What is the Henry's law constant of dissolved `O_(2) ` at `10^@C ` at 1 atmospheric pressure , if partial pressure of oxygen is 0.24 atm ?
the concentration of dissolved oxygen is `3.12xx10^(-4) mol dm ^(-3)`

Options

A. `2.5xx10^(-3) mol dm^(-3) atm^(-1)`
B. `1xx10 ^(-4) mol dm^(-3) atm^(-1)`
C. `1.3xx10^(-3) mol dm^(-3) atm^(-1)`
D. `1.3xx10^(-4) mol ^(-3) atm^(-1)`

Answer

To find the Henry's law constant (K_H) for dissolved oxygen at 10°C and 1 atm, we will use the formula derived from Henry's law:

\[ C = K_H \cdot P \]

Where:
- \( C \) is the concentration of the solute (in mol/dm³),
- \( K_H \) is the Henry's law constant (in mol/dm³·atm),
- \( P \) is the partial pressure of the solute (in atm).

### Step-by-Step Solution:

1. **Identify the given values:**
- Concentration of dissolved oxygen, \( C = 3.12 \times 10^{-4} \, \text{mol/dm}^3 \)
- Partial pressure of oxygen, \( P = 0.24 \, \text{atm} \)

2. **Rearrange the Henry's law equation to solve for \( K_H \):**
\[ K_H = \frac{C}{P} \]

3. **Substitute the known values into the equation:**
\[ K_H = \frac{3.12 \times 10^{-4} \, \text{mol/dm}^3}{0.24 \, \text{atm}} \]

4. **Perform the calculation:**
\[ K_H = \frac{3.12 \times 10^{-4}}{0.24} = 1.3 \times 10^{-3} \, \text{mol/dm}^3 \cdot \text{atm}^{-1} \]

5. **Final result:**
The Henry's law constant \( K_H \) for dissolved oxygen at 10°C and 1 atm is:
\[ K_H = 1.3 \times 10^{-3} \, \text{mol/dm}^3 \cdot \text{atm}^{-1} \]