CBSE Class 9 Science Atoms and Molecules Sure Shot Questions Set 02

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ASSIGNMENT QUESTIONS SET - 2

CHAPTER - 3

ATOMS AND MOLECULES

MOLECULAR MASS AND MOLE CALCULATION PROBLEMS

1. Calculate the number of moles in i) 4.6g sodium ii) 5.1g of Ammonia iii) 90g of water iv) 2g of NaOH [4 Marks]

Answer:
1. Number of moles = Given mass / Molar mass.
2. i) For 4.6g sodium (Na, atomic mass = 23 g/mol): Moles = 4.6 / 23 = 0.2 mol.
3. ii) For 5.1g of Ammonia (NH₃, molar mass = 14 + 3 = 17 g/mol): Moles = 5.1 / 17 = 0.3 mol.
4. iii) For 90g of water (H₂O, molar mass = 2 + 16 = 18 g/mol): Moles = 90 / 18 = 5 mol.
5. iv) For 2g of NaOH (molar mass = 23 + 16 + 1 = 40 g/mol): Moles = 2 / 40 = 0.05 mol.

Teacher's Note:
a) The fundamental formula relating mass and moles is n = given mass divided by molar mass.
b) Always calculate the correct molar mass by adding standard atomic masses of constituent atoms.
c) Mention units such as moles or mol in the final answer to avoid losing marks.
d) Verify each calculation carefully by cross-multiplying the moles and molar mass.

2. Calculate the number of molecules in 360g of glucose. [2 Marks]

Answer:
1. Chemical formula of glucose is C₆H₁₂O₆.
2. Molar mass of glucose = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 g/mol.
3. Number of moles = 360 / 180 = 2 moles.
4. Number of molecules = Moles × Avogadro constant = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules.

Teacher's Note:
a) Avogadro constant represents the number of particles in one mole of any substance.
b) Students often forget to multiply by the Avogadro number after finding moles.

3. Find the mass of 2.5 mole of oxygen atoms [2 Marks]

Answer:
1. Atomic mass of oxygen atoms (O) = 16 g/mol.
2. Number of moles = 2.5 mol.
3. Mass = Number of moles × Atomic mass.
4. Mass = 2.5 × 16 = 40 g.

Teacher's Note:
a) Distinguish carefully between oxygen atoms (O) and oxygen molecules (O₂).
b) The atomic mass is used here because the question specifies oxygen atoms.

4. Calculate the mass of 12.046 \(\times\) 10²³ molecules in CaO. [2 Marks]

Answer:
1. Molar mass of CaO = 40 + 16 = 56 g/mol.
2. Number of moles = Number of molecules / Avogadro number = (12.046 × 10²³) / (6.023 × 10²³) = 2 moles.
3. Mass = Number of moles × Molar mass.
4. Mass = 2 × 56 = 112 g.

Teacher's Note:
a) Convert the given number of particles to moles as an intermediate step.
b) Use 6.022 or 6.023 consistently for the Avogadro number as per school guidelines.

5. Calculate the number of moles in 24.092 \(\times\) 10²² molecules of water. [2 Marks]

Answer:
1. Given number of water molecules = 24.092 × 10²².
2. Avogadro number = 6.023 × 10²³ = 60.23 × 10²².
3. Number of moles = Given molecules / Avogadro number.
4. Number of moles = (24.092 × 10²²) / (6.023 × 10²³) = 0.4 mol.

Teacher's Note:
a) Adjust exponents carefully when dividing numbers with powers of 10.
b) The unit for the final calculated quantity is moles.

6. Calculate the number of moles in a) 12.046 \(\times\) 10²³ atoms of copper b) 27.95g of iron c) 1.51 \(\times\) 10²³ molecules of CO₂ [3 Marks]

Answer:
1. a) Moles of copper = (12.046 × 10²³) / (6.023 × 10²³) = 2 moles.
2. b) Moles of iron = Given mass / Atomic mass of iron (55.9 g/mol) = 27.95 / 55.85 = 0.5 mol.
3. c) Moles of CO₂ = (1.51 × 10²³) / (6.023 × 10²³) = 0.25 mol.

Teacher's Note:
a) Apply different formulas depending on whether atoms, molecules, or mass are provided.
b) Atomic mass of iron is commonly taken as 55.85 or 56 u.
c) Division involving decimal numbers requires careful rounding off.

