml of its 0.10 N solution is: NaOH having molality of 3 is: (1) 0.315 g (2) 3.150 g (3) 6.300 g (4) 63.00 g (1) 0.3 (2) 0.95 (3) 0.7 (4) 0.05 24 P NEET Dropper Module-1 CHEMISTRY W Itachi Uchiha [ @DRDIC ]  PRABAL EXERCISE-2 (LEARNING PLUS) 1. Which of the following is/are not affected by temperature? 12. The empirical formula of an organic compound is CH2O. Its (1) Molarity (2) Molality vapour density is 45. The molecular formula of the compound is: (3) Normality (4) None of these (1) CH2O (2) C2H4O2 2. Ferric sulphate on heating gives sulphur trioxide. The ratio (3) C3H6O3 (4) C6H12O6 between the weights of oxygen and sulphur present in SO3 13. 0.132 g of an organic compound gave 50 ml of N2 at STP. The obtained by heating 1 kg of ferric sulphate is: weight percentage of nitrogen in the compound is close to: (1) 2 : 3 (2) 1 : 3 (1) 15 (2) 20 (3) 48.9 (4) 47.34 (3) 3 : 1 (4) 3 : 2 14. 0.7 moles of potassium sulphate is allowed to react with 3. The number of atoms present in 4.25 grams of NH3 is 0.9 moles of barium chloride in aqueous solutions. The number approximately: of moles of the substance precipitated in the reaction is: (1) 1 × 1023 (2) 8 × 1020 (1) 1.4 moles of potassium chloride (3) 2 × 1023 (4) 6.02 × 1023 (2) 0.7 moles of barium sulphate 4. What will be the molarity of a solution, which contains 5.85 g (3) 1.6 moles of potassium chloride of NaCl (s) per 500 mL? (4) 1.6 moles of barium sulphate (1) 4 mol L–1 (2) 20 mol L–1 15. The number of moles of Fe2O3 formed when 0.5 moles of (3) 0.2 mol L –1 (4) 2 mol L–1 O2 and 0.5 moles of Fe are allowed to react are: 5. Number of atoms in 558.5 gram Fe (at. wt. of Fe = 55.85 g (1) 0.25 (2) 0.5 (3) 1/3 (4) 0.125 mol–1 is: 16. Amount of oxalic acid required to prepare 250ml of N/10 (1) Twice that 60 g carbon solution (MW of oxalic acid = 126) is: (2) 6.023 × 1022 (1) 1.5759 g (2) 3.15 g (3) Half that in 8g He (3) 15.75 g (4) 63.0 g (4) 5558.5 × 6.023 × 1023 17. The composition of compound A is 40% X and 60% Y. The 6. Neon has two isotpoes Ne20 and Ne22. If atomic weight of composition of compound B is 25% X and 75% Y. According Neon is 20.2, the ratio of the relative abundances of the to the law of multiple Proportions the ratio of the weight of isotopes is: element Y in compounds A and B is: (1) 1 : 9 (2) 9 : 1 (3) 70 % (4) 80 % (1) 1:2 (2) 2 : 1 (3) 2 : 3 (4) 3 : 4 7. The total weight of 1022 molecular units of CuSO4. 5H2O 18. If the concentration of glucose (C6H12O6) in blood is 0.9 g L–1, is nearly: what will be the molarity of glucose in blood? (1) 4.144 g (2) 5.5 g (1) 5 M (2) 50 M (3) 0.005 M (4) 0.5 M (3) 24.95 g (4) 41.45 g 19. What will be the molality of the solution containing 18.25 g 8. The number of Cl– and Ca+2 ions in 222g. of CaCl2 are: of HCl gas in 500 g of water? (1) 4NA, 2NA (2) 2NA, 4NA (1) 0.1 m (2) 10 m (3) 0.5 m (4) 1 m (3) 1NA, 2NA (4) 2NA, 1NA 20. Increasing order of number of moles of the species: 9. The empirical formula of a gaseous compound is ‘CH2’. (i) 3 grams of NO The density of the compound is 1.25 gm/lit. at S.T.P. The (ii) 8.5 grams of PH3 and molecular formula of the compound is ‘X’: (iii) 8 grams of methane is (1) C2H4 (2) C3H6 (1) (i) < (ii) < (iii) (2) (iii) < (ii) < (i) (3) C6H12 (4) C4H8 (3) (i) < (iii) < (ii) (4) (ii) < (iii) < (i) 10. If 500 mL of a 5 M solution is diluted to 1500 mL, what will be the molarity of the solution obtained? 