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 present in the average titrate:

1. Use the formula: $ext{Moles} = ext{Concentration (mol/L)} imes ext{Volume (L)}$

2. The concentration of I₃⁻ is given as 2.00 × 10⁻⁴ M, and the average volume is 15.65 mL, which needs to be converted to liters:

   $15.65 ext{ mL} = 0.01565 ext{ L}$

3. Now, substitute the values:

   $ext{Moles of I₃⁻} = 2.00 imes 10^{-4} ext{ mol/L} imes 0.01565 ext{ L} = 3.13 imes 10^{-6} ext{ moles}$

From the stoichiometry of the reaction, 1 mole of vitamin C reacts with 1 mole of I₃⁻. Therefore, the amount of vitamin C in each 25.00 mL aliquot is equal to the moles of I₃⁻ calculated above:

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

To find the concentration of vitamin C in the original fruit juice sample, first determine how many times the sample was diluted:

20.00 mL sample was diluted to 250.0 mL.

Dilution factor = ( \frac{250.0 , ext{mL}}{20.00 , ext{mL}} = 12.5 )

Next, since the amount of vitamin C in the diluted sample is 3.13 × 10⁻⁶ moles, we can find the amount in the original sample:

$\text{Moles of vitamin C in original} = 3.13 \times 10^{-6} ext{ moles} \times 12.5 = 3.91 \times 10^{-5} ext{ moles}$

Finally, find the concentration in the original sample:

$\text{Concentration} = \frac{3.91 \times 10^{-5} ext{ moles}}{0.020 ext{ L}} = 1.955 \times 10^{-3} ext{ mol/L}$