Chemistry Assignment For Revision (Carbon) - YAKEEN 4.0-2025 PDF

1 Chemistry Assignment for Revision (CARBON) YAKEEN 4.0-2025 Mole Concept 1. Calculate the number of atoms in 11.2 L of SO2 gas 11. 1 amu is equal to at STP :...

1 Chemistry Assignment for Revision (CARBON) YAKEEN 4.0-2025 Mole Concept 1. Calculate the number of atoms in 11.2 L of SO2 gas 11. 1 amu is equal to at STP : (1) 1/12 of C12 NA 3N A (2) 1/14 of O16 (1) (2) 2 2 (3) 1 g of H2 (3) 3NA (4) NA (4) 1.66 × 10–24 kg 2. Which of the following has maximum mass : 12. The actual molecular mass of chlorine is : (1) 0.1 gram atom of carbon (1) 58.93 × 10–24 g (2) 117.86 × 10–24 g (2) 0.1 mol f ammonia (3) 58.93 × 1024 g (4) 117.86 × 1024 g (3) 6.02 × 1022 molecules of hydrogen (4) 1120 cc of carbon dioxide at STP 13. The number of atoms present in 16 g of oxygen is 3. The total number of electrons present in 18 mL of (1) 6.02 × 1011.5 (2) 3.01 × 1023 water : (3) 3.01 × 1011.5 (4) 6.02 × 1023 (1) 6.02 × 1022 (2) 6.02 × 1023 24 (3) 6.02 × 10 (4) 6.02 × 1025 14. The number of atoms in 4.25 g of NH3 is approx. : (1) 1 × 1023 (2) 1.5 × 1023 4. The volume of 1.0 g of hydrogen at NTP is : (3) 2 × 1023 (4) 6 × 1023 (1) 2.24 L (2) 22.4 L (3) 1.12 L (4) 11.2 L 15. Which of the following contains maximum number 5. 11 grams of a gas occupy 5.6 litre of volume at STP. of oxygen atoms? The gas is : (1) 1 g of O2 (1) NO (2) N2O4 (2) 1 g of O2 (3) CO (4) CO2 (3) 1 g of O3 (4) All have the same number of atoms 6. At NTP, 5.6 L of a gas weight 8 grams. The vapour density of gas is : 16. The number of atoms present in 0.5 g atom of (1) 32 (2) 40 nitrogen is same as the atoms in - (3) 16 (4) 8 (1) 12 g of C (2) 32 g of S (3) 8 g of oxygen (4) 24 g of Mg 7. The vapour densities of two gases are in the ratio of 1 : 3. Their molecular masses are in the ratio of : (1) 1 : 3 (2) 1 : 2 17. Which of the following contains maximum number (3) 2 : 3 (4) 3 : 1 of atoms? (1) 4 g of H2 (2) 16 g of O2 8. The modern atomic weight scale is based on. (3) 28 g of N2 (4) 18 g of H2 (1) C12 (2) O16 1 (3) H (4) C13 18. Number of neutrons present in 1.7 g of ammonia is : (1) NA (2) NA/10 × 4 9. Gram atomic weight of oxygen is (3) (NA/10) × 7 (4) NA × 10 × 7 (1) 16 amu (2) 16 g (3) 32 amu (4) 32 g 19. 5.6 L of oxygen at STP contains - 10. Molecular weight of SO2 is : (1) 6.02 × 1023 atoms (1) 64 g (2) 64 amu (2) 3.01 × 1023 atoms (3) 32 g (4) 32 amu (3) 1.505 × 1023 atoms (4) 0.7525 × 1023 atoms  2 20. Number of oxygen atoms in 8 g of ozone is - 30. 4.4 g of an unknown gas occupies 2.24 L of volume 6.02 10 23 at STP. The gas may be : (1) 6.02 × 1023 (2) (1) N2O (2) CO 2 (3) CO2 (4) (1) & (3) both 6.02 1023 6.02 1023 (3) (4) 3 6 31. Which contains least number of molecules : (1) 1 g CO2 (2) 1 g N2 21. Sum of number of protons, electrons and neutrons in (3) 1 g O2 (4) 1 g H2 12g of 126C is : (1) 1.8 (2) 12.044 × 1023 32. If V mL of the vapours of substance at NTP weight (3) 1.084 × 10 25 (4) 10.84 × 1023 W g. Then molecular weight of substance is : (1) (W/V) × 22400 (2) V/W × 22.4 22. The weight of one atom of Uranium is 238 am. Its (3) (W – V) × 22400 actual weight is …… g. W 1 (1) 1.43 × 1026 (2) 3.94 × 10–22 (4) (3) 6.99 × 10–23 (4) 1.53 × 10–22 V  22400 33. If 3.01 × 1020 molecules are removed from 