AP Chemistry Quiz: Moles And Molar Mass
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Moles And Molar MassQuestion 1 of 20

A sample of pure aluminum, Al\mathrm{Al}, contains 3.01×10233.01\times 10^{23} atoms. How many moles of Al\mathrm{Al} are in the sample? (Avogadro's number: 6.022×1023 mol−16.022\times 10^{23}\ \mathrm{mol^{-1}}.)

0.250 mol0.250\ \mathrm{mol}
2.00 mol2.00\ \mathrm{mol}
6.02×1023 mol6.02\times 10^{23}\ \mathrm{mol}
0.500 mol0.500\ \mathrm{mol}
1.00 mol1.00\ \mathrm{mol}
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AP Chemistry Quiz

AP Chemistry Quiz: Moles And Molar Mass

Practice Moles And Molar Mass in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Moles And Molar Mass, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A sample of pure aluminum, Al\mathrm{Al}, contains 3.01×10233.01\times 10^{23} atoms. How many moles of Al\mathrm{Al} are in the sample? (Avogadro's number: 6.022×1023 mol−16.022\times 10^{23}\ \mathrm{mol^{-1}}.)

  1. 0.250 mol0.250\ \mathrm{mol}
  2. 2.00 mol2.00\ \mathrm{mol}
  3. 6.02×1023 mol6.02\times 10^{23}\ \mathrm{mol}
  4. 0.500 mol0.500\ \mathrm{mol} (correct answer)
  5. 1.00 mol1.00\ \mathrm{mol}

Explanation: This question tests the skill of moles and molar mass. Molar mass relates mass to moles, but this problem involves converting from particles to moles using Avogadro's number. Avogadro's number bridges moles to the number of atoms or molecules, allowing us to find moles by dividing the number of particles by 6.022 × 10^23 per mole. For elemental aluminum, this calculation reveals the amount in moles from a count of atoms. A common distractor is choice C, 6.02 × 10^23 mol, which comes from mistakenly multiplying instead of dividing by Avogadro's number, inverting the conversion factor. Ensure you divide when going from particles to moles. A transferable strategy is to write the unit you want (moles) and use Avogadro's number as the bridge by dividing particles by particles per mole.

Question 2

A sample of pure potassium bromide, KBr\mathrm{KBr}, has a mass of 11.9 g11.9\ \mathrm{g}. (Atomic masses: K=39.1\mathrm{K}=39.1, Br=79.9\mathrm{Br}=79.9.) How many moles of KBr\mathrm{KBr} are in the sample?

  1. 11.9 mol11.9\ \mathrm{mol}
  2. 0.0500 mol0.0500\ \mathrm{mol}
  3. 0.200 mol0.200\ \mathrm{mol}
  4. 1.00 mol1.00\ \mathrm{mol}
  5. 0.100 mol0.100\ \mathrm{mol} (correct answer)

Explanation: This question tests the skill of moles and molar mass. Molar mass links mass to amount through moles = mass / molar mass, with KBr's molar mass as 39.1 + 79.9 = 119 g/mol. This allows direct calculation from given mass. Avogadro's number, 6.02 × 10²³, links moles to particles for further conversions if needed. A tempting distractor is choice A, 0.0500 mol, from a common unit-conversion error of using half the molar mass, perhaps treating Br as 40 instead of 79.9. A transferable strategy is to write the unit you want and use molar mass as the bridge.

Question 3

A 23.0 g23.0\ \mathrm{g} sample of pure ethanol, C2H6O\mathrm{C_2H_6O}, is used. (Atomic masses: C=12.0\mathrm{C}=12.0, H=1.0\mathrm{H}=1.0, O=16.0\mathrm{O}=16.0.) How many moles of ethanol are present?

  1. 0.250 mol0.250\ \mathrm{mol}
  2. 0.500 mol0.500\ \mathrm{mol} (correct answer)
  3. 1.00 mol1.00\ \mathrm{mol}
  4. 2.00 mol2.00\ \mathrm{mol}
  5. 46.0 mol46.0\ \mathrm{mol}

Explanation: This question tests the skill of moles and molar mass. Molar mass links mass to amount as moles = mass / molar mass, with ethanol at 46.0 g/mol giving 0.500 mol. This conversion is fundamental. Avogadro's number, 6.02 × 10²³, links moles to particles. A tempting distractor is choice A, 0.250 mol, from a common unit-conversion error of halving the mass. A transferable strategy is to write the unit you want and use molar mass as the bridge.

