At 25 degrees celsius and 1.0 atm, 0.25g of a gas dissolves in 1.00L of water. What mass of the gas dissolves in 1.00 L of water at 25 degrees celsius and 3.0 atm?
Henry's Law.
P(gas) = kC
But in words, at the same T and the same volume of water, then 3 x pressure means 3 x the amount dissolved or 3 x 0.25 g.
0.75
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 pressure of the gas above the liquid.
First, let's calculate the solubility constant "k" using the given data:
P1 = 1.0 atm (initial pressure)
P2 = 3.0 atm (new pressure)
C1 = 0.25 g/L (initial concentration)
C2 = unknown (new concentration)
According to Henry's law:
C1 / P1 = C2 / P2
Plugging in the values:
0.25 g/L / 1.0 atm = C2 / 3.0 atm
Now, let's solve for C2:
C2 = (0.25 g/L / 1.0 atm) * 3.0 atm
C2 = 0.75 g/L
Therefore, the mass of the gas that dissolves in 1.00 L of water at 25 degrees Celsius and 3.0 atm is 0.75 grams.
To find the mass of the gas that dissolves in 1.00 L of water at 25 degrees Celsius and 3.0 atm, we can use Henry’s Law. Henry’s Law states that the concentration of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid.
Henry’s Law can be expressed as:
C = k × P
where:
C is the concentration of the dissolved gas in the liquid
k is a constant specific to the gas-solvent pair
P is the partial pressure of the gas
Since we know the concentration (0.25g/1.00L) at 25 degrees Celsius and 1.0 atm, we can rearrange Henry’s Law to solve for k:
k = C / P
Now, we can use this constant (k) along with the new partial pressure (3.0 atm) to find the new concentration (mass of the gas dissolved in 1.00 L of water). Rearrange Henry’s Law again:
C = k × P
Substitute the values into the equation:
C = k × P
C = (0.25g/1.00L) / (1.0 atm / 3.0 atm)
C = (0.25g/1.00L) / (1/3)
C = 0.25g * (3/1)
C = 0.75g
Therefore, the mass of the gas that dissolves in 1.00 L of water at 25 degrees Celsius and 3.0 atm is 0.75g.