The Henry's law constant for `O_(2)`, dissolved in water is `4.34 xx 10^(4)` atm at certain temperature. If the partial pressure of `O_(2)`, in a gas mixture that is in equilibrium with water is 0.434 atm, what is the mole fraction of `O_(2)`, in the solution?

Options

A. `1 xx 10^(-5)`
B. `1 xx 10^(-4)`
C. `2 xx 10^(-5)`
D. `1 xx 10^(-6)`

Explanation

To solve the problem, we will use Henry's Law, which states that the concentration of a gas dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid. The equation can be expressed as:

\[ P = k_H \cdot x \]

Where:
- \( P \) is the partial pressure of the gas (in atm),
- \( k_H \) is the Henry's law constant (in atm),
- \( x \) is the mole fraction of the gas in the solution.

Given:
- \( k_H = 4.34 \times 10^4 \) atm
- \( P = 0.434 \) atm

We need to find the mole fraction \( x_{O_2} \).

### Step 1: Rearranging the Henry's Law equation
We can rearrange the equation to solve for the mole fraction \( x \):

\[ x_{O_2} = \frac{P}{k_H} \]

### Step 2: Substituting the values
Now, we will substitute the known values into the equation:

\[ x_{O_2} = \frac{0.434 \text{ atm}}{4.34 \times 10^4 \text{ atm}} \]

### Step 3: Performing the calculation
Now, we will perform the division:

\[ x_{O_2} = \frac{0.434}{4.34 \times 10^4} \]

Calculating this gives:

\[ x_{O_2} = \frac{0.434}{43400} \]

\[ x_{O_2} = 1.00 \times 10^{-5} \]

### Final Answer
Thus, the mole fraction of \( O_2 \) in the solution is:

\[ x_{O_2} = 1.00 \times 10^{-5} \]

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