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

To calculate the amount of I3− in moles, we use the formula:

$ext{moles} = ext{concentration} imes ext{volume}$

Given the concentration of I3− is $2.00 imes 10^{-4} ext{ M}$ and the average titration volume is $15.65 ext{ mL}$ (or $0.01565 ext{ L}$), the calculation is as follows:

$ext{moles of I3−} = (2.00 imes 10^{-4} ext{ mol/L}) imes (0.01565 ext{ L}) = 3.1295 imes 10^{-6} ext{ mol}$

From the reaction stoichiometry, one mole of I3− reacts with one mole of vitamin C.

Thus, the amount of vitamin C will be equal to the moles of I3− in the aliquot taken:

Since the average titration yielded $3.1295 imes 10^{-6} ext{ mol}$ of I3−, the amount of vitamin C in the 25.00 mL aliquot is also:

$ext{moles of vitamin C} = 3.1295 imes 10^{-6} ext{ mol}$

The original fruit juice sample was diluted to 250.0 mL and the amount of vitamin C in each 25.00 mL aliquot is $3.1295 imes 10^{-6} ext{ mol}$.

To find the concentration in the original sample, we can scale up by the dilution factor:

ext{concentration of vitamin C} = rac{ ext{moles of vitamin C in 25.00 mL aliquot} imes ext{dilution factor}}{ ext{original volume}}

The dilution factor is 10 (since 250.0 mL / 25.0 mL = 10), and since the total volume of the original sample is 20.00 mL:

$ext{moles of vitamin C in original sample} = 3.1295 imes 10^{-6} ext{ mol} imes 10 = 3.1295 imes 10^{-5} ext{ mol}$

Since the original sample is 20.00 mL (or $0.020 ext{ L}$),

ext{concentration} = rac{3.1295 imes 10^{-5} ext{ mol}}{0.020 ext{ L}} = 1.56475 imes 10^{-3} ext{ mol/L}