One litre sea water (d = 1.03g/`cm^3` ) contains 10.3 mg `O_2` gas. Determine concentration of `O_2` in ppm.

Answer

To determine the concentration of \( O_2 \) in parts per million (ppm) in seawater, we can follow these steps:

### Step 1: Understand the definition of ppm
PPM stands for "parts per million." It is a way to express very dilute concentrations of substances. Specifically, it indicates how many parts of a substance are present in one million parts of a solution.

### Step 2: Gather the given data
- Volume of seawater = 1 liter = 1000 cm³
- Density of seawater = 1.03 g/cm³
- Mass of \( O_2 \) in seawater = 10.3 mg

### Step 3: Convert the mass of \( O_2 \) to grams
To convert milligrams to grams:
\[ 10.3 \text{ mg} = 10.3 \times 10^{-3} \text{ g} = 0.0103 \text{ g} \]

### Step 4: Calculate the mass of seawater
Using the density formula:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
We can rearrange this to find the mass of seawater:
\[ \text{Mass of seawater} = \text{Density} \times \text{Volume} \]
Substituting the values:
\[ \text{Mass of seawater} = 1.03 \text{ g/cm}^3 \times 1000 \text{ cm}^3 = 1030 \text{ g} \]

### Step 5: Calculate the concentration of \( O_2 \) in ppm
To find the concentration in ppm, we use the formula:
\[ \text{Concentration (ppm)} = \left( \frac{\text{mass of } O_2}{\text{mass of seawater}} \right) \times 10^6 \]
Substituting the values:
\[ \text{Concentration (ppm)} = \left( \frac{0.0103 \text{ g}}{1030 \text{ g}} \right) \times 10^6 \]
Calculating the fraction:
\[ \frac{0.0103}{1030} = 1.000 \times 10^{-5} \]
Now, multiplying by \( 10^6 \):
\[ \text{Concentration (ppm)} = 1.000 \times 10^{-5} \times 10^6 = 10.3 \text{ ppm} \]

### Final Answer
The concentration of \( O_2 \) in seawater is **10.3 ppm**.

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