To determine the molarity of a solution that contains 87.75 g of NaCl in 500 mL of solution, we can follow these steps:
1. Convert the volume of the solution from milliliters to liters:
[tex]\[
\text{Volume of solution in liters} = \frac{\text{Volume in mL}}{1000}
\][/tex]
Given the volume of the solution is 500 mL:
[tex]\[
\text{Volume in liters} = \frac{500 \text{ mL}}{1000} = 0.5 \text{ L}
\][/tex]
2. Calculate the number of moles of NaCl:
The number of moles ([tex]\(n\)[/tex]) is calculated by dividing the mass ([tex]\(m\)[/tex]) of the solute (NaCl) by its molar mass (M).
Given:
- The mass of NaCl is 87.75 g
- The molar mass of NaCl is 58.44 g/mol
[tex]\[
\text{Moles of NaCl} = \frac{\text{mass of NaCl}}{\text{molar mass of NaCl}} = \frac{87.75 \text{ g}}{58.44 \text{ g/mol}} \approx 1.501275 \text{ moles}
\][/tex]
3. Calculate the molarity of the solution:
Molarity (M) is defined as the number of moles of solute divided by the volume of the solution in liters.
[tex]\[
\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{1.501275 \text{ moles}}{0.5 \text{ L}} \approx 3.00255 \, \text{M}
\][/tex]
Given the possible choices and knowing our exact result is approximately 3.003, the correct choice can be verified:
- [tex]\(0.333 \, \text{M}\)[/tex]: Too low
- [tex]\(0.751 \, \text{M}\)[/tex]: No, also too low
- [tex]\(150 \, \text{M}\)[/tex]: Improbable, excessively high
- [tex]\(300 \, \text{M}\)[/tex]: Also too high
None of the given options closely match the calculated 3.003 M, but strictly speaking, [tex]\(300 \, \text{M}\)[/tex] is the closest by relative magnitude (though off by a factor of 100).
However, under realistic lab identification or classroom clarification, the code or value of [tex]\(0.751 \, \text{M}\)[/tex] indicated a miscalculated step or input. Verifying choices and methods correctly, we assume 300-3 recrunchg versions that align with fundamental principles_ `_initial 3.003 serves an anchoring correctness despite matching anomaly in answer keys.
