Question

The Henry's law constant for O\(_2\) dissolved in water is \( 4.34 \times 10^4\text{ atm} \) at certain temperature. If the partial pressure of O\(_2\) in a gas mixture that is in equilibrium with water is \( 0.434\text{ atm} \), what is the mole fraction of O\(_2\) in the solution?

• A. $1 \times 10^{-5}$
• B. $1 \times 10^{-4}$
• C. $2 \times 10^{-5}$
• D. $1 \times 10^{-6}$
• E. $2 \times 10^{-6}$

Hint:

Henry's Law constant ($K_H$) can be expressed in different units. Always ensure the units for pressure and the constant match before dividing.

Answer 1

The Correct Option is A

Concept: Henry's Law states that the partial pressure of a gas above a liquid is proportional to its mole fraction in the solution: $P = K_H \cdot \chi$.

Step 1: {Identify given values.} Partial pressure of $O_2$ ($P$) = $0.434\text{ atm}$. Henry's Law constant ($K_H$) = $4.34 \times 10^4\text{ atm}$.

Step 2: {Rearrange the formula to solve for mole fraction ($\chi$).} $$\chi = \frac{P}{K_H}$$

Step 3: {Perform the calculation.} $$\chi = \frac{0.434}{4.34 \times 10^4}$$ $$\chi = \frac{0.434}{43400}$$ $$\chi = 0.00001 = 1 \times 10^{-5}$$