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B) In this experiment an aspirin ta and dissolved in 10 mL of 1.0M Na mark...

Question

B) In this experiment an aspirin ta and dissolved in 10 mL of 1.0M Na mark...

b) In this experiment an aspirin ta and dissolved in 10 mL of 1.0M Na mark with deionized water. Using a into a 100-mL volume
b) In this experiment an aspirin ta and dissolved in 10 mL of 1.0M Na mark with deionized water. Using a into a 100-mL volumetric flask and make a diluted aspirin solution'. ent an aspirin tablet was placed in a 250-ml volumetric flask 10 mL of 1.OM NaOH solution, the flask was then filled to the zed water. Using a pipette, 5.0 mL of this solution was added IL Volumetric flask and diluted to the mark with FeCl solution to A = absorbance C = concentration y = 1442.7x + 0.0149 0 0.0002 0.0008 0.001 0.0004 0.0006 C(mol/L) Using the standard curve above determine the concentration of acetylsalicylic acid in the 'diluted aspirin solution'[Fe3+ SA) if the absorbance reading was 0.667. 0.504 c) Calculate the concentration of the initial solution prepared by the dissolving the aspirin tablet in the 250 ml volumetric flask. If the mass of ASA is 180 g/mole, how many milligrams were in the tablet?
Science Chemistry
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Answers

b.

Bestline equation is

y = 1442.7 x + 0.0149

Absorbance is plotted in Y axis and Concentration of acetylsalicylic acid is in  X axis.

hence

0.667 = 1442.7 \small \times Concentration + 0.0149

or, 1442.7 \small \times concentration = 0.667 - 0.0149 = 0.6521

or, concentration = ( 0.6521/1442.7) = 4.52 \small \times 10-4 M or mol/L

C.

The 5 mL solution was diluted to 100 mL

hence,

M1V1 = M2V2

or, M1\small \times 5 =  4.52 \small \times 10-4 \small \times 100

or, M1 =  (4.52 \small \times 10-4 \small \times 100) /5

or, M1 = 9.04 \small \times 10-3 mol/L

now ,in 250 ml solution moles of ASA =   9.04 \small \times 10-3 \small \times\small \frac{250}{1000} = 2.26 \small \times10-3 mol

mass of ASA = moles of ASA \small \times molar mass of ASA

= 2.26\small \times10-3 (mol) \small \times 180 (g/mol)

= 0.4068 g

= 0.4068 \small \times1000 mg 19

= 406.8 mg

hence 406.8 miligrams ASA were in the tablet.

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