Application Questions 1. Y ou wish to adapt the AA method to measure the amount

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Application Questions 1. Y ou wish to adapt the AA method to measure the amount of Iron in leaf tissues. The nimum amount of iron inthe tissues is expected to be about 0.002S%byt ass minimum concentration for AA measurements is 0.30 ppm (using a slightly more sensitiv method than the one you used). The Your plan is to weigh out 6.0 g leaf tissue samples, digest them in acid, filter them to 50 mL. This solution is your"Sample Stock Solution". You wil of this solution into a 25-mL volumetric flask and dilute to volume. This solution and dilute I then pipet a portion uted Sample Solution," and you,wil makeyourAAmeas urements on this solution. 4.0 g filter Into 50-m volumetrie flask pipet portion volume? original leaf sample Diluted Sample Solution acid digestion Sample Stock Solution a. How many milligrams of Fe are there in 6.025 g of a leaf tissue that is 0.0025% iron by mass? (Note: 0.0025% by mass = 0.0025 g Fe per 100 g sample.) b. If all of the iron from (a) is diluted in a 50-ml flask (Sample Stock Solution), what is the concentration, in ppm, of the resulting solution? c. What volume of the Sample Stock Solution (from (b)) is needed to prepare 25.0 ml of a Dilute Sample Solution with a concentration of 0.30 ppm Fe?

Explanation / Answer

a)

100 g leaf tissue contain 0.0025 g Fe.

So 6.025 g contain (0.0025*6.025)/100 = 1.506*10^-4 g = 0.1506 mg

b)

50 ml contains 1.506*10^-4 g

So 10^6 ml contains (1.506*10^4*10^6)/50 = 3.012 g

Thus 3.012 ppm.

C)

We take x ml from a solution which contain 1.5006*10^-4 g and make up to 25 ml.

This solution is with a concentration 0.30 ppm.

That is 10^6 ml contains 0.30 g Fe.

So 25 ml contains (0.30*25)/10^6= 7.5*10^-6 g

Thus (1.506*10^-4 * x)/25 = 7.5*10^-6

X = (7.5*10^-6*25)/1.506*10^-4 = 1.25 ml

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