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

The concentration of Na2S2O3 is given as 0.900 mol L$^{-1}$. Thus, the moles of Na2S2O3 used in titration can be calculated:

$\text{Volume (L)} = \frac{28.56667}{1000} = 0.02856667 \text{ L}$

$n(\text{Na}_2\text{S}_2\text{O}_3) = c \cdot V = 0.900 \text{ mol L}^{-1} \cdot 0.02856667 \text{ L} = 0.025771 mol$

From the reaction, 2 moles of Na2S2O3 react with 1 mole of I2:

$n(\text{I}_2) = \frac{0.025771 mol}{2} = 0.0128855 mol$

Considering the stoichiometry of the reaction, where 1 mole of Cr2O7^{2-} reacts with 3 moles of I2:

$n(\text{Cr}_2\text{O}_7^{2-}) = \frac{0.0128855 mol}{3} = 0.0042952 mol$

The initial moles of Cr2O7^{2-} calculated from the dilution:

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

Now we find the moles of Cr2O7^{2-} that reacted with ethanol:

$n(\text{Cr}_2O_7^{2-} \text{ reacted with ethanol}) = n(\text{initial } Cr_2O_7^{2-}) - n(\text{Cr}_2O_7^{2-} \text{ consumed})$

$= 0.0100 - 0.0042952 = 0.0057048 mol$

From the stoichiometry, the ratio is 2:3:

$n(\text{C}_2\text{H}_5\text{OH}) = \frac{3}{2} \cdot n(\text{Cr}_2O_7^{2-} \text{ reacted with ethanol})$

$= \frac{3}{2} \cdot 0.0057048 = 0.0085572 mol$

Using the molar mass of ethanol:

$m(\text{C}_2\text{H}_5\text{OH}) = n \cdot MM = 0.0085572 mol \cdot 46.068 g mol^{-1} = 0.394918 g$

Using the density of ethanol:

$V(\text{C}_2\text{H}_5\text{OH}) = \frac{m}{\text{density}} = \frac{0.394918 g}{0.789 g mL^{-1}} \approx 0.500 mL$

Finally, the concentration is:

$\text{Volume of undiluted sample} = 20.0 mL + 25.0 mL = 45.0 mL$

$\text{Concentration (v/v)} = \frac{0.500 mL}{45.0 mL} \times 100 \approx 1.11\% v/v$

Since the concentration of ethanol found is 1.11% v/v, which is less than the required 85% v/v, the sample does not meet the manufacturer’s requirements.