1. Determine the concentration of the pure ASA in the 10.0mL of 1 M NaOH in mg/m

Question

1. Determine the concentration of the pure ASA in the 10.0mL of 1 M NaOH in mg/mL using the Beer's law plot. Then calculate the mass (in mg) of pure ASA knowing that two consecutive dilutions were done to prepare the solution

yout Reference Mailings Review View * A Aa - E.F.EE| 21 | A bct Abccc A XA-A-===2--Normal I No Spac. Heading1 Heading 2 Title Subt Font Paragraph Styles Table 1: Qualitative Observations Part A: Microscale Production of Acetylsalicylic -salicylic acid (C,H,O3 was liquid, clear, Acid (ASA) colourless CHO3 (ag + CHO C6H,049+ -acetic anhydride (C4H6O3 (9) was solid, powder, CH3COOH (as) white -solution of “impure" ASA + acetic acid contained white solute/precipitate; cloudy solution Part B: Isolation of ASA Crystals -white, crystal powder formed on filter paper | Part C: Melting Point of Synthesized ASA vs. | -crystals took very long time to melt Standard ASA -when melted (liquid), it was clear, colourless Part D: Impure ASA + Iron (III) Chloride -solution turned dark brown, dark purple Table 2: Measurements from Synthesis and Isolation of Pure ASA Mass of salicylic acid (g) 0.14 Mass of filter paper (g) 0.38 Mass of “impure" ASA and filter paper (g) | 0.50 Mass of prepared “impure" ASA (g) | 0.12 Melting point of prepared ASA (*C) 132.5 Melting point of standard ASA (*C) Table 3: Absorbance Readings of Standard ASA Sample Absorbance Reading 0 0 e 60 M N O

Explanation / Answer

Use the regression equation to find out the concentration of the dilute ASA in the sample. The dilute solution of the prepared ASA recorded an absorbance of 0.601. Put y = 0.601 in the regression equation and obtain

0.601 = 8.935x + 0.0346

====> 8.935x = 0.601 – 0.0346 = 0.5664

====> x = 0.5664/8.935 = 0.06339 0.0634

The concentration of prepared ASA in 10.0 mL of 1 M NaOH solution is 0.0634 mg/mL (ans).

I need to know the dilutions to find out the mass of pure ASA (ans).

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