One kilogram of a sea water sample contains 6 mg of dissolved `O_(2)`. The concentration of `O_(2)` in the sample in ppm is

Options

A. 0.6
B. 6
C. 60
D. 16

Explanation

To find the concentration of dissolved oxygen (O₂) in the seawater sample in parts per million (ppm), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Data**:
- We have 1 kg of seawater which contains 6 mg of dissolved O₂.

2. **Convert Kilograms to Milligrams**:
- 1 kg = 1,000 grams = 1,000,000 mg (since 1 kg = 10^3 g and 1 g = 10^3 mg).
- Therefore, 1 kg of seawater = 10^6 mg of seawater.

3. **Set Up the Concentration Formula**:
- The concentration in ppm (parts per million) can be calculated using the formula:
\[ \text{ppm} = \left( \frac{\text{mass of solute (mg)}}{\text{mass of solution (mg)}} \right) \times 10^6 \]

4. **Substitute the Values**:
- Here, the mass of the solute (dissolved O₂) = 6 mg.
- The mass of the solution (seawater) = 10^6 mg.
- Now substituting these values into the formula:
\[ \text{ppm} = \left( \frac{6 \text{ mg}}{10^6 \text{ mg}} \right) \times 10^6 \]

5. **Calculate the Concentration**:
- The \(10^6\) mg in the numerator and denominator cancels out:
\[ \text{ppm} = 6 \]

6. **Final Answer**:
- The concentration of O₂ in the seawater sample is **6 ppm**.