Question

PLS HELP ! 5.00 mL of a stock solution, containing 0.200 M of Na3PO4 was pipetted into a 25.00 mL volumetric flask and made up to the calibration mark with deionised water. A serial dilution was carried out for another two times to obtain the final solution of desired concentration. Calculate the concentration of the final solution, express in ppm (by volume).

Answer 1

262 ppm of Na₃PO₄

Step-by-step explanation:

In a dilution, the concentration of the initial solution is decreased. When you take 5.00mL of the solution that is diluted to 25.0mL The solution is diluted 25/5 = 5 times

If you make another two serial dilutions the final solution wil decrease its concentration 5*5*5 = 125 times

As original solution containing 0.200 M of Na3PO4, the final solution will have a concentration of:

0.200M / 125 = 1.6x10⁻³M

Molarity is defined as the ratio between moles and liters. 1.6x10⁻³ moles of Na3PO4 in 1L are:

1.6x10⁻³mol ₓ (164g/mol) = 0.262g Na₃PO₄ / L

Assuming density of Na3PO4 as 1g/mL the concentration of the solution is:

0.262mL Na₃PO₄ / L

As 1mL = 1000μL:

262μL Na₃PO₄ / L

μL of solute per L of solution is equal to ppm, that means the solution has:

262 ppm of Na₃PO₄