Alcohol levels in blood can be determined by a redox titration with potassium di

Question

Alcohol levels in blood can be determined by a redox titration with potassium dichromate according to the balanced equation
C2H5OH(aq)+2Cr2O27(aq)+16H+(aq)2CO2(g)+4Cr3+(aq)+11H2O(l)

Part A

What is the blood alcohol level in mass percent if 8.83 mL of 0.04988 M K2Cr2O7 is required for titration of a 9.9950 g sample of blood?

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Alcohol levels in blood can be determined by a redox titration with potassium dichromate according to the balanced equation
C2H5OH(aq)+2Cr2O27(aq)+16H+(aq)2CO2(g)+4Cr3+(aq)+11H2O(l)

Part A

What is the blood alcohol level in mass percent if 8.83 mL of 0.04988 M K2Cr2O7 is required for titration of a 9.9950 g sample of blood?

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Explanation / Answer

Moles of K2Cr2O7 = 8.83 x 0.04988 / 1000 = 0.00044044 Moles

Moles of Ethanol was present in the sample =     0.00044044 / 2 = 0.00022022 Moles

The molecular weight of C2H5OH = 46.07 g/mol

Mass of ethanol present in the sample =    0.00022022 x 46.07 = 0.0101455 gm

Calculate the  the percent = 0.0101455 x 100 / 9.995 = 0.10151 %

Hence 0.10151 % of the blood alcohol was present

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