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) + 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?

Answer 1

0.1727%

Step-by-step 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%