Aperson'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 26.60 g of plasma, what is the mass percent of alcohol in the blood?

0.18 %

Explanation:

Considering:

Or,

Given :

For :

Molarity = 0.05961 M

Volume = 35.46 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 35.46×10⁻³ L

Thus, moles of :

Moles of  = 0.002114 moles

According to the given reaction:

2 moles of react with 1 mole of alcohol

Thus,

0.002114 moles  of react with 1/2*0.002114 mole of alcohol

Moles of alcohol = 0.001057 moles

Molar mass of ethanol = 46.07 g/mol

Mass = Moles * Molar mass = 0.001057 * 46.07 g = 0.048696 g

Mass of plasma = 26.60 g

Mass Percent is the percentage by the mass of compound present in the mixture.

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