Question

"Enter your answer in the provided box. 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 28.40 g of plasma, what is the mass percent of alcohol in the blood?"

Answer 1

Answer:

0.1714 (w/w) %

Explanation:

Using the equation:

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

2 moles of dichromate ion (Cr₂O₇²⁻) are used to titrate 1 mole of alcohol (C₂H₅OH)

35.46mL = 0.03546L of a 0.05961M Cr₂O₇²⁻ solution used to reach the equivalence point in the titration contains:

0.03546L ₓ (0.05961 moles Cr₂O₇²⁻ / L) = 2.114x10⁻³ moles Cr₂O₇²⁻

As 2 moles of dichromate reacts per mole of alcohol, moles of alcohol in the sample of plasma are:

2.114x10⁻³ moles Cr₂O₇²⁻ ₓ ( 1 mole C₂H₅OH / 2 moles Cr₂O₇²⁻) = 1.0569x10⁻³ moles of C₂H₅OH

As molar mass of alcohol is 46.07g/mol, mass of alcohol is:

1.0569x10⁻³ moles of C₂H₅OH ₓ (46.07g / mol) = 0.04869g of C₂H₅OH

Thus, mass percent of alcohol in the blood using the 28.40g of plasma is:

(0.04869g of C₂H₅OH / 28.40g) × 100 = 0.1714 (w/w) %