In the reaction $\mathrm{Cu}+4\mathrm{HNO}{}_3\to \mathrm{Cu}(\mathrm{NO}{}_3){}_2+2\mathrm{NO}{}_2+2\mathrm{H}{}_2\mathrm{O}$, what fraction of the nitric acid consumed acts as the oxidising agent?

Of the 4 $\mathrm{HNO}{}_3$ molecules, 2 end up as nitrate in $\mathrm{Cu}(\mathrm{NO}{}_3){}_2$ with N still at +5 (acid role). The other 2 are reduced to $\mathrm{NO}{}_2$ (N +5 to +4) and act as oxidant. The fraction is 2/4 = 1/2.