Redox Titration Question

[squance]

Need help with a Year 12 chem question;

The amount of ethanol (C2H6O) present in a cider drink was determined by treating a sample of the cider with acidified potassium permanganate. The KMnO4(aq) oxidises the ethanol to ethanoic acid (C2H4O2), while being itself reduced to Mn2+. It was found that 23.01 mL of 0.450 KMnO4(aq) was required to oxidise all the ethanol in a 10.00mL mole of cider.

Determine the concentration of ethanol in hte cider as a percentage volume per volume (%v/v), given that the density of ethanol is 0.785 g ML^-1.

[squance]

My sister does not have the answers because the teacher didn't give them to her. The question is from Chemistry Dimensions 2 .
So im not sure if thats the answer

[squance]

Ok.
I just got my sister to read this.
She doesn't understanding the "halving it to get moles of C6H6O" bit and she doesn't understand the "find out how many mL's it is using the density " bit

[squance]

Crap! it is C2H6O
Typing error. I will fix it right now.
My sister does get the mole is 5/4 but she doesn't know what to do next.

[pHysiX]

haha..that means we have to change the equation to:
MnO4- + 8H+ + 5e- ---> Mn2+ + 4H2O
C2H6O + H2O ---> C2H4O2 + 4H+ + 4e-

therefore the equation is:

4MnO4- + 12H+ + 5C2H6O --> 4Mn2+ + 5C2H4O2 + 11H20

So therefore (n luckily) the ratio stays the same:
n(C2H6O) = (5/4)n(MnO4-)

now we know that 23.01mL of .450M MnO4- ions was used,
which means that there was 1.03545x10^-2 mol of MnO4-

From that, n(C2H6O) = 5/4 x 1.03545x10^-2
=1.2943125 * 10^-2 mol of C2H6O

Next, we convert that to grams, so multiply by molar mass (46g mol-1)
=.59538375 grams of C2H6O

Now, the density: if 0.785g of C2H6O per mL
so for .59538375 grams, it is:
0.59538375/0.785
=0.7584506369 mL

Therefore %v/v is:
0.7584506369/10 x 100
=7.58% v/v

sorry for like being dodgy with the chem...obviously, it's not my forte but i'm pretty sure that's the correct answer. hope the concepts help =]

[squance]

Thanks physiX
My sister just had never done the density thing before.
THanks for helping out.

[pHysiX]

No probs. I'm glad that I could be of some help =]