Cycloalkanes 31 Atoms, Molecules and Ions 8 Type I. MCQ with only ONE Correct Alternate 1. A sample of CaCO 3 is 50% pure. On heating 1.12 L of CO 2 6. Consider following laws of chemical combination with (at STP) is obtained. Residue left (assuming non-volatile examples: impurity) is: I. Law of multiple proportion : N 2O, NO, NO 2 (a) 7.8 g (b) 3.8 g II. Law of reciprocal proportion : H 2O, SO 2, H 2S (c) 2.8 g (d) 8.9 g Which is correct with examples? 2. In the decomposition of 10 g of MgCO 3, 0.1 mole CO 2 and (a) I and II (b) I only 4.0 g MgO are obtained. Hence, percentage purity of (c) II only (d) None of these MgCO 3 is: 7. H 2S contains 94.11% sulphur; SO 2 contains 50% oxygen (a) 50% (b) 60% and H 2O contains 11.11% hydrogen. Thus: (c) 40% (d) 84% (a) law of multiple proportion is followed 3. Consider the following pairs: (b) law of reciprocal proportion is followed I. CH 4 , C 2H 6 II. CO, CO 2 (c) law of conservation of mass is followed III. NO, NO 2 IV. H 2O, H 2O 2 (d) all of the above 35 37 In which cases, law of multiple proportion is followed? 8. Sodium combines with 17 Cl and 17 Cl to give two samples of (a) I, II (b) I, II, III sodium chloride. Their formation follows the law of: (c) I, III, IV (d) I, II, III, IV (a) gaseous diffusion (b) conservation of mass 4. Two substances of carbon and oxygen have respectively (c) reciprocal proportion (d) none of these 72.73% and 47.06% oxygen. Hence, they follow: 9. According to Dalton’s atomic theory, the smallest particle (a) law of multiple proportion in which matter can exist, is called: (b) law of reciprocal proportion (a) an atom (b) an ion (c) law of definite proportion (c) an electron (d) a molecule (d) law of conservation of mass 10. The nucleus of an atom consists of: 5. In which case purity of the substance is 100%? (a) neutron (b) proton (a) 1 mol of CaCO 3 gave 11.2 L CO 2 (at STP) (c) electron (d) both (a) and (b) 35 37 (b) 1 mol of MgCO 3 gave 40.0 g MgO 11. 17 Cl and 17 Cl are two isotopes of chlorine. If average (c) 1 mol of NaHCO 3 gave 4 g H 2O atomic mass is 35.5 then ratio of these two isotopes is: (d) 1 mol of Ca(HCO 3 )2 gave 1 mol CO 2 (a) 35 : 37 (b) 1 : 3 (c) 3 : 1 (d) 2 : 1 (c) 50 : 50 (d) 75 : 25 Number of H 2O molecules in one drop is: 30. If Avogadro’s number would have been 1 × 10−10.8 g H 2O is: 27. volume of one (c) C 3H 8 (d) CH 2 equivalent of O 2 gas at STP is: (a) 22.5 : 12.02 × 1023 (a) 10 (b) 13 (d) none of the above (c) 7 (d) 17 24.4 L (b) 11.5 19. If it has 10 electrons. Thus. Mass of one atom of an element is 6. X . then number 32 × 2. Y and Z are isoelectronic of CO 2. molar mass of X is: each other most closely in chemical properties? (a) 16 g mol −1 (b) 32 g mol −1 (a) 11 H and 12 H (b) 168O and 168O 2– (c) 4 g mol −1 (d) 64 g mol −1 (c) 24 12 Mg 24 and 12 Mg 2+ (d) 14 7N and 147 N 3– −23 21. In which of the following pairs do the two species resemble moles. Hence.6 L (d) 44.02 × 3 × 102 (c) CH 3Cl (d) CCl 4 20.66 × 10–23 (c) mol of neutrons are: 6. Number of mol of 126C in 1 amu is: 13.64 × 10−23 amu (b) 40. This is 32.5 (b) 87. If each O-atom has two equivalents. they must have: increasing order of atomic number of X. The anion derived (a) 6. Mass of one 147 N-atom is: (a) X + = Y 2+ = Z − (b) X + < Y 2+ < Z − (a) 14 amu (b) 7 amu (c) Z − < X + < Y 2+ (d) Y 2+ < X + < Z − (c) 14 g (d) 7 g 26.5 mol Cl 2 (b) mol 2. A hydrocarbon has 3 g carbon per gram of hydrogen.02 × 1020 (a) C 6H 6Cl 6 (b) C 6H 5Cl (c) 22.10 mol KCl (d) 30. (c) 5.66 × 10−23 g.0 amu from the isotope has 54 electrons.2 L 29. (a) CH 4 (b) CuSO 4 ⋅ 5H 2O hence.02 × 1023 –23 (c) 0. Number of atoms in increasing order in 1. 1. simplest formula is: (c) H 2O 2 (d) H 2O (a) CH 4 (b) C 6H 6 18. Increasing order of protons in X + .4 × 10−3 (d) 6.66 × 10 × 6. ionic mass of M 2+ is: (a) (b) N 0 N0 (a) 70 (b) 66 1 (c) 68 (d) 64 (c) N 02 (d) + 2+ − N 0 × 12 14.8 L Hence. One isotope of a non-metallic element has mass number equal to: 127 and 74 neutrons in the nucleus.64 amu (a) 127 − (b) 127 − 40 54 X 53 X 74 − 74 22.4 Objective Chemistry 12. instead (c) 53 X (d) 54 X− of 6. Which of the following is the richest source of ammonia on (a) 1 amu (b) 1 × 1010 amu a mass percentage basis? (c) 6 amu (d) 6 × 1013 amu (a) NH 4 NO 3 (b) NH 2CONH 2 23. Which has maximum percentage of chlorine? (a) 1 × 10−3 (b) 6. Each drop of H 2O has 0.0 g MgCl 2 . Molar ratio of Na 2SO 3 and H 2O is 1 : 7 in Na 2SO 3 ⋅ xH 2O. M 2+ ion is isoelectronic of SO 2 and has ( Z + 2) neutrons (Z 1 is atomic number of M ). Y and Z is: (a) same viscosity (a) X < Y < Z (b) Z < Y < X (b) same vapour density (VD) (c) X = Y = Z (d) Z < X < Y (c) different molecular weight 16.6 gram CH 4 .0 g Cl 2 (b) 0.018 mL at room temperature. symbol for the 1 anion is: (c) amu (d) 6.7 (d) different percentage composition g NH 3 and 1. then the (a) H 2O = NH 3 = CH 4 (b) H 2O < NH 3 < CH 4 percentage of oxygen in its oxide is: (c) CH 4 < NH 3 < H 2O (d) CH 4 = NH 3 < H 2O (a) 16 (b) 40 17. Thus. Mass of one atom of X is 2. Thus. If two compounds have same empirical formula but 15.02 × 1023 then mass of one atom of H would be: 33. 1 g CH 4 and 4 g of compound X have equal number of 31.66 × 10−23 mol of chlorine? 32 (a) 5.5 : 87. If the equivalent weight of an element is 32. their mass percentage is: (a) 12. Y 2− and Z 3− are isotonic and isoelectronic. X − .64 × 10 g. Which has maximum number of H-atoms per gram of the (c) 32 (d) 20 substance? 28. different molecular formula. Which of the following substances contains greatest mass (a) 32 × 2. then its 32 g is (c) NH 3 (d) HNC(NH 2 )2 equal to: 34. Ionic mass of X 3− is 17. Y 2+ and Z − is: 25. 