A 50.02 gram sample of blood was taken from an individual who had been pulled ov

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A 50.02 gram sample of blood was taken from an individual who had been pulled over for a DUI. The sample was prepared and titrated with 0.05182 M solution of potassium dichromate. It took 35.48 mL of potassium dichromate to reach the titration a endpoint. Assuming the alcohol (C2H5OH) present in the individuals blood sample reacts only with the dichromate solution, determine if this person should be booked for a DUI or not. Reaction: C2H5OH + K2Cr2O7 ---> CO2 + K + Cr2 A 50.02 gram sample of blood was taken from an individual who had been pulled over for a DUI. The sample was prepared and titrated with 0.05182 M solution of potassium dichromate. It took 35.48 mL of potassium dichromate to reach the titration a endpoint. Assuming the alcohol (C2H5OH) present in the individuals blood sample reacts only with the dichromate solution, determine if this person should be booked for a DUI or not. Reaction: C2H5OH + K2Cr2O7 ---> CO2 + K + Cr2 A 50.02 gram sample of blood was taken from an individual who had been pulled over for a DUI. The sample was prepared and titrated with 0.05182 M solution of potassium dichromate. It took 35.48 mL of potassium dichromate to reach the titration a endpoint. Assuming the alcohol (C2H5OH) present in the individuals blood sample reacts only with the dichromate solution, determine if this person should be booked for a DUI or not. Reaction: C2H5OH + K2Cr2O7 ---> CO2 + K + Cr2 Reaction: C2H5OH + K2Cr2O7 ---> CO2 + K + Cr2

Explanation / Answer

The titration reaction is

2Cr2O72? + 16H+ + 3C2H5OH -----------> 4Cr3+ + 11H2O + 3CH3COOH

Therefore 2 mol of Cr2O72? is equivalent to 3 mol of C2H5OH

Now 35.48 mL (or 35.48 x 10-3 L) of 0.05182 M solution of potassium dichromate was consumed.

Therefore number of moles of potassium dichromate used = 0.05182 x 35.48 x 10-3 = 1.839 x 10-3 mole

so the number of moles of alcohol present in 50.02 g blood sample = (3/2) x 1.839 x 10-3 mole = 2.758 x 10-3 mole

Now mass of 2.758 x 10-3 mole of C2H5OH = 46 x 2.758 x 10-3 = 0.1269 g

therefore % of alcohol in blood = (0.1269 g x 100) / 50.02 g = 0.254 %

Hence the person should be booked for DUI as the threshold level for DUI is 0.08 %

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