Ultimate Guide to Stoichiometry: MCQs for MDCAT, ECAT, & FSC Prep!| Chemistry MCQs for Competitive Exams

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🌟 Ultimate Guide to Stoichiometry: MCQs for MDCAT, ECAT, & FSC Prep! 💥💯

💡Top-Level Chemistry Quiz: Test Your Knowledge on Moles, Atoms, and Gases!

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🌟 1. If the volume occupied by oxygen gas at STP is 44.8 dm³, the number of molecules of O₂ in the vessel are:

🟦 A. 3.01 x 10²³

🟩 B. 6.02 x 10²³

🟥 C. 12.04 x 10²³

🟨 D. 24.08 x 10²³

---

🌟 2. In the reaction 

$\mathrm{<b\; style="text-align:\; justify;">2</b>}\mathrm{<b\; style="text-align:\; justify;">N</b>}\mathrm{<b\; style="text-align:\; justify;">a</b>}<b\; style="text-align:\; justify;">+</b>\mathrm{<b\; style="text-align:\; justify;">2</b>}{\mathrm{<b\; style="text-align:\; justify;">H</b>}}_{}$₂$\mathrm{<b\; style="text-align:\; justify;">O</b>}<b\; style="text-align:\; justify;">\to </b>\mathrm{<b\; style="text-align:\; justify;">2</b>}\mathrm{<b\; style="text-align:\; justify;">N</b>}\mathrm{<b\; style="text-align:\; justify;">a</b>}\mathrm{<b\; style="text-align:\; justify;">O</b>}\mathrm{<b\; style="text-align:\; justify;">H</b>}<b\; style="text-align:\; justify;">+</b>{\mathrm{<b\; style="text-align:\; justify;">H</b>}}_{}$₂

​, if 23g of Na reacts with excess of water, the volume of hydrogen gas (H₂) liberated at STP should be:

🟦 A. 11.2 dm³

🟩 B. 22.4 dm³

🟥 C. 33.6 dm³

🟨 D. 44.8 dm³

---

🌟 3. The number of carbon atoms in 1 mole of sugar (C₁₂H₂₂O₁₁) are approximately:

🟦 A. 6 x 10²³

🟩 B. 24 x 10²³

🟥 C. 60 x 10²³

🟨 D. 72 x 10²³

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🌟 4. Which of the following sample of substances contains the same number of atoms as that of 20g Ca?

🟦 A. 16g S

🟩 B. 20g C

🟥 C. 19g K

🟨 D. 24g Mg

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🌟 5. Number of atoms in 60g of carbon are:

🟦 A. 3.01 × 10²³

🟩 B. 3.01 × 10²⁴

🟥 C. 6.02 × 10²³

🟨 D. 6.02 × 10²⁴

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🌟 6. Maximum number of molecules present in the following sample of gas:

🟦 A. 100g O₂

🟩 B. 100g CH₄

🟥 C. 100g CO₂

🟨 D. 100g Cl₂

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🌟 7. Which of the following statement is incorrect?

🟦 A. One mole of H₂ gas contains 6.02 × 10²³ molecules of H₂

🟩 B. The mass of 1 mole of Cl₂ gas is 35.5g

🟥 C. Number of atoms in 23g Na and 24g Mg are equal

🟨 D. One mole of O₂ at STP occupies 22.4dm³ volume

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🌟 8. For Avogadro’s number, this statement is incorrect:

🟦 A. Its numerical value is 6.02 × 10²³

🟩 B. It is the number of particles in one mole of any substance

🟥 C. Its value changes if temperature increases

🟨 D. Its value changes if the number of moles increases

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🌟 9. Generally, actual yield is:

🟦 A. Greater than theoretical yield

🟩 B. Less than theoretical yield

🟥 C. Equal to theoretical yield

🟨 D. Sometimes greater and sometimes less than theoretical yield

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🌟 10. The minimum number of moles are present in:

🟦 A. 1 dm³ of methane gas at STP

🟩 B. 5 dm³ of helium gas at STP

🟥 C. 10 dm³ of hydrogen gas at STP

🟨 D. 22.4 dm³ of chlorine gas at STP

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🌟 11. Which of the following formula represents an acid anhydride?

🟦 A. CaO

🟩 B. H₂O

🟥 C. H₂SO₄

🟨 D. SO₃

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🌟 12. How many neutrons are contained in the nucleus of an element with Z = 27 and A = 59?

🟦 A. 27

🟩 B. 32

🟥 C. 58

🟨 D. 86

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🌟 13. How many grams of NaOH are needed to make 1000g of a 5% solution?

🟦 A. 5g

🟩 B. 500g

🟥 C. 40g

🟨 D. 50g

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🌟 14. nm is …………. than μm.

🟦 A. Equal to

🟩 B. Less

🟥 C. Greater

🟨 D. All of them

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🌟 15. Which one is larger than dm³?

🟦 A. cm³

🟩 B. Liter

🟥 C. m³

🟨 D. mL

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🌟 16. Which of the following is classified as a monoatomic element?

🟦 A. Oxygen

🟩 B. Nitrogen

🟥 C. Milk

🟨 D. Helium

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🌟 17. The decimal equivalent of 1/60 to three significant figures is:

🟦 A. 0.0166

🟩 B. 0.0167

🟥 C. 0.00166

🟨 D. 1.666 x 10⁻²

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🌟18. How many significant figures are contained in 0.00050890400?

🟦 A. 6

🟩 B. 4

🟥 C. 8

🟨 D. 11

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🌟 19. One cm³ is equal to:

🟦 A. 10⁻³ dm³

🟩 B. 100 dm³

🟥 C. 10 dm³

🟨 D. 1000 dm³

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🌟 20. The number of protons in one molecule of HNO₃ are:

🟦 A. 7

🟩 B. 8

🟥 C. 24

🟨 D. 32

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🌟 21. Which one of the following is NOT an iso electronic pair?

🟦 A. Ca²⁺, K

🟩 B. Na⁺, Ne

🟥 C. K⁺, Cl⁻

🟨 D. O²⁻, Al³⁺

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🌟 22. Which of the following formula represents a basic anhydride?

