Answer:
Using the information provided in the question, the percentage by mass of ethanol in the blood is 0.1949%.
Explanation:
The balanced equation for the reaction can be written as:
3CH3CH2OH + 2K2Cr2O7+ 16H+ >>>> 3CH3COOH + 4Cr3+ + 4K+ + 11H2O. This will be used to solve the problem.
The no of moles of K2Cr2O7 used = concentration X volume in dm cube.
= 0.05 X 22.6/1000 = 1.13/1000 = 0.00113 moles of K2Cr2O7 was used.
From the balanced equation above, 3 moles of ethanol will react with 2 moles of K2Cr2O7.
Therefore, 0.00113 X 3/2 moles of ethanol would have been in the blood of the subject = 0.001695 moles of ethanol.
Remember that no of moles = mass/molar mass. Then mass = moles X molar mass.
The molar mass of ethanol is 46g/mole.
Then mass = 0.001695 X 46 = 0.07797gram of Ethanol was in the blood.
Percentage by mass in blood is: 0.07797/40 X 100 = 0.1949%