Answer: The solubility of oxygen at 682 torr is [tex]4.58\times 10^{-3}M[/tex]
Explanation:
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{A}=K_H\times p_{A}[/tex]
Or,
[tex]\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}[/tex]
where,
[tex]C_1\text{ and }p_1[/tex] are the initial concentration and partial pressure of oxygen gas
[tex]C_2\text{ and }p_2[/tex] are the final concentration and partial pressure of oxygen gas
We are given:
Conversion factor used: 1 atm = 760 torr
[tex]C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm[/tex]
Putting values in above equation, we get:
[tex]\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M[/tex]
Hence, the solubility of oxygen gas at 628 torr is [tex]4.58\times 10^{-3}M[/tex]
