$0.60$ mol of $\mathrm{Cl}{}_2$ reacts completely with hot concentrated NaOH according to $3\mathrm{Cl}{}_2+6\mathrm{NaOH}\to 5\mathrm{NaCl}+\mathrm{NaClO}{}_3+3\mathrm{H}{}_2\mathrm{O}$. The number of moles of electrons lost by the chlorine that is oxidised is

In the balanced equation, one Cl goes from 0 to $+5$ (loses 5 electrons) and five Cl go from 0 to $-1$ (gain 5 electrons) for every 3 mol of $\mathrm{Cl}{}_2$. Moles of electrons lost $=\frac{0.60}{3}\times 5=1.00$. The same amount is gained by the chlorine reduced to chloride.