Question

11. A gas has a solubility of 16.9 g/L at a

temperature of 20 °C and 5.00 atm of pressure.
What is the solubility at 20 °C and 3.00 atm
pressure?

Answer 1

Using Henry's law, we know that solubility is directly proportional to pressure:

S1/P1 = S2/P2

where S1 and P1 are the initial solubility and pressure, and S2 and P2 are the new solubility and pressure.

Plugging in the values given:

16.9 g/L / 5.00 atm = S2 / 3.00 atm

Solving for S2:

S2 = 10.14 g/L

Therefore, the solubility of the gas at 20 °C and 3.00 atm pressure is 10.14 g/L.