7. If 3.0115 \(\times\) 10²³ particles are present in CO₂. Find the number of moles. [1 Mark]

Answer:
1. Number of moles = Given particles / Avogadro number = (3.0115 × 10²³) / (6.023 × 10²³) = 0.5 mol.

Teacher's Note:
a) Half of the Avogadro number gives 0.5 moles directly.

8. Find the number of moles present in 24.088 \(\times\) 10²³ particles of carbon dioxide [1 Mark]

Answer:
1. Number of moles = (24.088 × 10²³) / (6.022 × 10²³) = 4 moles.

Teacher's Note:
a) Notice that 24.088 is exactly four times 6.022.

9. Calculate the number of atoms in 48g of Mg [2 Marks]

Answer:
1. Atomic mass of Mg = 24 g/mol.
2. Number of moles = 48 / 24 = 2 moles.
3. Number of atoms = Moles × Avogadro number.
4. Number of atoms = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ atoms.

Teacher's Note:
a) Magnesium is a monoatomic element, so moles of atoms equal moles of substance.
b) Always write the correct unit for the final particle count.

10. Calculate the number of molecules in 3.6 g of water [2 Marks]

Answer:
1. Molar mass of water (H₂O) = 18 g/mol.
2. Number of moles = 3.6 / 18 = 0.2 mol.
3. Number of molecules = 0.2 × 6.022 × 10²³.
4. Number of molecules = 1.2044 × 10²³ molecules.

Teacher's Note:
a) Multiplying decimals correctly is essential for accurate molecule counts.
b) Water is a molecular compound consisting of hydrogen and oxygen.

11. Calculate the number of atoms in 0.5 moles of carbon [1 Mark]

Answer:
1. Number of atoms = Moles × Avogadro number = 0.5 × 6.022 × 10²³ = 3.011 × 10²³ atoms.

Teacher's Note:
a) Carbon is treated as single atoms in elemental form.

12. Calculate the number of moles in 12g of oxygen gas [1 Mark]

Answer:
1. Oxygen gas is O₂, so its molar mass = 2 × 16 = 32 g/mol.
2. Number of moles = 12 / 32 = 0.375 mol.

Teacher's Note:
a) Gas name implies diatomic molecule O₂, not single oxygen atom O.

13. Calculate the number of moles present in 14g of carbon monoxide . [1 Mark]

Answer:
1. Formula of carbon monoxide is CO.
2. Molar mass of CO = 12 + 16 = 28 g/mol.
3. Number of moles = 14 / 28 = 0.5 mol.

Teacher's Note:
a) Carbon monoxide comprises one carbon and one oxygen atom.

14. Find the mass of 5 moles of aluminium atoms? [1 Mark]

Answer:
1. Atomic mass of aluminium = 27 g/mol.
2. Mass = Number of moles × Atomic mass = 5 × 27 = 135 g.

Teacher's Note:
a) Direct multiplication of moles by atomic mass yields the total mass.

15. Calculate the molar mass of sulphur. [1 Mark]

Answer:
1. Sulphur exists commonly as S₈ molecules in calculations unless specified otherwise, but atomic molar mass is 32 g/mol.
2. Standard molar mass of sulphur atom = 32 g/mol (and for S₈ molecule = 256 g/mol).

Teacher's Note:
a) Atomic mass of sulphur is 32 u, giving a molar mass of 32 g/mol.

16. Calculate the mass of 0.2 mole of water molecules. [1 Mark]

Answer:
1. Molar mass of water = 18 g/mol.
2. Mass = Number of moles × Molar mass = 0.2 × 18 = 3.6 g.

Teacher's Note:
a) Product of moles and molar mass gives the required mass in grams.

17. Which has greater number of atoms, 100g of sodium or 100g of iron? [2 Marks]

Answer:
1. Moles of sodium = 100 / 23 = 4.35 moles.
2. Moles of iron = 100 / 56 = 1.79 moles.
3. Since sodium has more moles, it contains a greater number of atoms.
4. Therefore, 100g of sodium has a greater number of atoms than 100g of iron.

Teacher's Note:
a) Lighter atoms yield more moles for the same given mass.
b) Comparing moles directly answers which substance has more atoms.

18. How many atoms of oxygen are present in 300 grams of CaCO₃? [3 Marks]

Answer:
1. Molar mass of CaCO₃ = 40 + 12 + (3 × 16) = 100 g/mol.
2. Number of moles of CaCO₃ = 300 / 100 = 3 moles.
3. Each molecule of CaCO₃ contains 3 atoms of oxygen.
4. Total oxygen atoms = 3 × 3 × 6.022 × 10²³ = 5.4198 × 10²⁴ oxygen atoms.

