Answer:
From the search results, I found the following question:
"The Ozone (O3) ambient air quality standard is 0.070ppmv express this concentration in mgO3/L−Air 4. Compute the molar volume of air at 25∘C and 1 atm pressure. Express your answer in L/mol."
To answer the first part of the question, we first need to convert the concentration of ozone from parts per million by volume (ppmv) to milligrams/m^3. We can do this using the molar mass of O3 and the ideal gas law:
Concentration of O3 (mg/m^3) = (0.070 ppmv) x (molecular weight of O3 / 24.45 L/mole) x (1 atm / 101.325 kPa) x (273.15 K / (273.15 + 25) K)
Using the molecular weight of O3 (48 g/mol) and simplifying the units, we get:
Concentration of O3 (mg/m^3) = 0.0277 mg/m^3
Therefore, the concentration of O3 in mgO3/L-Air is also 0.0277 mgO3/L-Air.
To compute the molar volume of air at 25°C and 1 atm pressure, we can use the ideal gas law:
PV = nRT
where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature in Kelvin.
At 25°C, which is 298 K, and 1 atm pressure, we have:
V = (nRT) / P
The molar volume of air will be the volume occupied by one mole of gas, so we can set n = 1 mole. The gas constant R is 0.08206 L·atm/(mol·K). Substituting the values we get:
V = (1 mole x 0.08206 L·atm/(mol·K) x 298 K) / 1 atm = 24.45 L/mol
Therefore, the molar volume of air at 25°C and 1 atm pressure is 24.45 L/mol.
Step-by-step explanation: