At 25 degrees celcius and 1.0 atm, 0.25 g of gas dissolves in 1.00 L of water. What mass of the gas dissolves in 1.00 L of water at 25 degrees celcius and 3.0 atm?
I think this describes Henry's Law very well.
http://www.800mainstreet.com/9/0009-006-henry.html
To solve this problem, 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 equation for Henry's Law can be written as:
C = k * P
Where:
C is the concentration of the gas in the liquid (in mol/L)
k is the proportionality constant (specific to each gas and solvent)
P is the partial pressure of the gas (in atm)
In this case, we are given the solubility of the gas at 1.0 atm, so we can use this information to find the proportionality constant (k). Once we have k, we can use it to calculate the solubility of the gas at 3.0 atm.
First, let's convert the given mass of gas (0.25 g) to moles.
The molar mass of the gas is needed to do this calculation. Let's assume we have the molar mass.
Next, we need to find the concentration (C) at 1.0 atm and 25 degrees Celsius using the given mass of the gas and the volume of water (1.00 L). To find the concentration, we divide the number of moles by the volume:
C = (mass of gas in moles) / volume of water (in L)
Now, we can use this concentration (C) and the known pressure (1.0 atm) to find the proportionality constant (k):
k = C / P
Once we have this k value, we can use it to find the concentration at 3.0 atm:
C = k * P
Finally, to find the mass of the gas that dissolves in 1.00 L of water at 25 degrees Celsius and 3.0 atm, we multiply the new concentration by the volume of water (1.00 L) and the molar mass of the gas to convert from moles to grams:
mass of gas = C * volume of water * molar mass
Note: To fully answer the question, we need additional information such as the molar mass of the gas.