Calculate the iodine number of linoleic acid.

\[\begin{array}{*{20}{l}} {{\text{C}}{{\text{H}}_{\text{3}}}{{{\text{(C}}{{\text{H}}_{\text{2}}}{\text{)}}}_{\text{4}}}{{{\text{(CH=CHC}}{{\text{H}}_{\text{2}}}{\text{)}}}_{\text{2}}}{{{\text{(C}}{{\text{H}}_{\text{2}}}{\text{)}}}_{\text{6}}}{\text{COOH}}}&{{M_{\text{r}}} = 280.4} \end{array}\]

(amount of linoleic acid in 100 g =) \(\frac{{100}}{{280.4}} = 0.357{\text{ mol}}\);

2 double bonds in molecule so 1 mole of lipid reacts with 2 moles of \({{\text{I}}_{\text{2}}}\);

Can be implied in working.

mass of iodine that reacts \(( = 253.8 \times 0.357 \times 2) = 181{\text{ g}}\);

Award [3] for correct final answer.

Award [2] for 90.5 g.

In (b) many candidates could not define iodine number, with many referring to moles instead of grams. This then meant that few were able to calculate the iodine number successfully.