Teacher's Note:
a) Account for the atomicity of the specific element within the compound formula.
b) Multiply moles by Avogadro number and then by the number of oxygen atoms per formula unit.

19. The mass of one atom of an element 'A' is 2.65 \(\times\) 10^-23 g. Calculate its atomic mass and name the element. [3 Marks]

Answer:
1. Atomic mass = Mass of one atom × Avogadro number.
2. Atomic mass = 2.65 × 10^-23 × 6.022 × 10²³ = 15.958 g/mol ≈ 16 u.
3. The element with atomic mass 16 u is Oxygen.
4. Hence, the element is Oxygen.

Teacher's Note:
a) Multiplying mass of a single atom by Avogadro constant gives the gram atomic mass.
b) Recognition of common atomic masses helps identify the element correctly.

20. Calculate the number moles of magnesium in 0.478g of magnesium ? [1 Mark]

Answer:
1. Atomic mass of magnesium = 24 g/mol.
2. Number of moles = 0.478 / 24 = 0.0199 mol.

Teacher's Note:
a) Divide given mass by atomic mass for monoatomic elements.

21. In which of the following cases the number of hydrogen atoms is more ? Two moles of HCl or one mole of NH₃. [2 Marks]

Answer:
1. In 2 moles of HCl, number of hydrogen moles = 2 × 1 = 2 moles of H atoms.
2. In 1 mole of NH₃, number of hydrogen moles = 1 × 3 = 3 moles of H atoms.
3. Comparing hydrogen moles: 3 moles in NH₃ is greater than 2 moles in HCl.
4. Therefore, one mole of NH₃ has more hydrogen atoms.

Teacher's Note:
a) Multiply the moles of the compound by the subscript of the element to find atom moles.
b) Direct comparison of moles of atoms determines which has more atoms.

22. Calculation of number of hydrogen atoms present in 1 mole of NH₃ [2 Marks]

Answer:
1. One mole of NH₃ contains 3 moles of hydrogen atoms.
2. Number of hydrogen atoms = Moles of H atoms × Avogadro number.
3. Number of hydrogen atoms = 3 × 6.022 × 10²³.
4. Number of hydrogen atoms = 1.8066 × 10²⁴ atoms.

Teacher's Note:
a) Ammonia has three hydrogen atoms per molecule.
b) Avogadro constant scales the mole quantity to actual particle count.

23. Find the number of oxygen atoms in 88g of CO₂? [2 Marks]

Answer:
1. Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol.
2. Number of moles of CO₂ = 88 / 44 = 2 moles.
3. Each CO₂ molecule contains 2 oxygen atoms, so moles of O atoms = 2 × 2 = 4 moles.
4. Number of oxygen atoms = 4 × 6.022 × 10²³ = 2.4088 × 10²⁴ atoms.

Teacher's Note:
a) Each carbon dioxide molecule features two oxygen atoms.
b) Double the mole value of the compound to find the moles of oxygen atoms.

24. Calculate the number of water molecules contained in a drop of water weighing 0.06g ? [2 Marks]

Answer:
1. Molar mass of water = 18 g/mol.
2. Number of moles = 0.06 / 18 = 0.00333 mol.
3. Number of molecules = 0.00333 × 6.022 × 10²³.
4. Number of molecules = 2.007 × 10²¹ molecules.

Teacher's Note:
a) Small mass values yield correspondingly smaller mole quantities.
b) Standard multiplication gives the final molecule count in scientific notation.

25. Find the number of aluminium ions present in 0.051g of aluminium oxide(Al₂O₃). (Atomic masses: Al= 27u; O= 16u) [3 Marks]

Answer:
1. Molar mass of Al₂O₃ = (2 × 27) + (3 × 16) = 54 + 48 = 102 g/mol.
2. Number of moles of Al₂O₃ = 0.051 / 102 = 0.0005 = 5 × 10^-4 mol.
3. Each formula unit of Al₂O₃ contains 2 aluminium ions (Al³⁺).
4. Number of aluminium ions = 2 × (5 × 10^-4) × 6.022 × 10²³ = 6.022 × 10²⁰ ions.

Teacher's Note:
a) Aluminium oxide dissociates or contains two aluminium ions per formula unit.
b) Handle decimal powers carefully during division and multiplication.
c) Check atomic masses provided in the question bracket.

