Final answer:
To prepare 325 mL of a 0.900 M Na+ solution, 15.98435 g of Na3PO4 is required after performing calculations using the molar mass of the compound and the desired concentration of Na+ ions.
Step-by-step explanation:
To find how many grams of Na3PO4 are needed to make 325 mL of a solution with a concentration of Na+ ions of 0.900 M, we perform the following calculations:
Calculate the number of moles of Na+ ions in the desired volume.
Since molarity (M) is moles per liter of solution, we have 0.900 moles of Na+ ions per liter.
For 0.325 liters (325 mL), we calculate:
0.900 moles/L × 0.325 L = 0.2925 moles of Na+ ions.
Since Na3PO4 contains three Na+ ions per formula unit,
we need a third of the moles of Na3PO4 as the number of moles of Na+ ions, so we divide by 3:
0.2925 moles Na+ / 3 = 0.0975 moles of Na3PO4.
To find the mass of Na3PO4 needed, we multiply by its molar mass.
However, the molar mass of Na3PO4 is not given directly here.
According to the periodic table, the molar mass can be calculated as follows:
Na (Sodium) ≈ 22.99 g/mole, P (Phosphorus) ≈ 30.97 g/mole, O (Oxygen) ≈ 16.00 g/mole.
The molar mass of Na3PO4 is (3 × 22.99) + 30.97 + (4 × 16.00) = 163.94 g/mole.
Thus, the mass of Na3PO4 needed is:
0.0975 moles × 163.94 g/mole = 15.98435 g.
15.98435 g of Na3PO4 will be needed to produce 325 mL of a 0.900 M Na+ solution.