98 mg of 23. The actual weight of a molecule of water is - H2SO4, then the number of mole of H2SO4 left are : (1) 18 g (1) 0.1 × 10–3 (2) 0.5 × 10–3 (2) 2.99 × 10–23 g (3) 1.66 × 10 –3 (4) 9.95 × 10–2 (3) both (1) & (2) are correct (4) 1.66 × 10–24 g 34. A gas is found to have the formula (CO)x. It’s VD is 70. The value of x must be : 24. What is the mass of a molecule of CH4 : (1) 7 (2) 4 (1) 16 g (2) 26.6 × 1022 g (3) 5 (4) 6 –23 (3) 2.66 × 10 g (4) 16 NA g 35. Vapour density of gas is 11.2. Volume occupied by 25. Which of the following has the highest mass ? 2.4 g of this at STP will be - (1) 1 g atm of C (1) 11.2 L (2) 2.24 L (2) 1/2 mole of CH4 (3) 22.4 L (4) 2.4 L (3) 10 mL of water (4) 3.011 × 1023 atoms of oxygen 36. The volume of a gas in discharge tube is 1.12 × 10–7 mL at STP. Then the number f molecule of gas in 26. Which of the following contains the least number of the tube is - molecules? (1) 3.01 × 104 (2) 3.01 × 1015 12 (1) 4.4 g CO2 (2) 3.4 g NH3 (3) 3.01 × 10 (4) 3.01 × 1016 (3) 1.6 g CH4 (4) 3.2 g SO2 37. The number of electron in 3.1 mg NO3– is - 27. The number of molecule in 4.25 g of NH3 is - (NA = 6 × 1023) (1) 1.505 × 1023 (2) 3.01 × 1023 (1) 32 (2) 1.6 × 10–3 (3) 6.02 × 1023 (4) None of these (3) 9.6 × 10 20 (4) 9.6 × 1023 28. Elements A and B form two compounds B2A3 and 38. Given that one mole of N2 at NTP occupies 22.4 L B2A. 0.5 moles of B2A3 weight 9.0 g and 0.10 mole the density of N2 is - of B2A weight 10 g. Calculate the atom weight of A (1) 1.25 g L–1 (2) 0.80 g L–1 and B : (3) 2.5 g L–1 (4) 1.60 g L–1 (1) 20 and 30 (2) 30 and 40 (3) 40 and 30 (4) 30 and 20 39. The number of carbon atoms present in a signature, if a signature written by carbon pencil, weighing 29. 5.6 L of oxygen at NTP is equivalent to - 1.2 × 10–3 g is : (1) 1 mol (2) 1/2 mol (1) 12.04 × 1020 (2) 6.02 × 1019 (3) 1/4 mol (4) 1/8 mol (3) 3.01 × 1019 (4) 6.02 × 1020  3 40. A compound of X and Y has equal mass of them. If 50. The number of atoms of Cr and O are 4.8 × 1010 and their atomic weights are 30 and 20 respectively. 9.6 × 1010 respectively. Its empirical formula is - Molecular formula of the compound is :- (1) Cr2O3 (2) CrO2 (1) X2Y2 (2) X3Y3 (3) Cr2O4 (4) CrO5 (3) X2Y3 (4) X3Y2 51. In solution contains 3.4% sulpjur ; the minimum 41. An oxide of sulphur contains 50% of sulphur in it. molecular weight of insulin is : Its empirical formula is - (1) 941.176 (2) 944 (1) SO2 (2) SO3 (3) 945.27 (4) None (3) SO (4) S2O 42. A hydrocarbon contains 80% of carbon, then the 52. Caffine has a molecular weight of 194. It contains hydrocarbon is - 28.93 by mass of nitrogen Number of atoms of (1) CH4 (2) C2H4 nitrogen in one molecule of it is :- (3) C2H6 (4) C2H2 (1) 2 (2) 3 (3) 4 (4) 5 44. Empirical formula of glucose is - (1) C6H12O6 (2) C3H6O3 53. An organic compound contains carbon, hydrogen (3) C2H4O2 (4) CH2O and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the 44. An oxide of metal M has 40% by mass of oxygen. compound would be? Metal M has atomic mass of 24. The empirical (1) CHO (2) CH4O formula of the oxide is :- (3) CH3O (4) CH2O2 (1) M2O (2) M2O3 (3) MO (4) M3O4 54. The amount of water of (g) produced by combustion of 286 g of propane is 45. A compound contains 38.8%C, 16.0% H and 45.2% (1) 168 g (2) 200 g