4 mol H atom 40 × 103 40 × 103 53. One equivalent of magnesium oxide weighs 20 g then one (a) 0. A spherical ball of radius 7 cm contains 56% iron. C 2 .80 g (a) 160 amu (b) 64 amu (c) 4. O −2 (b) NO + . CO. In a glass-tube.576 g (c) 40 amu (d) 96 amu 48. In the following final result is …0. many grams of oxygen are in this sample? (Cu = 63. To make 0.Atoms Molecules and Ions 5 35. The mass percentage of carbon in mass of 186 g.265 (b) 0. Which of the following has maximum number of C-atoms? 49.1 mol CH 4 + 3. which of the following has maximum 32 (a) 32 g mol −1 (b) g mol −1 mass? 6 (a) NaHCO 3 (b) Na 2CO 3 1 × 1023 (c) 32 × 1023 g mol −1 (d) g mol −1 (c) Na 2SO 4 (d) Na 2C 2O 4 32 44. Hence. N 2. Mg 2C 3( X ) is decomposed by H 2O forming a gaseous (c) C 3H 8 (d) C 4 H10 hydrocarbon ( Y ). is: cortisone is 69.2 (c) 500 g (d) 186 g (c) 287.8 g H 2O (b) 1. Its molar mass is: (a) 200 g (b) 372 g (a) 176. MCQ with ONE or more than ONE Correct Alternates 1.06 (d) 3. CN − . Rest mass of an electron is 9.0 g C 2H 6 hydrocarbon is: (c) 4. CO.1 (b) 0.5 × 1023 molecules of NH 3 collection of isoelectronic species? (a) NO.378 g of the compound is formed.4 g /cm 3.2 equivalent of magnesium chloride weighs: (c) 0.25 g NH 3? 2.25 mol of NH 3 (a) N 3− (b) F− (b) It contains 0.60 g (c) law of conservation of mass 45.4 g CO 2 (b) 3. MS 2.022 × 1023 of O 2 will be taken as: 43.08 mol of glucose 54. O 2 .5 × 10−7 kg mol −1 (d) 6.75 mol of H-atoms (c) Ti + (d) Na + (c) It contains total of 1.11 × 10−31 kg mol –1 (a) 1 : 1 (b) 1 : 2 (c) 5. Mass of one atom of X is 6.7 g NH 3 is 1.66 × 10−23 g.0 g (d) 20. A certain metal sulphide. Hence. is used extensively as a high 47.3 (d) 0. A sample of ammonium phosphate (NH 4 )3 PO 4 contains atmosphere. Which of the following are isoelectronic of O 2− ? (a) It contains 0.2 mol H atom (a) 103 mol (b) 10−3 mol (c) 0.06% by mass sulphur.795 (c) Mg 2N 3 (d) MgN (c) 1. their mol ratio is: (a) 1.372 mol of it has a carbon per molecule.3 g C 6H 6 (a) CH 4 (b) C 2H 6 39. CN − . (a) 4. The number of moles of oxygen compound formed is: atoms in the sample is: (a) Mg 3N 2 (b) Mg 3N (a) 0.6 (d) 360. Glucose left in the glass-tube is: (a) law of definite proportion (a) 0.952 g (b) 3.273 g of Mg is heated strongly in a nitrogen (N 2 ) 46. Na 2SO 3 ⋅ xH 2O has 50% H 2O by mass.4 g of X gives ………… mol of Y.11 × 10−31 kg. number of molecules CH 4 − 9. Among the following groupings which represents the (d) It contains 1.50 × 10−31 kg mol −1 (b) 9.5 g 40. Cortisone is a molecular substance containing 21 atoms of 37. CO 2. Molar mass of (d) all of the above the electron is: 55.0 g (a) 20 g Mg (b) 1. When 0.01 × 1023 42.10 g (b) 17.01 mol. 0.5) metal M has atomic mass: (a) 0. x is: (a) 10 (b) 15 (a) 4 (b) 5 (c) 20 (d) 25 (c) 6 (d) 7 52. Hence.0 mol of N and H atoms 3. number of mol of Fe present approximately is: 41.5 (b) 252.2 × 1023 mol −1. 0. If we assume that N 0 = 1. 8.1 38.4 g C 3H 8 (d) 1.98%. 