Question

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 was 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.

Answer 1

Answer:

533.33 mg quinine

Explanation:

First we calculate the concentration of the diluted sample, using the values obtained by the standard:

• 288 * 100 ppm / 180 = 160 ppm

Now we use the dilution factors to calculate the concentration of quinine in the original sample:

• 160 ppm * 100mL/15mL = 1066.67 ppm

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

500 mL solution ⇒ 500/1000 = 0.5 L

• 1066.67 ppm * 0.5L = 533.33 mg quinine