21. The number of molecules present in 1.12 × 10–7 cc of a gas at STP is: (1) 1.5 M (2) 1.66 M (3) 0.017 M (4) 1.59 M (1) 6.02 × 1023 (2) 3.01 × 1012 (3) 6.02 × 1012 (4) 3.01 × 1023 11. The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following element 22. From 320 mg. of O2, 6.023 ×1020 molecules are removed, contains the greatest number of atoms? the no. of moles remained are: (1) 4 g He (2) 46 g Na (1) 9 × 10–3 moles (2) 9 × 10–2 moles (3) 0.40 g Ca (4) 12 g He (3) Zero (4) 3 × 10–3 moles Some Basic Concepts of Chemistry 25 Itachi Uchiha [ @DRDIC ]  23. An oxide of nitrogen has a molecular weight 92. Find the conditions is: total number of electrons in one gram mole of that oxide. (1) 1 gram (2) 0.5 grams (1) 4.6 N (2) 46 N (3) 23 N (4) 2.3 N (3) 0.25 gms. (4) 0.125 gm 24. No. of moles of water in 488.6 gms of BaCl2.2H2O are 29. Which of the following solutions has the highest normality? (molecular weight of BaCl2.2H2O = 244.33): (1) 172 milli equivalents in 200 ml (1) 2 moles (2) 4 moles (2) 84 milli equivalents in 100 ml (3) 3 moles (4) 5 moles (3) 275 milli equivalents in 250 ml 25. A certain compound contains magnesium, carbon and (4) 43 milli equivalents in 60 ml Nitrogen in the mass ratio 12 : 12 : 14. The formula of the 30. What volume of 75 % H2SO4 by mass is required to prepare compound is: 1.5 litres of 0.2 M H 2SO 4? (Density of the sample is (1) MgCN (2) Mg2CN 1.8 g/cc): (3) MgCN2 (4) Mg(CN)2 (1) 14.2cc (2) 28.4cc 26. An oxide of nitrogen contains 36.8% by weight of nitrogen. (3) 21.7cc (4) 7.1 cc The formula of the compound is: 31. The crystalline salt of Na2SO4.xH2O on heating loses 55.9% (1) N2O (2) N2O3 (3) NO (4) NO2 of its mass and becomes anhydrous the formula of crystalline 27. 40 ml. of a hydrocarbon undergoes combustion in 260 ml salt is: of oxygen and gives 160 ml of carbon dioxide. If all gases (1) Na2SO4.5H2O (2) Na2SO4.7H2O are measured under similar conditions of temperature and (3) Na2SO4.2H2O (4) Na2SO4.10H2O pressure, the formula of hydrocarbon is: 32. 10 g of hydrogen and 64 g of oxygen were filled in a steel (1) C3H8 (2) C4H8 (3) C6H14 (4) C4H10 vessel and exploded. Volume of gaseous product after 28. The mass of Hydrogen at S.T.P. that is present in a reaction is: vessel which can hold 4 grams of oxygen under similar (1) 67.2L (2) 89.6L (3) 44.8L (4) 22.4L PARIKSHIT EXERCISE-3 (MULTICONCEPT) MATCH THE LIST MCQs C. H2 + Cl2 → 2HCl R. 1.125 g 1. Match the List-I with List-II. 1g 1g D. 2H2 + C → CH4 S. 1.214 g List-I List-II 1g 1g A. 1 mole of Na P. 6.02 × 1023 (1) A-(Q); B-(R); C-(P); D-(S) B. 1 mole of H2O Q. Atomic weight in gram (2) A-(P); B-(Q); C-(S); D-(R) C. 1 mole of NH3 R. Molecular weight in gram (3) A-(R); B-(S); C-(P); D-(Q) D. No. of molecules S. Avogadro’s number (4) A-(S); B-(Q); C-(P); D-(R) in 16 g CH4 3. Match the List-I with List-II. (1) A-(P,Q,S); B-(P,R,S); C-(P,R,S); D-(P,S) List-II (2) A-(P,S); B-(Q,R,S); C-(P,S); D-(P,S) List-I (No. of moles of (Amount of substance) particular atoms in (3) A-(P,S); B-(Q,S); C-(P,S); D-(P,S) the given substance) (4) A-(Q,S); B-(Q,R,S); C-(P,S); D-(P,S) A. 