I have a lot of trouble doing this chemistry exercise
It says:
The altitude of the lake affects the solubility of oxygen in lakes high in the alps. if the solubility of O2 from the air is 7.89 mg/L M at sea level and 25C, what is the solubility of O2 in moles/L at an elevation of 13,000 ft where the atmospheric pressure is 0.612atm and the temperature is also 25C? Assume that the moles fraction of O2 in air remains constant, at 0.209 at both elevations
Thanks a lot for the help!!!
To solve this problem, you can use the Ideal Gas Law equation, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L * atm / mol * K), and T is the temperature in Kelvin.
In this case, we are given the atmospheric pressure at an elevation of 13,000 ft (0.612 atm) and the temperature (25°C). We also know that the mole fraction of O2 in air remains constant at both elevations (0.209).
We need to find the solubility of O2 in moles/L at the given conditions.
Step 1: Convert the temperature to Kelvin
T = 25°C + 273 = 298 K
Step 2: Use the Ideal Gas Law equation to calculate the number of moles
PV = nRT
n = PV / RT
Substitute the given values:
P = 0.612 atm
V = 1 L (since we are looking for solubility in moles/L)
R = 0.0821 L * atm/mol * K
T = 298 K
n = (0.612 atm * 1 L) / (0.0821 L * atm/mol * K * 298 K)
n = 0.0249 mol
Step 3: Since the mole fraction of O2 in air is given as 0.209, we can multiply this by the total number of moles calculated in Step 2 to find the number of moles of O2.
Moles of O2 = 0.209 * 0.0249 mol
Moles of O2 = 0.00519 mol
Step 4: Calculate the solubility of O2 in moles/L by dividing the number of moles by the volume.
Solubility of O2 = 0.00519 mol / 1 L
Solubility of O2 = 0.00519 mol/L
Therefore, the solubility of O2 in moles/L at an elevation of 13,000 ft is 0.00519 mol/L.
To solve this chemistry exercise, we can use Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The formula for Henry's Law is:
S = k * P
Where:
S is the solubility of the gas in moles per liter (mol/L).
k is Henry's Law constant.
P is the partial pressure of the gas in atmospheres (atm).
To find the solubility of oxygen in moles per liter at an elevation of 13,000 ft, we need to know the partial pressure of oxygen at that altitude.
Given information:
Partial pressure of oxygen at sea level: 1 atm (since atmospheric pressure at sea level is 1 atm and the mole fraction of O2 is 0.209, which we assume remains constant).
Partial pressure of oxygen at an elevation of 13,000 ft: 0.612 atm.
To find the solubility of oxygen in moles per liter, we need to find the value of k (Henry's Law constant). We can rearrange the formula to solve for k:
k = S / P
Now, we need to convert the solubility of O2 from mg/L to moles/L.
Given information:
Solubility of O2 from the air at sea level: 7.89 mg/L
To convert to moles, we need to know the molar mass of O2, which is 32 g/mol.
Conversion:
7.89 mg/L * (1 g/1000 mg) * (1 mol/32 g) = 0.000246875 mol/L
Now we can substitute the values into the formula to find k:
k = 0.000246875 mol/L / 1 atm
Now, we have the value of k, which is approximately 0.000246875 mol/L * atm.
Finally, we can use this value of k and the partial pressure at an elevation of 13,000 ft (0.612 atm) to find the solubility of oxygen in moles per liter:
S = k * P
S = 0.000246875 mol/L * atm * 0.612 atm
S ≈ 0.00015120625 mol/L
So, the solubility of oxygen in moles per liter at an elevation of 13,000 ft is approximately 0.00015120625 mol/L.