Incorrect Question 7 of 11 Incorrect Incorrect Map em Sapling Learning macmillan
ID: 509321 • Letter: I
Question
Incorrect Question 7 of 11 Incorrect Incorrect Map em Sapling Learning macmillan learning A 25.0-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 150.0 mL. At 347.5 nm, the fluorescence intensity of the diluted sample was measured as 91.9 on an arbitrary scale. Under similar conditions, a 25.0 ppm standard quinine solution had a fluorescence intensity of 177. Calculate the mass of quinine in the original tonic water sample. Number 0.075 mgExplanation / Answer
It would be nice to have a standard calibration curve of quinine instead on one measurement.
Anyway, let's calculate;
The fluorescence intensity of the diluted sample = 91.9 a.u.
The standard 25 ppm solution intensity .......177.
The diluted sample concentration = (25 x 91.9)/177 = 12.98 ppm
The diluted sample is actually from the dilution of 25 (V1) mL of the original solution to 150 mL(V2) solution.
M2 = 12.98 ppm
So, we can use M1V1 = M2V2
M1 = M2V2/V1 = 12.98 ppm x 150mL/25mL = 77.88 ppm.
The unknown sample concentration = 77.88 ppm. = 77.88 mg/L
we need to calculate the mass of quinine present in 500mL solution. = 77.88 (mg/L) x (0.5 L) = 38.94 mg