The mass of 70% ${\mathrm{H}}_2{\mathrm{SO}}_4$ required for neutralization of one mole of NaOH is :

Since 70 g ${\mathrm{H}}_2{\mathrm{SO}}_4$ is in 100 gm of solution i.e. 70% ${\mathrm{H}}_2{\mathrm{SO}}_4$ it follows that:

49 g ${\mathrm{H}}_2{\mathrm{SO}}_4$ will be found in $=\frac{100}{70}\times 49=70 \mathrm{g}$