N. The formula of the compound would be (3) 468 g (4) 693 g (1) CH3NH2 (2) CH3CN (3) C2H5CN (4) CH2(NH)2 55. In a gaseous reaction of the type 46. The simplest formula of a compound containing aA + bB ⎯→ cC + dD 50% (w/w) of element X(at wt. = 10) and 50% of which statement is wrong ? element Y (at wt. = 20) is : (1) a litre of A combines with b litre of B to give C (1) XY (2) X2Y and D (3) XY2 (4) X3Y (2) a mole of A combines with b moles of B to 47. Which of the following compound has same give C and D empirical formula as that of glucose:- (3) a g of A combines with b g of B to give C and (1) CH3CHO (2) CH3COOH D (3) CH3OH (4) C2H6 (4) a molecules of A combines with b molecules of B to give C and D 48. 2.2 g of a compound of phosphorous and sulphur has 1.24 g of 'P' in it. Its empirical formula is - (1) P2S3 (2) P3S2 56. Assuming that petrol is octane (C8H18) and has (3) P3S4 (4) P4S3 density 0.8 g mL–1. 1.425 L of petrol on complete combustion will consume. 49. On analysis, a certain compound was found to (1) 50 mole of O2 (2) 100 mole of O2 contain iodine and oxygen in the mass ratio of (3) 125 mole of O2 (4) 200 mole of O2 254 : 80. The formula of the compound is : (At mass I = 127, O = 16) (1) IO (2) I2O 57. In a given reaction, 9 g of Al will react with (3) I5O2 (4) I2O5 2Al + 3/2 O2 → Al2O3 (1) 6 g O2 (2) 8 g O2 (3) 9 g O2 (4) 4 g O2  4 65. For the reaction A + 2B ⎯→ C, 58. The equation : 5 mol of A and 8 mol of B will produce 3 (1) 5 mole of C (2) 4 mole of C 2Al(s) + O2 ( g) → Al2O3 (s) shows that : 2 (3) 8 mole of C (4) 13 mole of C (1) 2 mol of Al reacts with 3/2 mol of O2 to produce 7/2 mol of Al2O3 66. If 1.6 g of SO2 and 1.5 × 1022 molecules of H2S are (2) 2 g of Al reacts with 3/2 g of O2 to produce one mixed and allowed to remain in contact in a closed mol of Al2O3 vessel until the reaction (3) 2 g of Al reacts with 3/2 L of O2 to produce 1 2H2S + SO2 ⎯→ 3S + 2H2O, proceeds to mol of Al2O3 completion. Which of the following statement is (4) 2 mol of Al reacts with 3/2 mol of O2 to true ? produce 1 mol of Al2O3 (1) Only ‘S’ and H2O’ remain in the reaction vessel. 59. 1 L of CO2 is passed over hot coke. When the (2) ‘H2S’ will remain in excess volume of reaction mixture becomes 1.4 L, the (3) ‘SO2’ will remain in excess composition of reaction mixture is- (4) None (1) 0.6 L CO (2) 0.8 L CO2 67. 12 L of H2 and 11.2 L of Cl2 are mixed and (3) 0.6 L CO2 and 0.8 L CO exploded. The composition by volume of mixture is- (4) None (1) 24 L of HCl (g) (2) 0.8 L Cl2 and 20.8 L HCl (g) 60. 26 cc of CO2 are passed over red hot coke. The volume of CO evolved is :- (3) 0.8 L H2 and 22.4 L HCl (g) (1) 15 cc (2) 10 cc (4) 22.4 L HCl (g) (3) 32 cc (4) 52 cc 68. 10 mL of gaseous hydrocarbon on combustion give 61. If 1/2 mol of oxygen combine with Aluminium to 40 mL of CO2(g) and 50 mL of H2O (vap.). The form Al2O3 then weight of Aluminium metal used in hydrocarbon is : the reaction is (Al = 27) - (1) C4H5 (2) C8H10 (1) 27 g (2) 18 g (3) C4H8 (4) C4H10 (3) 54 g (4) 40.5 g 69. The law of multiple proportion was proposed by : 62. If 0.5 mol of BaCl2 is mixed with 0.2 mol of (1) Lavoisier (2) Dalton Na3PO4, the maximum number of moles of moles of (3) Proust (4) Gaylussac Ba3(PO4)2 that can be formed is – 3BaCl2 + 2Na3 PO4 → Ba3(PO4)2 + 6NaCl 70. 