To find the molarity of the solution, we will follow these steps in detail:
1. Determine the molar mass of sodium hydrogen carbonate ([tex]\(NaHCO_3\)[/tex]):
- Sodium (Na): [tex]\( 22.99 \, \text{g/mol} \)[/tex]
- Hydrogen (H): [tex]\( 1.01 \, \text{g/mol} \)[/tex]
- Carbon (C): [tex]\( 12.01 \, \text{g/mol} \)[/tex]
- Oxygen (O): [tex]\( 16.00 \, \text{g/mol} \times 3 = 48.00 \, \text{g/mol} \)[/tex]
Therefore, the total molar mass of [tex]\(NaHCO_3\)[/tex] is:
[tex]\[
22.99 + 1.01 + 12.01 + 48.00 = 84.01 \, \text{g/mol}
\][/tex]
2. Convert the volume of the solution from milliliters to liters:
[tex]\[
\text{Volume of solution} = 250.0 \, \text{mL} = 250.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.25 \, \text{L}
\][/tex]
3. Calculate the number of moles of [tex]\(NaHCO_3\)[/tex] in the solution:
Using the formula:
[tex]\[
\text{Moles of } NaHCO_3 = \frac{\text{mass of } NaHCO_3}{\text{molar mass of } NaHCO_3}
\][/tex]
[tex]\[
\text{Moles of } NaHCO_3 = \frac{12.0 \, \text{g}}{84.01 \, \text{g/mol}} \approx 0.14284 \, \text{moles}
\][/tex]
4. Calculate the molarity of the solution:
Molarity (M) is given by the formula:
[tex]\[
\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}
\][/tex]
[tex]\[
\text{Molarity} = \frac{0.14284 \, \text{moles}}{0.25 \, \text{L}} \approx 0.571 \, M
\][/tex]
Therefore, the molarity of the solution is [tex]\(0.571 \, M\)[/tex]. The correct answer from the provided choices is:
[tex]\[
\boxed{0.571 \, M}
\][/tex]