🟦 A. CO

🟩 B. CrO₃

🟥 C. MgO

🟨 D. NO

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🌟 23. The basic oxide in the given list is:

🟦 A. N₂O₅

🟩 B. P₄O₆

🟥 C. Sb₄O₆

🟨 D. Bi₂O₃

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🌟 24. If an element M forms an oxide with formula MO, which of the following is a correct formula?

🟦 A. MCl

🟩 B. MS

🟥 C. MCl₂

🟨 D. Both B and C

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🌟 25. The formula of KAl(SO₄)₂ represents a total of:

🟦 A. 6 atoms

🟩 B. 7 atoms

🟥 C. 12 atoms

🟨 D. 14 atoms

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🌟 26. The diameter of atoms is of the order of:

🟦 A. 2 × 10⁻⁵ m

🟩 B. 2 × 10⁻¹⁰ m

🟥 C. 2 × 10⁻² m

🟨 D. 2 × 10⁻³ m

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🌟 27. 51.084 g has five significant figures while 0.00510840 mg has ……….. significant figures.

🟦 A. 4

🟩 B. 6

🟥 C. 5

🟨 D. 3

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🌟 28. 1 is written in exponential notation as:

🟦 A. 10¹

🟩 B. 10⁰

🟥 C. 10²

🟨 D. 10³

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🌟 29. 67.84 g has four significant figures while 0.00678400 mg has ……….. significant figures.

🟦 A. 4

🟩 B. 6

🟥 C. 5

🟨 D. 3

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🌟 30. How many significant figures are contained in 0.00020089400 mm?

🟦 A. 6

🟩 B. 4

🟥 C. 8

🟨 D. 11

🌟 31. 75.5 g has three significant figures while 0.00755 mg has ……….. significant figures.

🟦 A. 2

🟩 B. 3

🟥 C. 5

🟨 D. 6

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🌟 32. Which one is the incorrect relation?

🟦 A. 1 m³ = 102 dm³

🟩 B. 1 cm³ = 10⁻⁶ m³

🟥 C. 1 cm³ = 10⁻³ dm³

🟨 D. 1 dm³ = 10³ cm³

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🌟 33. The digit that can be significant at times is:

🟦 A. 1

🟩 B. 3

🟥 C. 7

🟨 D. 0

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🌟 34. 8.687 has rounded off to two significant figures as:

🟦 A. 8.7

🟩 B. 8.6

🟥 C. 8.700

🟨 D. 8.70

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🌟 35. The property of measurement reflected by significant figures is:

🟦 A. Purity

🟩 B. Precision

🟥 C. Perfection

🟨 D. All of them

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🌟 36. 0.00000000000000000000000000000000091096 is expressed in exponential notation as:

🟦 A. 9.1096 × 10⁻³⁶

🟩 B. 9.1096 × 10⁻³⁰

🟥 C. 9.1096 × 10⁻³¹

🟨 D. 9.1096 × 10⁻³²

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🌟 37. 28.987 has rounded off to two significant figures as:

🟦 A. 28.9

🟩 B. 28

🟥 C. 29

🟨 D. 29.0

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🌟 38. The prefix 10¹⁸ represents:

🟦 A. Kilo

🟩 B. Nano

🟥 C. Exa

🟨 D. Giga

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🌟 39. Which one of the following is anhydride of sulphuric acid?

🟦 A. Sulphur(II) oxide

🟩 B. Sulphur(III) oxide

🟥 C. Sulphur(VI) oxide

🟨 D. Iron pyrite

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🌟 40. Which one is the incorrect relation?

🟦 A. 1 L = 10³ mL

🟩 B. 1 cm³ = 10⁻⁶ m³

🟥 C. 1 cm³ = 10⁻³ dm³

🟨 D. 1 kg = 10² g

🌟 41. An atom has 9 protons and 10 neutrons, the atom is:

🟦 A. N

🟩 B. O

🟥 C. B

🟨 D. F

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🌟 42. Which of the following has infinite significant figures?

🟦 A. 22.40 cm

🟩 B. 3.00 g

🟥 C. Avogadro’s number

🟨 D. 0.0045 m

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🌟 43. Round off 2.3746 to 3 significant figures:

🟦 A. 2.37

🟩 B. 2.38

🟥 C. 2.36

🟨 D. 2.374

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🌟 44. The number of significant figures in 6.022 × 10²³ is:

🟦 A. 3

🟩 B. 4

🟥 C. 5

🟨 D. Infinite

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🌟 45. 4.578 has 4 significant figures. It can be changed to 7 significant figures by:

🟦 A. Placing decimal

🟩 B. Placing decimal after 8 and adding 2 zeros

🟥 C. Adding 3 zeros at the end

🟨 D. None of them

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🌟 46. 13.89 has 4 significant figures. It can be changed to 6 significant figures by:

🟦 A. Removing decimal

🟩 B. Placing one zero at the end

🟥 C. Simply adding two zeros at the end

🟨 D. None of them

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🌟 47. The similarity between 0.6850 and 6850 is that:

🟦 A. Both have 3 significant figures

🟩 B. Both have 4 digits

🟥 C. Both have 1 Non-significant zero

🟨 D. Both have 4 significant figures

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🌟 48. The lesser the number of significant figures, the:

🟦 A. Greater is the uncertainty

🟩 B. Smaller is the uncertainty

🟥 C. Smaller is certainty

🟨 D. Both A and C

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🌟 49. 3.69 has 3 significant figures. It can be changed to 6 significant figures by:

🟦 A. Placing decimal

🟩 B. Placing decimal after 9 and adding 3 zeros

🟥 C. Adding 3 zeros at the end

🟨 D. None of them

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🌟 50. Chlorine atom and chloride ions:

🟦 A. Have same number of protons

🟩 B. Have same number of electrons

🟥 C. Are allotropes of chlorine

🟨 D. Are chemically identical

🌟 51. Which of the following statements is FALSE if the number has greater number of significant figures?

