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) + 2Cr2O72 - (aq) + C2H5OH (aq) → 4Cr3 + (aq) + 2CO2 (g) + 11H2O (l) If 35.46 mL of 0.05961 M Cr2O72 - is required to titrate 28.20 g of plasma, what is the mass percent of alcohol in the blood?

0.1727%

Explanation:

Let's consider the following balanced equation.

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

The moles of Cr₂O₇²⁻ in 35.46 mL of 0.05961 M Cr₂O₇²⁻ are:

35.46 * 10⁻³ L * 0.05961 mol/L = 2.114 * 10⁻³ mol

The molar ratio of Cr₂O₇²⁻ to C₂H₅OH is 2:1. The moles of C₂H₅OH are 1/2 * 2.114 * 10⁻³ mol = 1.057 * 10⁻³ mol

The molar mass of C₂H₅OH is 46.07 g/mol. The mass corresponding to 1.057 * 10⁻³ moles is:

1.057 * 10⁻³ mol * (46.07 g/mol) = 0.04870 g

The mass percent of alcohol is 28.20 g of plasma is:

(0.04870 g/28.20g) * 100% = 0.1727%