26. Calculate the mass of 1.000 mole of CaCl₂ [1 Mark]

Answer:
1. Molar mass of CaCl₂ = 40 + (2 × 35.5) = 40 + 71 = 111 g/mol.
2. Mass of 1 mole = 1.000 × 111 = 111 g.

Teacher's Note:
a) Chlorine atomic mass is taken as 35.5 u for precision.

27. Calculate grams in 3.0000 moles of CO₂ [1 Mark]

Answer:
1. Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol.
2. Mass = 3.0000 × 44 = 132 g.

Teacher's Note:
a) Multiply the number of moles by the molar mass to find grams.

28. Calculate number of moles in 32.0 g of CH₄ [1 Mark]

Answer:
1. Molar mass of CH₄ = 12 + (4 × 1) = 16 g/mol.
2. Number of moles = 32.0 / 16 = 2 moles.

Teacher's Note:
a) Methane has one carbon and four hydrogen atoms.

29. Determine mass in grams of 40.0 moles of Na₂CO₃ [2 Marks]

Answer:
1. Molar mass of Na₂CO₃ = (2 × 23) + 12 + (3 × 16) = 46 + 12 + 48 = 106 g/mol.
2. Mass = Moles × Molar mass = 40.0 moles × 106 g/mol = 4240 g.

Teacher's Note:
a) Multiply the number of moles by the molar mass of the substance.
b) Ensure atomic masses of Na, C, and O are added correctly.

30. Calculate moles in 168.0 g of HgS [2 Marks]

Answer:
1. Molar mass of HgS = 200.6 + 32.1 = 232.7 g/mol (using Hg = 200.6, S = 32.1).
2. Moles = Given mass / Molar mass = 168.0 g / 232.7 g/mol = 0.722 moles.

Teacher's Note:
a) Use the standard atomic mass for Mercury (Hg ≈ 200.6) and Sulfur (S = 32.1).
b) Divide the given mass by the calculated molar mass.

31. Calculate moles in 510.0 g of Al₂S₃ [2 Marks]

Answer:
1. Molar mass of Al₂S₃ = (2 × 27) + (3 × 32) = 54 + 96 = 150 g/mol.
2. Moles = Given mass / Molar mass = 510.0 g / 150 g/mol = 3.4 moles.

Teacher's Note:
a) Atomic mass of Al is 27 and S is 32.
b) Total molar mass is 150 g/mol, giving 3.4 moles for 510 g.

32. How many moles are in 27.00 g of H₂O [2 Marks]

Answer:
1. Molar mass of H₂O = (2 × 1) + 16 = 18 g/mol.
2. Moles = 27.00 g / 18 g/mol = 1.5 moles.

Teacher's Note:
a) Water has a molar mass of 18 g/mol.
b) A simple division yields 1.5 moles.

33. Determine the mass in grams of Avogadro number of C₁₂H₂₂O₁₁ [2 Marks]

Answer:
1. Avogadro number of particles represents exactly one mole of the substance.
2. Molar mass of C₁₂H₂₂O₁₁ = (12 × 12) + (22 × 1) + (11 × 16) = 144 + 22 + 176 = 342 g/mol.
3. Hence, the mass of one mole (Avogadro number) is 342 g.

Teacher's Note:
a) Recognize that Avogadro number of molecules equals 1 mole.
b) Calculate the molar mass of sucrose (C₁₂H₂₂O₁₁) as 342 g/mol.

34. Find mass in grams of 9.03 moles of H₂S [2 Marks]

Answer:
1. Molar mass of H₂S = (2 × 1) + 32 = 34 g/mol.
2. Mass = Moles × Molar mass = 9.03 moles × 34 g/mol = 307.02 g.

Teacher's Note:
a) Hydrogen sulfide has a molar mass of 34 g/mol.
b) Multiply by the given number of moles to get the total mass.

35. Determine grams in 1.204 mole of NH₃ [2 Marks]

Answer:
1. Molar mass of NH₃ = 14 + (3 × 1) = 17 g/mol.
2. Mass = 1.204 moles × 17 g/mol = 20.468 g.

Teacher's Note:
a) Ammonia (NH₃) has a molar mass of 17 g/mol.
b) Product of moles and molar mass gives the mass in grams.

Consider the molecule CuNH₄Cl₃ as you answer 11 - 19. [Note: numbered 36-44 in sequence]
36. Name the elements present. [1 Mark]

Answer: Copper, Nitrogen, Hydrogen, and Chlorine.

Teacher's Note:
a) Identify the chemical symbols: Cu, N, H, Cl.

