A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4 will be needed to produce 650mL of a solution that has a concentration of Na+ ions of 1.50M ? Explain answer with three significant figures and include the appropriate units:)
To find the number of grams of Na3PO4 needed, we need to use the concentration of Na+ ions in the solution and the volume of the solution.
First, let's determine the number of moles of Na+ ions required:
Concentration (C) is given as 1.50 M, which represents 1.50 moles of Na+ ions per liter of solution.
To convert the volume from milliliters (mL) to liters (L), divide it by 1000:
650 mL ÷ 1000 = 0.650 L
Now, we can calculate the number of moles using the formula:
moles (n) = concentration (C) × volume (V)
moles (n) = 1.50 mol/L × 0.650 L
moles (n) = 0.975 mol
Next, we need to find the molar mass of Na3PO4 to convert moles to grams. The molar mass of Na3PO4 can be calculated by adding up the atomic masses of each element present in the formula:
Na3PO4 = 3(Na) + 1(P) + 4(O)
Na3PO4 = 3(22.99 g/mol) + 1(30.97 g/mol) + 4(16.00 g/mol)
Na3PO4 = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol
Na3PO4 = 163.94 g/mol
Finally, we can determine the mass of Na3PO4 using the formula:
mass (m) = moles (n) × molar mass (M)
mass (m) = 0.975 mol × 163.94 g/mol
mass (m) = 159.44 g
Therefore, approximately 159.44 grams of Na3PO4 will be needed to produce 650 mL of a solution with a concentration of Na+ ions of 1.50 M.
If Na^+ = 1.5M then Na3PO4 will be 1/3 that or 0.05M
Then how many mols do I need? That's M x L = 0.05 x 0.650 L = ?
Finally, mols = grams/molar mass. You know molar mass and mols, solve for grams.