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) %

Answer 2

The mass percent of ethanol in the plasma has been 0.171%.

The titration has been the neutralization reaction, that converts the acid and base and results in the formation of the salt and water.

From the balanced equation, there has been the requirement of 2 moles of chromate for the neutralization of 1 mole of alcohol.

The moles of chromate in 0.05961 M has been:

Molarity = $\rm \dfrac{Moles}{Volume}$

Moles = Molarity × Volume

The moles of chromate has been:

Moles of chromate = 0.05961 mol × 0.03546 L

Moles of chromate = 0.002113 mol.

From the balanced equation:

2 moles chromate neutralizes = 1 mole of ethanol

0.002113 moles of chromate neutralizes = 0.00105 mol of Ethanol.

The mass of 0.00105 mol ethanol has been:

Mass = Moles × Molecular weight

Mass of 0.00105 mol of ethanol = 0.00105 mol × 46.07 g/mol

Mass of 0.00105 mol of ethanol = 0.0486 grams.

The total mass of plasma has been 28.40 grams.

The mass percent of ethanol in 28.40 grams plasma has been:

The mass percent of ethanol =  $\rm \dfrac{0.0486}{28.40}$ × 100

The mass percent of ethanol has been 0.171%.

The mass percent of ethanol in the plasma has been 0.171%.

For more information about mass percent, refer to the link:

https://brainly.com/question/5394922