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

16H⁺(aq) + 2 Cr₂O₇ ²⁻(aq) + C₂H₅OH(aq)  → 4Cr³⁺(aq) + 2CO₂(g) + 11H₂O(l)

If 35.46 mL of 0.05961 M Cr₂O₇ ²⁻ is required to titrate 29.80 g of plasma, what is the mass percent of alcohol in the blood?

Answer: ________ mass %

Answer 1

Volume of Cr2O72- = 35.46 ml = 0.03546 L

Concentration of Cr2O72- = 0.05961 M = 0.05961 mol / L

Number of moles of Cr2O72- =( 0.05961 mol/L)(0.03546 L)

Number of moles of Cr2O72- = 0.002114 mol

From reaction, 2.0 mol of Cr2O72- reacts with 1.0 mol of C2H5OH, hence 0.002114 mol of Cr2O72- will react with (1/2)*0.002114 mol = 0.001057 mol

Molar mass of C2H5OH = 46.068 g/mol

Mass of ethanol = 0.001057 mol * 46.068 g/mol = 0.0487 g

Mass % of alcohol = (0.0487 g / 29.80 g)*100 = 0.1634 mass%