Question 4

A student measures out 44.0 g44.0\ \text{g} of pure carbon dioxide, CO2\text{CO}_2. (Molar mass of CO2=44.0 g/mol\text{CO}_2=44.0\ \text{g/mol}.) How many moles of CO2\text{CO}_2 are present?

  1. 2.00 mol2.00\ \text{mol}
  2. 0.0227 mol0.0227\ \text{mol}
  3. 44.0 mol44.0\ \text{mol}
  4. 1.00 mol1.00\ \text{mol} (correct answer)
  5. 88.0 mol88.0\ \text{mol}

Explanation: This question tests the skill of moles and molar mass. Molar mass provides the conversion factor between mass and moles through the relationship: moles = mass ÷ molar mass. With 44.0 g of CO₂ and a molar mass of 44.0 g/mol, we calculate: 44.0 g ÷ 44.0 g/mol = 1.00 mol. Avogadro's number would be necessary to find the number of molecules, but here we only need moles. A tempting wrong answer (choice A, 0.0227 mol) results from incorrectly inverting the calculation and dividing 1 by 44.0 instead of dividing 44.0 by 44.0. To solve mole conversions accurately, set up the calculation so that grams cancel out, leaving moles as your final unit.

Question 5

A sample contains 0.500 mol0.500\ \text{mol} of pure magnesium metal, Mg\text{Mg}. How many magnesium atoms are in the sample? (Use NA=6.02×1023 mol−1N_A = 6.02\times10^{23}\ \text{mol}^{-1}.)

  1. 1.20×1023 atoms1.20\times10^{23}\ \text{atoms}
  2. 3.01×1023 atoms3.01\times10^{23}\ \text{atoms} (correct answer)
  3. 6.02×1023 atoms6.02\times10^{23}\ \text{atoms}
  4. 0.500 atoms0.500\ \text{atoms}
  5. 1.00×1024 atoms1.00\times10^{24}\ \text{atoms}

Explanation: This question tests the skill of moles and molar mass. While molar mass links mass to moles, Avogadro's number (6.02×10²³ particles/mol) connects moles to the number of particles (atoms, molecules, or formula units). To find atoms from moles, we multiply: 0.500 mol × 6.02×10²³ atoms/mol = 3.01×10²³ atoms. The molar mass of Mg would only be needed if we were converting between mass and moles, but here we're converting between moles and atoms. A common error (choice C, 6.02×10²³ atoms) results from forgetting to multiply by the number of moles and just using Avogadro's number directly. To convert from moles to particles, always multiply the number of moles by Avogadro's number to get the total particle count.

Question 6

A 18.0 g18.0\ \text{g} sample of pure glucose, C6H12O6\text{C}_6\text{H}_{12}\text{O}_6, is placed in a container. How many moles of glucose are present? (Molar masses: C=12.0 g/mol\text{C}=12.0\ \text{g/mol}, H=1.0 g/mol\text{H}=1.0\ \text{g/mol}, O=16.0 g/mol\text{O}=16.0\ \text{g/mol}.)

  1. 0.300 mol0.300\ \text{mol}
  2. 0.100 mol0.100\ \text{mol} (correct answer)
  3. 0.600 mol0.600\ \text{mol}
  4. 1.00 mol1.00\ \text{mol}
  5. 180 mol180\ \text{mol}

Explanation: This problem tests moles and molar mass by converting mass to moles for a complex molecule. The molar mass of glucose (C₆H₁₂O₆) equals 6 carbon atoms (6 × 12.0 = 72.0 g/mol) plus 12 hydrogen atoms (12 × 1.0 = 12.0 g/mol) plus 6 oxygen atoms (6 × 16.0 = 96.0 g/mol), totaling 180.0 g/mol. To find moles: 18.0 g ÷ 180.0 g/mol = 0.100 mol. A common mistake would be to use 18.0 as the molar mass directly, giving 1.00 mol. The key strategy is to systematically calculate molar mass by counting each type of atom and multiplying by its atomic mass before adding them all together.

