Question

What is the blood alcohal level in mass percent if 8.05ml of .04988 M K2 Cr2 O7 is required for titration of a 10.000 g sample of blood?

Balanced Equation

C2H5OH + 2Cr2O7+ 16H yields 2CO2 + 4Cr + 11H2O

Answer 1

Molarity = Moles / Liter

Moles = Molarity x Liter

Now,

8.05 mL of 0.04988 M K2Cr2O7

So, moles of K2Cr2O7 = 0.04988 M x (0.00805 L)

= 0.0004 moles

Now, the reaction equation is

C2H5OH + 2Cr2O7 + 16H 2CO2 + 4Cr + 11H2O

In the reaction equation,;

2 moles of K2Cr2O7 required to react with 1 mole of C2H5OH

1 moles of K2Cr2O7 required to react with 1/2 mole of C2H5OH

0.0004 moles of K2Cr2O7 required to react with 0.0004/2 mole of C2H5OH

0.0004 moles of K2Cr2O7 required to react with 0.0002 mole of C2H5OH

Molar mass of C2H5OH = 46 g/mol

So, 1 mole of C2H5OH = 46 g

0.0002 mole of C2H5OH = 0.0002 x 46 g = 0.0092 g

Now, mass % of alcohol in 10 g of blood sample = (0.0092 g / 10 g) x 100

= 0.092 %