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

Answer 1

Answer:

0.18 %

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $Cr_2O_7^{2-}$ :

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 $Cr_2O_7^{2-}$ :

$Moles=0.05961 \times {35.46\times 10^{-3}}\ moles$

Moles of $Cr_2O_7^{2-}$  = 0.002114 moles

According to the given reaction:

$16H^+_{(aq)}+2Cr_2O_7^{2-}_{(aq)}+C_2H_5OH_{(aq)}\rightarrow 4Cr^{3+}_{(aq)}+2CO_2_{(g)}+11H_2O_{(l)}$

2 moles of $Cr_2O_7^{2-}$ react with 1 mole of alcohol

Thus,

0.002114 moles  of $Cr_2O_7^{2-}$ 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.

$Mass\ \%=\frac{Mass_{ethanol}}{Total\ mass\ of\ plasma}\times 100$

$Mass\ \%=\frac{0.048696}{26.60}\times 100=0.18\ \%$