37. How many atoms form the molecule? [1 Mark]

Answer:
1. Total atoms = 1 (Cu) + 1 (N) + 4 (H) + 3 (Cl) = 9 atoms.

Teacher's Note:
a) Sum up all the subscript numbers present in the molecular formula.

38. How many of each atom in the molecule? [2 Marks]

Answer:
1. Copper (Cu): 1 atom
2. Nitrogen (N): 1 atom
3. Hydrogen (H): 4 atoms
4. Chlorine (Cl): 3 atoms

Teacher's Note:
a) List each constituent element with its exact subscript count.
b) Elements without a written subscript have a count of 1.

39. How many hydrogen atoms in one mole of molecules? [2 Marks]

Answer:
1. One molecule contains 4 hydrogen atoms.
2. One mole of molecules contains \(4 \times 6.022 \times 10^{23}\) = \(2.4088 \times 10^{24}\) hydrogen atoms.

Teacher's Note:
a) Use Avogadro number to scale up from molecules to moles.
b) Multiply the number of hydrogen atoms per molecule by Avogadro constant.

40. How many chlorine atoms in six moles of molecules? [2 Marks]

Answer:
1. One mole of CuNH₄Cl₃ contains 3 moles of chlorine atoms.
2. Six moles of molecules contain 6 × 3 = 18 moles of chlorine atoms = \(18 \times 6.022 \times 10^{23}\) = \(1.084 \times 10^{25}\) chlorine atoms.

Teacher's Note:
a) Account for both the 6 moles factor and the 3 chlorine atoms per molecule.
b) Express the final count either in moles of atoms or absolute number of atoms.

41. What is the molar mass of this molecule? [2 Marks]

Answer:
1. Molar mass = 63.5 (Cu) + 14 (N) + (4 × 1) (H) + (3 × 35.5) (Cl)
2. Molar mass = 63.5 + 14 + 4 + 106.5 = 188 g/mol.

Teacher's Note:
a) Use standard atomic weights: Cu = 63.5, N = 14, H = 1, Cl = 35.5.
b) Sum all atomic masses to get 188 g/mol.

42. Name this molecule. [1 Mark]

Answer: Ammonium copper(II) trichloride (or Copper(II) ammonium chloride trichloride).

Teacher's Note:
a) Name the cation and anion components following standard chemical nomenclature rules.

43. What is the mass in grams of one molecule? [2 Marks]

Answer:
1. Mass of one molecule = Molar mass / Avogadro number
2. Mass = 188 g / \(6.022 \times 10^{23}\) = \(3.12 \times 10^{-22}\) g.

Teacher's Note:
a) Divide the molar mass by Avogadro constant to find the mass of a single entity.
b) Result is expressed in grams with proper exponent notation.

44. How many moles would be in 6.84 g of this substance? [2 Marks]

Answer:
1. Molar mass of CuNH₄Cl₃ = 188 g/mol.
2. Moles = Given mass / Molar mass = 6.84 g / 188 g/mol = 0.0364 moles.

">Teacher's Note:
a) Use the previously calculated molar mass of 188 g/mol.
b) Divide given mass by molar mass to determine moles.

45. You need 0.0100 mole of lead (II) chromate. How much should you weigh on the scale? [2 Marks]

Answer:
1. Formula of lead(II) chromate is PbCrO₄.
2. Molar mass of PbCrO₄ = 207.2 (Pb) + 52 (Cr) + (4 × 16) (O) = 207.2 + 52 + 64 = 323.2 g/mol.
3. Mass = 0.0100 mole × 323.2 g/mol = 3.232 g.

Teacher's Note:
a) Determine the correct chemical formula PbCrO₄.
b) Multiply desired moles by the calculated molar mass.

46. Given 6.40 g of HBr. How many moles is this? [2 Marks]

Answer:
1. Molar mass of HBr = 1 (H) + 79.9 (Br) = 80.9 g/mol.
2. Moles = 6.40 g / 80.9 g/mol = 0.0791 moles.

Teacher's Note:
a) Atomic mass of bromine is taken as 79.9 g/mol.
b) Divide given mass by molar mass.

47. Write the correct formula for calcium acetate and then answer 23 - 25 based on it. [1 Mark]

Answer: Ca(CH₃COO)₂

Teacher's Note:
a) Calcium has a valency of +2 and acetate ion has a charge of -1.
b) Cross valencies to obtain the formula Ca(CH₃COO)₂.

