B false i think, i may be wrong
Answer:
Explanation:
to show how precise your measurements were.
<u>Answer:</u> The solubility of oxygen at 682 torr is ![4.58\times 10^{-3}M](https://tex.z-dn.net/?f=4.58%5Ctimes%2010%5E%7B-3%7DM)
<u>Explanation:</u>
To calculate the molar solubility, we use the equation given by Henry's law, which is:
![C_{A}=K_H\times p_{A}](https://tex.z-dn.net/?f=C_%7BA%7D%3DK_H%5Ctimes%20p_%7BA%7D)
Or,
![\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}](https://tex.z-dn.net/?f=%5Cfrac%7BC_%7B1%7D%7D%7BC_%7B2%7D%7D%3D%5Cfrac%7Bp_%7B1%7D%7D%7Bp_2%7D)
where,
are the initial concentration and partial pressure of oxygen gas
are the final concentration and partial pressure of oxygen gas
We are given:
Conversion factor used: 1 atm = 760 torr
![C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm](https://tex.z-dn.net/?f=C_1%3D1.38%5Ctimes%2010%5E%7B-3%7DM%5C%5Cp_1%3D0.27atm%5C%5CC_2%3D%3F%5C%5Cp_2%3D682torr%3D0.897atm)
Putting values in above equation, we get:
![\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](https://tex.z-dn.net/?f=%5Cfrac%7B1.38%5Ctimes%2010%5E%7B-3%7D%7D%7BC_2%7D%3D%5Cfrac%7B0.27atm%7D%7B0.897atm%7D%5C%5C%5C%5CC_2%3D%5Cfrac%7B1.38%5Ctimes%2010%5E%7B-3%7D%5Ctimes%200.897atm%7D%7B0.27atm%7D%3D4.58%5Ctimes%2010%5E%7B-3%7DM)
Hence, the solubility of oxygen gas at 628 torr is ![4.58\times 10^{-3}M](https://tex.z-dn.net/?f=4.58%5Ctimes%2010%5E%7B-3%7DM)