Final answer:
To find the moles and grams of boron trifluoride (BF3), we use the Ideal Gas Law. The calculation yields 0.231 moles and 15.65 grams of BF3, but this does not match the given options, indicating a possible discrepancy.
Step-by-step explanation:
To determine the number of moles of gaseous boron trifluoride, BF3, in a 4.3410-L bulb at 788.0 K and 1.220 atm, we can use the Ideal Gas Law equation PV = nRT. Here, P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. First, convert the pressure to atmospheres (if necessary) and the volume to liters, which they already are in this case. The gas constant R is 0.0821 L·atm/(K·mol).
Plugging in the values we get:
n = PV / RT = (1.220 atm)(4.3410 L) / (0.0821 L·atm/(K·mol))(788.0 K). After calculating, the number of moles n is approximately 0.231 moles.
To find out how many grams of BF3 this corresponds to, we need to multiply the number of moles by the molar mass of BF3. The molar mass of BF3 is about 67.80 g/mol, so we multiply 0.231 moles by 67.80 g/mol to get approximately 15.65 g.
However, this calculated mass does not match any of the options given. We need to check the provided answer choices.
The closest matching option in terms of moles is option b) 0.231 moles.
Multiplying 0.231 moles by the molar mass of BF3: 0.231 moles × 67.80 g/mol, we get 15.65 g which does not correspond to option b).
Since non-integer answer choices are provided and none matches the calculation, it seems like there may be a mistake either in the answers given or in our calculation.
We base our answer on the data provided and the principles of chemistry, but we must note the discrepancy.