Question
Calculate the solubility of $\mathrm{O}_{2}$ in water at a partial pressure of $\mathrm{O}_{2}$ of 120 torr at $25^{\circ} \mathrm{C}$ . The Henry's law constant for $\mathrm{O}_{2}$ is $1.3 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot$ atm for Henry's law in the form $C=k P$ where $C$ is the gas concentration $(\mathrm{mol} / \mathrm{L})$
Step 1
We know that 1 atm is equal to 760 torr. Therefore, the pressure in atm is given by: \[ P = \frac{120 \, \text{torr}}{760 \, \text{torr/atm}} = 0.16 \, \text{atm} \] Show more…
Show all steps
Your feedback will help us improve your experience
James Irizarry and 63 other Chemistry 102 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Calculate the solubility of $\mathrm{O}_{2}$ in water at a partial pressure of $\mathrm{O}_{2}$ of 120 torr at $25^{\circ} \mathrm{C}$. The Henry's law constant for $\mathrm{O}_{2}$ is $1.3 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot \mathrm{atm}$ for Henry's law in the form $C=k P$ where $C$ is the gas concentration (mol/L).
The Henry's law constant for $\mathrm{N}_{2}$ in water at $25^{\circ} \mathrm{C}$ is $8.4 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{mmHg}^{-1}$. Calculate the solubility of $\mathrm{N}_{2}$ in $\mathrm{mol} / \mathrm{L}$ if its partial pressure is $1520 \mathrm{mmHg}$. Calculate the solubility when the $\mathrm{N}_{2}$ partial pressure is 20. mmHg.
Calculate the solubility of O2 in water at a partial pressure of O2 of 120 torr at 25°C. The Henry's law constant for O2 is 1.3×10-3 mol/L·atm for Henry's law in the form C = kP, where C is the gas concentration (mol/L).
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD