A test solution is prepared by pipetting 2.0 mL of an original sample into a 250 mL volumetric flask and then diluting to the
mark. The concentration of iron in the test solution is determined (from a calibration curve) to be 1.66 μ g / mL or 1.66 ppm. What is the iron
concentration in the original sample
1.66 μ g/mL*(250mL/2.0mL)=??????
2 Sig Figs
207.5 M Fe
How much total iron (in mg) was in the original 2mL sample?
1.66 μ g/mL*(250mL/2.0mL)=?????? <==== Answer will be in μg/mL. Use the mw to convert it to mole and convert mL to L.
To find the iron concentration in the original sample, we need to consider the dilution factor of the test solution. Here is the step-by-step explanation:
1. Determine the dilution factor:
The dilution factor is the ratio of the final volume to the initial volume. In this case, the initial volume is 2.0 mL, and the final volume is 250 mL. So, the dilution factor is 250 mL / 2.0 mL = 125.
2. Consider the iron concentration in the test solution:
The iron concentration in the test solution is given as 1.66 μg/mL or 1.66 ppm (parts per million).
3. Calculate the iron concentration in the original sample:
To calculate the iron concentration in the original sample, we need to divide the concentration of iron in the test solution by the dilution factor.
If the iron concentration in the test solution is given in μg/mL:
Iron concentration in original sample = (Concentration in test solution) / (Dilution factor)
= (1.66 μg/mL) / 125
= 0.0133 μg/mL
If the iron concentration in the test solution is given in ppm:
Iron concentration in original sample = (Concentration in test solution) / (Dilution factor)
= (1.66 ppm) / 125
= 0.0133 ppm
So, the iron concentration in the original sample is 0.0133 μg/mL or 0.0133 ppm, depending on the units provided for the test solution concentration.