The blood alcohol (C2H5OH) level can bedetermined by titrating a sample of blood plasma with an acidicpotassium dichromate solution, resulting in the production ofCr3 + (aq) and carbon dioxide. The reaction canbe monitored because the dichromate ion (Cr2O72-) is orange in solution, and the Cr3 + ion is green. The unbalanced redox equationis shown below.

Cr2O72 - (aq) + C2H5OH (aq) → Cr3 + (aq) + CO2 (g)

If 31.05 mL of 0.0600 M potassium dichromate solution isrequired to titrate 30.0 g of blood plasma, determine the masspercent of alcohol in the blood.

Question

Mathematics

Jenny Burgess

31 August, 14:08

The blood alcohol (C2H5OH) level can bedetermined by titrating a sample of blood plasma with an acidicpotassium dichromate solution, resulting in the production ofCr3 + (aq) and carbon dioxide. The reaction canbe monitored because the dichromate ion (Cr2O72-) is orange in solution, and the Cr3 + ion is green. The unbalanced redox equationis shown below.

Cr2O72 - (aq) + C2H5OH (aq) → Cr3 + (aq) + CO2 (g)

If 31.05 mL of 0.0600 M potassium dichromate solution isrequired to titrate 30.0 g of blood plasma, determine the masspercent of alcohol in the blood.

+3

Answers (1)

Know the Answer?

Eli Combs · Answer 1

Step-by-step explanation:

In acidic reaction, equivalent weight of potassium dichromate is

mol weight / 6

= 294 / 6

49 g

equivalent weight of alcohol = 46 g

31.05 cc of. 06M of potassium dichromate

= 31.05 x. 06 cc of M potassium dichromate

= 1.863 cc of M potassium dichromate

= 1.863 x 6 cc of N potassium dichromate

= 11.178 cc of N potassium dichromate

= = 11.178 cc of N alcohol

= (11.178 / 1000) x equivalent weight of alcohol

= (11.178 / 1000) x 46 g

=.514 g

30 g of blood contains. 514 g of alcohol

100 g of blood contains (0.514 / 30) x 100

= 1.71 % of alcohol.