PV = nRT and solve for n = number of mols.
Then n = grams/molar mass. You know n and molar mass, solve for grams.
Then n = grams/molar mass. You know n and molar mass, solve for grams.
PV = nRT
Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, let's convert the given values to the proper units:
- Pressure: 475 mm Hg = (475/760) atm ≈ 0.625 atm
- Volume: 1.25 L
- Temperature: -22 degrees Celsius = -22 + 273.15 Kelvin ≈ 251.15 K
Now we can rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Substituting the values we have:
n = (0.625 atm * 1.25 L) / (0.0821 L·atm/(mol·K) * 251.15 K)
n ≈ 0.0397 moles
In order to determine the mass of oxygen gas, we need to know the molar mass of oxygen, which is approximately 32 g/mol.
Finally, we can calculate the mass (m) of oxygen gas using the number of moles (n) and the molar mass of oxygen (M):
m = n * M
m ≈ 0.0397 moles * 32 g/mol
m ≈ 1.27 grams
Therefore, the mass of oxygen gas that exerts a pressure of 475 mm Hg in a volume of 1.25 L at a temperature of -22 degrees Celsius is approximately 1.27 grams.