Answer:
S= 1.40x10⁻⁵mol/L
Explanation:
The Henry's Law is given by the next expression:
[tex] S = k_{H} \cdot p [/tex] (1)
where S: is the solubility or concentration of Ar in water, [tex]k_{H}[/tex]: 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:
[tex] S = 1.5 \cdot 10^{-3} \frac{mol}{L\cdot atm} \cdot 1.0atm \cdot \frac{0.93}{100} = 1.40 \cdot 10^{-5} \frac{mol}{L} [/tex]
Therefore, the argon solubility in water is 1.40x10⁻⁵mol/L.
Have a nice day!
