The blood alcohol () level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of and carbon dioxide. The reaction can be monitored because the dichromate ion () is orange in solution, and the ion is green. The unbalanced redox equation is If 26.70 mL of 0.0600 M potassium dichromate solution is required to titrate 30.0 g blood plasma, determine the mass percent of alcohol in the blood.

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30-09-2019
Biology

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The blood alcohol () level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of and carbon dioxide. The reaction can be monitored because the dichromate ion () is orange in solution, and the ion is green. The unbalanced redox equation is If 26.70 mL of 0.0600 M potassium dichromate solution is required to titrate 30.0 g blood plasma, determine the mass percent of alcohol in the blood.

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fcoxavierhs13 fcoxavierhs13 · Answer 1 · 2019-10-06T07:55:18+02:00

Answer

0.184 % of ethanol

Explanation:

The balanced equation for this reaction is attached.

The statement indicates that 26.70 ml of a 0.06 M solution potassium dichromate titrates the ethanol present in 30 grams of blood plasma.

First you need to calculate the moles of potassium dichromate requiered to titrate the ethanol. The calculus is described below:

(0.06 mol potassium dichromate)/1000 ml * 26.70 ml = 0.002403 moles of dichromate titrate the ethanol present in 30 g of blood plasma.

Based on the balanced equation, 2 molecules of potassium dichromate react with one molecule of ethanol. So the moles of ethanol present in 30 grams of blood plasma are 0.0012015 mol ( 0.002403 moles of potassium dichromate / 2)

The you need to calculate the grams of ethanol present in 0.0012015 moles. The calculus is described below:

Molecular weight of ethanol 46.07 g/mol

(0.0012015 moles of ethanol) * 46.07 g/mol = 0.0553 g

Finally you need to calculate the % of ethanol presente in 30 g of blood:

(0.0553g/30g) * 100 = 0.184 %