Chemistry High School

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).

Answers

Answer 1

Answer:

262 ppm of Na₃PO₄

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

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