Determine the molarity of in an aqueous solution at equilibrium with air at normal atmospheric pressure. The volume percent of in air is 20.65

Under an pressure of 1.00 , 28.31 of dissolves in 1.00 at 25.

To determine the molarity of a gas in an aqueous solution, we need to use Henry's Law, which states that the concentration of a gas in a solution is directly proportional to its partial pressure.

Step 1: Convert the volume percent of nitrogen in air to partial pressure.
The volume percent of nitrogen in air is given as 20.65%. This means that 20.65 mL of nitrogen is present in every 100 mL of air. Since the volume of air is not specified, we can assume it to be 1 L (1000 mL) to simplify the calculation.
So, the volume of nitrogen in air is (20.65 mL / 100 mL) * 1000 mL = 206.5 mL.

To convert this volume to moles, we need to use the ideal gas law, PV=nRT.
Assuming the temperature is consistent at 25 °C (298 K) and the pressure is normal atmospheric pressure (1 atm), we can calculate the number of moles using the formula:
n = PV / RT, where:
P = pressure in atm
V = volume in L
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in K

Substituting the values into the formula:
n = (1 atm) * (0.2065 L) / (0.0821 L·atm/mol·K * 298 K)
n ≈ 0.008146 mol

Step 2: Determine the molarity of nitrogen in the aqueous solution.
The molarity is defined as the moles of solute (nitrogen gas) divided by the volume of the solution (in liters).
Given that 28.31 g of nitrogen dissolves in 1.00 L of water, we can calculate the molarity using the formula:
Molarity (M) = moles of solute (nitrogen gas) / volume of solution (in liters)

First, we need to convert the mass of nitrogen to moles using the molar mass of nitrogen (N2 = 28.02 g/mol):
moles of nitrogen = mass of nitrogen / molar mass of nitrogen
moles of nitrogen = 28.31 g / 28.02 g/mol
moles of nitrogen ≈ 1.010 mol

Finally, we can calculate the molarity using the formula:
Molarity (M) = 1.010 mol / 1.00 L
Molarity (M) ≈ 1.010 M

Therefore, the molarity of nitrogen in the aqueous solution at equilibrium with air at normal atmospheric pressure is approximately 1.010 M.

To determine the molarity of CO2 in an aqueous solution at equilibrium with air at normal atmospheric pressure, we need to follow these steps:

Step 1: Calculate the moles of CO2 dissolved in 1.00 L of water.
Since we know that 28.31 g of CO2 dissolves in 1.00 L of water, we can convert this mass to moles using the molar mass of CO2. The molar mass of CO2 is 44.01 g/mol, so:

Moles of CO2 = 28.31 g / 44.01 g/mol = 0.643 mol

Step 2: Determine the volume of the aqueous solution at equilibrium.
We are given that the volume percent of CO2 in air is 20.65%. This means that 20.65% of the total volume of air is occupied by CO2. Since air is mostly nitrogen and oxygen, we can assume the total volume to be 100.00%. Therefore:

Volume of CO2 = 20.65% of 100.00% = 20.65/100 * 1.00 L = 0.2065 L

Step 3: Calculate the molarity of CO2 in the aqueous solution.
Molarity is defined as moles of solute (CO2) divided by liters of solution. In this case, the solute is CO2 and the solution is the aqueous solution. Therefore:

Molarity of CO2 = Moles of CO2 / Volume of aqueous solution
= 0.643 mol / 0.2065 L ≈ 3.11 M

So, the molarity of CO2 in the aqueous solution at equilibrium with air at normal atmospheric pressure is approximately 3.11 M.