if air has an average density of 1.29 g/L and CO is reported at 2.0 ppm, what volume of air will contain 1.00 mol of CO ?
To answer this question, we need to calculate the volume of air required to contain 1.00 mol of CO.
Here's how you can do it:
1. Determine the molar mass of CO.
- Carbon (C) has a molar mass of 12.01 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- So, the molar mass of CO = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol.
2. Use the given information to calculate the volume of air required to contain 1.00 mol of CO.
- The density of air is given as 1.29 g/L, which means 1 L of air has a mass of 1.29 g.
- Now, we need to convert the mass of CO (1.00 mol) into grams. The molar mass of CO is 28.01 g/mol, so 1.00 mol of CO is equal to 28.01 g.
- Since the density of air is given in grams per liter, we divide the mass of CO by the density of air to find the corresponding volume.
- Volume = (Mass of CO) / (Density of air) = 28.01 g / 1.29 g/L = 21.7 L
Therefore, to contain 1.00 mol of CO, you would need a volume of approximately 21.7 liters of air.