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Q.
The mass of $70\%\,\,H_{2}SO_{4}$ required for neutralisation of $1$ mol of NaOH.
Some Basic Concepts of Chemistry
Solution:
$ H_{2}SO_{4}+2NaOH \to Na_{2} SO_{4}+2H_{2}O$
for $1$ mole $NaOH \frac{1}{2}$ mole $H_{2}SO_{4}$ required
Molar mass of $H_{2}SO_{4} = 98$ g
$70$ g $H_{2}SO_{4}$ in $100$ g solution
$\frac{98}{2}\,g\,H_{2}SO_{4}\,\, in =\frac{100}{70} \times 49=70 $g solution.