Quinine in a 0.8320 g antimalarial tablet was dissolved into sufficient 0.10 M H

Question

Quinine in a 0.8320 g antimalarial tablet was dissolved into sufficient 0.10 M HCl to give 500 mL of solution. A 15 ml. aliquot was then diluted to 100 mL with the acid. The fluorescence intensity of this diluted solution (that had a 0.05 absorbance) at 345.7 nm was 144 (eps). Under identical instrumental conditions, the fluorescence intensity was 180 for a standard 100 ppm quinine solution. Calculate the mass in mg for the quinine tablet. What is the KCN used for in the limestone experiment? Why must Keep this solution basic?

Explanation / Answer

We have dissolved 0.8320 g antimalarial tablet into 500 mL solution = 1.664 g / L

Now we have taken 15mL aliquot so mass of tablet in aliquot = 1.664 X 0.015 = 0.02496 grams
Now this is diluted to 100mL so the final concentration is

0.02496 grams / 100mL or 0.2496 grams of tablet / Litre

One standard solution of concentration 100 ppm is given

100ppM = 100 mg quinine/L = 0.1 grams of quinine in one litre

It shows an intensity of 180
We know that absorption is directly proportional to concentration

Absorption = K X concentration

Absroption 1 / concentration 1 = Absorption 2 / concentration 2

180 / 0.1 = 144 / concentration 2

concentration 2 = 0.08 g quinine / L

This amount is from the intial concentration of tablet = 0.2496 g of tablet / Litre

So in a gram of tablet the amount of quinine = 0.08 / 0.2496 = 0.3205 grams per gram of tablet

So in 0.8320 grams of tablet the amount of quinine = 0.2667 grams quinine = 266.7 mg / 0.8320 grams of tablet

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