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:

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