How many grams of K3PO4 would be needed to prepare a 1.73 M solution in a total volume of 427 mL?
Change 427 ml to liters
Multiply liters and the molarity (1.73) and ur answer will be in moles
Find the molar mass of k3po4 then convert your moles to grams using the molar mass
To determine the number of grams of K3PO4 needed to prepare a 1.73 M solution in a total volume of 427 mL, you will need to use the formula:
Molarity (M) = moles of solute (K3PO4) / volume of solution (in liters)
First, convert the volume of the solution from milliliters (mL) to liters (L) by dividing it by 1000:
Volume of solution (L) = 427 mL / 1000 = 0.427 L
Now, rearrange the formula to solve for moles of solute:
Moles of solute (K3PO4) = Molarity (M) * Volume of solution (in liters)
= 1.73 mol/L * 0.427 L
≈ 0.73971 moles
To calculate the mass of K3PO4, you need to know its molar mass, which can be found by adding up the atomic masses of its constituent elements. The molar mass of K3PO4 is:
Molar mass of K = 39.1 g/mol
Molar mass of P = 31.0 g/mol
Molar mass of O = 16.0 g/mol
Total molar mass of K3PO4 = (3 * Molar mass of K) + Molar mass of P + (4 * Molar mass of O)
= (3 * 39.1) + 31.0 + (4 * 16.0)
= 122.3 g/mol
Now, multiply the number of moles (0.73971 moles) by the molar mass (122.3 g/mol) to get the mass of K3PO4:
Mass of K3PO4 = Moles of K3PO4 * Molar mass of K3PO4
= 0.73971 moles * 122.3 g/mol
≈ 90.51 grams
Therefore, approximately 90.51 grams of K3PO4 would be needed to prepare a 1.73 M solution in a total volume of 427 mL.