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

Answer 1

Answer: The mass percent of alcohol in blood is 0.183 %

Step-by-step explanation:

To calculate the moles for given molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}`

Molarity of
`Cr_2O_7^(2-)` solution = 0.05961 M

Volume of solution = 35.46 mL = 0.03546 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

`0.05961M=\frac{\text{Moles of }Cr_2O_7^(2-)}{0.03546L}\\\\\text{Moles of }Cr_2O_7^(2-)=(0.05961mol/L* 0.03546L)=2.11* 10^(-3)mol`

The given chemical reaction follows:

`16H^+(aq.)+2Cr_2O_7^(2-)(aq.)+C_2H_5OH(aq.)\rightarrow 4Cr^(3+)(aq.)+2CO_2(g)+11H_2O(l)`

By Stoichiometry of the reaction:

2 moles of dichromate ions react with 1 mole of ethanol.

So,
`2.11* 10^(-3)mol` of dichromate ions will react with =
`(1)/(2)* 2.11* 10^(-3)=1.06* 10^(-3)mol`

• To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Moles of ethanol =
`1.06* 10^(-3)mol`

Molar mass of ethanol = 46.07 g/mol

Putting values in above equation, we get:

`1.06* 10^(-3)mol=\frac{\text{Mass of ethanol}}{46.07g/mol}\\\\\text{Mass of ethanol}=(1.06* 10^(-3)mol* 46.07g/mol)=0.0488g`

• To calculate the mass percent of ethanol in blood, we use the equation:

`\text{Mass percent of ethanol}=\frac{\text{Mass of ethanol}}{\text{Mass of blood plasma}}* 100`

Mass of blood plasma = 26.60 g

Mass of ethanol = 0.0488 g

Putting values in above equation, we get:

`\text{Mass percent of ethanol}=(0.0488g)/(26.60g)* 100=0.183\%`

Hence, the mass percent of alcohol in blood is 0.183 %