A gas has a solubility of 2.45 g/L at a pressure of 0.750 atm. What pressure wold be required to produce an aqueous solution containing 6.25 g/L of this gas at constant temperature
Answer:
1.91 atm
Explanation:
Step 1: Calculate Henry's constant (k)
A gas has a solubility (C) of 2.45 g/L at a pressure (P) of 0.750 atm. These two variables are related to each other through Henry's law.
C = k × P
K = C/P
K = (2.45 g/L)/0.750 atm = 3.27 g/L.atm
Step 2: Calculate the pressure required to produce an aqueous solution containing 6.25 g/L of this gas at constant temperature.
We have C = 6.25 g/L and k = 3.27 g/L.atm. The required pressure is:
C = k × P
P = C/k
P = (6.25 g/L)/(3.27 g/L.atm) = 1.91 atm