The mass of 80% pure H₂SO₄ required to completely neutralise 60g of NaOH is

$$\begin{gathered} {\mathrm{H}}_2{\mathrm{SO}}_4+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O} \\ 98\mathrm{g} 2\times 40\mathrm{g} \\ ? 60\mathrm{g} \end{gathered}$$

The mass of pure(100%) H₂SO₄ required = $\frac{60}{80}\times 98=73.5\mathrm{g}$

The mass of 80% pure H₂SO₄ required = $73.5\times \frac{100}{80}=92\mathrm{g}$