6.022×1024 formula units P. 15-mole O-atoms 2. Match the List-I with List-II. of Al2(SO4)3.3H2O List-I List-II B. 90 gm C6H12O6 Q. 3-mole O-atoms A. 2H2 + O2 → 2H2O P. 1.028 g 1g 1g C. 112 litre SO3(g) at 1 atm R. 2.5 mole O-atoms and 0°C B. 3H2 + N2 → 2NH3 Q. 1.333 g 1g 1g D. 54 gram N2O5 (g) S. 150-mole O-atoms 26 P NEET Dropper Module-1 CHEMISTRY W Itachi Uchiha [ @DRDIC ]  (1) A-(S); B-(Q); C-(P); D-(R) (1) A-(Q); B-(P); C-(S); D-(R) (2) A-(Q); B-(R); C-(P); D-(S) (2) A-(Q); B-(R); C-(P); D-(S) (3) A-(S); B-(P); C-(R); D-(Q) (3) A-(S); B-(P); C-(R); D-(Q) (4) A-(P); B-(R); C-(S); D-(Q) (4) A-(P); B-(R); C-(S); D-(Q) 4. Match the List-I with List-II. 7. Where MA, MB are molar masses, nA, nB are no of moles & List-I List-II XA, XB is mole fractions of solute and solvent respectively. (Reaction) (At the end) Match the List-I with List-II. 50% yield List-I List-II A. 2A + 2B  → 3C P. 3 moles C formed 4 mol 6 mol A. Molarity P. Dependent on temperature 1 80% yield M A × nA B. A + 2B  →C Q. 3.2 moles C formed B. Molality Q. × 100 2 8 mol nA M A + nB M B 4 mol 60% yield C. Mole fraction R. Independent of temperature C. 3A + 2B  →C R. A is limiting reagent 15 mol 20 mol XA D. Mass % S. × 1000 A + 3B  20% yield → 2C S. B is limiting reagent XBMB D. 5 mol 12 mol (1) A-(Q); B-(P,R); C-(S); D-(R,S) T. 1.6 moles C formed (2) A-(Q,S); B-(R); C-(P,R); D-(S) (1) A-(S,R); B-(Q,S); C-(P,R); D-(R,T) (3) A-(P); B-(R,S); C-(R); D-(Q,R) (2) A-(S,T); B-(P,R); C-(Q,S); D-(P,R) (4) A-(P,R); B-(R); C-(S); D-(Q,S) (3) A-(S,R); B-(P,S); C-(R,P); D-(Q,T) 8. Match the List-I with List-II. (4) A-(P,R); B-(Q,S); C-(P,R); D-(S,T) List-I List-II 5. Match the List-I with List-II. A. 88g of CO2 P. 0.25 mol List-II List-I B. 6.022 × 1023 molecules of H2O Q. 2 mol (Relative amounts of products on (Compound) C. 5.6 litres of O2 at STP R. 1 mol complete combustion) A. CH4 P. Moles of CO2 < Moles of H2O D. 96 g of O2 S. 6.022 × 1023 Molecules B. C2H4 Q. Moles of CO2 = Moles of H2O E. 1 mol of any gas T. 3 mol C. C2H2 R. Moles of CO2 > Moles of H2O (1) A-(T); B-(Q); C-(P); D-(R); E-(S) D. C3H8 S. Moles of CO2 < Moles of O2 (2) A-(Q); B-(R); C-(P); D-(T); E-(S) (3) A-(S); B-(P); C-(R); D-(Q); E-(T) (1) A-(S,R); B-(Q,S); C-(P,R); D-(R,T) (4) A-(P); B-(T); C-(S); D-(Q); E-(R) (2) A-(S,T); B-(P,R); C-(Q,S); D-(P,R) 9. Match the List-I with List-II. (3) A-(S,R); B-(P,S); C-(R,P); D-(Q,T) (4) A-(P,S); B-(Q,S); C-(R,S); D-(P,S) List-I List-II 6. Match the List-I with List-II. A. 32 gm each of O2 and S P. 2 moles of Fe B. 2 gram molecule of Q. 3 moles of ozone List-I List-II K3[Fe(CN)6] molecule A. 20% (w/w) solution of KOH P. 8.6 M C. 144 gm of oxygen atom R. 1 mole (density of solution = 1.02 gm/mL) D. From 168 g of iron 6.023 × S. 12 moles of B. Solution containing 954.6 gm of Q. 3.64 M 1023 atoms of iron are carbon atoms CaCl2 in 1 L removed the iron left C. Volume of 1.204 × 1024 molecules R. 5 mL of water at 4°C (1) A-(R); B-(P,S); C-(Q); D-(P) (2) A-(Q); B-(R); C-(P,S); D-(S) D. Volume of 0.2 M NaOH solution S. 36 mL containing 40 mg of NaOH (3) A-(S); B-(P); C-(R); D-(Q,R) (4) A-(P,Q); B-(R); C-(S); D-(Q) Some Basic Concepts of Chemistry 27 Itachi Uchiha [ @DRDIC ]  10. Match the List-I with List-II. 16. In a gaseous reaction of the