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 Formula for PPM
The formula for calculating parts per million (ppm) is given by:

\[ \text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6 \]

### Step 2: Convert the Volume of Seawater to Mass
We know that the density of seawater is given as \( 1.03 \, \text{g/cm}^3 \) and the volume of seawater is \( 1 \, \text{L} \).

1 liter is equivalent to \( 1000 \, \text{cm}^3 \).

Using the formula for density:

\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]

We can rearrange this to find the mass:

\[ \text{Mass} = \text{Density} \times \text{Volume} = 1.03 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 1030 \, \text{g} \]

### Step 3: Convert the Mass of \( O_2 \) to Grams
The mass of \( O_2 \) gas in the seawater is given as \( 10.3 \, \text{mg} \). We need to convert this to grams:

\[ 10.3 \, \text{mg} = 10.3 \times 10^{-3} \, \text{g} = 0.0103 \, \text{g} \]

### Step 4: Substitute Values into the PPM Formula
Now we can substitute the values we have into the ppm formula:

\[ \text{ppm} = \frac{0.0103 \, \text{g}}{1030 \, \text{g}} \times 10^6 \]

### Step 5: Calculate the PPM
Now we perform the calculation:

\[ \text{ppm} = \frac{0.0103}{1030} \times 10^6 \]

Calculating the fraction:

\[ \frac{0.0103}{1030} \approx 0.00001 \]

Now multiplying by \( 10^6 \):

\[ \text{ppm} \approx 0.00001 \times 10^6 = 10.0 \, \text{ppm} \]

### Final Answer
The concentration of \( O_2 \) in seawater is approximately **10 ppm**.

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