What is the mass in grams of a 15.8 L sample of ozone (O3) at 1.67 atm and 30.0 °C?
Use PV = nRT and solve for n = mols.
Then mols = grams/molar mass. You know mols and molar mass, solve for grams.
To determine the mass in grams of the ozone sample, we need to use the ideal gas law equation:
PV = nRT
where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L atm/mol K)
T = temperature (in Kelvin)
We are given:
P = 1.67 atm
V = 15.8 L
T = 30.0 °C = 30.0 + 273.15 = 303.15 K
First, we need to calculate the number of moles (n) of ozone using the ideal gas law equation and rearranging it to solve for n:
n = PV / RT
Substituting the given values into the equation:
n = (1.67 atm * 15.8 L) / (0.0821 L atm/mol K * 303.15 K)
Simplifying the equation:
n = 0.8289 mol
Next, we convert the moles of ozone to grams using the molar mass of ozone.
The molar mass of ozone (O3) is:
Oxygen (O) has a molar mass of 16.00 g/mol
Ozone (O3) consists of three oxygen atoms, so the molar mass of ozone is:
3 * 16.00 g/mol = 48.00 g/mol
Finally, we calculate the mass of the ozone sample using the formula:
Mass (in grams) = number of moles * molar mass
Mass = 0.8289 mol * 48.00 g/mol
Calculating the mass:
Mass ≈ 39.77 grams
Therefore, the mass of the 15.8 L sample of ozone is approximately 39.77 grams.