Question

The solubility of oxygen gas at 35.1 °C and a oxygen pressure of 649 mmHg is 8.06 × 10−3 g/L. What is the Henry's Law constant in mol⋅L−1⋅atm−1?

Answer 1

The answer is 2.95 × 10^(-4) mol.L-1.atm-1 (three significant digits)

The process of the calculation is given below.

Answer 2

Henry's Law relates the solubility of a gas in a liquid to the partial pressure of the gas above the liquid. It is expressed as:

S = kH P

where: S is the solubility of the gas in the liquid (in mol/L) P is the partial pressure of the gas (in atm) kH is Henry's Law constant (in mol/L·atm)

To find kH, we need to convert the given solubility from grams per liter to moles per liter:

8.06 × 10−3 g/L = 8.06 × 10−3/32 g/mol = 2.52 × 10−4 mol/L

Now we can use the given oxygen pressure to find kH:

kH = S/P = (2.52 × 10−4 mol/L) / (649 mmHg / 760 mmHg/atm) = 1.97 × 10−5 mol/L·atm−1

Therefore, the Henry's Law constant for oxygen gas at 35.1 °C is 1.97 × 10−5 mol/L·atm−1.