Answers
Answer:
B B
W BW BW
W BW BW
Explanation:
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Related Questions
Quinine in a 1.664-g antimalarial tablet was dissolved in sufficient 0.10 M 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 327 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.
Answers
Answer:
605.6mg of quinine
Explanation:
Based on Lambert-Beer's law, the intensity of an optical measurement is directely proportional to its concentration.
Unknown concentration gives a reading of 327
100 ppm gives 180 of intensity
The concentration of the diluted quinine tablet is:
327 * (100ppm / 180) = 181.67 ppm. The final dilution
The concentration of the first diluted solution is:
181.67 ppm * (100.0mL / 15mL) = 1211ppm
ppm could be defined as mass of solute (In this case of quinine) in mg per liter of solution. That is:
1211mg / L
As the tablet was diluted to 500mL = 0.5L, the mass of the quinine is:
0.5L * (1211mg / L) =
605.6mg of quinine