2. Which one of the following pairs of compound (1) 0.7 (2) 0.5 illustrate the law of multiple proportions? (3) 0.3 (4) 0.1 (1) H2O, Na2O (2) MgO, Na2O 63. If 8 mL of uncombined O2 remain after exploding (3) Na2O, BaO (4) SnCl2, SnCl4 O2 with 4mL of hydrogen, the number of mL of O2 originally were - 71. In the reaction N2 + 3H2 ⎯→ 2NH3, ratio by (1) 12 (2) 2 volume of N2, H2 and NH3 is 1 : 3 : 2. This (3) 10 (4) 4 illustrates laws of - (1) Definite proportion 64. 4 g of hydrogen are ignited with 4 g of oxygen. The (2) Multiple proportion weight of water formed is - (3) Law of conservation of mass (1) 0.5 g (2) 3.5 g (4) Gaseous volumes (3) 4.5 g (4) 2.5 g  5 72. Different proportions of oxygen in the various 77. Four one litre flasks are seperately filled with the oxides of nitrogen prove the law of - gases hydrogen, helium, oxygen and ozone at same (1) Equivalent proportion room temperature and pressure. The ratio of total (2) Multiple proportion number of atoms of these gases present in the (3) Constant proportion different flasks would be - (4) Conservation of matter (1) 1 : 1 : 1 : 1 (2) 1 : 2 : 2 : 3 (3) 2 : 1 : 2 : 3 (4) 2 : 1 : 3 : 2 73. Oxygen combines with two isotopes of carbon 12C and 14C to form two sample of carbon dioxide. The 78. A container of volume V, contains 0.28 g of N2 gas. If same volume of an unknown gas under similar data illustrates - condition of temperature and pressure weighs (1) Law of conservation of mass 0.44 g, the molecular mass of the gas is (2) Law of multiple proportions (1) 22 (2) 44 (3) Law of gaseous volume (3) 66 (4) 88 (4) None of these 79. When 100 g of ethylene polymerizes to 74. The law of conservation of mass holds good for all polyethylene according to equation of the following except - nCH2 = CH2 → –(–CH2 – CH2 –)n–. The weight of (1) All chemical reactions polyethylene produced will be : (2) Nuclear reactions (1) n/2 g (2) 100 g (3) Endothermic reactions (3) 100 / n g (4) 100 n g (4) Exothermic reactions 80. A chemical equation is balanced according to the 75. Number of molecules in 100 mL of each of O2, NH3 law of - and CO2 at STP are - (1) Multiple proportion (1) In the order CO2 < O2 < NH3 (2) Constant composition (2) In the order NH3 < O2 < CO2 (3) Gaseous volume (3) The same (4) Conservation of mass (4) NH3 = CO2 < O2 76. The empirical formula of an organic compound containing carbon and hydrogen is CH2. The mass of one litre of this organic gas is exactly equal to that of one litre of N2 at same temperature and pressure. Therefore, the molecular formula of the organic gas is - (1) C2H4 (2) C3H6 (3) C6H12 (4) C4H8  6 Answer Key 1. (2) 41. (1) 2. (4) 42. (3) 3. (3) 43. (4) 4. (4) 44. (3) 5. (4) 45. (1) 6. (3) 46. (2) 7. (1) 47. (2) 8. (1) 48. (4) 9. (2) 49. (4) 10. (2) 50. (2) 11. (1) 51. (1) 12. (2) 52. (3) 13. (4) 53. (3) 14. (4) 54. (3) 15. (4) 55. (3) 16. (3) 56. (3) 17. (1) 57. (2) 18. (3) 58. (4) 19. (2) 59. (3) 20. (2) 60. (4) 21. (3) 61. (2) 22. (2) 62. (4) 23. (2) 63. (3) 24. (3) 64. (3) 25. (1) 65. (2) 26. (4) 66. (3) 27. (1) 67. (3) 28. (3) 68. (4) 29. (3) 69. (2) 30. (4) 70. (4) 31. (1) 71. (4) 32. (1) 72. (2) 33. (2) 73. (4) 34. (3) 74. (2) 35. (4) 75. (3) 36. (3) 76. (1) 37. (3) 77. (3) 38. (1) 78. (2) 39. (2) 79. (2) 40. (3) 80. (4)