Question

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, in x $\times$ 10$^{-6}$ m. x = _______. (Nearest integer)
Given: Molar mass of NaCl is 58.5 g mol$^{-1}$Molar mass of O$_2$ is 32 g mol$^{-1}$.

Hint:

When dealing with ppm and molarity, remember that ppm represents parts per million, and it can be converted to mass per volume. Once you have the mass of a solute, converting it to moles using the molar mass will give you the molarity.

Answer 1

Correct Answer: 2

We are given that sea water is a 6 molar (6 M) solution of NaCl and has a density of 2 g mL$^{-1}$. We also know that the concentration of dissolved oxygen (O$_2$) in sea water is 5.8 ppm. 
First, we need to calculate the concentration of O$_2$ in mol/L using the given data: 
1. Molar mass of NaCl = 58.5 g/mol 
- Sea water has a molarity of 6 M, meaning each liter of sea water contains 6 moles of NaCl. 
- Therefore, in 1 liter of sea water, the mass of NaCl is: \[ \text{mass of NaCl} = 6 \times 58.5 = 351 \text{ grams} \] 
2. Given that the density of the solution is 2 g/mL, the mass of 1000 mL (1 L) of sea water is: \[ \text{mass of solution} = 2 \times 1000 = 2000 \text{ grams} \] 
3. Now, we can calculate the mass of O$_2$ dissolved in 1 liter of sea water using the given ppm value of 5.8 ppm: \[ \text{ppm of O}_2 = \frac{\text{mass of O}_2}{\text{mass of solution}} \times 10^6 \] \[ \text{mass of O}_2 = 5.8 \times 10^{-3} \text{ g} \] \[ \text{mass of O}_2 = 5.8 \text{ mg} = 5.8 \times 10^{-3} \text{ grams} \] 
4. To calculate the molarity (mol/L) of O$_2$ in the solution, we use the molar mass of O$_2$: \[ \text{molality for O}_2 = \frac{5.8 \times 10^{-3}}{32} = 1.81 \times 10^{-4} \text{ moles} \] This corresponds to a concentration of O$_2$ as: \[ = 2.19 \times 10^{-4} \text{ mol/L} \] 
Therefore, the concentration of O$_2$ in sea water is \( \mathbf{2.19 \times 10^{-4}} \) mol/L, which is approximately \( 2.19 \times 10^{-6} \).