Unit 1 · Problem 1
apchem-u01-q001
9 points
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A sample contains 0.125 mol Al2O3. Use M = 101.96 g mol−1, NA = 6.022 × 1023 mol−1 and 1 u ≈ 1.66054 × 10−24 g.
- Determine the sample mass and number of oxygen atoms. Explain why the oxygen counting factor is 3.
- State the average mass represented by one Al2O3 formula unit in u, and the mass of one mole of formula units in g. Show how the supplied u-to-gram factor and Avogadro’s constant connect these quantities. Evaluate the claim “101.96 u and 101.96 g mol−1 are the same units.”
Part a 5 points
| Criterion | Points |
|---|---|
a1 Sets up the mass conversion as 0.125 mol × 101.96 g mol⁻¹, with units that cancel to grams. | 1 |
a2 Calculates the sample mass as 12.7 g Al₂O₃ to three significant figures. | 1 |
a3 Sets up the formula-unit count as 0.125 mol × 6.022 × 10²³ formula units mol⁻¹. Do not award for
| 1 |
a4 Explains that the subscript 3 represents exactly three O atoms per Al₂O₃ formula unit. | 1 |
a5 Calculates 2.26 × 10²³ O atoms to three significant figures. | 1 |
Part b 4 points
| Criterion | Points |
|---|---|
b1 States that one formula unit represents an average mass of 101.96 u, whereas one mole of formula units has mass 101.96 g. | 1 |
b2 Sets up the connection using 101.96 u × 1.66054 × 10⁻²⁴ g u⁻¹, then multiplying by 6.022 × 10²³ formula units mol⁻¹. | 1 |
b3 Calculates approximately 102.0 g mol⁻¹ from the supplied rounded constants, verifying the numerical correspondence with the supplied molar mass. Accept
| 1 |
b4 Rejects the claim that the units are identical: u measures particle-scale mass, while g mol⁻¹ measures mass per mole, despite their corresponding numerical values. | 1 |