Question

If a blood sample contains (m/m) 0.335% iron determine how much blood (mu L) should be diluted in a 250 mL volumetric flask with the appropriate solvent to give an absorbance reading of 0.2500. Assume that the solution being analyzed has a density of 1000 g/cc but blood has a density of 1.070 g/cc.

Hint: first use the correct calibration curve from 2c to calculate the concentration of the iron sample that yields 0.2500 absorbance. then apply the dilution factor to the calculate the volume (uL) of blood necessary to yield this result.

My calibration equation: y = 0.1133x + 0.0031

R² = 0.99988

The completed answer is 152(micro)L

This is as far as I get...

.2500=0.1133x+.0031
I get conc Fe 2.179ppm?

Answer 1

NOTE: I am assuming that the density of blood mentioned above is 1070 g/cc, and not 1.070 g/cc

Putting y = 0.25 in the calibration plot equation and solving we get:

x = (0.25-0.0031)/0.1133 = 2.179 ppm or 2.179 mg/kg

Final solution volume = 250 mL = 250 cc

So, Mass of solution = 250*1000 = 250 kg

Mass of Fe in this solution = Solution mass*Conc. = 250*2.179 = 544.75 mg

All this Fe comes from the blood sample which is diluted.

Assume that the required blood volume is 'p' L = p*10^-3 mL = p*10^-3 cc

Mass of blood sample = 1070*p*10^-3 = 1.07*p*10^-3 kg

Given:

Mass of Fe/Mass of blood sample = 0.00335

so, Mass of Fe = 0.00335*Mass of blood sample = 0.00335*1.07*p*10^-3 kg = 3.5845*p mg

Equating this to the calculating volume above, we get:

3.5845*p = 544.75

Thus,

p = 151.97 L