Henry's law constant of oxygen is `1.4 xx 10^(-3)` mol `"lit"^(-1) "atm"^(-1)` at 298 K. How much of oxygen is dissolved in 100 mL at 298 K when the partial pressure of oxygen is 0.5 atm?

Options

A. 22.4 mg
B. 22.4 g
C. 2.24 g
D. 2.24 mg

Explanation

To solve the problem, we will use Henry's Law, which states that the amount of gas that dissolves in a liquid is directly proportional to the partial pressure of the gas above the liquid. The formula for Henry's Law is given by:

\[ C = k_H \times P \]

where:
- \( C \) is the concentration of the dissolved gas (in moles per liter),
- \( k_H \) is Henry's law constant (in mol/(L·atm)),
- \( P \) is the partial pressure of the gas (in atm).

### Step-by-Step Solution:

1. **Identify the given values:**
- Henry's law constant, \( k_H = 1.4 \times 10^{-3} \, \text{mol/(L·atm)} \)
- Partial pressure of oxygen, \( P = 0.5 \, \text{atm} \)
- Volume of the solution, \( V = 100 \, \text{mL} = 0.1 \, \text{L} \)

2. **Calculate the concentration of oxygen dissolved using Henry's Law:**
\[ C = k_H \times P \]
Substituting the values:
\[ C = (1.4 \times 10^{-3} \, \text{mol/(L·atm)}) \times (0.5 \, \text{atm}) \]
\[ C = 0.7 \times 10^{-3} \, \text{mol/L} \]

3. **Calculate the number of moles of oxygen in 100 mL:**
Since we have the concentration in moles per liter, we can find the number of moles in 0.1 L (100 mL):
\[ \text{Moles of oxygen} = C \times V \]
\[ \text{Moles of oxygen} = (0.7 \times 10^{-3} \, \text{mol/L}) \times (0.1 \, \text{L}) \]
\[ \text{Moles of oxygen} = 0.07 \times 10^{-3} \, \text{mol} = 7 \times 10^{-5} \, \text{mol} \]

4. **Calculate the mass of oxygen dissolved:**
The molecular mass of oxygen (O₂) is approximately 32 g/mol. To find the mass, we use:
\[ \text{Mass} = \text{Moles} \times \text{Molecular mass} \]
\[ \text{Mass} = (7 \times 10^{-5} \, \text{mol}) \times (32 \, \text{g/mol}) \]
\[ \text{Mass} = 2.24 \times 10^{-3} \, \text{g} \]

5. **Convert the mass from grams to milligrams:**
Since \( 1 \, \text{g} = 1000 \, \text{mg} \):
\[ \text{Mass in mg} = 2.24 \times 10^{-3} \, \text{g} \times 1000 \, \text{mg/g} \]
\[ \text{Mass in mg} = 2.24 \, \text{mg} \]

### Final Answer:
The amount of oxygen dissolved in 100 mL at 298 K when the partial pressure of oxygen is 0.5 atm is **2.24 mg**.