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

0.18 %

Step-by-step explanation:

Considering:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

Or,

`Moles =Molarity * {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 * {35.46* 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\ \%=(Mass_(ethanol))/(Total\ mass\ of\ plasma)* 100`

`Mass\ \%=(0.048696)/(26.60)* 100=0.18\ \%`