Question 7

A student has a 18.0 g18.0\text{ g} sample of pure water, H2O\text{H}_2\text{O}. What amount of H2O\text{H}_2\text{O}, in moles, is present? (Molar mass of H2O=18.0 g/mol\text{H}_2\text{O}=18.0\text{ g/mol}.)​​

  1. 0.500 mol0.500\text{ mol}
  2. 18.0 mol18.0\text{ mol}
  3. 36.0 mol36.0\text{ mol}
  4. 1.00 mol1.00\text{ mol} (correct answer)
  5. 0.0556 mol0.0556\text{ mol}

Explanation: This problem tests the skill of moles and molar mass. Molar mass serves as the conversion factor between mass (in grams) and amount (in moles), allowing us to determine how many moles are in a given mass of substance. To convert from grams to moles, we divide the mass by the molar mass: moles = mass ÷ molar mass. For this water sample: moles = 18.0 g ÷ 18.0 g/mol = 1.00 mol. A common error would be multiplying instead of dividing (18.0 × 18.0 = 324), which doesn't appear here but would give an incorrect large value. When converting mass to moles, always divide by molar mass—think of it as 'how many molar mass units fit into your sample mass.'

Question 8

A student has 0.500 mol0.500\ \mathrm{mol} of pure ammonia, NH3(s)\mathrm{NH_3(s)}. The molar mass of NH3\mathrm{NH_3} is 17.0 g mol−117.0\ \mathrm{g\ mol^{-1}}. What is the mass of the ammonia sample?

  1. 34.0 g34.0\ \mathrm{g}
  2. 8.50 g8.50\ \mathrm{g} (correct answer)
  3. 17.0 g17.0\ \mathrm{g}
  4. 0.0294 g0.0294\ \mathrm{g}
  5. 0.500 g0.500\ \mathrm{g}

Explanation: This problem requires using moles and molar mass to convert from moles to mass. The molar mass of NH₃ is 17.0 g/mol, which tells us the mass of one mole. To find the mass of 0.500 mol, we multiply: 0.500 mol × 17.0 g/mol = 8.50 g. A student might mistakenly divide the molar mass by moles (17.0 ÷ 0.500 = 34.0), which gives twice the molar mass rather than half. The reliable method is dimensional analysis: start with moles, multiply by molar mass (g/mol), and verify that mol cancels to leave grams.

Question 9

A student measures out 0.100 mol0.100\text{ mol} of pure magnesium chloride, MgCl2\text{MgCl}_2. How many formula units of MgCl2\text{MgCl}_2 does this sample contain? (Use NA=6.02×1023 mol−1N_A=6.02\times10^{23}\text{ mol}^{-1}.)​​

  1. 6.02×10226.02\times10^{22} formula units (correct answer)
  2. 6.02×10236.02\times10^{23} formula units
  3. 3.01×10233.01\times10^{23} formula units
  4. 1.00×10−11.00\times10^{-1} formula units
  5. 1.20×10241.20\times10^{24} formula units

Explanation: This problem involves moles and molar mass concepts. Avogadro's number bridges between moles and particles, with one mole containing 6.02×10²³ formula units regardless of the substance. To find the number of formula units from moles, we multiply: formula units = moles × NA. For this MgCl₂ sample: formula units = 0.100 mol × 6.02×10²³ mol⁻¹ = 6.02×10²² formula units. A tempting error would be forgetting to account for the 0.100 coefficient and selecting 6.02×10²³ (choice B), which represents one full mole. When converting moles to particles, multiply by Avogadro's number and carefully track your decimal places or powers of ten.

Question 10

A sample contains 0.100 mol0.100\ \mathrm{mol} of pure aluminum metal, Al(s)\mathrm{Al(s)}. The molar mass of aluminum is 27.0 g mol−127.0\ \mathrm{g\ mol^{-1}}. What is the mass of the aluminum sample?

  1. 270 g270\ \mathrm{g}
  2. 2.70 g2.70\ \mathrm{g} (correct answer)
  3. 0.00370 g0.00370\ \mathrm{g}
  4. 27.0 g27.0\ \mathrm{g}
  5. 0.100 g0.100\ \mathrm{g}

Explanation: This problem tests converting moles to mass using moles and molar mass. The molar mass of aluminum (27.0 g/mol) tells us that one mole of aluminum has a mass of 27.0 grams. To find the mass of 0.100 mol, we multiply: 0.100 mol × 27.0 g/mol = 2.70 g. A typical mistake would be dividing (0.100 ÷ 27.0 = 0.00370), which gives the wrong value and units (mol²/g instead of g). Always check your units: when multiplying moles by g/mol, the mol units cancel, leaving grams as desired.

