Question

Argon makes up 0.93% by volume of air. Calculate its solubility (in mol/L) in water at 20°C and 1.0 atm. The Henry's law constant for Ar under these conditions is 1.5 × 10−3 mol/L·atm. Enter your answer in scientific notation.

Answer 1

S= 1.40x10⁻⁵mol/L

Step-by-step explanation:

The Henry's Law is given by the next expression:

`S = k_(H) \cdot p` (1)

where S: is the solubility or concentration of Ar in water,
`k_(H)`: is Henry's law constant and p: is the pressure of the Ar

Since the argon is 0.93%, we need to multiply the equation (1) by this percent:

`S = 1.5 \cdot 10^(-3) (mol)/(L\cdot atm) \cdot 1.0atm \cdot (0.93)/(100) = 1.40 \cdot 10^(-5) (mol)/(L)`

Therefore, the argon solubility in water is 1.40x10⁻⁵mol/L.

Have a nice day!