The solubility of pure oxygen in water at ` 20 ^(@) C ` and one atmosphere pressure is ` 1.38 xx 10 ^(-3)` mol ` L^(-1)` Calculate the concentration of oxygen at `20 ^(@) C ` and partial pressure of 0.21 atm.

Answer

To solve the problem, we will use Henry's Law, which states that the solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of that gas above the liquid.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Solubility of pure oxygen at 1 atm: \( M = 1.38 \times 10^{-3} \, \text{mol L}^{-1} \)
- Partial pressure of oxygen: \( P = 0.21 \, \text{atm} \)

2. **Write Henry's Law Equation:**
\[ M = K_H \times P \]
Where:
- \( M \) = concentration of the gas (mol L\(^{-1}\))
- \( K_H \) = Henry's constant (mol L\(^{-1}\) atm\(^{-1}\))
- \( P \) = partial pressure of the gas (atm)

3. **Calculate Henry's Constant \( K_H \):**
Using the solubility at 1 atm:
\[ 1.38 \times 10^{-3} = K_H \times 1 \]
Thus,
\[ K_H = 1.38 \times 10^{-3} \, \text{mol L}^{-1} \text{atm}^{-1} \]

4. **Calculate the Concentration of Oxygen at 0.21 atm:**
Now, we can use the value of \( K_H \) to find the concentration at the new partial pressure:
\[ M_A = K_H \times P_A \]
Substituting the values:
\[ M_A = (1.38 \times 10^{-3}) \times (0.21) \]

5. **Perform the Calculation:**
\[ M_A = 1.38 \times 10^{-3} \times 0.21 = 2.898 \times 10^{-4} \, \text{mol L}^{-1} \]
Rounding this to two significant figures, we get:
\[ M_A \approx 2.9 \times 10^{-4} \, \text{mol L}^{-1} \]

6. **Final Answer:**
The concentration of oxygen at 20 °C and a partial pressure of 0.21 atm is approximately \( 2.9 \times 10^{-4} \, \text{mol L}^{-1} \).