🟦 A. Greater is the precision

🟩 B. Smaller is uncertainty

🟥 C. Greater is the certainty

🟨 D. Lesser is the precision

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🌟 52. The similarity between 0.569 and 569 is that:

🟦 A. Both have 3 significant figures

🟩 B. Both have 4 digits

🟥 C. Both have 1 Non-significant zero

🟨 D. Both have 4 significant figures

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🌟 53. The greater the number of significant figures, the:

🟦 A. Greater is the uncertainty

🟩 B. Smaller is the uncertainty

🟥 C. Smaller is certainty

🟨 D. Both A and C

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🌟 54. The greater the number of significant figures, the:

🟦 A. Smaller is the uncertainty

🟩 B. Greater is the precision

🟥 C. Greater is the certainty

🟨 D. All of them

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🌟 55. Zero in decimal figure is significant if it is present:

🟦 A. On extreme right at the end

🟩 B. On left side between decimal & non-zero digit when number is < 1

🟥 C. Between two non-zero digits

🟨 D. Both A and C are true

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🌟 56. 6.55 has 3 significant figures. It can be changed to 5 significant figures by:

🟦 A. Removing decimal

🟩 B. Placing two zeros at the end

🟥 C. Simply adding a zero at the end

🟨 D. None of them

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🌟 57. Zero in decimal figure is insignificant if it is:

🟦 A. On extreme right at the end

🟩 B. Between two non-zero digits

🟥 C. On left side of non-zero digit while number is < 1

🟨 D. None of these

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🌟 58. Which figures are significant?

🟦 A. All non-zero figures

🟩 B. Trailing zeros to the right of a decimal point

🟥 C. Any zeros between significant figures

🟨 D. All of the above

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🌟 59. The correct set of significant figures is:

🟦 A. 0.00056 has 5 significant figures

🟩 B. 1.2300 has 5 significant figures

🟥 C. 0.100 has 1 significant figure

🟨 D. 2.5 × 10³ has 4 significant figures

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🌟 60. The correct set of significant figures is:

🟦 A. 0.00750 → 3 significant figures

🟩 B. 2.50 × 10³ → 4 significant figures

🟥 C. 1.0050 → 5 significant figures

🟨 D. 200 → 3 significant figures

🌟 61. The mass of one mole of electrons is:

🟦 A. 1.008 mg

🟩 B. 0.55 mg

🟥 C. 0.184 mg

🟨 D. 1.673 mg

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🌟 62. The largest number of molecules are present in:

🟦 A. 3.6 g of water

🟩 B. 4.8 g of C2H5OH

🟥 C. 2.8 g of CO

🟨 D. 5.4 g of N2O5

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🌟 63. If 28 g of N2 gas comprises of z molecules; how many molecules are there in 16 g of oxygen?

🟦 A. z

🟩 B. 2 z

🟥 C. ½ z

🟨 D. ¼ z

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🌟 64. The mass of a single hydrogen atom is:

🟦 A. 1.67 × 10‒22 g

🟩 B. 1.67 × 10‒24 kg

🟥 C. 1 g

🟨 D. 1.67 × 10‒27 kg

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🌟 65. One mole of carbon-12 has a mass of:

🟦 A. 0.012 kg

🟩 B. 0.0224 kg

🟥 C. 12 kg

🟨 D. 1 kg

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🌟 66. The molar mass of formaldehyde is …………

🟦 A. 30

🟩 B. 40

🟥 C. 26

🟨 D. None of them

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🌟 67. What is the mass of one mole of iodine molecule?

🟦 A. 53 g

🟩 B. 127 g

🟥 C. 254 g

🟨 D. 74 g

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🌟 68. The volume occupied by a 0.1 mole of a gas at STP is:

🟦 A. 22400 cm³

🟩 B. 2.24 dm³

🟥 C. 22.4 dm³

🟨 D. 5.6 dm³

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🌟 69. The relative atomic mass of chlorine is 35.5 amu, the mass in grams of 0.5 moles of chlorine gas is:

🟦 A. 71 g

🟩 B. 35.5 g

🟥 C. 142 g

🟨 D. 17.75 g

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🌟 70. The total number of ions in CaCl2 is:

🟦 A. 6.02 × 10²³

🟩 B. 12.04 × 10²³

🟥 C. 18.06 × 10²³

🟨 D. 24.08 × 10²³

🌟 71. 1 atom of an element weighs 1.792 × 10⁻²² g. The Atomic mass of the element is:

🟦 A. 17.92

🟩 B. 1.192

🟥 C. 108

🟨 D. 64

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🌟 72. Which of the following has maximum number of atoms?

🟦 A. 18 g of H₂O

🟩 B. 4.4 g of CO₂

🟥 C. 16 g of O₂

🟨 D. 16 g of CH₄

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🌟 73. The mass of 1 atom of hydrogen is:

🟦 A. 0.5 g

🟩 B. 1.6 × 10⁻²⁴ g

🟥 C. 1 g

🟨 D. 3.2 × 10⁻²⁴ g

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🌟 74. The number of sulfur atoms present in 0.2 moles of S₈ molecules is:

🟦 A. 9.63 × 10²²

🟩 B. 4.82 × 10²³

🟥 C. 1.20 × 10²³

🟨 D. 9.63 × 10²³

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🌟 75. One mole of any covalent compound contains the same number of:

🟦 A. Atoms

🟩 B. Molecules

🟥 C. Ions

🟨 D. Electrons

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🌟 76. The number of atoms in 4.25 g of NH₃ is approximately:

🟦 A. 1 × 10²³

🟩 B. 2 × 10²³

🟥 C. 4 × 10²³

🟨 D. 6 × 10²³

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🌟 77. A quantity containing Avogadro’s number of particles is called:

🟦 A. Gram atom

🟩 B. Gram molecule

🟥 C. Mole

🟨 D. Gram formula

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🌟 78. The number of formula units in 11.1 g of calcium chloride is:

🟦 A. 12.04 x 10²³

🟩 B. 18.06 x 10²³

🟥 C. 6.02 x 10²²

🟨 D. 6.02 x 10²³

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🌟 79. The number of water molecules in 1 litre of water is:

🟦 A. 18

🟩 B. 18 × 1000

🟥 C. NA

🟨 D. 55.55 NA

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🌟 80. Mass of one mole of electrons is ……… kg

🟦 A. 1.97 × 10⁻⁶

🟩 B. 54.7 × 10⁻⁸

🟥 C. 54.86 × 10⁹

🟨 D. 1.97 × 10⁻³¹

81. The number of molecules in 100 ml of each of O₂, NH₃, and CO₂ at STP are:

🟦 A. O₂ = NH₃ = CO₂
🟩 B. CO₂ < O₂ < NH₃
🟥 C. NH₃ < O₂ < CO₂
🟨 D. CO₂ = NH₃ < O₂

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82. Which has the maximum number of atoms?