type List-I List-II aA + bB → cC + dD A. 49 g H2SO4 P. 0.5 mole Which statement is wrong? (1) a litre of A combines with b litre of B to give C and D B. 20 g NaOH Q. 1.5 NA atoms (2) a mole of A combines with b moles of B to give C and D C. 11.2 L of CO2 at STP R. 0.5 NA molecules (3) a gram of A combines with b gram of B to give C and D D. 6.023 × 1023 atoms of S. 2 mole of ‘O’ atom (4) a molecules of A combines with b molecules of B to Oxygen give C and D (1) A-(P,Q,R); B-(P,S,R); C-(P,R); D-(P,Q,R) 17. Which statement is false for the balanced equation given (2) A-(P,R); B-(P,Q,R); C-(P,S); D-(P,Q,R) below? (3) A-(P,Q,R); B-(P,Q,S); C-(P,S,R); D-(Q,R) CS2 + 3O2 → 2SO2 + CO2 (4) A-(P,S,R); B-(P,R); C-(P,Q,R); D-(P,R) (1) One mole of CS2 will produce one mole of CO2 (2) The reaction of 16 g of oxygen produces 7.33 g of CO2 CORRECT-INCORRECT STATEMENT MCQs (3) The reaction of one mole of O2 will produce 2/3 mole 11. Which of the following is correct? of SO2 (1) The sum of mole fractions of all the components in a (4) Six molecules of oxygen require three molecules of CS2 solution is always unity 18. 8 g H2 and 32 g O2 is allowed to react to form water then (2) Mole fraction depends upon temperature which of the following statement is correct? (3) Mole fraction is always negative (1) O2 is a limiting reagent (2) O2 is reagent in excess (4) Mole fraction is independent of the content of solute in (3) H2 is limiting reagent (4) 40 g of water is formed the solution 19. In an ionic compound, molar ratio of cation to anion is 12. Equal masses of SO2 and O2 are placed in a flask at STP choose the incorrect statement: 1 : 2. If atomic masses of metal and non-metal respectively are 138 and 19, then the correct statement is: (1) The number of molecules of O2 is more than SO2 (1) The molecular mass of the compound is 176 (2) Volume occupied at STP is more for O2 than SO2 (2) Formula mass of the compound is 176 (3) The ratio of the number of atoms of SO2 and O2 is 3 : 4 (4) Moles of SO2 is greater than the moles of O2 (3) Formula mass of the compound is 157 (4) The molecular mass of the compound is 157 13. For the reaction: 2Fe2S3 + 6H2O + 3O2 —→ 4Fe(OH)3 + 6S STATEMENT BASED MCQs If 4 moles of Fe2S3 are combined with 2 mole of H2O and 3 moles of O2, then which of the following statements (1) Both Statement-I and Statement-II are correct. incorrect: (2) Both Statement-I and Statement-II are incorrect. (1) H2O limiting reagent (3) Statement-I is correct and Statement-II is incorrect. (2) Moles of Fe(OH)3 formed is 4/3 (4) Statement-I is incorrect and Statement-II is correct. (3) Moles of Fe2S3 remaining is 10/3 20. Statement-I: 10,000 molecules of CO2 have the same (4) Mass of O2 remaining is 32 gm volume at STP as 10,000 molecules of CO at STP. 14. Select the incorrect statements: Statement-II: Both CO and CO2 are formed by combustion (1) Ratio of gm/litre & % w/v of a solution is independent of carbon in presence of oxygen. of solute substance. 21. Statement-I: Molality and mole fraction are not affected by (2) Ratio of % w/v and molarity of a solution depends on temperature. the solute substance. W 1 (3) Ratio of % w/v and molarity of a solution depends on Statement-II: Molality (m) = × (where, b = mass of solvent). GMM b(Kg) the solvent substance. (4) Ratio of % w/v & ppm for any solution is same 22. Statement-I: The percentage of nitrogen in urea is 46.6%. 