The alcohol content in a 10.0 g sample of blood from a driver required 4.23 mL of 0.7654 M K2Cr2O7?

Question

The alcohol content in a 10.0 g sample of blood from a driver required 4.23 mL of 0.7654 M K2Cr2O7 for titration. What is the percent alcohol by mass in the driver's blood? The reaction equation is given by:
3CH3CH2OH + 2K2Cr2o7 + 8H2SO4 ----> 3CH3COOH + 2Cr2(SO4)3 + 2K2SO4 + 11H2O ethanol

Answer 1

From the given reaction, we know that two moles of K2Cr2O7 would react with 3 moles of ethanol CH3CH2OH. Hence: The 10.0 g sample of blood contains:
(3/2)*(4.23ml)*(0.7654 M) = 4.86 mMole alcohol = 0.224g alcohol, which is 2.24% by mass.