Question

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?

Answer 1

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 \times$10^-3 mol

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

= 2.26$\small \times$10^-3 (mol) $\small \times$ 180 (g/mol)

= 0.4068 g

= 0.4068 $\small \times$$\small \frac{1000 \, mg}{1\: g }$

= 406.8 mg

hence 406.8 miligrams ASA were in the tablet.