5. The organic chemical aniline, CHsNH2, was to be determined instrumentally. 5.

Question

5. The organic chemical aniline, CHsNH2, was to be determined instrumentally. 5.0 milligrams of pure aniline was dissolved in 1.00 liter of water. The instrument aralyzed this solution, and gave an output reading of 0.655. Assume the instrument responds linearly. An impure unknown sample of aniline weighing 0.200 grams was then dissolved in 500.0 ml of water. A 25.00 ml aliquot of this was then diluted to 250.0 ml, and mixed well. A 10.0 ml aliquot of this was then diluted to 100.0 ml, and measured on the same instrument under the same conditions, giving a reading of 0.425 units. What was the % purity of the unknown aniline sample? (10 pts)

Explanation / Answer

Concentration of pure aniline=mol of aniline/volume of solution

mol of aniline=mass/molar mass=5.0*10^-3g/93.13g/mol=0.0537*10^-3 mol

Thus,Concentration of pure aniline=Cp=0.0537*10^-3 mol/1.0 L=5.377*10^-5 mol/Lor M

As the output reading is linear to the concentration value ,so Cp=k*O   [O=output,k=constant]

concentration of stock of impure anilne=( 0.2g/93.13g/mol) /0.5L=0.00429 mol/L

After first Dilution ,the concentration=(25ml/250ml)*0.00429 mol/L=0.000429 mol/L

After second dilution ,concentration: (10ml/100ml)*0.000429 mol/L=0.0000429 mol/L

So, the final concentration of unknown=4.29*10^-5 mol/L

Calculated value for concentration of impure aniline:

Cp/Cu=Op/Ou [where Ou = output for the impure aniline)

Cu=(Ou/OP)*Cp=(0.425/0.655)* 5.377*10^-5 mol/L=3.489*10^-5 mol/L

% purity=(actual concentration/expected concentration)*100=(3.489*10^-5 mol/L/4.29*10^-5 mol/L)*100=81.328%

Answer:81.3%

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