Question

Use Henry's law to determine the molar solubility of Helium at a pressure of 1.5 atm and 25⁰C. Henry's law constant for helium gas in water at 25⁰C is 3.70×10⁻⁴ M/atm.

Answer 1

The molar solubility of Helium at a pressure of 1.5 atm and 25°C, using Henry's law and a Henry's law constant of 3.70×10⁻⁴ M/atm, is calculated to be 5.55×10⁻⁴ M.

Step-by-step explanation:

To determine the molar solubility of Helium (He) at a pressure of 1.5 atm using Henry's law, we'll apply the following formula which relates pressure and concentration for a gas in solution:

S = kH × P

Where:

• S is the solubility of the gas in moles per liter (M).
• kH is the Henry's law constant for the gas in M/atm.
• P is the partial pressure of the gas above the solution in atm.

Using the given values:

1. Henry's law constant for helium, kH = 3.70 × 10⁻⁴ M/atm.
2. Pressure, P = 1.5 atm.

Plugging these values into the equation gives us the solubility (S):

S = (3.70 × 10⁻⁴ M/atm) × (1.5 atm) = 5.55 × 10⁻⁴ M

Therefore, the molar solubility of Helium at 1.5 atm and 25°C is 5.55 × 10⁻⁴ M.