(a) Sulphur Reason: An atom of aluminium is 27 times heavier than 1 / 12th of the mass of the C – 12 atoms. Gram molar mass of Ca3(PO4)2 = (40 à 3) + [30 + (16 à 4)] à 2 (c) 34.2 g of C12H22O11 Does the solubility of a substance change with temperature? Solution: (a) One gram of C – 12 contains Avogadroâs number of atoms. Density of gaseous element = 16 à 5 = 80 u (vi) By comparing the definition of relative molecular mass and vapour density we can write as follows. It determines the relation between molecular mass and vapour density. Which of the following contains the largest number of molecules? Mass of SO2 = \(\frac{64 \times 18.069 \times 10^{23}}{6.023 \times 10^{23}}\) Number of moles = \(\frac{\text { S.T.P. Answer: (b) 18 g (b) 22.4 litres, Question 25. Question 2. Answer: Answer: Relative molecular mass is equal to _____. (c) 320 g of Sulphur The number of molecules of CO2 present in 44 g of CO2 is _____. Mass of Ca = 40 g Eg: Isobars 18Ar40, 20Ca40. Verify by calculating that Answer: Question 5. Question 2. Question 32. \mathrm{T.P}}{\text { Molar volume } \mathrm{SO}_{2} \text { at } \mathrm{S} . Others are less. (iv) 8 g of calcium Mass of NH4Cl = 53.5 g. (iii) NH3 (Ammonia) gas will remain after the completion of the reaction. }}{\text { Molar volume at } \mathrm{S} . Answer: Question 21. (a) F2 (a) \(_{1} \mathrm{H}^{2}\) Atoms of different elements having ______ mass number, but ______ atomic numbers are called isobars. (d) Atomicity. 18 g of water contains = 6.02 2 à 1023 molecules (b) 23 g Hint: (i) 2 moles of water Answer: Question 1. Calculate the number of moles in 16 g of Calcium. (b) 100 g (iii) 52 g of He False. Hint: Atomic mass of 1(Na) = 1 à 23 = 23 Question 1. [NCERT Exemplar] Question 18. Answer: Calculate the number of moles present in: 6.022 à 1022 molecules of oxygen = 1 mol à \(\frac{6.022 \times 10^{22}}{6.022 \times 10^{23}}\) = 0.1 mol. Avogadro’s law. Answer: Let the fraction of relative abundance of B – 10 = x = 98 g. (d) 18. (f) 4 = = 24.3202 amu Which of the following has the smallest mass? = 1.51 à 1023 molecules, (c) 1 mole of carbon atoms contains 6.022 à 1023 atoms Answer: 1 mole of water contain molecules = 6.022 à 1023 2.5 mole of CO2 at S.T.P. = \(\frac{1 \mathrm{mol}}{(22.4 \mathrm{L})} \times(44.8 \mathrm{L})\) = 2.0 mol. Solution: (Atomic mass of Na = 23, C = 12, H = 1, O = 16) (d) 8 (b) 1 g of H2 (d) C2H5OH (e) 2 \(\text { Relative atomic mass }=\frac{\text { Average mass of the isotopes of the element }}{1 / 12^{\text {th }} \text { of the mass of one Carbon- } 12 \text { atom }}\). Which one of the following represent Avogadro’s law? (b) 8.0 g of H2 Applications of Avogadro’s law: Question 4. Mass of 1 atom of He = \(\frac{0.004}{6.023 \times 10^{23}}=6.6423 \times 10^{-27} \mathrm{kg}\). Number of moles = \(\frac { 14 }{ 28 }\) = 0.5 mole Answer: Computing Molecular Mass for a Covalent Compound Ibuprofen, C 13 H 18 O 2, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Advil and Motrin.What is the molecular mass (amu) for this compound? Gram molar mass of H2O = (1 à 2) + (16 à 1) = 2 + 16 (i) A and R are correct, R explains the A. Which of the following has the highest number of molecules? (d) Atomicity. Calcium chloride when dissolved in water dissociates into its ions according to the following equation. Question 4. 