Sea water, which can be considered as a 6 molar (6 M) solution of NaCl, has a density of 2 g mL$$^{-1}$$. The concentration of dissolved oxygen (O$$_2$$) in sea water is 5.8 ppm. Then the concentration of dissolved oxygen (O$$_2$$) in sea water, is $$x \times 10^{-4}$$ m. (Nearest integer)
Given: Molar mass of $$NaCl$$ is 58.5 g $$mol^{-1}$$
          Molar mass of $$O_{2}$$ is 32 g $$mol^{-1}$$

For dissolved oxygen we must convert ppm (mass ratio) to molality (moles of $$\mathrm{O_2}$$ per kg of water).

Case 1: Work with exactly $$1\ \text{L}$$ of sea water (any convenient volume can be chosen).

Density is $$2\ \text{g mL}^{-1}$$, so mass of this $$1\ \text{L} = 1000\ \text{mL}$$ of solution is
$$1000 \times 2 = 2000\ \text{g}$$.

Case 2: Separate solute (NaCl) from solvent (water).

The solution is $$6\ \text{M}$$ in NaCl, i.e. $$6$$ moles NaCl per litre.
Molar mass of NaCl is $$58.5\ \text{g mol}^{-1}$$, therefore mass of NaCl in $$1\ \text{L}$$ is
$$6 \times 58.5 = 351\ \text{g}$$.

Mass of water (solvent) present = total mass − mass of NaCl:
$$2000 - 351 = 1649\ \text{g} = 1.649\ \text{kg}$$.

Case 3: Find moles of dissolved $$\mathbf{O_2}$$ present in the same litre.

Concentration of $$\mathrm{O_2}$$ is $$5.8\ \text{ppm}$$. For mass ratios, $$1\ \text{ppm} = 1\ \text{mg}$$ solute per $$1\ \text{kg}$$ solution.

Hence in $$1\ \text{kg}$$ solution, mass of $$\mathrm{O_2} = 5.8\ \text{mg} = 0.0058\ \text{g}$$.
Our sample is $$2000\ \text{g} = 2\ \text{kg}$$ of solution, so

Mass of $$\mathrm{O_2} = 2 \times 0.0058 = 0.0116\ \text{g}$$.

Molar mass of $$\mathrm{O_2} = 32\ \text{g mol}^{-1}$$, therefore moles of $$\mathrm{O_2}$$ are
$$\frac{0.0116}{32} = 3.625 \times 10^{-4}\ \text{mol}$$.

Case 4: Convert to molality.

Molality $$m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}$$ = $$\frac{3.625 \times 10^{-4}}{1.649} = 2.20 \times 10^{-4}\ m$$.

Expressed as $$x \times 10^{-4}\ m$$, the value of $$x \approx 2.20$$, which to the nearest integer is $$2$$.

Therefore the concentration of dissolved oxygen is $$\mathbf{2 \times 10^{-4}\ m}$$ (nearest integer $$x = 2$$).