What is the mass of 2.31 x 1023 formula units of SrF2?
The Correct Answer and Explanation is :
The correct answer is: 48.19 grams
To find the mass of 2.31 × 10²³ formula units of SrF₂ (strontium fluoride), follow these steps:
Step 1: Calculate the molar mass of SrF₂.
SrF₂ consists of one strontium (Sr) atom and two fluorine (F) atoms.
- The atomic mass of Sr (strontium) is 87.62 g/mol.
- The atomic mass of F (fluorine) is 18.998 g/mol.
The molar mass of SrF₂ is calculated by adding the mass of one Sr atom and two F atoms:
[
Molar\ mass\ of\ SrF_2 = (87.62\ g/mol) + 2 \times (18.998\ g/mol) = 125.616\ g/mol
]
Step 2: Convert the number of formula units to moles.
Use Avogadro’s number (6.022 × 10²³ formula units/mol) to convert from formula units to moles. This step is important because molar mass relates grams to moles.
[
\frac{2.31 \times 10^{23}\ formula\ units}{6.022 \times 10^{23}\ formula\ units/mol} = 0.3837\ mol
]
Step 3: Calculate the mass of SrF₂.
Now, multiply the number of moles by the molar mass to get the mass in grams:
[
mass\ of\ SrF_2 = 0.3837\ mol \times 125.616\ g/mol = 48.19\ g
]
Explanation:
The key to solving this problem is converting the given number of formula units into moles using Avogadro’s number, which provides the connection between the number of particles and the amount of substance in moles. From there, using the molar mass of SrF₂, we can calculate the mass. The molar mass represents the mass of one mole of SrF₂ formula units, and since the number of formula units given is a fraction of a mole, we scale the mass accordingly.
In summary, 2.31 × 10²³ formula units of SrF₂ corresponds to approximately 48.19 grams.