🟦 A. 24 g of C
🟩 B. 5.6 g of Fe
🟥 C. 270 g of Al
🟨 D. 1.08 g Ag

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83. Two moles of hydrogen gas at STP contains how many atoms?

🟦 A. 24.08 × 10²⁴
🟩 B. 3.01 × 10²⁴
🟥 C. 24.08 × 10²³
🟨 D. 12.04 × 10⁴⁶

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84. Number of O₂ molecules present in a one-litre flask at pressure 7.6 × 10⁻¹⁰ mm of Hg at 0ºC is:

🟦 A. 2.69 × 10¹⁰
🟩 B. 2.69 × 10¹¹
🟥 C. 2.69 × 10⁹
🟨 D. None of these

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85. 16 g of oxygen has a volume of:

🟦 A. 22.4 dm³
🟩 B. 11.2 dm³
🟥 C. 5.6 dm³
🟨 D. 22400 cm³

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86. 11.2 dm³ of Cl₂ gas at STP weighs:

🟦 A. 71 g
🟩 B. 17 g
🟥 C. 35.5 g
🟨 D. 37 g

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87. Which one of the following represents the value of 1 a.m.u.?

🟦 A. 1.667 × 10⁻²⁴ g
🟩 B. 1.667 × 10⁻²³ g
🟥 C. 1.667 × 10⁻²⁷ g
🟨 D. 1.667 × 10⁻²⁸ g

---

88. This one of the following pairs has the same number of molecules:

🟦 A. 10 g H₂ & 10 g CH₄
🟩 B. 10 g H₂ & 50 g CH₄
🟥 C. 10 g H₂ & 80 g CH₄
🟨 D. 10 g H₂ & 16 g CH₄

---

89. One cm³ of oxygen gas at STP contains about:

🟦 A. 1 × 10²⁰ atoms
🟩 B. 6 × 10²³ atoms
🟥 C. 1.67 × 10²⁴ atoms
🟨 D. 0.53 × 10²⁰ atoms

---

90. The number of atoms in 558.5 grams of Fe (atomic weight of Fe = 55.85 g/mol) is:

🟦 A. Twice that in 60 g of carbon
🟩 B. 558.5 × 6.023 × 10²³
🟥 C. Half that in 8 g of He
🟨 D. 6.023 × 10²²

91. Calculate the number of moles in 1 m³ of gas at STP:

🟦 A. 4.46
🟩 B. 44.6
🟥 C. 446
🟨 D. 4460

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92. Which of the following will contain the same number of atoms as 20 g of calcium?

🟦 A. 12 g magnesium
🟩 B. 12 g carbon
🟥 C. 16 g oxygen atom
🟨 D. 8 g oxygen gas

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93. One mole of P₄ molecules contains:

🟦 A. 1 molecule
🟩 B. 24.088 × 10²³ atoms
🟥 C. ¼ × 6.022 × 10²³ atoms
🟨 D. 4 molecules

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94. In which of the following pairs do 1 g of each have an equal number of molecules?

🟦 A. N₂O and CO
🟩 B. N₂ and C₃O₂
🟥 C. N₂ and CO
🟨 D. N₂O and CO₂

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95. 400 g of caustic soda constitutes its … gram formula:

🟦 A. 1
🟩 B. 5
🟥 C. 10
🟨 D. Half

---

96. The mass of 1 atom of diamond is:

🟦 A. 2.99 × 10⁻²³ g
🟩 B. 1.99 × 10⁻²³ g
🟥 C. 3.99 × 10⁻²³ g
🟨 D. 2.99 × 10⁻²⁴ g

---

97. The mass of 1 dm³ of nitrogen gas at STP is:

🟦 A. 1.25
🟩 B. 1.43
🟥 C. 0.089
🟨 D. None of them

---

98. 11200 ml of argon gas at STP weighs:

🟦 A. 40
🟩 B. 20
🟥 C. 30
🟨 D. 10

---

99. 840 g of baking soda constitutes its … gram formula:

🟦 A. 5
🟩 B. Half
🟥 C. 10
🟨 D. 2

---

100. The mass of 1 atom of uranium is:

🟦 A. 3.95 × 10⁻²³ g
🟩 B. 3.95 × 10⁻²⁴ g
🟥 C. 3.95 × 10⁻²² g
🟨 D. None of them

📝 Answer Key:

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1. 🟥 C. 12.04 x 10²³

2. 🟦 A. 11.2 dm³

3. 🟨 D. 72 x 10²³

4. 🟦 A. 16g S

5. 🟩 B. 3.01 × 10²⁴

6. 🟩 B. 100g CH₄

7. 🟩 B. The mass of 1 mole of Cl₂ gas is 35.5g

8. 🟥 C. Its value changes if temperature increases

9. 🟩 B. Less than theoretical yield

10. 🟦 A. 1 dm³ of methane gas at STP

11. 🟨 D. SO₃

12. 🟩 B. 32

13. 🟨 D. 50g

14. 🟩 B. Less

15. 🟥 C. m³

16. 🟨 D. Helium

17. 🟩 B. 0.0167

18. 🟥 C. 8

19. 🟦 A. 10⁻³ dm³

20. 🟨 D. 32

21. 🟦 A. Ca²⁺, K

22. 🟥 C. MgO

23. 🟨 D. Bi₂O₃

24. 🟨 D. Both B and C

25. 🟥 C. 12 atoms

26. 🟩 B. 2 × 10⁻¹⁰ m

27. 🟥 B. B

28. 🟩 B. 10⁰

29. 🟩 B. 6

30. 🟥 C. 8

31. 🟥 C. 5

32. 🟦 A. 1 m³ = 102 dm³

33. 🟨 D. 0

34. 🟦 A. 8.7

35. 🟩 B. Precision

36. 🟥 C. 9.1096 × 10⁻³¹

37. 🟥 C. 29

38. 🟩 C. Exa

39. 🟥 C. Sulphur(VI) oxide

40. 🟨 D. 1 kg = 10² g

41. 🟨 D. F

42. 🟥 C. Avogadro’s number

43. 🟩 B. 2.38

44. 🟩 B. 4

45. 🟥 C. Adding 3 zeros at the end

46. 🟥 C. Simply adding two zeros at the end

47. 🟩 B. Both have 4 digits

48. 🟨 D. Both A and C

49. 🟥 C. Adding 3 zeros at the end

50. 🟦 A. Have same number of protons

51. 🟨 D. Lesser is the precision

52. 🟦 A. Both have 3 significant figures

53. 🟩 B. Smaller is the uncertainty

54. 🟨 D. All of them

55. 🟨 D. Both A and C are true

56. 🟩 B. Placing two zeros at the end

57. 🟥 C. On left side of non-zero digit while number is < 1

58. 🟨 D. All of the above

59. 🟩 B. 1.2300 has 5 significant figures

60. 🟥 C. 1.0050 → 5 significant figures

61. 🟩 B. 0.55 mg

62. 🟦 A. 3.6 g of water

63. 🟥 C. ½ z

64. 🟨 D. 1.67 × 10‒27 kg

65. 🟦 A. 0.012 kg

66. 🟦 A. 30

67. 🟥 C. 254 g

68. 🟩 B. 2.24 dm³

69. 🟩 B. 35.5 g

70. 🟥 C. 18.06 × 10²³

71. 🟥 C. 108

72. 🟨 D. 16 g of CH₄

73. 🟩 B. 1.6 × 10⁻²⁴ g

74. 🟨 D. 9.63 × 10²³

75. 🟩 B. Molecules

76. 🟨 D. 6 × 10²³

77. 🟩 C. Mole

78. 🟥 C. 6.02 x 10²²

79. 🟨 D. 55.55 NA

80. 🟦 A. 1.97 × 10⁻⁶

81. 🟦 A. O₂ = NH₃ = CO₂

82. 🟩 C. 270 g of Al

83. 🟩 C. 24.08 × 10²³

84. 🟩 B. 2.69 × 10¹¹

85. 🟩 B. 11.2 dm³

86. 🟥 C. 35.5 g

87. 🟦 A. 1.667 × 10⁻²⁴ g

88.  🟥 C. 10 g H₂ & 80 g CH₄

89. 🟨 D. 0.53 × 10²⁰ atoms

90. 🟦 A. Twice that in 60 g of carbon

91. 🟩 B. 44.6

92. 🟦 A. 12 g magnesium

93.  🟩 B. 24.088 × 10²³ atoms

94. 🟩 C. N₂ and CO

95.  🟥 C. 10

96.  🟩 B. 1.99 × 10⁻²³ g

97. 🟦 A. 1.25

98.  🟩 B. 20

99.  🟥 C. 10

100.  🟥 C. 3.95 × 10⁻²² g

---

📝 Explanations:

---

1. 🟥 C. 12.04 x 10²³
   Explanation:
   No. of molecules = 
   $\frac{V_{\mathrm{g/}}}{V_{\mathrm{M }}}\times N_{\mathrm{A }}= \frac{\mathrm{44.8/}}{\mathrm{22.4 }}\times 6.02\times {10}^{\mathrm{23 }}=12.04\times {10}^{23}$

---

2. 🟦 A. 11.2 dm³
   Explanation:
   

---

3. 🟨 D. 72 x 10²³
   Explanation:
   1 mole of sugar contains 12 moles of carbon.
   No. of carbon atoms = 
   $\mathrm{12 }\times N_{\mathrm{A }}= \mathrm{72 }\times {10}^{23}$

---

4. 🟦 A. 16g S
   Explanation:
   
   
   

---

5. 🟩 B. 3.01 × 10²⁴
   Explanation:
   Number of atoms = 
   $\frac{60}{12}\times N_A=5N_A=5\times 6.02\times {10}^{23}=3.01\times {10}^{24}$

---

6. 🟩 B. 100g CH₄
   Explanation:
   CH₄ has the least molar mass of 16. For the same mass, CH₄ contains the most molecules.
   Moles in 100g CH₄ = 100/16 = 6.25
   

---

7. 🟩 B. The mass of 1 mole of Cl₂ gas is 35.5g
   Explanation:
   Chlorine gas (Cl₂) is diatomic and its molar mass is 71g/mol, not 35.5g.

---

8. 🟥 C. Its value changes if temperature increases
   Explanation:
   Avogadro’s number is independent of temperature and pressure. It depends only on the number of moles.

---

9. 🟩 B. Less than theoretical yield
   Explanation:
   Actual yield is always less than the theoretical yield due to losses during reactions, side reactions, or incomplete reactions.

---

10. 🟦 A. 1 dm³ of methane gas at STP
    Explanation:
    According to Avogadro's law, volume is directly proportional to the number of moles. Therefore, 1 dm³ of methane has the least number of moles compared to larger volumes. 

11. 🟨 D. SO₃
    Explanation:
    Anhydrides are oxides of non-metals. Basic anhydrides (like CaO) are oxides of metals, while acid anhydrides (like SO₃) are oxides of non-metals. SO₃ dissolves in water to form H₂SO₄, making it an acid anhydride.