15. Identify the incorrect statement from the following. Statement-II: Urea is an ionic compound. (1) The multiple of prefix femto is 10–15 23. Statement-I: Equal moles of different substances contain (2) The multiple of prefix pico is 10–12 same number of constituent particles. (3) The multiple of prefix nano is 10–18 Statement-II: Equal weights of different substances contain (4) The multiple of prefix micro is 10–6 the same number of constituent particles. 28 P NEET Dropper Module-1 CHEMISTRY W Itachi Uchiha [ @DRDIC ]  24. Statement-I: Equivalent of K2Cr2O7 has 1 equivalent of K, Reason (R): The mole fraction of component A is given by, Cr and O each. No. of moles of A Mole fraction of A = Statement-II: A species contains same number of Total no. of moles of all components equivalents of its components. 32. Assertion (A): One molal solution prepared at 20°C will 25. Statement-I: The molality of the solution does not change retain the same molality at 100°C, provided there is no loss with change in temperature. of solute or solvent on heating. Statement-II: The molality of the solution is expressed in Reason (R): Molality is independent of temperature. units of moles per 1000 g of solvent. 33. Assertion (A): Laboratory reagents are usually made up to 26. Statement-I: Equivalent weight of ozone in the change a specific molarity rather than a given molality. O3→O2 is 8. Reason (R): The volume of a liquid is more easily measured 3 than its mass. Statement-II: 1 mole of O3 on decomposition gives moles of O2. 2 34. Assertion (A): Molality and mole fraction concentration 27. Statement-I: A solution which contains one gram equivalent units do not change with temperature. of solute per litre of solutions is known as molar solution. Reason (R): These units are not defined in terms of any mol. wt. of solute volume. Statement-II: Molarity = normality × eq. wt. of solute 35. Assertion (A): The molality and molarity of very dilute 28. Statement-I: Weight of 1 molecule of O2 = 32 u aqueous solutions differ very little. Statement-II: 1 g molecule = 6.023 × 1023 molecules. Reason (R): The density of water is about 1.0 g cm–3 at room temperature. 29. Statement-I: Normality and molarity can be calculated from each other. 36. Assertion (A): For calculating the molality or the mole fraction of solute, if the molarity is known, it is necessary Statement-II: Normality is equal to the product of molarity to know the density of the solution. and n. Reason (R): Molality, molarity and the mole fraction of solute can be calculated from the weight percentage and the ASSERTION & REASON MCQs density of the solution. (1) If both Assertion (A) and Reason (R) are True and the 37. Assertion (A): The ratio of the mass of 100 billion atoms Reason (R) is a correct explanation of the Assertion (A). of magnesium to the mass of 100 billion atoms of lead can (2) If both Assertion (A) and Reason (R) are True but Reason (R) 24 be expressed as. is not a correct explanation of the Assertion (A). 