Question 11

A sample contains 0.250 mol0.250\text{ mol} of pure calcium carbonate, CaCO3\text{CaCO}_3. What is the mass of the sample? (Molar mass of CaCO3=100.0 g/mol\text{CaCO}_3=100.0\text{ g/mol}.)​​

  1. 0.250 g0.250\text{ g}
  2. 25.0 g25.0\text{ g} (correct answer)
  3. 100.0 g100.0\text{ g}
  4. 40.0 g40.0\text{ g}
  5. 250 g250\text{ g}

Explanation: This problem requires understanding moles and molar mass. Molar mass connects the amount of substance in moles to its mass in grams, with one mole of any substance having a mass equal to its molar mass. To find mass from moles, we multiply: mass = moles × molar mass. For calcium carbonate: mass = 0.250 mol × 100.0 g/mol = 25.0 g. A tempting error would be dividing instead (100.0 ÷ 0.250 = 400), but this reverses the relationship and gives an incorrect answer. Remember: when going from moles to grams, multiply by molar mass—you're scaling up from the number of mole units to find total mass.

Question 12

A sample of pure sulfur, S\mathrm{S}, has a mass of 6.40 g6.40\ \mathrm{g}. Given that the molar mass of sulfur is 32.0 g mol−132.0\ \mathrm{g\ mol^{-1}}, how many moles of S\mathrm{S} are present?

  1. 0.200 mol0.200\ \mathrm{mol} (correct answer)
  2. 2.00 mol2.00\ \mathrm{mol}
  3. 0.0500 mol0.0500\ \mathrm{mol}
  4. 32.0 mol32.0\ \mathrm{mol}
  5. 6.40 mol6.40\ \mathrm{mol}

Explanation: This question tests the skill of moles and molar mass. Molar mass connects mass in grams to moles by dividing for elemental sulfur at 32.0 g/mol. This value is the atomic mass of sulfur. Avogadro's number links moles to atoms, but this focuses on mass-to-moles. A tempting distractor is choice E, 6.40 mol, from multiplying 6.40 by 32.0 instead of dividing, a unit-conversion mix-up. Always divide mass by molar mass. A transferable strategy is to write the unit you want (moles) and use molar mass as the bridge by dividing grams by grams per mole.

Question 13

A student has 0.125 mol0.125\ \mathrm{mol} of pure potassium bromide, KBr\mathrm{KBr}. Given that the molar mass of KBr\mathrm{KBr} is 119.0 g mol−1119.0\ \mathrm{g\ mol^{-1}}, what is the mass of the sample?

  1. 14.9 g14.9\ \mathrm{g} (correct answer)
  2. 119 g119\ \mathrm{g}
  3. 0.125 g0.125\ \mathrm{g}
  4. 952 g952\ \mathrm{g}
  5. 1.49 g1.49\ \mathrm{g}

Explanation: This question tests the skill of moles and molar mass. Molar mass connects the amount in moles to the mass in grams by multiplying moles by grams per mole for compounds like KBr. The molar mass of 119.0 g/mol comes from potassium and bromine atomic masses, allowing mass calculations from moles. Avogadro's number links moles to particles, but here the focus is moles-to-mass. A common distractor is choice B, 119 g, resulting from using 1 mole instead of 0.125, forgetting to multiply by the given moles—a frequent oversight in unit conversion. Remember to incorporate the exact mole quantity. A transferable strategy is to write the unit you want (grams) and use molar mass as the bridge by multiplying moles by grams per mole.

Question 14

A sample of pure magnesium chloride, MgCl2\mathrm{MgCl_2}, has a mass of 9.50 g9.50\ \mathrm{g}. Given that the molar mass of MgCl2\mathrm{MgCl_2} is 95.0 g mol−195.0\ \mathrm{g\ mol^{-1}}, how many moles of MgCl2\mathrm{MgCl_2} are in the sample?

