"Enter your answer in the provided box. A person's blood alcohol (C2H5OH) level can be determined by titrating a sample of blood

Question

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"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?"

Chemistry

1 answer:

Tems11 [2.7K] · Answer 1 · 2026-02-27T07:46:55+03:00

Answer:

0.1714 (w/w) %

Explanation:

Utilizando la ecuación:

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

Se emplean 2 moles de ion dicromato (Cr₂O₇²⁻) para titular 1 mol de alcohol (C₂H₅OH)

35.46mL = 0.03546L de una solución de Cr₂O₇²⁻ a 0.05961M utilizada para alcanzar el punto de equivalencia en la titulación contiene:

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

Dado que 2 moles de dicromato reaccionan por cada mol de alcohol, los moles de alcohol en la muestra de plasma son:

2.114x10⁻³ moles Cr₂O₇²⁻ ₓ ( 1 mol C₂H₅OH / 2 moles Cr₂O₇²⁻) = 1.0569x10⁻³ moles de C₂H₅OH

Como la masa molar del alcohol es 46.07g/mol, la masa de alcohol es:

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

Por lo tanto, el porcentaje en masa de alcohol en sangre utilizando los 28.40g de plasma es:

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

"Enter your answer in the provided box. A person's blood alcohol (C2H5OH) level can be determined by titrating a sample of blood

= 84.34 grams