207 (3) If Assertion (A) is True but the Reason (R) is False. Reason (R): Atomic weights are relative masses. (4) Assertion (A) is False but Reason (R) is True. 38. Assertion (A): A molecule of butane, C4H10 has a mass of 58.12 amu. 30. Assertion (A): One molal aqueous solution of glucose contains 180 g of glucose in 1 kg of water. Reason (R): One mole of butane contains 6.022 × 1023 molecules and has a mass of 58.12 g. Reason (R): A solution containing one mole of solute in 1000 g of solvent is called one molal solution. 39. Assertion (A): Volume of a gas is inversely proportional to 31. Assertion (A): The weight percentage of compound A in a the number of moles of gas. solution is given by Reason (R): The ratio by volume of gaseous reactants and products is in agreement with their mole ratio. Mass A % of A = × 100 Total mass of solution 40. Assertion (A): 1 Avogram is equal to 10 amu Reason (R): Avogram is reciprocal of Avogadro number. PYQ's EXERCISE-4 (NEET PAST YEAR QUESTIONS) 1. The right option for the mass of CO2 produced by heating 2. What mass of 95% pure CaCO3 will be required to neutralise 20 g of 20% pure limestone is (Atomic mass of Ca = 40) 50 mL of 0.5 M HCl solution according to the following  (2023) reaction? (2022) CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + 2H2O(l) [CaCO3 1200K CaO + CO2] [Calculate upto second place of decimal point] (1) 1.76 g (2) 2.64 g (3) 1.32 g (4) 1.12 g (1) 9.50 g (2) 1.25 g (3) 1.32 g (4) 3.66 g Some Basic Concepts of Chemistry 29 Itachi Uchiha [ @DRDIC ]  3. An organic compound contains 78% (by wt.) carbon and 11. The number of water molecules is maximum in: (2015 Re) remaining percentage of hydrogen. The right option for the (1) 18 moles of water empirical formula of this compound is: [Atomic wt. of C is (2) 18 molecules of water 12, H is 1] (2021) (3) 1.8 gram of water (1) CH2 (2) CH3 (3) CH4 (4) CH (4) 18 gram of water 4. Which one of the followings has maximum number of atoms? 12. If Avogadro number NA, is changed from 6.022 × 1023 mol–1 to (1) 1 g of Mg(s) [Atomic mass of Mg = 24] (2020) 6.022 × 1020 mol–1, this would change: (2015 Re) (2) 1 g of O2(g) [Atomic mass of O = 16] (1) The ratio of elements to each other in a compound (3) 1 g of Li(s) [Atomic mass of Li = 7] (2) The definition of mass in units of grams (4) 1 g of Ag(s) [Atomic mass of Ag = 108] (3) The mass of one mole of carbon 5. One mole of carbon atom weighs 12g, the number of atoms (4) The ratio of chemical species to each other in a balanced in it is equal to.  (2020 Covid) equation (Mass of carbon- 12 is 1.9926 × 10–23 g) 13. What is the mass of the precipitate formed when 50 mL of (1) 6.022 × 1022 (2) 12 × 1022 16.9% solution of AgNO3 is mixed with 50 mL of 5.8% NaCl (3) 6.022 × 10 23 (4) 12 × 1023 solution? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5) 6. The number of moles of hydrogen molecules required to  (2015 Re) produce 20 moles of ammonia through Haber’s process is:  (2019) (1) 3.5 g (2) 7 g (3) 14 g (4) 28 g (1) 10 (2) 20 (3) 30 (4) 40 14. 20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will 7. A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated be the percentage purity of magnesium carbonate in the sample? with concentration H2SO4. