Solution Found!
A person's blood alcohol (C2H5OH) level can be determined
Chapter 4, Problem 62P(choose chapter or problem)
A person’s blood alcohol (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)) level can be determined by titrating a sample of blood plasma with a potassium dichromate solution. The balanced equation is
\(16 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2}(\mathrm{aq})+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq}) \rightarrow 4 \mathrm{Cr}^{3+}(\mathrm{aq})+2 \mathrm{CO}_{2}(\mathrm{~g})+11 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)
If \(35.46 mL\) of \(0.05961 \mathrm{M} \ \mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2}\) is required to titrate \(28.00 g\) of plasma, what is the mass percent of alcohol in the blood?
Equation Transcription:
C2H5OH
16H+(aq) + 2Cr2O72-(aq) + C2H5OH(aq) 4Cr3+(aq) + 2CO2(g) + 11H2O(l)
0.05961 M Cr2O72-
Text Transcription:
C_2H_5OH
16H^+(aq) + 2Cr_2O_7^2-(aq) + C_2H_5OH(aq) rightarrow 4Cr^3+(aq) + 2CO_2(g) + 11H_2O(l)
35.46 mL
0.05961 M Cr_2O_7^2-
28.00 g
Questions & Answers
QUESTION:
A person’s blood alcohol (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)) level can be determined by titrating a sample of blood plasma with a potassium dichromate solution. The balanced equation is
\(16 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2}(\mathrm{aq})+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq}) \rightarrow 4 \mathrm{Cr}^{3+}(\mathrm{aq})+2 \mathrm{CO}_{2}(\mathrm{~g})+11 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)
If \(35.46 mL\) of \(0.05961 \mathrm{M} \ \mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2}\) is required to titrate \(28.00 g\) of plasma, what is the mass percent of alcohol in the blood?
Equation Transcription:
C2H5OH
16H+(aq) + 2Cr2O72-(aq) + C2H5OH(aq) 4Cr3+(aq) + 2CO2(g) + 11H2O(l)
0.05961 M Cr2O72-
Text Transcription:
C_2H_5OH
16H^+(aq) + 2Cr_2O_7^2-(aq) + C_2H_5OH(aq) rightarrow 4Cr^3+(aq) + 2CO_2(g) + 11H_2O(l)
35.46 mL
0.05961 M Cr_2O_7^2-
28.00 g
ANSWER:
Solution 62P:
Here we have to calculate the mass percent of alcohol in the blood.
Step 1
The balanced equation for the redox reaction is
16H+ (aq) + 2 Cr2O72- (aq) + C2H5OH (aq) → 4 Cr3+ (aq) + 2 CO2 (g) + 11H2O (l)
Given:
Volume (V) of Cr2O72- = 35.46 mL
Concentration (C) of Cr2O72- = 0.05961 M
Mass of plasma = 28.00 g
1st we have to calculate how many moles of Cr2O72- is present as follows,
Moles = (C V) / 1000
Where C = concentration in mol dm-3
V is the volume in mL
Now substituting the values, moles can be calculated as,
Moles = (C V) / 1000
=
= 2.11 10-3 mol of dichromate