48. What is the mass of exactly one mole of calcium acetate? [2 Marks]

Answer:
1. Formula: Ca(CH₃COO)₂ (C₄H₆O₄Ca).
2. Molar mass = 40 (Ca) + 4 × 12 (C) + 6 × 1 (H) + 4 × 16 (O) = 40 + 48 + 6 + 64 = 158 g/mol.

Teacher's Note:
a) Count total atoms: 1 Ca, 4 C, 6 H, 4 O.
b) Sum their atomic masses to get 158 g/mol.

49. How many moles are contained in 1.58 g of the substance in #23? [2 Marks]

Answer:
1. Molar mass of calcium acetate = 158 g/mol.
2. Moles = 1.58 g / 158 g/mol = 0.01 moles.

Teacher's Note:
a) Refer to the molar mass calculated in question 48.
b) Divide 1.58 g by 158 g/mol to get 0.01 moles.

50. How much does 0.400 mole of #23 weigh? [2 Marks]

Answer:
1. Molar mass of calcium acetate = 158 g/mol.
2. Mass = 0.400 mole × 158 g/mol = 63.2 g.

Teacher's Note:
a) Multiply the number of moles (0.400) by the molar mass (158 g/mol).
b) Final weight is 63.2 g.

51. Write the formula for oxygen gas. [1 Mark]

Answer: O₂

Teacher's Note:
a) Oxygen gas exists naturally as a diatomic molecule.

52. How many atoms (and moles) are represented by the formula in #26? [2 Marks]

Answer:
1. The formula O₂ represents 2 atoms of oxygen per molecule (or 2 moles of oxygen atoms per mole of molecules).

Teacher's Note:
a) Subscript 2 indicates 2 atoms in a single molecule of oxygen gas.

53. What is the mass of Avogadro Number of oxygen molecules? [2 Marks]

Answer:
1. Avogadro number of O₂ molecules equals 1 mole of O₂ molecules.
2. Mass = Molar mass of O₂ = 2 × 16 = 32 g.

Teacher's Note:
a) Avogadro number of any molecular substance equals its gram molecular mass.
b) Molar mass of O₂ is 32 g/mol.

54. Calculate the molar mass of HNO₃. [N = 14, O = 16, H = 1] [2 Marks]

Answer:
1. Molar mass of HNO₃ = Atomic mass of H + Atomic mass of N + (3 × Atomic mass of O)
2. Molar mass = 1 + 14 + (3 × 16) = 1 + 14 + 48 = 63 g/mol.

Teacher's Note:
a) Substitute the given atomic masses into the formula sum.
b) Final molar mass is 63 g/mol.

55. Calculate the formula mass of CaCl₂. [Ca = 40, Cl = 35.5] [2 Marks]

Answer:
1. Formula mass of CaCl₂ = Atomic mass of Ca + (2 × Atomic mass of Cl)
2. Formula mass = 40 + (2 × 35.5) = 40 + 71 = 111 u (or g/mol).

Teacher's Note:
a) Ionic compounds use the term formula mass instead of molecular mass.
b) Sum 40 and two times 35.5 to get 111.

56. A certain non-metal X forms two oxides I and II. The mass percentage of oxygen in oxide I (X₄O₆) is 43.7, which is same as that of X in oxide II. Find the formula of the second oxide. [4 Marks]

Answer:
1. In oxide I (X₄O₆), mass percentage of oxygen is given as 43.7%, so the mass percentage of X is 100 - 43.7 = 56.3%.
2. In oxide II, the mass percentage of X is 43.7% and oxygen is 56.3%.
3. Let the atomic mass of X be \(M_X\). For X₄O₆, mass ratio of X to O = \(\frac{4 \times M_X}{6 \times 16} = \frac{56.3}{43.7}\).
4. Solving for \(M_X\): \(\frac{4 M_X}{96} = \frac{56.3}{43.7} \implies M_X = \frac{96 \times 56.3}{43.7 \times 4} \approx 30.9 \approx 31\) (Phosphorus).
5. For oxide II, ratio of mass of X to O = \(\frac{43.7}{56.3}\). Let formula be X_aO_b, then \(\frac{a \times 31}{b \times 16} = \frac{43.7}{56.3} \implies \frac{31a}{16b} = 0.776 \implies \frac{a}{b} = \frac{0.776 \times 16}{31} \approx 0.4 \implies \frac{2}{5}\).
6. Thus, the formula of the second oxide is X₂O₅.

