Alcohol levels in blood can be determined by a redox titration with potassium dichromate according to the balanced equation:

C2H5OH(aq) + 2Cr2O7(-2) (aq) + 16H+(aq)---> 2CO2 (g) + 4 Cr3(+)(aq) + 11H2O (l)
What is the blood alcohol level in mass perecent if 8.76 mL of 0.04988 M K2Cr2O7 is required for titration of a 10.0002 g sample of blood?

moles dichromate = M x L = xx.

moles alcohol = 1/2 of xx.
grams alcohol = moles alcohol x molar mass.
% alcohol = (mass alcohol/total mass)*100

Given 10.0 L of Nitrogen at -78 deg C. What volume will the Nitrogen occupy at 25 deg C (P is constant)?

If you wish to post a question of your own, I suggest you go to the top of the page, click on Post a new question, and type in your question. Posting piggy back style often goes unanswered. In this case, use PV = nRT.

which part is the alcohol

To determine the blood alcohol level in mass percent, we need to calculate the amount of ethanol reacted with potassium dichromate during the titration.

First, calculate the amount in moles of K2Cr2O7 used:
moles of K2Cr2O7 = volume (in L) x molarity
Convert the volume from mL to L:
8.76 mL = 8.76/1000 L = 0.00876 L
moles of K2Cr2O7 = 0.00876 L x 0.04988 M = 4.363 x 10^-4 mol

Since the balanced equation shows that 1 mole of C2H5OH reacts with 2 moles of K2Cr2O7, we can use stoichiometry to find the moles of ethanol reacted:
moles of C2H5OH = (moles of K2Cr2O7) / 2 = 4.363 x 10^-4 mol / 2 = 2.1815 x 10^-4 mol

Next, we calculate the mass of ethanol using the molar mass of C2H5OH (ethyl alcohol):
molar mass of C2H5OH = (2 x 12.01 g/mol) + (6 x 1.01 g/mol) + 16.00 g/mol = 46.07 g/mol
mass of C2H5OH = moles of C2H5OH x molar mass = 2.1815 x 10^-4 mol x 46.07 g/mol = 0.01006 g

Finally, we can calculate the blood alcohol level in mass percent:
mass percent = (mass of ethanol / mass of blood sample) x 100%
mass percent = (0.01006 g / 10.0002 g) x 100% = 0.10062%

Therefore, the blood alcohol level is approximately 0.10062% (mass percent).