This question is confusing; can anyone help me solve it?
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
16 H+(aq) + 2 Cr2O72-(aq) + C2H5OH(aq) 4 Cr3+(aq) + 2 CO2(g) + 11 H2O(l)
If 31.45 mL of 0.05971 M Cr2O72- is required to titrate 28.67 g plasma, what is the mass percent of alcohol in the blood?
1 Answer
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moles Cr2O72- = 0.03145 L x 0.05971 = 0.001878
Moles ethanol = 0.001878 / 2 =0.0009390
Mass ethanol = 0.0009390 mol x 46.0694 g/mol = 0.04326 g
% = 0.04326 x 100 / 28.67 = 0.151
