Question

1028 grams of seawater sample contains 7 mL of dissolved oxygen ($ \text{O}_2 $). What is the concentration of oxygen in parts per million (ppm)?

• A. 0.6 ppm
• B. 6 ppm
• C. 6.8 ppm
• D. 60 ppm

Hint:

For dissolved gases in water like oxygen, the approximation \[ \text{1 mL} \approx \text{1 mg} \] is commonly used, unless otherwise specified.

Answer 1

The Correct Option is C

Step 1: Understand the definition of ppm (mass basis)
In environmental chemistry, for dilute aqueous solutions, ppm is often calculated using mass: \[ \text{ppm} = \frac{\text{Mass of solute (in mg)}}{\text{Mass of solution (in kg)}} \] Step 2: Convert volume of $\text{O_2$ to mass}
We are given volume of oxygen gas, so we must convert it to mass using density. Assume the density of oxygen gas at STP is: \[ \text{Density of } \text{O}_2 = 1.429 \text{ g/L} = 1.429 \times 10^{-3} \text{ g/mL} \] Given:
Volume of $\text{O}_2 = 7 \text{ mL}$ \[ \text{Mass of } \text{O}_2 = 7 \times 1.429 \times 10^{-3} \text{ g} = 0.010003 \text{ g} = 10.003 \text{ mg} \] Step 3: Convert mass of seawater to kg
Given:
Mass of seawater = 1028 g = 1.028 kg Step 4: Calculate ppm
Now apply the formula: \[ \text{ppm} = \frac{10.003}{1.028} \approx 9.73 \text{ ppm} \] More Accurate Approach:
For dissolved oxygen in water, volume is usually expressed in mg/L, where: \[ 1 \text{ mg/L} = 1 \text{ ppm} \] So if 7 mL of oxygen corresponds to 7 mg (a standard assumption), then: \[ \text{ppm} = \frac{7}{1.028} \approx 6.8 \text{ ppm} \] Step 5: Conclusion
Using standard assumptions for dissolved oxygen in water: \[ (C) \quad 6.8 \text{ ppm} \]