What volume of water, in mL, is needed to dissolve 8.7 mg of carbon monoxide gas when the partial pressure of carbon monoxide above the water is 1.6 atm at 25 °C? The Henry's constant for carbon monoxide gas in water at 25 °C is 9.5×10-4 M/atm.
C = KcP
Substitute and solve for C.
Then C, in mol/L. You convert 8.7 mg to mols. That gives you mols and M, solve for L and convert to mL.
To solve this problem, we need to use Henry's law, which states that at a constant temperature, the amount of gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid.
Henry's law equation is written as:
C = k * P
Where:
C is the concentration of the gas in the liquid (in Molarity, M)
k is Henry's constant (in M/atm)
P is the partial pressure of the gas (in atm)
First, let's convert the given mass of carbon monoxide gas (8.7 mg) to moles. The molar mass of carbon monoxide (CO) is 28.01 g/mol:
8.7 mg * (1 g/1000 mg) * (1 mol/28.01 g) = 0.000310 Mols of CO
Next, we will convert the given partial pressure of carbon monoxide (1.6 atm) to concentration using Henry's constant.
Concentration (C) = k * P
C = (9.5×10^-4 M/atm) * (1.6 atm) = 1.52×10^-3 M
Since concentration is defined as moles of solute per liter of solution, we need to convert this to moles per milliliter.
1.52×10^-3 M * (1 mol/1000 mL) = 1.52×10^-6 moles/mL
Now, we have the concentration of carbon monoxide gas (CO) in terms of moles/mL. To find the volume of water needed to dissolve the given mass of carbon monoxide gas (8.7 mg), we will use the formula:
Volume (V) = Mass (m) / Concentration (C)
V = 8.7 mg * (1 g/1000 mg) / (1.52×10^-6 moles/mL) = 5723 mL
Therefore, approximately 5723 mL of water is needed to dissolve 8.7 mg of carbon monoxide gas when the partial pressure of carbon monoxide above the water is 1.6 atm at 25 °C.