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining (Atomic weight of Mg = 24) (2015 Re) product at STP will be: (2018) (1) 96 (2) 60 (3) 84 (4) 75 (1) 1.4 (2) 3.0 (3) 4.4 (4) 2.8 15. When 22.4 litres of H2(g) is mixed with 11.2 litres of Cl2(g), 8. In which case number of molecules of water is maximum? each at STP, the moles of HCl(g) formed is equal to:  (2014) (1) 18 mL of water (2018) (1) 2 mol of HCl(g) (2) 0.5 mol of HCl(g) (2) 0.18 g of water (3) 1.5 mol of HCl(g) (4) 1 mol of HCl(g) (3) 10–3 mol of water 16. 1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel. (4) 0.00224 L of water vapours at 1 atm and 273 K Which reactant is left in excess and how much? (2014) 9. A hydrocarbon contains 85.7% of Carbon and 14.3% of (Atomic weight Mg = 24; O = 16) Hydrogen. If 42 mg of the compound contains 3.01 × 1020 (1) O2, 0.16 g (2) Mg, 0.44 g molecules, the molecular formula of the compound will be:  (2017-Gujarat) (3) O2, 0.28 g (4) Mg, 0.16 g (1) C2H4 (2) C3H6 (3) C6H12 (4) C12H24 17. Equal masses of H2, O2 and methane have been taken in a container of volume V at temperature 27°C in identical 10. Suppose the elements X and Y combine to form two conditions. The ratio of the volumes of gases H2 : O2 : methane compounds XY2 and X3Y2. When 0.1 mole of XY2 weighs would be: (2014) 10 g and 0.05 mole of X3Y2 weighs 9 g, the atomic weights of X and Y are: (2016-II) (1) 8 : 16 : 1 (2) 16 : 8 : 1 (1) 20, 30 (2) 30, 20 (3) 40, 30 (4) 60, 40 (3) 16 : 1 : 2 (4) 8 : 1 : 2 30 P NEET Dropper Module-1 CHEMISTRY W Itachi Uchiha [ @DRDIC ]  Answer Key CONCEPT APPLICATION 1. (2) 2. (4) 3. (3) 4. (4) 5. (2) 6. (3) 7. (2) 8. (2) 9. (1) 10. (3) 11. (1) 12. (2) 13. (3) 14. (4) 15. (1) 16. (1) 17. (1) 18. (2) 19. (4) 20. (4) 21. (3) 22. (1) 23. (4) 24. (1) PRARAMBH EXERCISE-1 (TOPICWISE) 1. (2) 2. (3) 3. (2) 4. (3) 5. (2) 6. (1) 7. (3) 8. (3) 9. (1) 10. (4) 11. (2) 12. (3) 13. (3) 14. (4) 15. (3) 16. (4) 17. (3) 18. (3) 19. (2) 20. (2) 21. (2) 22. (1) 23. (2) 24. (3) 25. (4) 26. (3) 27. (2) 28. (1) 29. (2) 30. (4) 31. (4) 32. (4) 33. (3) 34. (4) 35. (3) 36. (3) 37. (4) 38. (3) 39. (3) 40. (4) 41. (1) 42. (1) 43. (4) 44. (2) 45. (2) 46. (1) 47. (3) 48. (2) 49. (1) 50. (3) 51. (1) 52. (4) 53. (1) 54. (1) 55. (2) 56. (4) 57. (2) 58. (2) 59. (1) 60. (2) 61. (1) 62. (1) 63. (2) 64. (4) 65. (2) 66. (2) 67. (1) 68. (1) 69. (2) 70. (2) 71. (2) 72. (2) 73. (3) 74. (1) 75. (3) 76. (4) 77. (4) 78. (2) 79. (3) 80. (2) 81. (1) 82. (1) 83. (2) 84. (2) 85. (2) PRABAL EXERCISE-2 (LEARNING PLUS) 1. (2) 2. (4) 3. (4) 4. (3) 5. (1) 6. (2) 7. (1) 8. (1) 9. (1) 10. (2) 11. (4) 12. (3) 13. (4) 14. (2) 15. (1) 16. (1) 17. (1) 18. (3) 19. (4) 20. (1) 21. (2) 22. (1) 23. (2) 24. (2) 25. (4) 26. (2) 27. (4) 28. (3) 29. (3) 30. (3) 31. (4) 32. (4) PARIKSHIT EXERCISE-3 (MULTICONCEPT) 1. (1) 2. (3) 3. (1) 4. (4) 5. (4) 6. (1) 7. (3) 8. (2) 9. (1) 10. (4) 11. (1) 12. (4) 13. (4) 14. (3) 15. (3) 16. (3) 17. (4) 18. (1) 19. (2) 20. (1) 21. (1) 22. (3) 23. (3) 24. (1) 25. (1) 26. (1) 27. (2) 28. (1) 29. (1) 30. (1) 31. (2) 32. (1) 33. (1) 34. (1) 35. (1) 36. (2) 37. (1) 38. (1) 39. (4) 40. (4) PYQ's EXERCISE-4 (NEET PAST YEAR QUESTIONS) 1. (1) 2. (3) 3. (2) 4. (3) 5. (3) 6. (3) 7. (4) 8. (1) 9. (3) 10. (3) 11. (1) 12. (3) 13. (2) 14. (3) 15. (4) 16. (4) 17. (3) Some Basic Concepts of Chemistry 31 Itachi Uchiha [ @DRDIC ]