Use Henry's law to determine the molar solubility of helium at a pressure of 1.5 atm and 25deg C.

To determine the molar solubility of helium at a given pressure and temperature using Henry's law, we need to know the Henry's law constant, which represents the solubility of a gas in a solvent at a particular temperature.

Henry's law states that the solubility of a gas is directly proportional to the partial pressure of the gas above the solvent:

C = k * P

Where:
C is the molar concentration (molarity) of the gas in the solution,
k is the Henry's law constant, and
P is the partial pressure of the gas.

In this case, we need to find k, the Henry's law constant for helium at 25°C. The Henry's law constant varies depending on the solvent and temperature, so a specific value needs to be given or looked up.

Let's assume the given Henry's law constant for helium in water at 25°C is 4.5 x 10^(-4) mol/(L·atm).

Now, we can substitute the values into Henry's law equation to find the molar solubility of helium (C):

C = (k * P) / R * T

Where:
R is the ideal gas constant (0.0821 L·atm/(mol·K)),
T is the temperature in Kelvin (25°C = 298 K). Please note that temperatures should be in Kelvin when working with gas laws.

C = (4.5 x 10^(-4) mol/(L·atm) * 1.5 atm) / (0.0821 L·atm/(mol·K) * 298 K)

C ≈ 9.1 x 10^(-5) mol/L

Therefore, the molar solubility of helium at a pressure of 1.5 atm and 25°C is approximately 9.1 x 10^(-5) mol/L.