The solubility of air in water is approximately 2.1x10^-3 M at 20 C and 1 atm. What is the constant for air.?
The constant for air, denoted as K, can be calculated using Henry's law. Henry's law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas.
The equation for Henry's law is:
C = K * P
Where:
C is the concentration of the gas in the liquid (in this case, air) in units of M (mol/L)
K is the Henry's law constant
P is the partial pressure of the gas (in this case, 1 atm)
Given that the solubility of air in water is approximately 2.1×10^-3 M at 20 °C and 1 atm, we can substitute these values into the Henry's law equation:
2.1×10^-3 M = K * 1 atm
To solve for K, divide both sides of the equation by 1 atm:
K = 2.1×10^-3 M / 1 atm
Therefore, the Henry's law constant for air in water is approximately 2.1×10^-3 M/atm.
To determine the constant for air in this context, we need to use Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
In this case, the solubility of air in water is given as approximately 2.1x10^-3 M at 20°C and 1 atm. This means that the concentration of air (in moles per liter) dissolved in water is 2.1x10^-3 M when the pressure of air above the water is 1 atm.
To find the constant for air, we can rearrange Henry's Law equation and solve for it:
C = k * P
Where:
C is the concentration of air in moles per liter (2.1x10^-3 M)
k is the constant for air (what we're trying to find)
P is the partial pressure of air above the water (1 atm)
Plugging in the values we have:
2.1x10^-3 M = k * 1 atm
Now, solving for k:
k = (2.1x10^-3 M) / (1 atm)
k ≈ 2.1x10^-3 M/atm
Therefore, the constant for air in this context is approximately 2.1x10^-3 M/atm.
Which formula do you use?
p = Kc*molarity
Plug and chug.