Question

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

16H+ (aq) + 2Cr2O72- (aq) + C2H5OH (aq) 4Cr3+ (aq) + 2CO2 (g) + 11H2O (l)

If 35.46 mL of 0.05961 M Cr2O72- is required to titrate 30.50 g of plasma, What is the mass percent of alcohol in the blood?

__________ mass %

40 mL mass % Alcohol in blood = mass alcohol/ mass blood
0.06 M

Answer 1

mols Cr2O7^2- = M x L = ?
Convert to mols C2H5OH by
?mols Cr2O7^2- x (1 mol C2H5OH/2 mol Cr2O6^2-) = ?mols Cr2O7^2- x 1/2 = ?
Then grams ethanol = mols ethanol x molar mass ethanol
%w/w = (mass ethanol/mass sample)*100 = ?
mass ethanol from above.
mass sample is 30.50 g from the problem.