How many grams of Na3PO4 are needed to produce 250mL solution that has an Na concentration of 1.04M?

1.04 M Na = 1.04/3 M Na3PO4 = 0.347

M Na3PO4
M Na3PO4 = mols Na3PO4/L Na3PO4. You know L and M, solve for mols.
Then mol Na3PO4 = grams Na3PO4/molar mass Na3PO4.
You know M

To calculate the number of grams of Na3PO4 needed, we can use the formula:

Number of moles = Concentration x Volume

First, let's convert the volume from milliliters (mL) to liters (L):

250 mL = 250/1000 L = 0.25 L

Next, we can calculate the number of moles of Na+ ions needed:

Number of moles of Na+ = Concentration x Volume

Number of moles of Na+ = 1.04 mol/L x 0.25 L = 0.26 mol

Since Na3PO4 contains three Na+ ions per formula unit, we need to multiply the number of moles of Na+ by the conversion factor of 1 mol Na3PO4 / 3 mol Na+:

Number of moles of Na3PO4 = (0.26 mol Na+) / (3 mol Na3PO4 / 1 mol Na+) = 0.087 mol Na3PO4

Finally, we can calculate the number of grams of Na3PO4 using its molar mass. The molar mass of Na3PO4 is approximately 163.94 g/mol:

Mass of Na3PO4 = number of moles x molar mass

Mass of Na3PO4 = 0.087 mol x 163.94 g/mol ≈ 14.238 g

Therefore, approximately 14.238 grams of Na3PO4 are needed to produce a 250 mL solution with an Na concentration of 1.04M.

To determine the number of grams of Na3PO4 needed to produce a 250 mL solution with an Na concentration of 1.04 M, we can use the formula:

Molarity (M) = moles of solute / volume of solution (L)

First, we need to convert the volume from milliliters (mL) to liters (L):

Volume = 250 mL = 250 / 1000 = 0.25 L

Next, rearrange the formula to solve for moles of solute:

moles of solute = Molarity × volume of solution (L)

Substitute the values into the equation:

moles of solute = 1.04 M × 0.25 L = 0.26 moles

Since Na3PO4 dissociates into four sodium ions (Na+), we can calculate the number of moles of Na+ ions as:

moles of Na+ ions = 0.26 moles × 4 = 1.04 moles

Finally, to determine the mass of Na3PO4 needed, we need to multiply the number of moles of Na+ ions by the molar mass of Na3PO4:

Molar mass of Na3PO4 = (22.99 g/mol × 3) + (15.999 g/mol × 1) + (31.9988 g/mol × 4) = 163.9408 g/mol

Mass of Na3PO4 = moles of Na+ ions × molar mass of Na3PO4
= 1.04 moles × 163.9408 g/mol
≈ 170.464 g

Therefore, approximately 170.464 grams of Na3PO4 are needed to produce a 250 mL solution with an Na concentration of 1.04 M.