Question

10.30 mg of O₂ is dissolved into a liter of sea water of density 1.03 g/mL. The concentration of O₂ in ppm is _______.

Answer 1

10

Answer 2

The concentration in parts per million (ppm) is defined as the mass of solute per mass of solution multiplied by $10^6$.

$$\text{Concentration (ppm)} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 10^6$$

Given:

• Mass of solute (O₂) = $10.30 \text{ mg}$
• Volume of sea water = $1 \text{ liter}$
• Density of sea water = $1.03 \text{ g/mL}$

First, convert the mass of solute from milligrams to grams:

Mass of O₂ = $10.30 \text{ mg} = 10.30 \times 10^{-3} \text{ g} = 0.01030 \text{ g}$

Next, convert the volume of sea water from liters to milliliters:

Volume of sea water = $1 \text{ liter} = 1000 \text{ mL}$

Now, calculate the mass of the sea water (solution) using its volume and density:

Mass of sea water = Density $\times$ Volume

Mass of sea water = $1.03 \text{ g/mL} \times 1000 \text{ mL} = 1030 \text{ g}$

Finally, calculate the concentration of O₂ in ppm:

$$\text{Concentration (ppm)} = \frac{\text{Mass of O₂}}{\text{Mass of sea water}} \times 10^6$$ $$\text{Concentration (ppm)} = \frac{0.01030 \text{ g}}{1030 \text{ g}} \times 10^6$$ $$\text{Concentration (ppm)} = \frac{10.30 \times 10^{-3}}{1030} \times 10^6$$ $$\text{Concentration (ppm)} = \frac{10.30}{1030} \times 10^{-3} \times 10^6$$ $$\text{Concentration (ppm)} = 0.01 \times 10^3$$ $$\text{Concentration (ppm)} = 10$$

The concentration of O₂ in ppm is 10.

Explanation:

Convert the mass of O₂ from mg to g (10.30 mg = 0.01030 g). Convert the volume of sea water from L to mL (1 L = 1000 mL). Calculate the mass of sea water using its density and volume (1.03 g/mL * 1000 mL = 1030 g). Calculate the concentration in ppm using the formula: (mass of solute / mass of solution) * 10⁶ = (0.01030 g / 1030 g) * 10⁶ = 10 ppm.