Teacher's Note:
a) Use mass percentage data to set up proportional atomic mass relations.
b) Determine the atomic mass of element X (phosphorus in this case).
c) Use the percentage composition of oxide II to find the simplest whole number ratio of X to O.
d) The resulting formula is X₂O₅.

57. Calculate the mass of 0.2 moles of water (O=16, H=1). [2 Marks]

Answer:
1. Molar mass of water (H₂O) = (2 × 1) + 16 = 18 g/mol.
2. Mass = Moles × Molar mass = 0.2 moles × 18 g/mol = 3.6 g.

Teacher's Note:
a) Molar mass of water is 18 g/mol.
b) Multiply by 0.2 to obtain 3.6 grams.

58. What is the volume of 7.1 g of chlorine (Cl=35.5) at S.T.P. [3 Marks]

Answer:
1. Chlorine gas exists as Cl₂. Molar mass of Cl₂ = 2 × 35.5 = 71 g/mol.
2. Moles of Cl₂ = Given mass / Molar mass = 7.1 g / 71 g/mol = 0.1 moles.
3. Volume at S.T.P. = Moles × 22.4 L = 0.1 × 22.4 L = 2.24 L.

Teacher's Note:
a) Remember that chlorine is diatomic (Cl₂), so its molar mass is 71 g/mol.
b) One mole of any gas occupies 22.4 L at S.T.P.
c) Multiply the number of moles (0.1) by 22.4 L to get 2.24 L.

59. The reaction between aluminium carbide and water takes place according to the following equation: Al₄C₃ + 12H₂O → 3CH₄ + 4Al(OH)₃. Calculate the volume of CH₄ released from 14.4 g of Al₄C₃ by excess water at S.T.P. (C = 12, Al = 27) [4 Marks]

Answer:
1. Molar mass of Al₄C₃ = (4 × 27) + (3 × 12) = 108 + 36 = 144 g/mol.
2. Moles of Al₄C₃ = 14.4 g / 144 g/mol = 0.1 moles.
3. According to the balanced equation, 1 mole of Al₄C₃ produces 3 moles of CH₄.
4. Moles of CH₄ produced = 0.1 × 3 = 0.3 moles.
5. Volume of CH₄ at S.T.P. = Moles × 22.4 L = 0.3 × 22.4 = 6.72 L.

Teacher's Note:
a) Calculate the molar mass of reactant Al₄C₃ correctly as 144 g/mol.
b) Use stoichiometry from the balanced chemical equation (1:3 ratio).
c) Convert moles of methane gas to volume at S.T.P. using molar volume (22.4 L).
d) Final calculated volume is 6.72 L.

60. A compound of sodium, sulphur and oxygen has the following percentage composition. Na=29.11%, S=40.51%, O=30.38%. Find its empirical formula (O=16, Na=23, S=32). [3 Marks]

Answer:
1. Percentage of Na = 29.11%, Atomic mass = 23, Relative moles = \(\frac{29.11}{23}\) = 1.266
2. Percentage of S = 40.51%, Atomic mass = 32, Relative moles = \(\frac{40.51}{32}\) = 1.266
3. Percentage of O = 30.38%, Atomic mass = 16, Relative moles = \(\frac{30.38}{16}\) = 1.898
4. Simplifying the mole ratio (dividing by smallest value 1.266): Na = 1, S = 1, O = 1.5
5. Multiplying by 2 to get whole numbers: Na = 2, S = 2, O = 3
6. Hence, the empirical formula is Na₂S₂O₃.

Teacher's Note:
a) The empirical formula represents the simplest whole-number ratio of atoms of each element in a compound.
b) Always divide the percentage of each element by its respective atomic mass to find the relative number of moles.
c) If the mole ratio contains a fraction like 1.5, multiply all values by the smallest integer that converts it to a whole number.

61. Solid ammonium dichromate with relative molecular mass of 252 g decomposes according to the equation: (NH₄)₂Cr₂O₇ → N₂ + Cr₂O₃ + 4H₂O. (i) What volume of nitrogen at S.T.P will be evolved when 63 g of (NH₄)₂Cr₂O₇ is decomposed? (ii) If 63 g of (NH₄)₂Cr₂O₇ is heated above 100°C, what will be the loss of mass? (H=1, N=14, O=16, Cr=52). [5 Marks]

