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 $1.3 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot \mathrm{atm}$ for Henry's law in the form $C=k P$where $C$ is the gas concentration (mol/L).
Step 1
We know that 1 atmosphere is equivalent to 760 torr. So, we can use this conversion factor to convert 120 torr to atmosphere: \[ P = 120 \, \text{torr} \times \frac{1 \, \text{atm}}{760 \, \text{torr}} \] Show more…
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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 $(\mathrm{mol} / \mathrm{L})$
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.
The solubility of $\mathrm{O}_{2}$ in water is $6.5 \mathrm{mg} / \mathrm{L}$ at an atmospheric pressure of 1 atm and temperature of $40^{\circ} \mathrm{C}$. Calculate the Henry's law constant of $\mathrm{O}_{2}$ at $40^{\circ} \mathrm{C}$. The mole fraction of $\mathrm{O}_{2}$ in air is 0.209.
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