For quality control, a chemist analyses the vitamin C (molecular formula C6H8O6) content of a new brand of fruit juice - VCE - SSCE Chemistry - Question 2 - 2005 - Paper 1

First, calculate the number of moles of I3- using the concentration and volume:

$ext{Moles of } I3^- = ext{Concentration} imes ext{Volume}$

Using the given concentration:

$ext{Moles of } I3^- = (2.00 imes 10^{-4} ext{ mol/L}) imes (15.65 imes 10^{-3} ext{ L}) = 3.13 imes 10^{-6} ext{ moles}$

Therefore, the amount of I3- present in the average titre is 3.13 × 10^-6 moles.

From the reaction, we see that 1 mole of vitamin C reacts with 1 mole of I3-. Therefore, the amount of vitamin C is equal to the amount of I3- present.

$ext{Moles of Vitamin C} = Moles of I3^- = 3.13 imes 10^{-6} ext{ moles}$

Thus, the amount of vitamin C in each 25.00 mL aliquot is also 3.13 × 10^-6 moles.

To find the concentration of vitamin C in the original sample, we first note that the original sample of fruit juice was diluted before titration:

Total dilution factor = Total volume / Aliquot volume = 250.0 mL / 25.0 mL = 10.

Now, the concentration of vitamin C in the 25.00 mL aliquot:

C = rac{ ext{Moles}}{ ext{Volume}} = rac{3.13 imes 10^{-6} ext{ moles}}{25.0 imes 10^{-3} ext{ L}} = 1.252 imes 10^{-4} ext{ mol/L}

To find the concentration in the undiluted sample:

$ext{Concentration in original} = 10 imes 1.252 imes 10^{-4} ext{ mol/L} = 1.252 imes 10^{-3} ext{ mol/L}$

Therefore, the concentration of vitamin C in the original sample of fruit juice is 1.252 × 10^-3 mol/L.