50.3 mol H atom (d) 0.60 mol (d) 3.761 g (d) 8. there is 18 g of glucose.18 36.4 (a) 29. NO (d) CO.92 g (b) law of multiple proportion (c) 3. In a gas S and O are 50% by mass. How temperature lubricant.66 × 10 6.02 × 1023 kg mol −1 (c) 2 : 1 (d) 3 : 1 8 Type II. then molar mass (c) −23 mol (d) mol 6. A sample of CuSO 4 ⋅ 5H 2O contains 3. hence. C 2O 3 follow: is taken.6 g CH 4 51. Hence. C 2− + 2 . In a hydrocarbon C is 3 g per gram of hydrogen. Number of atoms in 20 g Ca is equal to number of atoms in: (c) 95. If density (c) 1.6 g CH 4 = x mol H atoms: moles of atom X in 40 kg is: (a) 0 mol H atom (b) 0. The molar mass of a compound if 0.18 mol of H-atoms. NO . C 2− − 2 .782 g of Cu. CO 2− (c) N 2. 3. If MS 2 is 40. Which is/are correct about 4.75 g (b) 47. When stearic acid is added to water.82 kg −1 (d) Rs. Potash production is often reported as the potassium oxide (K 2O) to form a monolayer over water in a dish of diameter 20 cm. 2. 1021 molecules are removed from 200 mg of CO 2. XHCO 3 and YCO 3 are two pure substances of equal molar masses decomposing as shown 8 Type IV Comprehension Based MCQ Example 1. If a 2. Which of the following may contain one proton and one (d) atomic number of X − is 14 neutron? (a) H +2 (b) 42 He 8 Type III. 50 per kg. collect at the surface and form a monolayer. What is the cost of K per mol of the KCl sample? number is: (a) Rs.01 × 1012 (b) 3.62 Å.20 per cent Al by mass. What is equivalent of 1g H in amu for this value of same amount of potassium as in KCl? Avogadro’s number? (a) Rs. (CO 2 ). 4. molecules is thus equal to: (a) ionic mass of X − is 28 (a) 3. 42.01 × 1018 (b) ionic mass of X − is 30 (c) 3. Testing of Numerical Skill 1. Volume of a gas at NTP is 1. I. F2 (d) O −2 . (the equivalent or the amount of K 2O that could be made from a area of a circle of radius r is πr 2. Study following observations and answer the (a) 1.5 × 10−24 g (d) 1 × 10−23 g The following is a crude but effective method for estimating the order of magnitude of Avogadro’s number using stearic acid Example 2.00 kg −1 .2 g of mixture of H 2O and CO 2. 35 37 and 17 Cl differ in: 17 Cl (b) half of the atomic weight (c) electrical charge of the nucleus (a) atomic number (b) number of neutrons (d) number of protons (c) number of electrons (d) atomic mass 5. Thus. its molecules K 2 O and answer the questions at the end of it. 58. O 2− (b) CO. 3. Then: 9.4 × 10−4 g of stearic acid is needed chloride (KCl) and potassium sulphate (K 2SO 4 ). Read the following passage regarding fertiliser (C18H 36O 2 ). Most of the potash produced in the United States goes acid molecule has been measured to be 0.00 mol −1 (c) 4 × 1023 (d) 1 × 1023 Q. A bivalent metal ion has equivalent mass of 12. In one into fertilizer. Cesium atoms are the largest naturally occurring atoms.21 nm 2. What is possible formula of carbon cesium atoms 2.73 mol −1 (a) 3 × 1023 (b) 6 × 1023 (c) Re. I.12 × 10−7 cc.32 g of C and 1. show that the law of multiple atoms would have to be laid side by side to give a row of proportions is followed. ∆ 2XHCO 3 → H 2O + CO 2 + X 2CO 3 number of moles of CO 2 left is ∆ 2.01 × 1024 (d) 3. 