1 mole of any substance contains ______ molecules. = 1.368 à 1022 molecules of CO2. Which has the highest number of molecules? Atomic mass of 4(H) = 4 à 1 = 4 g = 3.764 à 1022 molecules of CH4 1.12 à 10-7 cc contains = \(\frac{6.023 \times 10^{23}}{22400} \times 1.12 \times 10^{-7}\) Number of molecules of CO2 = Number of moles of CO2 à Avogadroâs number = 24 à \(\frac{78.99}{100}\) + 25 à \(\frac{10}{100}\) + 26 à \(\frac{11.01}{100}\) Which of the following is a triatomic molecule? Gram molar mass of Ca3(PO4)2 = 308 g. You can Download Samacheer Kalvi 10th Science Guide Pdf Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. Question 1. Question 2. 4.4 g of CO2 at S.T.P will occupy \(\frac{22.4}{44}\) à 4.4 = 2.24 L. Question 41. (b) MgCl2 Answer: Question 12. Find the percentage of nitrogen in ammonia. Percentage abundance of B – 11 = 80.4 %. (c) 1 g of H2 (c) 17 g of H2O (c) 4 dm³ (a) 1 litre of N2 Question 24. (a) 8 g of CH4 We know that one mole of sodium contains 6.022 à 1023 atoms. Assertion (A): Nitrogen, oxygen and hydrogen are diatomic molecules. Homo. 2.24 à 10-3 c molecules 6.023 à 1023 molecules Gram molar mass of CO2 = 44 g. (iii) Ca3(PO4)2 (i) 8 Solution: Reason (R): 1 molecule of ozone contains 3 atoms of oxygen. (iii) 40 g of calcium = 40 + 12 + 3(16) Mass of 2 moles of S8 = Atomic mass à Number of moles = 256 à 2 = 512 g. (iii) 4 moles of ozone molecule O3 Answer: (i) 1 mole of chlorine molecule, Cl2 Answer: (a) Both (A) and (R) are correct One mole of gold contains NA atoms = 6.022 à 1023 â 11 – x = 10.804 amu (ii) 1 g atom of Ag = Gram atomic mass of Ag = 108 g Question 37. Question 37. (d) 0.1 L. _____ is one-twelfth of the mass of C – 12 atom, an isotope of carbon which contains _____ protons and ____ neutrons. Assertion: The Relative Molecular Mass of Chlorine is 35.5 a.m.u. State Avogadro hypothesis (or) Avogadro’s Law. Question 38. Give one example. Answer: Answer: = 4 à 124 = 496 g. Question 2. 16 g of methane contain molecules = 6.022 à 1023 Assertion (A): \(_{17} \mathrm{Cl}^{35}\) and \(_{17} \mathrm{Cl}^{37}\) are isotones. (v) Atoms may not always combine in a simple whole number ratio. Reason (R): Atom is the ultimate particle of an element which may or may not have an independent existence. An equal volume of all gases under similar conditions of temperature and pressure contain a different number of molecules. Answer: Question 11. Write the different types of isotopes of oxygen and its percentage abundance. Answer: \(_{20} \mathrm{Ca}^{40}\) contains 20 protons, 20 electrons and 20 neutrons. Mass of 1 mole of NH3 = 17 g Question 3. Answer: It is helpful in determining the molecular formula of gaseous compounds. Solution: (d) 1 g of Fe Atomicity = \(\frac{\text { Molecular mass }}{\text { Atomic mass }}\) = 1.51 à 1023 CO2 molecules. Atomicity of sulphur is _____. (a) 9 The molecular formula of gases can be derived using Avogadro’s law. Chemical bonds. (i) Sulphur Question 14. The volume occupied by NH3 in glass bulb A is three times more than the volume occupied by HCl in glass bulb B at STP. (c) C – 12, Question 8. Mass of oxygen = 56 – 40 = 16 g. (ii) \(\frac{\text { No of moles of oxygen atom }}{\text { Mole }=\text { mass/atomic mass }}=\frac{16}{16}=1 \text { mole }\) Answer: Question 29. Correct statement: Atom is the smallest particle that takes part in a chemical reaction. Atoms of different elements having the same number of _____ are called isotones. GMM = 23 + 16 + 1 6.022 à 1022 molecules of oxygen = 1 mol Answer: Which one of the following element is used as the standard for measuring the relative atomic mass of an element in now a days? (b) 3.011 à 1012 False. No. Question 7. Atomic masses of Ca = 40, P = 30, O = 16. Question 5. (b) Protium \(_{1} \mathrm{H}^{1}\), Question 6. Correct statement: Equal volume of all gases under similar conditions of temperature and pressure contain the same number of molecules. (i) 1 mol of He = 6.022 à 1023 atoms = 2 à 27 u + 3 à 16 u of molecules = mole à Avogadro number = 4 à 6.023 à 1023 = 2.409 à 1024, Question 35. Monoatomic. Number of molecules = \(\frac{6.023 \times 10^{23} \times 360}{180}\) (d) Protium. (iii) A is wrong, R is correct. An example of a triatomic molecule is _____. (d) 16.2 %. Number of moles = \(\frac{\text { Number of atoms of iron }}{\text { Avogadro’s number }}\) (R) explain (A) So, mass of Al = 0.3 à 27 = 8.1 g. Question 2. = 3.011 à 1023 molecules. Question 10. (d) 2 dm³ 44 g (molar mass) of CO2. (a) Atomic masses of the elements A and B. Question 16. Answer: Answer: (a) 0.125 mole (c) Sodium Answer: (iv) 2 moles of nitrogen molecules, N2 of atoms }}{\text { Avogadro number }}} \\ {=\frac{12.046 \times 10^{22}}{6.023 \times 10^{23}}=2 \times 10^{-1}} \\ {=0.2 \text { mole. Question 1. Homo atomic molecule. = 0.5 mole. 8.0 g of calcium = 1 mol à \(\frac{(8.0 \mathrm{g})}{(40 \mathrm{g})}\) = 0.2 mol. 44 g of CO2 at S.T.P occupies 22.4 L = 0.4 mole. Answer: The number of _____ present in the molecule is called its atomicity. All noble gases are _____ molecules. Question 10. Now, Molar mass of H2O = 2(1) + 16 = 18 g (ii) Mass of sodium bicarbonate in this equation is mass of 2 moles of NaHCO3. Atoms of different elements with the same atomic mass but a different atomic number are called _____. (c) \(_{6} \mathrm{C}^{13},_{7} \mathrm{N}^{14}\) (b) 6.023 à 1023 Answer: of moles of sodium = \(\frac{46}{23}\) = 2 moles Solution: Thus, masses in the decreasing order are: 0.1 g atom of Ag > 0.1 mole of H2SO4 > 1023 molecules of CO2 > 1023 atoms of Ca > 1 g of carbon, Question 2. (b) atomic number For example, sodium chloride, calcium oxide. Answer: (c) zero Answer: â´ 0.1764 moles weigh = 80 à 0.1764 g Answer: Answer: Hint: (a) Hydrogen chloride (a) 6.023 à 1023 Question 1. Your email address will not be published. Classify each of the following on the basis of their atomicity. in 1961 for a universally accepted atomic mass unit, carbon-12 isotope was chosen as the standard reference for measuring atomic masses. Mass of oxygen = 16 à 2.5 = 40 g. Question 22. Answer: (a) CO2 has molar mass = 44 g mol-1 (b) triatomic Question 1. Its value is equal to 22.4 litre or 22400 ml or 22400 cm³ or 2.24 à 10-2 m³. (d) 18 g of CH4. (c) (A) is wrong but (R) is correct One mole of water weighs 18 g. (b) 5.6 litre (a) He of molecules present Molecules, volume. = 0.5 à 22.4 Answer: (c) HI, Question 16. Assertion (A): NH3, H2O, HCl are heteroatomic molecules. Answer: The number of atoms present in the molecule is called atomicity. Protons and neutrons have considerable mass, but _____ don’t have considerable mass. (iii) 6.023 à 1023 molecules of water (a) 5 moles of CO2 and 5 moles of H20 do not have the same mass. Write the applications of Avogadroâs Law. \mathrm{TP}}{22.4}}\end{array}\) (R) explain (A), Question 12. Question 6. Answer: Solution: Give two examples. (b) Both (A) and(R) are wrong Question 7. Correct statement: The value of Gram molar volume at STP is 22.4 litres. Which one of the two is heavier and by how many times? Question 2. }}{\text { Mass of } 2 \text { atoms of hydrogen }}\). Calculate the formula mass of sodium carbonate (Na2CO3.10H2O). Question 15. Calculate the number of moles in 24.092 à 1022 molecules of water. (ii) 4. (d) V2 â \(\frac{1}{n}\). Calculate the mass of glucose necessary to prepare a 500 mL pouch of D5W. We know that one mol of alumina contains 2 mol of Al3+ ions. The volume occupied by 1 mole of a diatomic gas at S.T.P is _____. (d) (ii) and (iv). Answer: â´Number of Al3+ ions in 0.056 g = 2 à 5.49 à 10-4 à 6.022 à 1023 (a) 0.5 e.g., \(_{1} \mathrm{H}^{1},_{1} \mathrm{H}^{2},_{1} \mathrm{H}^{3}\). 1998-2013 Professor of Chemistry, Iowa State University. (c) Oxygen Question 22. â´ 2 moles of sodium contain = 2 à 6.022 à 1023 atoms (b) 1 atom of He Gram atomic mass of Ca = 40 g (b) Gold (i) 3.011 à 1023 number of oxygen atoms. Isotopes of oxygen: (d) CH4. Assertion (A): Atoms of the same element may not be similar in all respects. Answer: Question 28. (vii) The mass of an atom can be converted into energy (E = mc²). Calculate the mass of 18.069 à 1023 molecules of SO2? (c) 18 g Solution: Gram molecular mass of CaO = ⦠= (2 à atomic mass of Na) + (1 à atomic mass of C) + (3 à atomic mass of O) + 10 [(2 à atomic mass of H) + (1 à atomic mass of O)] = 1 à 40u + 1 à 12u + 3 à 16u = 100u Asked for: mass of solute. Answer: (b) Hydrogen chloride Question 3. The gram atomic mass of an element has no unit? Solution: (c) isotopes, Question 13. Which one of the following is a hetero diatomic molecule? Antoine L. Lavoiser and Joseph L. Proust. (d) 3.011 à 1023 Calculate the number of molecules in 360 g of glucose. Mass = Mole à Molecular mass = 0.5 à 18 = 9 g. Question 55. No of moles = \(\frac{\text { Given mass }}{\text { Atomic mass }}\). Nitroglycerine is used as an explosive. Solution: In flask Q: (c) 3 litres of O2 Assertion (A): C12H22O11 is not a simple ratio. The actual number of atoms in 12 g of carbon – 12 is called the Avogadro number. (d) 12. Answer: (b) 6.023 à 10-23 Consider the following and classify them on the basis of their atomicity. = 4.456 à 1o24 electrons, RD Sharma Class 11 Solutions Free PDF Download, NCERT Solutions for Class 12 Computer Science (Python), NCERT Solutions for Class 12 Computer Science (C++), NCERT Solutions for Class 12 Business Studies, NCERT Solutions for Class 12 Micro Economics, NCERT Solutions for Class 12 Macro Economics, NCERT Solutions for Class 12 Entrepreneurship, NCERT Solutions for Class 12 Political Science, NCERT Solutions for Class 11 Computer Science (Python), NCERT Solutions for Class 11 Business Studies, NCERT Solutions for Class 11 Entrepreneurship, NCERT Solutions for Class 11 Political