Question

A person’s blood alcohol (C2H5OH) level can be determined by titrating a sample of blood plasma

with potassium dichromate solution. The balanced equation is:

16 H+ (aq) + 2 Cr2O7 2-(aq) + C2H5OH (aq) ? 4 Cr3+ (aq) + 2 CO2 (g) + 11 H2O (l)

If 35.46 mL of 0.05961 M Cr2O7 2-
is required to titrate 28.00 g of plasma, what is the mass percent
of alcohol in the blood?

Answer 1

0.17%

Step-by-step explanation:

With the equation:

2Cr2O7 2- + C2H5OH + H2O --> 4Cr3+ + 2CO2 + 11H2O

We can assume that every mole of ethanol needs 2 moles of Dichromate to react.

So if in 1L we have 0.05961 moles of dichromate we can discover how many moles we have in 35.46mL

1000 mL - 0.05962 moles

35.46 mL - x

x =
`(0.05962 * 35.46)/(1000)`

x = 2,11* 10^-3 moles

As we said earlier, 1 mole of ethanol needs 2 mole of dichromate, so in the solution we have 1,055*10^-3 moles of ethanol. We can discover the mass of ethanol present in the solution.

1 mole - 46g

1.055*10^-3 - y

y = 46 * 1.055*10^-3

y = 0.048 g

To discover the percent of alchol we can use a simple relation

28 g - 100%

0.048 - z

z =
`(0.048 * 100)/(28)`

z = 0.17%