A person\'s blood alcohol (C 2 H 5 OH) level can be determined by titrating a sa

Question

A person's blood alcohol (C2H5OH) level can be determined by titrating a sample of blood plasma with a potassium dichromate solution. The balanced equation is

If 32.94 mL of 0.05928 M Cr2O72- is required to titrate 28.95 g plasma, what is the mass percent of alcohol in the blood?

A person's blood alcohol (C2H5OH) level can be determined by titrating a sample of blood plasma with a potassium dichromate solution. The balanced equation is 6 H+(aq) + 2 Cr2O72-(aq) + C2H5OH(aq) 4 Cr3+(aq) + 2 CO2(g) + 11 H2O(l) If 32.94 mL of 0.05928 M Cr2O72- is required to titrate 28.95 g plasma, what is the mass percent of alcohol in the blood?

Explanation / Answer

moles K2Cr2O7 = 8.08 x 10^-3 L x 0.04988 M=0.0004030
the ratio between ethanol and Cr2O72- is 1 : 2
moles ethanol = 0.0004030/2=0.0002015
mass ethanol = 0.0002015 x 46.069 g/mol=0.009283
% = 0.009283 x 100/ 9.9950=0.09288

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