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17 March, 09:57

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.

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  1. 17 March, 10:20
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    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
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