Answer:
1. Given equation: (NH₄)₂Cr₂O₇ → N₂ + Cr₂O₃ + 4H₂O
2. Molar mass of (NH₄)₂Cr₂O₇ = 252 g/mol. Number of moles in 63 g = \(\frac{63}{252}\) = 0.25 moles.
3. (i) From the balanced equation, 1 mole of (NH₄)₂Cr₂O₇ gives 1 mole of N₂ gas. Therefore, 0.25 moles of (NH₄)₂Cr₂O₇ gives 0.25 moles of N₂.
4. Volume of N₂ at S.T.P = 0.25 × 22.4 L = 5.6 litres.
5. (ii) When heated above 100°C, water (H₂O) formed will exist as steam and escape along with nitrogen gas. Total gaseous products = 1 mole of N₂ + 4 moles of H₂O = 5 moles of gases per mole of reactant.
6. Total mass of gases evolved for 0.25 moles = 0.25 × (mass of N₂ + 4 × mass of H₂O) = 0.25 × (28 + 4 × 18) = 0.25 × 100 = 25 g. Hence, the loss of mass is 25 g.

Teacher's Note:
a) Stoichiometric coefficients determine the molar relationships between reactants and products in a balanced chemical equation.
b) At standard temperature and pressure (S.T.P), 1 mole of any ideal gas occupies 22.4 litres of volume.
c) Water is produced as a gas (steam) above 100°C, meaning both nitrogen and water vapour contribute to the total loss of solid mass.
d) Always verify that the chemical equation is completely balanced before applying mole ratios to mass or volume calculations.

62. How many litres of ammonia are present in 3.4 kg of it? (N = 14, H = 1) [3 Marks]

Answer:
1. Formula of ammonia = NH₃
2. Molar mass of NH₃ = 14 + (3 × 1) = 17 g/mol
3. Mass of ammonia given = 3.4 kg = 3400 g
4. Number of moles of ammonia = \(\frac{3400}{17}\) = 200 moles
5. Volume of ammonia at S.T.P = Number of moles × 22.4 L = 200 × 22.4 = 4480 litres.

Teacher's Note:
a) Convert kilograms to grams before calculating the number of moles to maintain uniform units.
b) One mole of any gas at S.T.P occupies a standard volume of 22.4 litres.
c) Ammonia is a common gas compound composed of one nitrogen atom and three hydrogen atoms.

63. About 640 mL of carbon monoxide is mixed with 800 mL of oxygen and ignited in an enclosed vessel. Calculate the total volume of gases after the burning is completed. All volumes are measured at S.T.P. [4 Marks]

Answer:
1. Balanced chemical equation for the combustion of carbon monoxide: 2CO + O₂ → 2CO₂
2. According to Gay-Lussac's law of combining volumes, 2 volumes of CO react with 1 volume of O₂ to form 2 volumes of CO₂.
3. Given volumes: CO = 640 mL, O₂ = 800 mL. Since 2 volumes of CO require 1 volume of O₂, 640 mL of CO will require \(\frac{640}{2}\) = 320 mL of O₂.
4. Carbon monoxide is the limiting reactant and is completely consumed, while unreacted oxygen = 800 mL - 320 mL = 480 mL.
5. Volume of carbon dioxide (CO₂) formed = Volume of CO consumed = 640 mL.
6. Total volume of gases remaining after burning = Volume of unreacted O₂ + Volume of CO₂ formed = 480 mL + 640 mL = 1120 mL.

Teacher's Note:
a) Gay-Lussac's law states that gases react in simple whole-number volume ratios under similar conditions of temperature and pressure.
b) Identify the limiting reactant correctly by comparing the mole or volume ratio with the stoichiometric coefficients.
c) The final gas mixture includes both the newly formed product gas and any excess unreacted reactant gas.
d) Always check if any gaseous reactant remains unconsumed at the end of the reaction.

64. Calculate the number of moles of ammonium sulphate present in 15.84 kg of it. (H=1, N=14, O=16, S=32) [3 Marks]

Answer:
1. Formula of ammonium sulphate = (NH₄)₂SO₄
2. Molar mass of (NH₄)₂SO₄ = 2 × (14 + 4 × 1) + 32 + 4 × 16 = 2 × (18) + 32 + 64 = 36 + 32 + 64 = 132 g/mol
3. Given mass of ammonium sulphate = 15.84 kg = 15,840 g
4. Number of moles = \(\frac{15840}{132}\) = 120 moles.

Teacher's Note:
a) Accurately determine the molecular mass by summing the atomic masses of all constituent atoms in the chemical formula.
b) Pay close attention to unit conversions, changing kilograms to grams by multiplying by 1,000.
c) Number of moles is calculated as the ratio of given mass to the molar mass of the substance.