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NNADVOKAT [17]
3 years ago
8

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 t

o 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).
Chemistry
1 answer:
JulijaS [17]3 years ago
4 0

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 = <em>5 times</em>

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 = <em>1.6x10⁻³M</em>

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:

<h3>262 ppm of Na₃PO₄</h3>

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⇒ with T = the temperature = 273.15 Kelvin

n(total) = p*V/RT = (3.32 atm*2.5 L)/(0.08206 L*atm/mol•K*273.15) = 0.3703 mol

Step 4: Calculate moles of Ne and F2

For one mole heated at constant volume,  

∆S = Cv*ln(288.15/273.15) = 0.05346*Cv

⇒ ∆S for 0.3703 mol,  

∆S = (0.3703 mol)(0.05346)Cv = 0.345 J/K

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For pure Ne, Cv = (3/2)R = 1.5*8.314 J/mol*K = 12.471 J/mol*K

For pure F₂, Cv = (5/2)R = 2.5 * 8.314 J/mol*K = 20.785 J/mol*K

if X is the mole fraction of Ne, we can find X by:

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moles Ne = (0.415)(0.3703 mol) = 0.154 mol

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