  1. 0.0100 mol0.0100\ \mathrm{mol}
  2. 0.100 mol0.100\ \mathrm{mol} (correct answer)
  3. 1.00 mol1.00\ \mathrm{mol}
  4. 9.50 mol9.50\ \mathrm{mol}
  5. 95.0 mol95.0\ \mathrm{mol}

Explanation: This question tests the skill of moles and molar mass. Molar mass serves as the conversion factor that links the mass of a substance in grams to the amount in moles, allowing us to calculate moles from mass by dividing the given mass by the molar mass. For compounds like MgCl2, the molar mass is the sum of the atomic masses of its constituent elements, providing a direct bridge between macroscopic measurements and microscopic quantities. Avogadro's number, 6.022 × 10^23 particles per mole, further connects the amount in moles to the number of individual particles, though this question focuses on mass-to-moles conversion. A tempting distractor might be choice A, 0.0100 mol, which results from incorrectly dividing 9.50 by 950 instead of 95.0, a common error in handling decimal places during unit conversion. Always double-check the placement of decimals when performing divisions with molar mass. To solve these problems reliably, write the unit you want (moles) and use molar mass as the bridge by dividing mass by grams per mole.

Question 15

A 18.0 g18.0\ \mathrm{g} sample of pure water, H2O\mathrm{H_2O}, is analyzed. Given that the molar mass of H2O\mathrm{H_2O} is 18.0 g mol−118.0\ \mathrm{g\ mol^{-1}}, how many moles of H2O\mathrm{H_2O} are present?

  1. 36.0 mol36.0\ \mathrm{mol}
  2. 0.0556 mol0.0556\ \mathrm{mol}
  3. 18.0 mol18.0\ \mathrm{mol}
  4. 1.00 mol1.00\ \mathrm{mol} (correct answer)
  5. 0.500 mol0.500\ \mathrm{mol}

Explanation: This question tests the skill of moles and molar mass. Molar mass links the mass in grams to the amount in moles, calculated by dividing mass by molar mass for substances like H2O. The molar mass of water is 18.0 g/mol, derived from hydrogen and oxygen atomic masses, enabling precise conversions. Avogadro's number connects moles to the number of molecules, though this question focuses on mass-to-moles. A tempting distractor is choice C, 18.0 mol, from incorrectly multiplying 18.0 by 18.0 instead of dividing, a unit-conversion error when mixing up operations. Always divide mass by molar mass for moles. A transferable strategy is to write the unit you want (moles) and use molar mass as the bridge by dividing grams by grams per mole.

Question 16

A sample contains 2.00 mol2.00\ \mathrm{mol} of pure carbon dioxide, CO2\mathrm{CO_2}. How many molecules of CO2\mathrm{CO_2} are in the sample? (Avogadro's number: 6.022×1023 mol−16.022\times 10^{23}\ \mathrm{mol^{-1}}.)

  1. 1.20×1023 molecules1.20\times 10^{23}\ \text{molecules}
  2. 1.20×1025 molecules1.20\times 10^{25}\ \text{molecules}
  3. 6.02×1023 molecules6.02\times 10^{23}\ \text{molecules}
  4. 3.01×1023 molecules3.01\times 10^{23}\ \text{molecules}
  5. 1.20×1024 molecules1.20\times 10^{24}\ \text{molecules} (correct answer)

Explanation: This question tests the skill of moles and molar mass. Molar mass relates mass to moles, but this involves converting moles to molecules using Avogadro's number. Avogadro's number bridges moles to particles by multiplying moles by 6.022 × 10^23 molecules per mole for CO2. This quantifies molecules from a mole amount. A tempting distractor is choice C, 6.02 × 10^23 molecules, from using 1 mole instead of 2.00, a common error in not scaling by the given moles. Always multiply by the specific mole value. A transferable strategy is to write the unit you want (molecules) and use Avogadro's number as the bridge by multiplying moles by molecules per mole.

Question 17

A 8.00 g8.00\ \text{g} sample of pure methane, CH4\text{CH}_4, is collected. How many molecules of CH4\text{CH}_4 are in the sample? (Molar masses: C=12.0 g/mol\text{C}=12.0\ \text{g/mol}, H=1.0 g/mol\text{H}=1.0\ \text{g/mol}; use NA=6.02×1023 mol−1N_A=6.02\times10^{23}\ \text{mol}^{-1}.)

