A 50.0-mL aliquot was taken from a 500.0 mL sample of tonic water containing an unknown amount of quinine and diluted to a volume of 200.0 mL. At 347.5 nm, the fluorescence intensity of the diluted sample was measured as 127 on an arbitrary scale. Under similar conditions, a 25.0 ppm standard quinine solution had a fluorescence intensity of 193.
(a) Calculate the mass of quinine in the original tonic water sample.

0.0329g of quinine

Explanation:

If 25.0ppm of quinine solution had a fluorescence of 193, the solution that had a fluorescence of 127 has as concentration:

127 × (25.0ppm / 193) = 16.45ppm

As the solution was diluted from 50.0mL to 200.0mL, the sample tonic of water contains:

16.45ppm × (200.0mL / 50.0mL) = 65.8 ppm

As the original solution contains 500.0mL, mass of quinine is:

0.5000L × (65.8mg / L) = 32.9mg ≡ 0.0329g of quinine

- lead + phosphoric acid = hydrogen + lead(ii) phosphate