Science, NCERT Solutions for Class 11 Indian Economic Development, NCERT Solutions for Class 10 Social Science, NCERT Solutions For Class 10 Hindi Sanchayan, NCERT Solutions For Class 10 Hindi Sparsh, NCERT Solutions For Class 10 Hindi Kshitiz, NCERT Solutions For Class 10 Hindi Kritika, NCERT Solutions for Class 10 Foundation of Information Technology, NCERT Solutions for Class 9 Social Science, NCERT Solutions for Class 9 Foundation of IT, PS Verma and VK Agarwal Biology Class 9 Solutions, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, Periodic Classification of Elements Class 10, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, CBSE Previous Year Question Papers Class 12, CBSE Previous Year Question Papers Class 10, One mole of an element contains 6.022 à 10. Question 30. (e) Incorrect, the correct symbol of sodium is Na. False. = 100 g. (iii) How many moles of CO2 are there in this equation? = 6.022 à 1023 à 0.4 à 3 atoms (c) (A) is correct but (R) is wrong, Question 7. Answer: 0.5 mole of oxygen gas = 6.022 à 1023 à 0.5 molecules Answer: [Given mass of a dot = 10-18 g] Al3+. (b) 2.24 litre Question 27. (d) 12.5 mole. The difference between activity and other measures of composition arises because molecules in non-ideal gases or solutions interact with each other, either to attract or to repel each other. Atomic mass of 3(O) = 3 à 16 = 48 g (a) 18 g 2NaHCO3 â Na2CO3 + H2O + CO2 â´ 0.1 g atom of Ag = 0.1 à 108 g = 10.8 g The atomicity of K2Cr2O7 is _____. True. Answer: â´ 0.0536 g of Mg will have = \(\frac{6.023Ã10^{23}}{24}\) à 0.536 Answer: Heterodiatomic molecules. Answer: (a) CaCO3 = 3.011 à 1012. Question 1. (c) 20.60 False. â´ 5.49 à 10-4 mol of Al2O3 contains 2 à 5.49 à 10-4 mol of Al3+ ions (ii) 23 g of sodium Calculate the mass of CO2 which contains the same number of molecules as are contained in 40 g of SO2. Question 7. (b) 4.4 g of CO2 (c) 47.05 % Deadline. Question 1. (b) isobars (j)5 = 3 à 71 = 213 g, (iii) 5 moles of sulphur molecule, S2 Analyse the table and fill in the blanks. One mole of screw weighs = 2.475 à 1024 g = 2.475 à 1021 kg The molecule that consists of atoms of different elements is called _____. = 120 u. Question 4. 12 à 2 : 1 à 6 : 16 1 atom of oxygen = \(\frac{1}{6.022 \times 10^{23}}\) mol (c) 0.5 mole Solution: Mass = No. (c) Atomic mass = 4 à 1022 à 10-23 The atomicity of hydrogen chloride is _____. (a) nucleus What are the applications of Avogadro’s Law? The average atomic mass of an element is calculated by adding the masses of its isotopes, each multiplied by their natural abundance on the Earth. (b) Cl– Na+ Answer: What is wrong in saying âone mole of nitrogenâ? Answer: Electronic configuration of A : 2, 7 Mass = Atomic mass à number of moles Which one is heavier? = 149.43 g. IV. Which one of the following represents the mass of 0.5 moles of water molecules? Question 6. (it) 60 g of calcium NH3 (Ammonia) + HCl (Hydrochloric acid) â NH4Cl (Ammonium chloride), Question 11. (c) \(\frac { 1 }{ 2 }\)th of the mass of a C – 12 atom \mathrm{T} \cdot \mathrm{P}}\) Give an example to show law of conservation of mass applies to physical changes also. (Atomic mass of N = 14, H = 1, Cl = 35.5) Number of molecules in a certain mass (i) H2O % of O = \(\frac{16}{18}\) à 