Laws passed in some states define a drunk driver as one who drives with a blood alcohol level of 0.08% by mass or higher. the level of alocohol can be determinded byt titrating blood plasma with pottasium dichromate according to the unbalanced redox reaction.
igot the balanced equation--
C2O72- + 14 H + 3H2O + C2H5OH ---> 2cr3+ + 7H2O + 2CO2 + 6H
-- assuming that the only substance that reacts with the dichromate in blood plasma is alcohol, is a person legally drunk if 38.94 ML of 0.0723M potassium dichromate is required to titrate 50.0 g sample of blood plasma?
HELP ME PLZ.. THIS IS FOR A GRADE.. THANKYOU! :)
Check your equation. I don't think it is balanced. I see 26 H atoms on the left and 20 on the right.
Step 1. moles dichromate = M x L
Step 2. Convert moles dichromate to moles alcohol using the coefficients in the balanced equation.(I think you need to start with 2Cr2O7^-2.
Step 3. Convert moles alcohol to grams alcohol.
Step 4. % ethanol = (g alcohol/50)*100=??
To determine if a person is legally drunk, we need to calculate whether the amount of alcohol present in the blood plasma sample exceeds the legal limit.
First, let's calculate the number of moles of potassium dichromate (K2Cr2O7) used in the titration:
Moles of K2Cr2O7 = Molarity × Volume
= 0.0723 mol/L × 38.94 mL
= 0.002818 mol
According to the balanced equation, 1 mole of alcohol (C2H5OH) reacts with 1 mole of K2Cr2O7. Therefore, the number of moles of alcohol in the blood plasma sample is also 0.002818 mol.
Next, we need to calculate the mass of alcohol in the blood plasma:
Mass = Moles × Molar mass
= 0.002818 mol × 46.07 g/mol (molar mass of C2H5OH)
= 0.1299 g
Now, we can compare the mass of alcohol in the blood plasma to the mass of the sample, which is given as 50.0 g.
Alcohol concentration by mass = (Mass of alcohol / Mass of sample) × 100%
= (0.1299 g / 50.0 g) × 100%
= 0.2598%
The legal limit for blood alcohol concentration is 0.08% (equivalent to 0.08 g alcohol per 100 g of blood plasma).
Since the calculated alcohol concentration (0.2598%) is higher than the legal limit, the person would be considered legally drunk based on the given information.
To determine whether a person is legally drunk based on the given information, we need to calculate the amount of alcohol (C2H5OH) present in the 50.0 g sample of blood plasma and compare it to the legal limit of 0.08% by mass.
First, we need to find the number of moles of potassium dichromate (K2Cr2O7) used in the titration. We can do this using the formula:
moles = concentration (M) x volume (L)
Given:
Concentration of K2Cr2O7 = 0.0723 M
Volume of K2Cr2O7 = 38.94 mL = 0.03894 L
moles of K2Cr2O7 = 0.0723 M x 0.03894 L = 0.002818 moles
Now, let's use the balanced equation to determine the molar ratio between K2Cr2O7 and C2H5OH. From the equation, we see that the ratio is 1:1.
Therefore, the moles of C2H5OH = 0.002818 moles.
Next, we'll calculate the mass of C2H5OH using its molar mass, which is 46.07 g/mol:
mass = moles x molar mass
mass of C2H5OH = 0.002818 moles x 46.07 g/mol = 0.1299 g
Now, we compare the mass of C2H5OH to the mass of the blood plasma sample:
mass percent = (mass of C2H5OH / mass of sample) x 100%
mass percent = (0.1299 g / 50.0 g) x 100% = 0.2598%
The calculated mass percent of alcohol is 0.2598%. Since this is below the legal limit of 0.08%, the person would not be considered legally drunk.
Please note that this calculation assumes that alcohol is the only substance reacting with the potassium dichromate in the blood plasma sample. In reality, there could be other substances that react with the dichromate, which might affect the accuracy of the result.