7. a - VCE - SSCE Chemistry - Question 7 - 2010 - Paper 1

First, we will calculate the moles of POP using its molar mass:

$n(\text{POP}) = \frac{\text{mass}}{\text{molar mass}} = \frac{10000 \, \text{g}}{833 \, \text{g mol}^{-1}} = 12.00 \, \text{mol}$

Since 3 moles of methanol are required for every mole of POP reacted:

$n(\text{methanol}) = 3 \times n(\text{POP}) = 3 \times 12.00 \, \text{mol} = 36.00 \, \text{mol}$

Now, we convert moles of methanol to volume using its density:

Using the formula:

$\text{Volume (L)} = \frac{n(\text{methanol})}{\text{density}}$

$\text{Volume} = \frac{36.00 \, \text{mol}}{0.79 \, \text{g mL}^{-1}} \times 1000 \approx 45641.77 \, \text{mL} \approx 45.64 \, \text{L}$

Therefore, the volume of methanol required is approximately 45.64 litres.

To find the number of double bonds, we first calculate the number of moles of cervonic acid reacted:

$n(\text{cervonic acid}) = \frac{0.328 \, \text{g}}{328 \, \text{g mol}^{-1}} = 1.00 \, \text{mol}$

The reaction with iodine solution indicates that each mole of cervonic acid has 6 double bonds based on the ratio determined from titration:

$\text{Ratio} = \frac{n(\text{cervonic acid})}{n(\text{I}_2)} = 1 : 6$

This means there are 6 double bonds in each molecule of cervonic acid.

Given that there are 22 carbon atoms in a molecule of cervonic acid and considering the presence of 6 double bonds, the general formula for polyunsaturated fatty acids can be applied:

$C_nH_{2n-2m}O_2$

where $n$ is the number of carbon atoms and $m$ is the number of double bonds. Thus:

$m = 6, n = 22$

The formula simplifies to:

$C_{22}H_{2(22)-2(6)}O_2 = C_{22}H_{34}O_2$

Hence, the formula of cervonic acid is $C_{22}H_{34}O_2$.