1. CN − (c) equivalent mass of its hydride is 13 (d) molar mass of its hydride is 14 10.40 kg −1 (b) Rs.00 kg −1 (c) Rs. 25. For what price must K 2SO 4 be sold in order to supply the Q.500 g sample of carbon suboxide contains The radius of cesium atom is 2. CN − (b) molar mass of its oxide is 40 (c) O 2− 2 . In addition to carbon monoxide (CO) and carbon dioxide Identify the substances.01 × 1030 (c) atomic number of X − is 13 7. that is. there is a third compound of carbon called carbon 5.66 × 10−24 g (b) 3.6 Objective Chemistry 2 3 4. 50. II. II. 13. The major sources of potash are potassium experiment it is found that 1. the layer is Potash is any potassium mineral that is used for its potassium only one molecule thick. suboxide.00 mol −1 (d) Rs.18 g of O. Q.42 mol −1 (b) Rs. The cross-sectional area of each stearic content. 3. Isoelectronic species are represented by pairs: (a) equivalent mass of its oxide is 28 (a) N 3− . X − is isoelectronic of CO and has ( Z + 2) neutrons 6. Al 2(SO 4 )3 ⋅ xH 2O has 8. Based on these measurements value of Avogadro’s Q. The number of (Z = atomic number of X − ).8 g of XHCO 3 gave 6. The atomic number of an element is always equal to: (c) 1D (d) 1T (a) number of neutrons in the nucleus 8. 16. ) given mineral. Thus. How many cesium 1. (c) 2.33 × 10−24 g questions at the end of it. Calculate YCO 3 → YO + CO 2 value of x.50 cm long? Assume that the atoms are suboxide? spherical. KCl costs Rs. 2 (d) 25. (d) A is false but R is true. 2.6 g of residue is left and 4. Number of atoms in 1. Miscellaneous Exercise ➥ A. ➥ B. Hence. Questions given below are based on this mass spectrum.7 g NH 3 and 1. Number of atoms in one mole of each isotope is placed in increasing order: (a) 24 Mg < 25Mg < 26Mg Relative abundance 80 26 (b) Mg < 25Mg < 24 Mg 60 26 25 24 (c) Mg = Mg = Mg (d) given data is insufficient 40 Q. Assertion (A) : SO 2 and SO 3 obey “law of multiple proportion”. molecular formula of hydrated salt is ………… and proton = electrons in neutral species 7. 1 ppm is equal to 1 mg per kg.66 × 10−24 g. Reason (R) : Ionic mass = neutron + proton 6. The mass spectrum (given below) of magnesium has three peaks.Atoms Molecules and Ions 7 Q. H 2S and SO 2 follow law of ………… ionic mass is 18. True/False 8. .8 g H 2O are (b) Both A and R are true but R is not the correct equal. 5.0 (b) 24. Out of CH 3Cl. Assertion (A) : When 10 g of CaCO 3 is decomposed. 3. Reason (R) : Law of mass conservation is followed. Concentration 1 ppm = ………… ppb. there are 24 electrons and 30 5.6 g CH 4 . 6. (a) Both A and R are true and R is the correct explanation 4. Concentration 1 microgram per gram = ………… ppm. 10. Equivalent mass of NH 3 is 17.315 kg (a) Mg-24 (b) Mg-25 (c) 1. (c) A is true but R is false. phosphate form ………… mol of barium phosphate. Percentage of sodium sulphite (Na 2SO 3 ) in hydrated salt is 50. 