Question

What is the molar solubility of helium in water at a pressure of 1.0 atm and 25 °C according to Henry's law? (Henry’s law constant for helium gas in water at 25 °C is 3.70⋅10−4M/atm)

a) 2.5 x 10⁻4 M
b) 3.7 x 10⁻4 M
c) 4.8 x 10⁻4 M
d) 5.6 x 10⁻4 M

Answer 1

The molar solubility of helium in water at a pressure of 1.0 atm and 25 °C according to Henry's law is 3.70⋅10−4 M.

Step-by-step explanation:

To find the molar solubility of helium in water at a pressure of 1.0 atm and 25 °C according to Henry's law, we can use the equation:

S = k * P

Where S is the molar solubility, k is the Henry's law constant, and P is the pressure of the gas. In this case, the Henry's law constant for helium gas in water at 25 °C is 3.70⋅10−4 M/atm. Plugging in the values, we get:

S = (3.70⋅10−4 M/atm) * (1.0 atm) = 3.70⋅10−4 M

Therefore, the molar solubility of helium in water at a pressure of 1.0 atm and 25 °C is 3.70⋅10−4 M.