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vaieri [72.5K]
3 years ago
10

Quinine in a 1.664-g antimalarial tablet was dissolved in sufficient 0.10 M HCl to give 500 mL of solution. A 15.00-mL aliquot w

as then diluted to 100.0 mL with the acid. The fluorescence intensity for the diluted sample at 347.5 nm provided a reading of 288 on an arbitrary scale. A standard 100 ppm quinine solution registered 180 when measured under conditions identical to those for the diluted sample. Calculate the mass of quinine in milligrams in the tablet.
Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
4 0

Answer:

533.33 mg quinine

Explanation:

First we<u> calculate the concentration of the diluted sample</u>, using the values obtained by the standard:

  • 288 * 100 ppm / 180 = 160 ppm

Now we use the dilution factors to <u>calculate the concentration of quinine in the original sample</u>:

  • 160 ppm * 100mL/15mL = 1066.67 ppm

ppm can be defined as <u>mg of quinine</u>/L solvent:

500 mL solution ⇒ 500/1000 = 0.5 L

  • 1066.67 ppm * 0.5L = 533.33 mg quinine
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To find - Identify what kind of ligand (weak or strong), what kind

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