1 amu is equal to 1. ➥ C. III.3 (c) 25. H 2O.262 kg (d) 0. I. Hydrocarbon containing 3 g carbon per gram hydrogen is ………… Reason (R) : Different samples of a pure chemical 4. and CCl 3CHO. 20 g CaCO 3 (50% pure) on strongly heating gives 15. percentage of chlorine is in C 6H 6Cl 6. Assertion (A) : In Fe2+ . 2. II. 8. 8 Type V. then its 9. NO and NO 2 follow the law of ………… sulphur and 1 part oxygen by mass. of A. Atomic mass unit is also called ………… 2.158 kg (b) 0. What mass (in kg) of K 2O contains the same number of Q. Average atomic mass of Mg is approximately: 20 (a) 25.631 kg (c) Mg-26 (d) equal 100 Q. 1 g H 2 is left. SO 3 contains sulphur and oxygen in the mass ratio of 3.48 × 10−7 kg. contains 10 electrons then ionic mass of X 2+ is ………… 4.8 24 25 26 Atomic mass (amu) Example 3. Assertion/Reason Type Questions 3. 7. Which isotope has maximum number of atoms per gram moles of K atoms as 1. 5.6 g Codes : residue. Fill in the Blanks Reason (R) : Every sample of SO 2 contains 1 part 1. which indicates that magnesium has three isotopes. If 1 mol of H 2 is taken out from 2 g H 2. 10. III. ionic mass is 56. 1. If X 2+ is isotonic (same number of neutrons) of H 2O and substance always contain the same proportions of elements by mass. Assertion (A) : SO 2 from oxidation of sulphur or H 2S ………… contains sulphur and oxygen in the mass ratio of 1 : 2. maximum 1. explanation of A. 1 mol of barium nitrate and 1 mol of ammonium neutrons and thus. If there are 10 electrons and 8 neutrons in X 2− . 1.4 g of CO 2 escapes. 9.00 kg KCl? of each? (a) 0. SO 2 contains sulphur and oxygen in 1 : 1 mass ratio. Mass of 1 mol of electrons is 5. C 6H 6Cl 6. 3% 4. MCQ with only ONE Correct Alternate ∆ Thus. S and O 2 combine to List I List II proportion form SO 2 and SO 3. 27. 12% 5. (D)—(4). (b) 17.0 g 0. (a). F 7. 1 Example 1 Q. CH 4 has carbon and mass hydrogen in 3 : 1 mass ratio. (d) 2. (b) D. Match % of carbon (in List I) with the compound proportion with 30 mL of H 2 to form 20 mL of NH 3.2% 2. 20 Type IV. Testing of Numerical Skill 1. (d) 11.r. (in List II). (d) Type II. (d) . 103 8. mass ratio of H D. CH 4 D. Law of multiple 2. (C)—(1). II. (b) 24. (d) 25.8 Objective Chemistry ➥ D.0 g residue on heating. Law of reciprocal 4. (c) 6.05 mol = 2. CO 2 proportion ratio of H and O w. (a) 5. (c) 27. (b) 20. Matching 1. II. (a) 26. (d) 21. (b) 9. sulphur is 1 : 16. I. Dalton (Da) Example 2 Q. (A)—(5). (c) 30. Law of definite 3. (d) 4. (b) 8. CaCO 3 (s ) → CaO(s ) ↓ + CO 2 (g ) ↑ ∆ 1 mol =100 g 1 mol = 56 g 22. C 2H 6O in H 2O. Matching B. (c) 19. (b) 3. (a) 29. 2 : 3 3. (a). (a) 2. (a) Q. (a) 2. Comprehension Based MCQ 5.2 g MgCO 3 gives 2. Reciprocal proportion 10. % of pure MgCO 3 = 84% Total residue = 7. (c) 12. Type I. (a) 14. Assertion/Reason Type Questions 5. E. (a) 10. CN 2OH 4 and O is 1 : 8.77 × 107 atoms C.8 g Impurity = 5. T 9.88 × 10−3 2. (d) 4. CaCO 3 E. XHCO 3 = NaHCO 3 . (c) 1. 