Question

The mass of 70% $H_{2}S{O}_4$ required for the neutralization of $1$ mol of $NaOH$ is:

• A. $49\text{g}$
• B. $98\text{g}$
• C. $70\text{g}$
• D. $34.3\text{g}$

Answer 1

The correct option is C $70\text{g}$

$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$
0.5 mole of $H_{2}S{O}_4$ or 49 g of $H_{2}S{O}_4$ reacts with 1 mole of NaOH.
In 100 g solution, 70g of $H_{2}S{O}_4$ is involved in the reaction.
49 g of $H_{2}S{O}_{4}in=\frac{100\times 49}{70}=70g$
Using 70 g of the acid solution,neutralises the given base