- #1

- 5

- 0

volumes:

Diluted vodka = 5ml

6.0M H2SO4 = 3ml

K2Cr2O7 = 5ml

In the back titration of excess dichromate with iron(ll) solution, the volume "delivered" of iron(ll) solution in order to titrate the excess dichromate is 10.35mL.

There were also three stock solutions that were made initially.

For the K2CrO7 solution, 4 grams of K2CrO7 was used and the concentration is 0.136mol/L.

For the Iron solution, 4 grams were used and the concentration is 0.102 mol/L.

And for the vodka solution (2 ml was used!)

So here's what I've done and I know its wrong T_T:

First I found mole of iron(ll) solution

n = MxL = (.102)(0.01035L) = 0.0010557 mol Fe

From the chemical equation, there's a 6:1 ratio when it reacts w/ dichromate. So 0.0010557 mol/6 = 0.00017595 mol dichromate (left in excess)

Initial dichromate reacted w/ ethanol

n = (0.005)(.136) = 0.00068 mol

0.00068 mol - 0.00017595 = 0.0005 mol dichromate that reacted w/ the ethanol

From the chemical equation with the reaction b/w ethanol and dichromate, there is a 3:2 ratio

so ... 0.0005 mol x 1.5 = 0.000756 mol ethanol

0.000756 mol x 46g/mol(molar mass) = 0.03477grams ethanol

volume = m/P = 0.03477g/.789g/ml = 0.04408 mL

... So basically in this experiment, it says "the quality of a shipment of vodka has been questioned, so quality contro ltests have been requested. The label on the vodka indicated that ethanol content is 40% by volume, where:

%ethanol content by volume = (volume ethanol)/(total volume)*100%"

When I plug the volume ethanol = 0.04408 and divide by 2ml (the total volume) and multiply it by a 100 .. I don't get 40% or above that or less that... i get a small percent like 2% which is obviously wrong. What am I doing wrong? I'm so confused!

I would APPRECIATE any sort of help.