Question

The mass of $70\%H_{2}S{O}_4$ required for neutralisation of 1 mole of $NaOH$ is:

• A. 49 g
• B. 98 g
• C. 70 g
• D. 34.3 g

Answer 1

The correct option is C 70 g

$H_{2}S{O}_4+2NaOH⟶N{a}_{2}S{O}_4+2H_{2}O$
for neutralizing 1 mole NaOH,$\frac{1}{2}$ mole (49 g)$H_{2}S{O}_4$ is required.

Since, $70g{H}_{2}S{O}_4$ is in 100 gm of solution i.e. 70% $H_{2}S{O}_4$, it follows that:

$49g{H}_{2}S{O}_4$ will be found in $=\frac{100}{70}\times 49=70g$ solution.