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- present in the average titre, use the molarity formula:

$n = C \times V$

where:

• $n$ is the number of moles,
• $C$ is the concentration in moles per litre,
• $V$ is the volume in litres.

Given:

• Concentration, $C = 2.00 \times 10^{-4} \text{ M}$
• Average titre volume, $V = 15.65 \text{ mL} = 0.01565 \text{ L}$

Now calculate:

$n = (2.00 \times 10^{-4}) \times (0.01565) = 3.130 \times 10^{-6} \text{ moles of } I3^-$

From the balanced equation:

$C6H8O6 + I3^- \rightarrow C6H8O7 + 3I^-$

1 mole of I3- reacts with 1 mole of vitamin C. Thus, the amount of vitamin C is the same as the amount of I3- that reacted:

• The amount of vitamin C in each 25.00 mL aliquot is:

$n(C6H8O6) = 3.130 \times 10^{-6} \text{ moles}$

Since the original 20.00 mL sample was diluted to 250.0 mL, we can calculate the concentration of vitamin C in the undiluted sample.

Using:

$C_1 \times V_1 = C_2 \times V_2$

where:

• $C_1$ is the concentration of the undiluted sample,
• $V_1 = 20.00 \text{ mL}$,
• $C_2$ is the concentration of diluted sample (the amount of vitamin C we calculated in part b ii),
• $V_2 = 250.0 \text{ mL}$.

To find $C_1$:

$C_1 = \frac{C_2 \times V_2}{V_1}$

Calculating: $C_2 = \frac{3.130 \times 10^{-6}}{0.025} = 1.252 \times 10^{-4} M$

Thus, $C_1 = \frac{1.252 \times 10^{-4} \times 250.0}{20.00} = 1.565 \times 10^{-3} \text{ M}$