---

12. 🟩 B. 32
    Explanation:
    The number of neutrons is calculated by subtracting the atomic number (Z) from the mass number (A).
    $\text{No of neutrons}=A-Z=59-27=32$

---

13. 🟨 D. 50g
    Explanation:
    The percentage concentration is given by:

$$\text{Mass of NaOH}=\frac{\text{Percent}\times \text{Total mass}}{100}=\frac{5\times 1000}{100}=50g$$

---

14. 🟩 B. Less
    Explanation:
    1 nm = 10⁻⁹ m, and 1 μm = 10⁻⁶ m. So, 1 nm is smaller than 1 μm.

---

15. 🟥 C. m³
    Explanation:
    1 m³ = 1000 dm³. So, cubic meter (m³) is larger than cubic decimeter (dm³).

    

---

16. 🟨 D. Helium
    Explanation:
    Helium (He) is a noble gas and is a monoatomic element. Other monoatomic gases include Ne, Ar, Kr, and Xe.

---

17. 🟩 B. 0.0167
    Explanation:
    1/60 = 0.016666... which rounds to 0.0167 when rounded to three significant figures.

---

18. 🟥 C. 8
    Explanation:
    0.00050890400 has 8 significant figures, as the leading zeros are not counted but the trailing zeros after the decimal point are significant.

    

    
    

---

19. 🟦 A. 10⁻³ dm³
    Explanation:
    1 dm³ = 1000 cm³, so 1 cm³ = 10⁻³ dm³.

---

20. 🟨 D. 32
    Explanation:
    In HNO₃, the number of protons is the sum of the atomic numbers of hydrogen (H), nitrogen (N), and oxygen (O).
    $\text{Protons}=1+7+(8\times 3)=32$

21. 🟦 A. Ca²⁺, K
    Explanation:
    Isoelectronic species have the same number of electrons. Ca²⁺ has 18 electrons (Z = 20 - 2), and K has 19 electrons (Z = 19), so they are not isoelectronic.

---

22. 🟥 C. MgO
    Explanation:
    MgO is a basic anhydride because it is a metal oxide that forms a basic solution when dissolved in water. MgO reacts with water to form Mg(OH)₂, a strong base.

---

23. 🟨 D. Bi₂O₃
    Explanation:
    Bi₂O₃ is a basic oxide because bismuth, being a metal, forms basic oxides. Other listed oxides are acidic.

---

24. 🟨 D. Both B and C
    Explanation:
    For the oxide MO, the oxidation state of M is +2. MS and MCl₂ both have M in the +2 oxidation state, making both formulas correct.

---

25. 🟥 C. 12 atoms
    Explanation:
    In KAl(SO₄)₂:
    1 K + 1 Al + 2 S + (2 × 4) O = 1 + 1 + 2 + 8 = 12 atoms.

---

26. 🟩 B. 2 × 10⁻¹⁰ m
    Explanation:
    The diameter of an atom is generally in the order of 10⁻¹⁰ meters, which is about the size of an atomic radius.

---

27. 🟥 B. 6
    Explanation:
    0.00510840 mg has 6 significant figures. Leading zeros do not count, but all non-zero digits and zeros between significant digits are counted.       
    
    
    

---

28. 🟩 B. 10⁰
    Explanation:
    1 in exponential notation is written as 10⁰, because any number to the power of zero equals 1.

---

29. 🟩 B. 6
    Explanation:
    0.00678400 mg has 6 significant figures. The trailing zeros after the decimal point are considered significant.

---

30. 🟥 C. 8
    Explanation:
    0.00020089400 mm has 8 significant figures. The zeros at the end after the decimal are significant.

31. 🟥 C. 5
    Explanation:
    0.00755 mg has 5 significant figures. The leading zeros do not count, but all non-zero digits and zeros between significant digits are considered significant.

---

32. 🟦 A. 1 m³ = 102 dm³
    Explanation:
    This is incorrect. 1 m³ = 10³ dm³, not 102 dm³. 1 cubic meter is equal to 1000 cubic decimeters.

---

33. 🟨 D. 0
    Explanation:
    Zeros may or may not be significant, depending on their position. Zeros at the beginning of a number are not significant, but zeros at the end after a decimal point are significant.

---

34. 🟦 A. 8.7
    Explanation:
    Rounding 8.687 to two significant figures gives 8.7. The digits 8 and 6 are the significant figures, and the last digit is dropped.

---

35. 🟩 B. Precision
    Explanation:
    Significant figures indicate the precision of a measurement, i.e., how accurately a value has been measured.

---

36. 🟥 C. 9.1096 × 10⁻³¹
    Explanation:
    The number 0.00000000000000000000000000000000091096 is written in scientific notation as 9.1096 × 10⁻³¹.
    
    
    

---

37. 🟥 C. 29
    Explanation:
    28.987 rounded to two significant figures gives 29. The digit 8 is rounded up, and the 7 becomes 9.

---

38. 🟩 C. Exa
    Explanation:
    The prefix 10¹⁸ is called "exa" (E) in the metric system, representing a factor of 10¹⁸.
    

---

39. 🟥 C. Sulphur(VI) oxide
    Explanation:
    Sulphur(VI) oxide (SO₃) is the anhydride of sulfuric acid. It dissolves in water to form sulfuric acid (H₂SO₄).

---

40. 🟨 D. 1 kg = 10² g
    Explanation:
    This is incorrect. 1 kg = 10³ g, not 10² g. Kilograms and grams differ by a factor of 1000.

41. 🟨 D. F
    Explanation:
    An atom with 9 protons has an atomic number of 9, which corresponds to fluorine (F).

---

42. 🟥 C. Avogadro’s number
    Explanation:
    Avogadro's number (6.022 × 10²³) is a defined constant, which means it has infinite significant figures by definition.

---

43. 🟩 B. 2.38
    Explanation:
    Rounding 2.3746 to 3 significant figures gives 2.38.

---

44. 🟩 B. 4
    Explanation:
    6.022 × 10²³ has 4 significant figures. The scientific notation does not affect the number of significant figures.

---

45. 🟥 C. Adding 3 zeros at the end
    Explanation:
    4.578 has 4 significant figures. To make it 7 significant figures, add 3 zeros to the right of the decimal point.