100 = 88.89% (i) 2 g of nitrogen Answer: Ca : C : O à 3 Find the mass of one mole of these steel screws. (b) 22.4 litres Which one of the following will have largest number of atoms? of moles of NaOH = \(\frac { 2 }{ 40 }\) = 0.05 moles of NaOH, 1 g Au = \(\frac{1}{197}\) mol = \(\frac{1}{197}\) à 6.02 à 10, 1 g Na = \(\frac{1}{23}\) mol = \(\frac{1}{23}\) à 6.02 à 10, 1 g Li = \(\frac{1}{7}\) mol = \(\frac{1}{7}\) à 6.02 à 10. (i) H2O True. True. C à 2 : H à 6 : O = 20 moles. (a) atomic mass Gram molar mass of glucose (C6H12O6) = (6 à 12) + (12 à 1) + (6 à 16) (c) relative molecular mass (iv) What mass of calcium will be obtained from 1000 g of calcium oxide? Question 5. â´ 0.5 mol of water contains \(\frac{2 \times 6.022 \times 10^{23}}{2}\) atoms of hydrogen Alternatively, the powder may be dissolved in water and checked for its conduction of electricity. Except for noble gases, atoms of most of the elements are found in the combined form. Question 1. (d) Sodium. (c) H2SO4 1 mole of oxygen atom. Question 16. = 12.046 à 1023 / 6.023 à 1023 (f) Ca(OH)2 Eg: isotopes 17Cl35, 17Cl37. of moles of 1 g of CO2 = \(\frac{1}{44}\) or 6.022 à 1023 atoms of hydrogen weigh = 1 g N2(g) + 3H2(g) â 2NH3(g) (iii) 1 g of carbon = 1 g = 32.65 %. i.e., 6.02 à 1023 atoms of Ca have mass = 40 g (c) 22400 litres Calculate the mass of 12.046 à 1023 molecules of CaO. 67.2 litre of NH3 = \(\frac{1}{22.4} \times 67.2\) = 3 moles of NH3. Question 4. Question 1. A big drop of water has volume 1.0 mL. Mass of 2 moles of H2 molecule Answer: The _____ is useful to determine the empirical formula and molecular formula. _____ determines the relation between molecular mass and vapour density. (d) 2. Question 1. (a) 1 litre of N2 (b) BCl3 (i) 4 moles of Nitroglycerine, (ii) 4 moles of Nitroglycerine produce 19 moles of gas molecules 52 mol of He = 52 à 6.022 à 1023 atoms = 3.131 à 1025 atoms. (b) \(_{18} \mathrm{Ar}^{40},_{20} \mathrm{Ca}^{40}\) Mass of 2 moles of NaHCO3 = 84 à 2 = 168 g. (iii) Number of moles of CO2 in this equation = 1 mole. (i) 52 moles of He 2. â´ 0.18 g of water contains 24 : 71, (c) H2SO4 Question 5. (c) 28 g. Question 5. Answer: 1.51 à 1023 molecules of water â´ 0.1 mole of carbon atoms contains = 6.022 à 1023 à 0.1 atoms The mass of 6.023 à 1023 atoms of Gold = 197 g Calculation based on number of moles from mass and volume: Question 1. Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules. (c) 18 g of CO2 Glucose has a molar mass of 180.16 g/mol. Answer: Question 13. Number of moles = \(\frac{\text { Mass }}{\text { atomic mass }}=\frac{32}{32}\) = 1 mole of sulphur. (iv) 56 g calcium oxide gives 40 g of calcium So (b) is the smallest mass as 6.6423 à 10-27 kg. = \(\frac{\text { Mass of a given volume of gas or vapour at S.T.P }}{\text { Mass of same volume of hydrogen }}\). NH3, H2O are _____ molecules whereas N2, O2 are _____ molecules. = 24.75 à 1023 g = 2.475 à 1024 g A sample of ethane (C2H6) gas has the same mass as 1.5 à 1020 molecules of methane (CH4). = 2 à 22.4 = 44.8 litres at S.T.P. (c) 1022 molecules of CO2 Molar mass of H2SO4 = 2 à 1 + 32 + 4 à 16 = 98 g Answer: Isotopes. (b) 240 g of calcium and 240 g of magnesium elements have a mole ratio of 3 : 5. Which one of the following is an example of a polyatomic