A manufacturer requires that its product contains at least 85% v/v ethanol - HSC - SSCE Chemistry - Question 35 - 2021 - Paper 1

Using the average volume obtained, calculate the moles of Na₂S₂O₃:

$n(S_2O_3^{2-}) = c \times V = 0.900 \text{ mol L}^{-1} \times 0.0285667 \text{ L} = 0.02571 \text{ mol}$

From the equation:

$I_2 + 2S_2O_3^{2-} \rightarrow 2I^-$

We know that:

$n(I_2) = \frac{1}{2} \times n(S_2O_3^{2-}) = \frac{1}{2} \times 0.02571 = 0.012855 \text{ mol}$

Using the stoichiometry from the reaction:

$Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O$

We determine:

$n(I_2) = 3 \times n(Cr_2O_7^{2-})$

Thus:

$n(excess Cr_2O_7^{2-}) = \frac{1}{3} \times n(I_2) = \frac{1}{3} \times 0.012855 = 0.004285 \text{ mol}$

Using the initial concentration of Cr₂O₇^2-:

$n(initial Cr_2O_7^{2-}) = c \times V = 0.500 \text{ mol L}^{-1} \times 0.0200 \text{ L} = 0.0100 \text{ mol}$

Now calculate the moles of Cr₂O₇^2- that reacted with ethanol:

$n(Cr_2O_7^{2-}) = n(initial Cr_2O_7^{2-}) - n(excess Cr_2O_7^{2-})$

So:

$n(Cr_2O_7^{2-}) = 0.0100 - 0.004285 = 0.005715 \text{ mol}$

Using the stoichiometry of the reaction:

$3C_2H_5OH + Cr_2O_7^{2-} \rightarrow products$

We have:

$\frac{1}{3} \times n(Cr_2O_7^{2-}) = n(C_2H_5OH)$

Thus:

$n(C_2H_5OH) = \frac{1}{3} \times 0.005715 = 0.001905 \text{ mol}$

Using the molar mass of ethanol (46.068 g mol^-1):

$m(C_2H_5OH) = n \times MM = 0.001905 \text{ mol} \times 46.068 \text{ g mol}^{-1} = 0.08757 \text{ g}$

Now, using the volume of the diluted sample:

$V(C_2H_5OH) = \frac{m}{\text{density}} = \frac{0.08757 \text{ g}}{0.789 \text{ g mL}^{-1}} = 0.111 ext{ mL}$

Finally, calculate the percent:

$\% \text{ v/v} = \left(\frac{20.02 \text{ mL}}{25.0 \text{ mL}}\right) \times 100 = 80 \text{ v/v}$

Since the concentration found is under 85%, the product does not meet the manufacturer's requirement.