---

46. 🟥 C. Simply adding two zeros at the end
    Explanation:
    13.89 has 4 significant figures. To make it 6 significant figures, add two zeros at the end (13.8900).

---

47. 🟩 B. Both have 4 digits
    Explanation:
    Both 0.6850 and 6850 have 4 digits, even though the position of the decimal point differs.

---

48. 🟨 D. Both A and C
    Explanation:
    The number of significant figures is inversely related to uncertainty. Fewer significant figures mean more uncertainty, which also means less certainty.

---

49. 🟥 C. Adding 3 zeros at the end
    Explanation:
    3.69 has 3 significant figures. To make it 6 significant figures, add 3 zeros to the right of the decimal point (3.690000).

---

50. 🟦 A. Have same number of protons
    Explanation:
    A chlorine atom and chloride ion both have the same number of protons (9), but chloride ions have gained one electron, while atoms of chlorine have no charge.

51. 🟨 D. Lesser is the precision
    Explanation:
    Greater significant figures imply greater precision. Therefore, "Lesser is the precision" is false.

---

52. 🟦 A. Both have 3 significant figures
    Explanation:
    0.569 and 569 both have 3 significant figures. The positioning of the decimal point doesn't affect the significant figures.

---

53. 🟩 B. Smaller is the uncertainty
    Explanation:
    The greater the number of significant figures, the smaller the uncertainty in the measurement.

---

54. 🟨 D. All of them
    Explanation:
    With more significant figures, you have greater precision, greater certainty, and smaller uncertainty.

---

55. 🟨 D. Both A and C are true
    Explanation:
    Zeros at the end of a decimal number and between non-zero digits are considered significant.

---

56. 🟩 B. Placing two zeros at the end
    Explanation:
    6.55 with 3 significant figures can be changed to 5 significant figures by adding two zeros at the end (6.5500).

---

57. 🟥 C. On left side of non-zero digit while number is < 1
    Explanation:
    Leading zeros (zeros before the first non-zero digit in a number less than 1) are not significant.

---

58. 🟨 D. All of the above
    Explanation:
    All of the listed scenarios involve significant figures: non-zero digits, trailing zeros in decimals, and zeros between significant digits.

---

59. 🟩 B. 1.2300 has 5 significant figures
    Explanation:
    1.2300 has 5 significant figures because the trailing zeros after the decimal point are significant.

---

60. 🟥 C. 1.0050 → 5 significant figures
    Explanation:
    1.0050 has 5 significant figures because the zeros between the decimal point and non-zero digits are significant.

61. 🟩 B. 0.55 mg
    Explanation:
    Mass of 1 electron = 
    $9.1 \times 10^{-31} \, \text{kg}$
    Mass of 1 mole of electrons = 
    $(9.1 \times 10^{-25} \, \text{mg}) \times (6.02 \times 10^{23}) = 0.55 \, \text{mg}$

---

62. 🟦 A. 3.6 g of water
    Explanation:
    The largest number of molecules corresponds to the largest number of moles, and water (H₂O) has the greatest number of molecules for the given mass.
    
    
    

---

63. 🟥 C. ½ z
    Explanation:
    28 g of N₂ = 1 mole = z molecules.
    16 g of O₂ = 0.5 moles = ½ z molecules.

---

64. 🟨 D. 1.67×10⁻²⁷ kg
    Explanation:
    Mass of a hydrogen atom is 
    $1.67 \times 10^{-27} \, \text{kg}$
    
    
    

---

65. 🟦 A. 0.012 kg
    Explanation:
    1 mole of carbon-12 weighs 12 grams, which equals 0.012 kg (since 1000 grams = 1 kilogram).

---

66. 🟦 A. 30
    Explanation:
    The molar mass of formaldehyde (HCHO) = 
    $2(1) + 12 + 16 = 30 \, \text{g/mol}$

---

67. 🟥 C. 254 g
    Explanation:
    The molar mass of iodine (I₂) is 
    $127\times 2=254\text{g}$

---

68. 🟩 B. 2.24 dm³
    Explanation:
    1 mole of any gas at STP occupies 22.4 dm³.
    0.1 mole of any gas at STP occupies 
    $22.4\times 0.1=2.24{\text{dm}}^3$

---

69. 🟩 B. 35.5 g
    Explanation:
    The molar mass of chlorine gas (Cl₂) is 71 g.
    0.5 moles of Cl₂ weighs 
    $0.5\times 71=35.5\text{g}$

---

70. 🟥 C. 18.06 × 10²³
    Explanation:
    1 mole of CaCl₂ contains 3 moles of ions: 1 Ca²⁺ and 2 Cl⁻.
    Thus, the total number of ions in 1 mole of CaCl₂ is 
    $3\times 6.02\times {10}^{23}=18.06\times {10}^{23}$

71. 🟥 C. 108
    Explanation:
    To find the atomic mass, use the formula:
    Atomic mass = mass of atom × Avogadro’s number.
    Thus, the atomic mass = 
    
    
    

---

72. 🟨 D. 16 g of CH₄
    Explanation:
    The number of atoms is proportional to the number of moles and the atomicity of the compound. Since CH₄ has the greatest atomicity (4 atoms per molecule), it contains the most atoms.
    
    
    

---

73. 🟩 B. 1.6 × 10⁻²⁴ g
    Explanation:
    Mass of 1 hydrogen atom = 
    $\frac{1}{6.02 \times 10^{23}}$

---

74. 🟨 D. 9.63 × 10²³
    Explanation:
    Number of atoms in 0.2 moles of S₈ = 
    $0.2\times 8\times 6.02\times {10}^{23}=9.63\times {10}^{23}$

---

75. 🟩 B. Molecules
    Explanation:
    Covalent compounds consist of molecules. Therefore, one mole of any covalent compound contains the same number of molecules, which is 6.02 × 10²³.

---

76. 🟨 D. 6 × 10²³
    Explanation:
    For NH₃:
    Moles of NH₃ = 
    $\frac{4.25}{17} = 0.25$
    Number of atoms = 
    $0.25 \times 4 \times 6.02 \times 10^{23} = 6.02 \times 10^{23}$

---

77. 🟩 C. Mole
    Explanation:
    A mole is the quantity containing Avogadro’s number of particles (atoms, molecules, ions).

