Final answer:
To find the weight in grams of 2.701 x 10^24 formula units of BeI2, calculate the molar mass of BeI2, use Avogadro's number to determine the number of moles, and then multiply the number of moles by the molar mass in grams per mole, resulting in 1179.41 grams, which does not match any of the provided answer choices.
Step-by-step explanation:
To calculate how many grams are in 2.701 x 10^24 formula units of BeI2, we must first determine the molar mass of BeI2, and then use Avogadro's number to find the number of moles of BeI2 represented by the given number of formula units.
The molar mass of BeI2 (beryllium iodide) can be found by adding the atomic mass of beryllium (Be, approximately 9.012 u) and the atomic mass of iodine (I, approximately 126.904 u) multiplied by 2, since there are two iodine atoms in one formula unit of BeI2. This gives us:
Molar mass of BeI2 = 9.012 u + 2(126.904 u) = 262.820 u
Then, convert the molar mass to grams per mole, since u is essentially grams per mole:
Molar mass of BeI2 = 262.820 g/mol
Next, we use Avogadro's number, which is approximately 6.022 x 10^23 particles per mole, to find the number of moles:
Number of moles = (2.701 x 10^24 formula units) ÷ (6.022 x 10^23 formula units/mol) = 4.487 moles
Finally, calculate the weight in grams:
Weight in grams = (Number of moles) × (Molar mass of BeI2) = 4.487 moles × 262.820 g/mol ≈ 1179.41 grams
However, as none of the answer choices match this result, it's possible that there may be a typo in the question or in the answer choices provided. Therefore, based on the calculation, none of the options (a) 97.73 grams (b) 5.62 grams (c) 10.27 grams (d) 27.01 grams are correct.