Question

For neutralization of one mol of $NaOH$ the mass of $70\%$H_{2}SO_{4}$ required is :

• A. 48 g
• B. 70 g
• C. 49 g
• D. 35 g

Answer 1

Answer: (B)

The neutralization of $NaOH$ by $H_{2}S{O}_4$ takes place as follows
$H_{2}S{O}_4+2NaOH⟶N{a}_{2}S{O}_4+H_{2}O$
For complete neutralization
Equivalents of acid = equivalents of base
Equivalents of $NaOH=$ moles $\times$ acidity
$=1\times 1=1$
Equivalents of $H_{2}S{O}_4=\frac{x}{98}\times 2=\frac{x}{49}$
(Mol. mass of $H_{2}S{O}_4=98$ )
Putting the values
$1\times 1=\frac{x}{49}$
$\Rightarrow x=49g$
but $H_{2}S{O}_4$ is $70\%$
let $yg70\%H_{2}S{O}_4$ is required
$\frac{70}{100}\times y=49$
$\Rightarrow y=70g$