Given that the molar mass of [tex]$NaNO_3[tex]$[/tex] is [tex]$[/tex]85.00 \, g/mol[tex]$[/tex], what mass of [tex]$[/tex]NaNO_3[tex]$[/tex] is needed to make [tex]$[/tex]4.50 \, L[tex]$[/tex] of a [tex]$[/tex]1.50 \, M[tex]$[/tex] [tex]$[/tex]NaNO_3$[/tex] solution?

Use the formula for molarity: [tex]\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}[/tex].

A. [tex]6.75 \, g[/tex]
B. [tex]18.9 \, g[/tex]
C. [tex]255 \, g[/tex]
D. [tex]574 \, g[/tex]



Answer :

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]