10 mL N 2 combines 1. III. (a). CH4 4. x = 18 3. (d) 9.05 mol = 5 g pure 0. Law of conservation of 1. (a) 7. (a) Q. MCQ with only ONE Correct Alternate 1. (a) 10. Match Laws of chemical combinations (in List I) with the examples followed (in List II). (c) Q. (a). I. MgCO 3 (s ) → MgO(s )+ CO 2 (g ) 0. MCQ with ONE or more than ONE Correct Alternates Type V. List I List II A. T 10. (b). (b).8 g Thus.33 6. F 2.1 mol in 10 g sample CaO(s ) left = 2. (d) 3. Na 2SO 3 ⋅ 7H2O 7. (D)—(2). (b). 2. (c) B. (d) 4. 2. 4. True/False Type III. A. F 4. (d) 28. (b). YCO 3 = MgCO 3 5. Miscellaneous Exercise 1. 52. (d) A. 0. II. (a) 13. (B)—(3). Fill in the Blanks 1. (a) 18. F 6. (E)—(2) H ints and Solutions 8 Type I. (c). (a) 23.4 g (pure) 4. (c) Q.t. (b). I. (A)—(4). (a) 7. T 5. T 4. C 3O 2 8. (d) Example 3 Q. (B)—(3). 4. In H 2S and SO 2 mass B. (C)—(1). C. 20% 1. (d) 3. Multiple proportion 2.12 L at STP 1 mol = 84 g 40 g 1 mol Impure CaCO 3 taken = 10 g (5 g pure CaCO 3 + 5 g impurity) 8. (b) Q. Gay Lussac’s law 5. (E)—(5) 2.8 g 1. hence C.0 g Thus. (c) 8. T 3. (c). (b) 22. 75% 3. (b) 6. (d) 9. III. (b) 16.4 L at STP 2. (c) 15. T 1. (a) 1. 1 N 0 0. M 2+ has electrons = 32 (isoelectronic of SO 2 ) 45. (b) (b) MgCO 3 → MgO + CO 2 1 mol 40 g 23. 100% = atomic mass ∆ (c) 2NaHCO 3 → Na 2CO 3 + H2O + CO 2 Thus.1 0.66 × 10−23 × 6. (a) CaCO 3 → CaO + CO 2 1 mol 22.4 g 3 H2O 0.48 × 10–7 kg mol −1 Thus. X 3− has 10 electrons.1 N 0 0. 0.02 mol = 3. (b) Percentage of oxygen in oxide = × 100 40 A1 X 1 + A2 X 2 11.02 × 1023 g mol −1 Thus. Rest mass of electron = 9. Na 2SO 3 : H2O X1 + X 2 Moles 1 : 7 X1 3 On solving = Mass 126 126 X2 1 % 50 50 Thus (c) Thus.10 − 0. (b) 5. 44.5 N 0 3 4 NH3 0. 44. (c) 12. Mass of one atom = 2.1 0. (b) ∆ 27.66 × 10–23 × 6. Mass of one atom = 6. one equivalent of oxide = 40 g Thus.64 × 10–23 × 6.02 × 1020 molecules Thus.5 = 29.001 mol 18 = 20 mol = 6. 18 g glucose = 0.11 × 10−31 kg protons = 34 = Z mass of one mole of electrons = 9. 50% = 40 ∆ Thus.02 × 1023 Thus. ionic mass =70 Thus.4 N 0 Mass of spherical ball = π (7 )3 × 1.02 × 1023 Thus. Number of moles in Molecules Atoms 4 = π (7 )3 CH4 0.08 mol neutron + proton = ionic mass =17 left = 0.018 mL = 0. (d) 13. (c) .02 × 1023 neutrons = 36 = 5.64 × 10−23 g actual = 11. (a) 51. protons (= atomic number) in X 3 − = 7 withdrawn = 0.4% Thus. number of moles in 32 g 2 mol 18 g 1 mol 9g 32 = actual = 4 g 2. Volume of spherical ball = πr3 3 16.018 3 × 56 × 100 = = 0. (c) 4 Thus.018 g (density of water = 1g/mL) = mol 0. (d) 35X 1 + 37 X 2 35.08 = 0.66 × 10–23 g Mass of N 0 atoms = 2. (a) Thus.6 g Thus.1 N 0 0. 50% 8 Thus.4 L 21. atomic mass = 6. neutrons =17 – 7 =10 Thus.3 N 0 4 56 Iron in spherical ball = π (7 )3 × 1. (b) 4π (7 )3 × 1.2 L Thus.1 0.4 × g 3 100 Thus. Equivalent weight of element = 32 g (d) Ca(HCO 3 )2 → CaO + H2O + 2CO 2 1 mol 2 mol and that of oxygen = 8 g actual = 1 mol Thus. Average atomic mass A = = 20% X1 + X 2 Thus.4 × 56 19.11 × 10−31 × 6.10 mol glucose Thus.Atoms Molecules and Ions 9 ∆ Thus.