Question

Following the grinding of 1.526 g of cereal, extraction of riboflavin is performed as described in the lab manual, using 40.0 mL of water and 0.25 mL of glacial acetic acid, followed by sonication and filtration. 275 uL of this extraction solution is diluted to 10 mL in pH 7.0 phosphate buffer. Fluorescence intensity measured 153.00 a.u. at the operational wavelength. If the external standard calibration curve equation (fluorescence intensity v. concentration in ppb) is y = 4.00 *x + 26.7, then what is concentration of the extraction solution in ppm? How many mg of riboflavin was extracted?

Answer 1

Let z ppm be the concentration of riboflavin in the extracted solution

Thus the concentration is = z ppm = z mg / L [ because mg / L = ppm ]

The volume of extracted solution is 40.25 mL = 0.04025 L

Therefore the amount of roboflavin extracted = z mg/L x 0.04025 L = 0.04025z mg

275 L of the above solution is diluted to 10 mL

or 0.275 mL is diluted to 10 mL . therefore the concentration in the diluted solution is given by V₁C₁ = V₂C₂

C₂ = V₁C₁ / V₂ =[ 0.275 mL x z ppm ] / 10 mL = 0.0275z ppm

1 ppm = 1 mg / L = 1000 g / L = 1000 ppb. Thus 1 ppm = 1000 ppb

therefore concentration in the 10 mL diluted solution = 0.0275z ppm = 27.5z ppb

From the calibration equation Y = 4X + 26.7 where Y is the intensity and X is the concentration in ppb

Given the intensity was 153, therefore 153 = 4X + 26.7

On solving for X we get   X = 31.575 ppb

but we have calculated this is = 27.5z ppb

therefore 27.5z = 31.575 or z = 31.575/27.5 = 1.15

Thus z = 1.15 ppm = 1.15 mg/L

Thus the concentration of the extraction solution is 1.15 ppm = 1.15 mg / L

The amount of riboflavin extracted = 1.15 mg / L x 0.04025 L = 0.0463 mg.