Exactly 5.00 m aliquots of a solution containing Phenobarbital were measured int

Question

Exactly 5.00 m aliquots of a solution containing Phenobarbital were measured into 50.00 mL volumetric flasks and made basic with KOH 0, 0.1, 0.15, 0.2, 0.25 and 0.3 mL volumes of 2.0 mu g/mL of standard solution of Phenobarbital were then introduced into each flask and the mixture was diluted to volume of 50 mL. The fluorescence of each of these solutions was measured with a fluorometer, which gave values of 1.30, 3.26, 4.80, 6.41, 8.02, and 9.56, respectively. Plot the data. Using the plot from (a), calculate the concentration of phenobarbital in the unknown. Find the concentration of phenobarbital from the equation in (c). Calculate Signal to Noise Ratio for the audio signal series (50, 25, 22, 35, 28, ) The following data were collected for fluorescence measurements on a noisy system: 0.101, 0.134, 0.109, 0.122, 0.143, 0.105. Assuming the noise random. Calculate the signal-to-noise ratio. How many measurements must be done to have S/N = 20. Calculate the frequency in Hertz, wavenumber, energy (J) and kJ/mol for electromagnetic radiation of 330 nm 2400 nm. Data is collected for a chemical analysis

Explanation / Answer

3)

Concentration you can calculate with Beer-Lamberts Law.

Absorbance A= E(molar absorption constant) C(Concentration) l(Path length)

2) SNR= Psig/Pnoise

SNR in desibell is= Psig db - Pnoise db

4)

For 330 nm

Speed of light = v = 3 x 108 m/s

Wave length = = 330 nm

Frequency = f = v/lamnda

                        = 90.90X104 Hz

Wavenumber =n=

1/ lambda= 1/330 = 0.003030303

Energy (J)

=E = h = (6.626 x 10¯34 J s) (0.003030303)

=2.0078X 10-36 J

For 2400 nm

Frequency f= 125000 Hz

Wavenumber =n= 0.0004166666

Energy = E= 2.7608X10-37 J

Phenobarbital ml KOH ml Fluroscence 5 0 1.30 5 0.1 3.26 5 0.15 4.80 5 0.2 6.4 5 0.25 8.02 5 0.3 9.56

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