Answer:
option, b Henry's constant = [tex]1.27\times 10^_{-3} [/tex] mol/L atm
Explanation:
Given:
Solubility of [tex]O_2 = 0.590\;g/L[/tex]
Molecular mass of [tex]O_2[/tex] = 16 g/mol
Solubility [tex]O_2[/tex] in mol/L = [tex]\frac{Solubility\;in\;g/L}{Molecular\ mass}[/tex]
Solubility [tex]O_2[/tex] in mol/L = [tex]\frac{0.590}{16} =0.0184\ mol/L[/tex]
According to Henry's Law,
[tex]C = K \times P_{gas}[/tex]
Where,
C = Solubility
K = Henry's constant
P = Pressure in atm
[tex]C = K \times P_{gas}[/tex]
[tex]K = \frac{C}{P}[/tex]
[tex]=\frac{0.0184}{14.5} = 1.27\times 10^_{-3}[/tex] mol/L atm
