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Q.
For neutralization of one mol of $NaOH$ the mass of $70\% $H_{2}SO_{4}$ required is :
AFMCAFMC 2002
Solution:
The neutralization of $NaOH$ by $H _{2} SO _{4}$ takes place as follows
$H _{2} SO _{4}+2 NaOH \longrightarrow Na _{2} SO _{4}+ H _{2} O$
For complete neutralization
Equivalents of acid = equivalents of base
Equivalents of $NaOH =$ moles $\times$ acidity
$=1 \times 1=1$
Equivalents of $H _{2} SO _{4}=\frac{x}{98} \times 2=\frac{x}{49}$
(Mol. mass of $H _{2} SO _{4}=98$ )
Putting the values
$1 \times 1 =\frac{x}{49}$
$\Rightarrow x =49\, g$
but $H _{2} SO _{4}$ is $70 \%$
let $y g 70 \% H _{2} SO _{4}$ is required
$\frac{70}{100} \times y =49$
$\Rightarrow y =70\, g$