---

78. 🟥 C. 6.02 x 10²²
    Explanation:
    Molar mass of CaCl₂ = 111 g/mol.
    Total formula units in 11.1 g of CaCl₂ = 
    $\frac{11.1}{111}\times 6.02\times {10}^{23}=6.02\times {10}^{22}$

---

79. 🟨 D. 55.55 NA
    Explanation:
    Molar mass of water = 18 g/mol.
    The number of water molecules in 1 L = 
    $\frac{1000}{18}\times 6.02\times {10}^{23}=55.55\times 6.02\times {10}^{23}=55.55NA$

---

80. 🟦 A. 1.97 × 10⁻⁶
    Explanation:
    The mass of one mole of electrons = 
    $9.1\times {10}^{-31}\times 6.02\times {10}^{23}=1.97\times {10}^{-6}\text{kg}$

81. 🟦 A. O₂ = NH₃ = CO₂
Explanation:
According to Avogadro's law, at the same temperature and pressure conditions, equal volumes of all gases contain the same number of molecules. Since the volume is the same (100 mL), the number of molecules of O₂, NH₃, and CO₂ will be the same.

---

82. 🟩 C. 270 g of Al
Explanation:
To find the largest number of atoms, we need to calculate the number of moles for each substance:

• 24 g of C = 2 moles

• 5.6 g of Fe = 0.1 mole

• 270 g of Al = 10 moles

• 1.08 g of Ag = 0.01 mole
  Therefore, Al has the greatest number of atoms.

---

83. 🟩 C. 24.08 × 10²³
Explanation:
For 2 moles of hydrogen gas (H₂), the number of atoms = atomicity × moles × Avogadro’s number = 2 × 2 × $6.02 \times 10^{23}$ = 24.08 × 10²³ atoms.

---

84. 🟩 B. 2.69 × 10¹¹
Explanation:
Use the ideal gas law to calculate the number of molecules in 1L of oxygen gas at the given pressure and temperature, yielding approximately 2.69 × 10¹¹ molecules.

---

85. 🟩 B. 11.2 dm³
Explanation:
At STP, 32 g of O₂ equals 22.4 dm³. Therefore, 16 g of O₂ = 11.2 dm³.

---

86. 🟥 C. 35.5 g
Explanation:
At STP, 22.4 dm³ of Cl₂ weighs 71 g. Therefore, 11.2 dm³ weighs 35.5 g.

---

87. 🟦 A. 1.667 × 10⁻²⁴ g
Explanation:
1 atomic mass unit (amu) is defined as 1.667 × 10⁻²⁴ g, which is the mass of one hydrogen atom.

---

88. 🟥 C. 10 g H₂ & 80 g CH₄
Explanation:
For equal moles (5 moles), you will have the same number of molecules:

10 g H₂ = 5 mol

80 g CH₄ = 5 mol

Thus, they contain the same number of molecules.

---

89. 🟨 D. 0.53 × 10²⁰ atoms
Explanation:
At STP, one cm³ of oxygen gas contains approximately 0.53 × 10²⁰ atoms. This is derived from the molar volume of gases at STP and the atomicity of O₂.

---

90. 🟦 A. Twice that in 60 g of carbon
Explanation:
The number of atoms in 558.5 g of Fe = $\frac{558.5}{55.85} \times 6.023 \times 10^{23}$
This gives twice the number of atoms as in 60 g of carbon (which is 1 mole of carbon, or $6.023 \times 10^{23}$ atoms).

91. 🟩 B. 44.6
Explanation:
1 m³ = 1000 L
At STP, 22.4 L = 1 mole, so:
1000 L = $\frac{1000}{22.4} = 44.6$

---

92. 🟦 A. 12 g magnesium
Explanation:
20 g of calcium = 0.5 moles (since atomic mass of calcium = 40).
12 g of magnesium (atomic mass 24) also constitutes 0.5 moles.
Thus, both contain the same number of atoms.

---

93. 🟩 B. 24.088 × 10²³ atoms
Explanation:
1 mole of P₄ molecules contains 6.022 × 10²³ molecules.
Each molecule contains 4 phosphorus atoms, so 1 mole of P₄ molecules contains $4 \times 6.022 \times 10²³ = 24.088 \times 10²³$ atoms of phosphorus.

---

94. 🟩 C. N₂ and CO
Explanation:
For 1 g of each to have the same number of molecules, their molar masses must be the same.
The molar mass of N₂ (28 g/mol) is the same as CO (28 g/mol), so 1 g of each will contain the same number of molecules.

---

95. 🟥 C. 10
Explanation:
1 mole (or gram formula) of NaOH (caustic soda) weighs 40 g.
Therefore, 400 g of NaOH contains 10 moles, i.e., 10 gram formulas.

---

96. 🟩 B. 1.99 × 10⁻²³ g
Explanation:
Mass of 1 atom of diamond = $\frac{\text{Atomic mass of carbon}}{\text{Avogadro's number}} = \frac{12}{6.022 \times 10^{23}} = 1.99 \times 10^{-23}$

---

97. 🟦 A. 1.25
Explanation:
For 1 dm³ of nitrogen gas at STP, the molar volume is 22.4 L (or 0.0224 m³).
Thus, the mass = $\frac{1 \text{ dm}^3}{22.4} \times 28$ g = 1.25 g.

---

98. 🟩 B. 20
Explanation:
At STP, molar volume of argon = 22.4 L (22400 mL).
Thus, 11200 mL = half the molar volume, which is 20 g.

---

99. 🟥 C. 10
Explanation:
1 mole of baking soda (NaHCO₃) weighs 84 g.
Thus, 840 g = 10 moles or 10 gram formulas.

---

100. 🟥 C. 3.95 × 10⁻²² g
Explanation:
Mass of 1 atom of uranium = $\frac{238}{6.022 \times 10^{23}} = 3.95 \times 10^{-22}$

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