To determine the mass of sodium nitrate (\(NaNO_3\)) needed to make a 4.50 L solution with a molarity of 1.50 M, we will follow these steps:
1. Calculate the moles of \(NaNO_3\) using the molarity formula:
[tex]\[
\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}
\][/tex]
Rearranging this formula to solve for moles of solute, we get:
[tex]\[
\text{moles of solute} = \text{Molarity (M)} \times \text{liters of solution}
\][/tex]
Given:
- Molarity (\(M\)) = 1.50 \(M\)
- Volume of solution (liters) = 4.50 \(L\)
Plugging in the values:
[tex]\[
\text{moles of } NaNO_3 = 1.50 \, M \times 4.50 \, L = 6.75 \, \text{moles}
\][/tex]
2. Calculate the mass of \(NaNO_3\) using its molar mass:
[tex]\[
\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}
\][/tex]
Given:
- Moles of \(NaNO_3\) = 6.75 moles
- Molar mass of \(NaNO_3\) = 85.00 \(g/mol\)
Plugging in the values:
[tex]\[
\text{mass of } NaNO_3 = 6.75 \, \text{moles} \times 85.00 \, \frac{\text{g}}{\text{mol}} = 573.75 \, \text{g}
\][/tex]
Therefore, the mass of \(NaNO_3\) needed is approximately \(574 \, g\). The correct answer from the given options is:
[tex]\(574 \, g\)[/tex]
