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$^{-4}$ m. x = _______. (Nearest integer)
Given: Molar mass of NaCl is 58.5 g mol$^{-1}$Molar mass of O$_2$ is 32 g mol$^{-1}$.
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$^{-4}$ m. x = _______. (Nearest integer)
Given: Molar mass of NaCl is 58.5 g mol$^{-1}$Molar mass of O$_2$ is 32 g mol$^{-1}$.
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Correct Answer: 2
Approach Solution - 1
Sea water is 6 Molar in NaCl, which means that 1000 mL of sea water contains 6 mol of NaCl.
Step 1: Calculate the mass of the solution
\[ \text{Mass of solution} = \text{Volume} \times \text{Density} \]
\[ = 1000 \times 2 = 2000 \, g \]
Step 2: Calculate the mass of O2 from ppm
\[ \text{ppm} = \frac{\text{mass of O}_2}{2000} \times 10^6 \]
Given that the O2 concentration is \( 5.8 \times 2 \times 10^{-3} \):
\[ \text{Mass of O}_2 = 1.16 \times 10^{-2} \, g \]
Step 3: Calculate molality of O2
\[ \text{Molality of O}_2 = \frac{1.16 \times 10^{-2} / 32}{(2000 - 6 \times 58.5)} \times 1000 \]
Simplifying,
\[ = \frac{1.16 \times 10^{-2}}{32 \times 1649} \]
\[ = 0.000219 = 2.19 \times 10^{-4} \]
Final Answer:
\[ \boxed{\text{Correct answer } = 2} \]
Approach Solution -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 \( 2.19 \times 10^{-6} \).
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