The question asks “What is the density of PF3 at STP”?
Molar mass of PF3 is 87.9680 so just divide that by 22.4? Is the correct answer 3.92 g/l?
The Correct Answer and Explanation is :
Yes, your approach is correct! At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 L of volume. To calculate the density of phosphorus trifluoride (PF₃) at STP, we use the formula:
[
\text{Density} = \frac{\text{Molar Mass}}{\text{Molar Volume at STP}}
]
Given:
- Molar mass of PF₃ = 87.9680 g/mol
- Molar volume at STP = 22.4 L/mol
Now, performing the calculation:
[
\text{Density} = \frac{87.9680 \text{ g/mol}}{22.4 \text{ L/mol}}
]
[
\text{Density} = 3.93 \text{ g/L}
]
So, the correct answer is 3.93 g/L, not 3.92 g/L (a minor rounding difference).
Explanation:
The concept of gas density at STP is based on the ideal gas law, which states:
[
PV = nRT
]
At STP (Standard Temperature = 273.15 K, Pressure = 1 atm), 1 mole of an ideal gas occupies 22.4 L. The density of a gas is given by:
[
\text{Density} = \frac{\text{Mass of Gas}}{\text{Volume of Gas}}
]
Since 1 mole of PF₃ weighs 87.9680 g and occupies 22.4 L, its density is simply the molar mass divided by the molar volume.
This method works for any ideal gas at STP and provides a quick way to determine its density. The slight difference in your calculation (3.92 g/L) is likely due to rounding. Using more precise values, 3.93 g/L is the most accurate answer.