[undergroundboy]

You are analyzing riboflavin in wheat flour by fluorescence spectroscopy (?ex = 450 nm, ?em =530 nm). The extraction procedure is fairly complex. You obtain a 100g sample of the flour. You then weigh out 4.542 g of the flour, add it to a 250 mL volumetric flask, and dilute to volume with 1 N boiling HCl. After thoroughly mixing this solution, you combine 2 mL with 7 mL 1 N HCl. You then freeze-dry 4 mL of the diluted solution, and redissolve the residue to a total volume of 1.5 mL. You combine 0.5 mL of this solution with 0.5 mL ddH2O and measure the fluorescence. The following fluorescence values are obtained for the sample and standards. Prepare a blanked standard curve, and determine the riboflavin concentration in the flour (ppm) (10 points)

Fluorescence (?ex=450 nm, ?em=530 nm).

Sample

1.706
Riboflavin (20 mg/L)

0.822
Riboflavin (10 mg/L)

0.368
Riboflavin (5 mg/L)

0.121
Riboflavin (2 mg/L)

0.004
Riboflavin (1 mg/L)

1.706
Riboflavin (0 mg/L)

0.476
Flour

[undergroundboy]

You work for a dairy that produces specialty butters. You use the Mohr titration to measure the salt content of
the butter. You titrate 288.2 mg KCl in 100 mL ddH2O with aqueous AgNO3 (28.11 mL). The “blank” (100 mL
ddH2O) with no KCl registers the endpoint when 1.77 mL AgNO3 is added. You then disperse 3.78 g butter in
100 mL boiling ddH2O and titrate to equivalence (31.9 mL AgNO3). The serving size for butter is 1 tablespoon
(14 g).
1. What is the Cl− content (mg/serving) of the butter?
2. What is the NaCL content (mg/serving) of the butter?
3. What is the Na+ content (mg/serving) of the butter?

[undergroundboy]

You work in a sports drink bottling facility. You are measuring the iron content of the water coming into the
plant using an Fe2+ ISE. You decide to use the “standard addition” method, as you are concerned that there
may unknown interferences present. You a spike a 0.5 L sample of water with a 12,000 ppm solution of
Fe(II)Cl2 (25 µL).You measure the electrode potential in unspiked and spiked water (112.5 and 99.4 mV,
respectively).
4. What is the concentration of Fe2+ in the water (ppm)?
5. What is the concentration of Fe2+ in the water (g/L)?

[Borek]

You have to show your attempts at solving the question to receive help. This is a forum policy.

Please read the forum rules.

I am locking other threads.

[Arkcon]

I hope you also don't mind, undergroundboy:, that I've merged your different posts into one thread.  You essentially ask the same question in each, just with different units.  When you start to solve one, and with some hints from us, you may have what you need to solve the others.  As it stands now however, it seems all you can do is type out assignment problems, and that's not how this forum produces results.

[undergroundboy]

I'm sorry guys I did not know, I should have read the forum rules. Makes total sense, this is not chegg. Haha.

Anyways, can anyone help me with my attempt at this question?

You are analyzing riboflavin in wheat flour by fluorescence spectroscopy (?ex = 450 nm, ?em =530 nm). The extraction procedure is fairly complex. You obtain a 100g sample of the flour. You then weigh out 4.542 g of the flour, add it to a 250 mL volumetric flask, and dilute to volume with 1 N boiling HCl. After thoroughly mixing this solution, you combine 2 mL with 7 mL 1 N HCl. You then freeze-dry 4 mL of the diluted solution, and redissolve the residue to a total volume of 1.5 mL. You combine 0.5 mL of this solution with 0.5 mL ddH2O and measure the fluorescence. The following fluorescence values are obtained for the sample and standards.

1.) Prepare a blanked standard curve, and determine the riboflavin concentration in the flour (ppm) (10 points) (ONE QUESTION)

Fluorescence (?ex=450 nm, ?em=530 nm).
Sample

1.706
Riboflavin (20 mg/L)

0.822
Riboflavin (10 mg/L)

0.368
Riboflavin (5 mg/L)

0.121
Riboflavin (2 mg/L)

0.004
Riboflavin (1 mg/L)

1.706
Riboflavin (0 mg/L)

0.476
Flour

What is the concentration of Ca2+ in the diluted drinking water (?g/L)?

Based on the equation of the calibration curve (y = 0.088x - 0.0575), the riboflavin in the diluted flour sample will be determined:

[tex]Riboflavin (mg/L)= \frac{0.476+0.0575}{0.088}= 6.0625 mg/L[/tex]

The amount of riboflavin in the flour sample will be: [itex]6.0625 (\frac{mg}{L})\times 1.5 (mL)\times (\frac{1 L}{1000 mL})=9.09375 \times 10^{-3} mg[/itex]

[tex]\frac{250}{2}\times 9.09375 \times 10^{-3} mg=1.137 mg[/tex]

4.542 g contains 1.137 mg of riboflavin

1000 g contains [itex]\frac{1.137 mg\times 1000 g}{4.542 g}=250.3 mg/kg=250.3 ppm [/itex]

Can anyone check if I did this right?

[Borek]

In general you are on the right track, but seems to me like you have ignored part of the dilution information. Unless I am reading something wrong your final sample (the measured one) was prepared by combining 0.5 mL (out of 1.5 mL) with 0.5 mL of distilled water, for freeze drying you have used 4 mL out of 9 mL.