The solubility of a gas is 0.890 8/1 at a pressure of 121 kPa. What is the solubility of the gas if the pressure is increased to 150 kPa,
given that the temperature is held constant?

1.10 g/L

Explanation:

Step 1: Calculate Henry's constant (k)

The solubility of a gas (C) is 0.890 g/L at a pressure (P) of 121 kPa. Solubility and pressure are related through Henry's law.

C = k × P

k = C / P

k = (0.890 g/L) / 121 kPa = 7.36 × 10⁻³ g/L.kPa

Step 2: Calculate the solubility of the gas if the pressure is increased to 150 kPa

We will use Henry's law.

C = k × P

C = (7.36 × 10⁻³ g/L.kPa) × 150 kPa = 1.10 g/L

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