(a)     Define the term iodine number.

(b)     Linoleic acid (\({M_{\text{r}}} = 281\)) has the following formula.

\({\text{C}}{{\text{H}}_{\text{3}}}{{\text{(C}}{{\text{H}}_{\text{2}}}{\text{)}}_{\text{4}}}{\text{CH=CHC}}{{\text{H}}_{\text{2}}}{\text{CH=CH(C}}{{\text{H}}_{\text{2}}}{{\text{)}}_{\text{7}}}{\text{COOH}}\)

Calculate the volume of \({\text{1.00 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) iodine solution required to react exactly with 1.00 g of linoleic acid.

(a)     the number of grams/mass of iodine that add to/react with 100 g of the fat/lipid/oil;

(b)     amount of linoleic acid \( = \frac{1}{{281}} = 0.00356{\text{ (mol)}}\);

amount of \({{\text{I}}_{\text{2}}}\) required \( = 2 \times 0.00356 = 0.00712{\text{ (mol)}}\);

volume of \({\text{1.00 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) solution \( = 7.12{\text{ c}}{{\text{m}}^{\text{3}}}{\text{/0.00712 d}}{{\text{m}}^{\text{3}}}\);

OR

281 g of acid require 507.6 g of iodine;

1 g of acid requires 1.806 g/0.00712 mol of iodine;

volume of \({\text{1.00 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) solution \( = 7.12{\text{ c}}{{\text{m}}^{\text{3}}}{\text{/0.00712 d}}{{\text{m}}^{\text{3}}}\);

Award [3] for correct final answer.

In part (a), many candidates failed to score a mark for defining iodine number as they stated it is the amount of iodine, rather than the mass of iodine. The Chemistry guide clearly states in 1.1.2 that amount means the number of moles of a substance. Few candidates could calculate the volume of iodine solution required in part (b). Few recognised that 2 moles of iodine reacts with each mole of linoleic acid. Many tried to use the volume of one mole of gas to find the volume of solution. This is a standard question that is clearly in the Chemistry Guide in B.4.5.