  1. 3.01×1023 molecules3.01\times10^{23}\ \text{molecules} (correct answer)
  2. 6.02×1023 molecules6.02\times10^{23}\ \text{molecules}
  3. 1.20×1024 molecules1.20\times10^{24}\ \text{molecules}
  4. 0.500 molecules0.500\ \text{molecules}
  5. 8.00×1023 molecules8.00\times10^{23}\ \text{molecules}

Explanation: This question requires using moles and molar mass to convert mass to molecules through moles. First, find the molar mass of CH₄: one carbon (12.0 g/mol) plus four hydrogens (4 × 1.0 = 4.0 g/mol) equals 16.0 g/mol. Next, convert mass to moles: 8.00 g ÷ 16.0 g/mol = 0.500 mol. Finally, convert moles to molecules: 0.500 mol × 6.02×10²³ molecules/mol = 3.01×10²³ molecules. A common error would be to skip the moles step and try to go directly from grams to molecules. The strategy is to use a two-step conversion: grams → moles (divide by molar mass), then moles → molecules (multiply by Avogadro's number).

Question 18

A sample contains 3.01×10233.01\times10^{23} molecules of pure water, H2O\text{H}_2\text{O}. What amount of water (in moles) is present? (Use NA=6.02×1023 mol−1N_A=6.02\times10^{23}\ \text{mol}^{-1}.)

  1. 0.250 mol0.250\ \text{mol}
  2. 0.500 mol0.500\ \text{mol} (correct answer)
  3. 2.00 mol2.00\ \text{mol}
  4. 1.00 mol1.00\ \text{mol}
  5. 6.02×1023 mol6.02\times10^{23}\ \text{mol}

Explanation: This problem requires using moles and molar mass concepts, specifically converting molecules to moles using Avogadro's number. One mole contains 6.02×10²³ particles (molecules, atoms, or formula units). To find moles from molecules, we divide: 3.01×10²³ molecules ÷ 6.02×10²³ molecules/mol = 0.500 mol. A common error would be to multiply instead of divide, giving 1.81×10⁴⁷, which makes no physical sense. The strategy to remember is: when going from particles to moles, divide by Avogadro's number; when going from moles to particles, multiply.

Question 19

A sample contains 1.204×10241.204\times10^{24} formula units of pure sodium fluoride, NaF\text{NaF}. What amount of NaF\text{NaF} (in moles) is present? (Use NA=6.02×1023 mol−1N_A=6.02\times10^{23}\ \text{mol}^{-1}.)

  1. 1.00 mol1.00\ \text{mol}
  2. 0.500 mol0.500\ \text{mol}
  3. 2.00 mol2.00\ \text{mol} (correct answer)
  4. 6.02×1023 mol6.02\times10^{23}\ \text{mol}
  5. 3.01×1023 mol3.01\times10^{23}\ \text{mol}

Explanation: This problem tests moles and molar mass concepts by converting formula units to moles using Avogadro's number. One mole contains 6.02×10²³ formula units (the particle count for ionic compounds). To find moles: 1.204×10²⁴ formula units ÷ 6.02×10²³ formula units/mol = 2.00 mol. A common mistake would be to use 1.204×10²³ instead of 1.204×10²⁴, which would give 0.200 mol. The key strategy is to recognize that "formula units" is used for ionic compounds and works exactly like molecules in calculations with Avogadro's number.

Question 20

A student has a sample of pure calcium chloride, CaCl2\mathrm{CaCl_2}, with a mass of 111 g111\ \mathrm{g}. (Atomic masses: Ca=40.1\mathrm{Ca}=40.1, Cl=35.5\mathrm{Cl}=35.5.) How many moles of CaCl2\mathrm{CaCl_2} are in the sample?

  1. 1.00 mol1.00\ \mathrm{mol} (correct answer)
  2. 111 mol111\ \mathrm{mol}
  3. 0.500 mol0.500\ \mathrm{mol}
  4. 3.00 mol3.00\ \mathrm{mol}
  5. 2.00 mol2.00\ \mathrm{mol}

Explanation: This question tests the skill of moles and molar mass. Molar mass is the mass of one mole of a substance, linking the mass in grams to the amount in moles through the equation moles = mass / molar mass. For compounds like CaCl2, the molar mass is calculated by adding the atomic masses considering the subscripts, such as 40.1 for Ca and 2 × 35.5 for Cl, totaling approximately 111 g/mol. Avogadro's number, 6.02 × 10²³, links the number of moles to the number of particles by multiplying moles by this constant to find the particle count. A tempting distractor is choice E, 111 mol, which results from a common unit-conversion error of mistaking the mass directly for moles without dividing by molar mass. A transferable strategy is to write the unit you want, such as moles, and use molar mass as the bridge by dividing the given mass by it.