molecule? (a) (A) is correct and (R) explains (A) (c) (A) is wrong but (R) is correct = 3.6132 à 1024 ions. (c) (A) is correct but (R) is wrong. Atoms of the same element with same atomic number but having different mass number are called _____. Find the atomicity of ozone if its atomic mass is 16 and its molecular mass is 48. False. (a) isotopes (i) (A) and (R) are correct. Answer: Volume of drop of water = 1.0 mL (a) 1 g of CO2 â´ 222 g of CaCl2 is equivalent to 2 moles of CaCl2 (ii) How many moles of oxygen atoms are there in this? (c) V â \(\frac{1}{n^{2}}\) Question 9. (b) 44 g Which of the following has the largest number of particles? (c) 22.4 The value of Avogadro’s number is _____. â -x = 10.804 – 11 The atomic number of an element is the number of protons or number of neutrons and electrons present in it. Answer: We should say âone mole of nitrogen atomsâ or âone mole of nitrogen moleculeâ. of moles of atoms Actual mass of a molecule is obtained by dividing the molar mass by Avogadroâs number whereas gram molecular mass represents the molecular mass expressed in grams, i.e., it is the mass of 1 mole of molecules, i.e., Avogadroâs number of molecules. What is a mole? (b) Glucose (C6H12O6), Question 22. Question 9. Atomic mass of 2(O) = 2 à 16 = 32 Gram molecular mass of CaO = 40 + 16 = 56 g (v) 1 mole of Ca = 40 g = 6.02 à 1023 atoms of Ca = 2.151 à 1022 molecules of N2. Question 1. \(\frac{2.68}{100}\) à 20 (ii) 44.8 litres of sulphur dioxide at N.T.P. (a) Mass of a C – 12 atom (b) Gay – Lussacâs law In the formula of C12H22O11 the carbon, hydrogen and oxygen combine in whole number ratio but the ratio is not simple. Find the gram molecular mass of the following from the data given: It is used to determine the atomicity of gases. Question 6. Write the relationship between. 5 moles of CO2 have molar mass = 44 à 5 = 220 g (i) 27 g of Al 5 moles of H2O = 5 à 18 = 90 g, (ii) 1 mole of Glucose (C6H12O6) (iii) How many moles of carbon dioxide are there in this equation? (c) Noble gases, Question 14. Answer: (iii) 1 gram of carbon Question 1. = 40 g Give an example of a triatomic molecule of an element. Mass of 4 moles of ozone = 48 à 4 = 192 g. (iv) 2 moles of Nitrogen molecule N2 = 1.66058 à 10-24g, Question 7. Answer: (a) 1 mol of water contains 6.022 à 1023 molecules Answer: Answer: (b) C2H5OH Carbon dioxide produced by action of dilute hydrochloric acid on potassium hydrogen carbonate is moist whereas that produced by heating potassium hydrogen carbonate is dry. Called _____ of O2 to produce 18 g of calcium atoms are there in a reaction! ÂOne mole of H atoms = 6.022 à 1023 ( d ) 12.5 mole chloride ( b ) hydrogen not! Glucose ( b ) 8 ( c ) 22.4 ( d ) 44 g of Li the... It establishes the relationship between relative molecular mass of calculate the mass of 100 molecules of sucrose in 2 × 10 24.... ) allotropes ( d ) Both ( a ) atomic mass à number of molecules in 360 of. React with five volumes of water at room temperature is 1.0 g/mL mole Hint: No Both ( a 6.023... Using Avogadro ’ s hypothesis is used as the standard for measuring the relative molecular mass ( d relative! The human body but a different atomic number of ions actual number moles... 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